chemical in the laboratory.pdf

Chemistry in the Laboratory

John }. Alexander University of Cincinnati

Margaret }. Steffel

The Ohio State University Marion Campus

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HARCOURT BRACE JOVANOVICH, PUBLISHERS San Diego New York Chicago Austin London Sydney Toronto

Preface

The collection of experiments, some classical, some new, in Chemistry in the Laboratory emphasizes the accumulation and interpretation of numerical data. Even in the qualitative experiments, students are often requested to draw semiquantitative conclusions. Most of the experiments in this manual have been class tested for several years at both the University of Cincinnati and The Ohio State University Marion Campus.

Many experiments—especially those in the qualitative analysis section—require more than one three-hour laboratory period to complete. In addi t ion, more than one experiment pertaining to a specific subject (for example, equi l ibr ium) is often presented.

We have designed the experiments in Chemistry in the Laboratory in the belief that an instructive experience in the laboratory is possible only if students carefully prepare for their work and keep complete records of their data in a laboratory notebook.

We are grateful for the useful suggestions contr ibuted by many teaching assistants and students. We especially wish to thank Sam Lucas, T i m Becker, Ralph Brewer, and Ed Schroer, all of the University of Cincinnati . And to Kathy Bailey, who accurately and cheerfully typed several early drafts of the manual. Finally, we wish to acknowledge w i t h gratitude the encouragement of Thomas B. Cameron, late of the University of Cincinnati .

The authors welcome comments and suggestions from the users of this manual.

John Alexander Margaret J. Steffel

Contents

Introduction

Experiment 1

Experiment 2

Experiment 3

Experiment 4

Experiment 5

Experiment 6

Experiment 7

Experiment 8

Experiment 9

Experiment 10

Experiment 11

Experiment 12

Experiment 13

Experiment 14

Experiment 15

Experiment 16

Experiment 17

Experiment 18

Experiment 19

Experiment 20

Experiment 21

Experiment 22

Experiment 23

Experiment 24

Experiment 25

Experiment 26

1

The Use of Balances: Hydrated Compounds 29

Properties of Substances 39

Compound Stoichiometry 49

Stoichiometry of a Reaction 55

Synthesis of Co(NH3 ) 6 C/ 3 61

Line Spectra of Elements 69

A Simple Covalent Compound: Synthesis and Properties 79

Salt Identification Using Visible and Infrared Spectra 87

Synthesis of Coordination Compounds and Their Identification Using Spectrophotometry 105

Vaporization of a Liquid: Vapor Pressure and Heat of Vaporization 119

Molar Weight by Vapor Density 127

"Molecular Interactions in Solution 135

Freezing Point Lowering 143

Chemical Kinetics: Oxidation of Ethanol by Chromium( VI) 151

A Complete Rate Law 163

Chemical Equilibrium 169

Spectropliotometric Determination of an Equilibrium Constant 181

Enthalpy of Reaction 191

Solubility of an Electrolyte 201

Gravimetric Determination of Iron 213

Acid-Base Titration 221

Antacids 229

Acidic and Basic Salts 235

Electrochemistry: Voltaic Cells 243

Electrochemistry: An Electrolytic Cell 253

An Oxidation-Reduction Titration 259

vii

viii Contents

Experiment 27 Metal Chelate Complexes 267

Experiment 28 Synthesis of an Organo metallic Compound 277

Introduction II Inorganic Qualitative Analysis 285

Experiment 29 Qualitative Analysis of the Group I Cations 293

Experiment 30 Qualitative Analysis of the Croup II Cations 301

Experiment 3 1 Qualitative Analysis of the Group III Cations 319

Experiment 32 Qualitative Analysis of the Group IV Cations 331

Experiment 33 Qualitative Analysis of the Group V Cations 339

Experiment 34 Qualitative Analysis of Anions 345

Experiment 35 Analysis of a General Unknown 351

Appendix I Sample Laboratory Report: Density Determination 365

Appendix II Graphing Experimental Data 367

Appendix I I I Infrared Spectrophotometers 373

Appendix IV Visible and Ultraviolet Spectrophotometers 375

Appendix V Reagents and Constants for Qualitative Analysis 378

Introduction

Why Have a Chemistry Laboratory?

Your faculty requires you to spend time each week working in a chemistry laboratory. If you have no intention of becoming a practicing chemist, you may wonder why there is this insistence on performing laboratory experiments. After all, you don' t have to write a novel when you take a course in literature. Why isn't just learning about chemistry sufficient' ' Why must you actually carry out experiments in a laboratory?

There are at least three important reasons why students who are preparing for careers in medicine, biology, physics, or chemistry must have actual experience wi th scientific experimentation.

First, chemistry (like other natural sciences) is based on experiment. Scientific conclusions arc reached by carefully studying the results of experimental observations—not by making plausible conjectures. To illustrate this point , consider a simple example. Someone might wonder whether iron chloride wi l l react w i th iodine. A balanced equation for a possible reaction can easily be wr i t ten

FeCl 2 + I 2 • F e l 2 + C l 2

But does this plausible equation represent the way nature really behaves? The way to find out is to / r y the reaction to see if it works! ( I t doesn't.) Information in your text has resulte"d from experiments of many scientists. What you do in the laboratory should make you personally aware that scientific knowledge is based as solidly on experimentation as it is on reasoning.

Second, you should begin to develop some manipulative skills in performing quantitative experiments. Good laboratory technique becomes increasingly essential as you progress to more advanced scientific courses and w i l l , o f course, be mandatory when you ul t imately practice your profession.

Th i rd , you need to develop the abili ty to keep careful records of your experimental observations and to communicate, both verbally and in wr i t ing , wi th others about these observations and the conclusions you draw from them. These functions are served by the laboratory notebook and the Report Sheets described later in the In t roduct ion.

W H A T Y O U ' L L D O I N T H E L A B O R A T O R Y

Even though one purpose of the laboratory is to make you aware that chemistry is an experimental science, it would be impossible for you to repeat all of the important chemical experiments ever conducted. Obviously, we can select only a reasonable number of projects to complete in the laboratory this year. Each of these has been chosen w i t h one (or all) of the fol lowing criteria in mind :

1. To show the relationship between experimental measurement and chemical theory. Sometimes this involves confirming the results that you have already

l

2 I n t r o d u c t i o n

read about in the text ; at other times, you w i l l be asked to design simple experiments to test various theoretical possibilities.

2. To teach basic laboratory technique. 3. To help you learn to manipulate and interpret numerical data. 4. To give you an idea of what a chemist's actual work entails.

) In all of your laboratory work , you should make a special effort to develop the

abili ty to observe what is actually happening in an experiment -not what you th ink should be happening. Most successful scientific experiments are usually designed wi th some not ion o f what wi l l occur. However, i f preconceived expectations are not realized, it does not always mean that the experiment has failed. It may only mean that the scientist overlooked or did not eompletely understand some theoretical principle or some aspect of experimental design. The scientist's task then is to try to explain what happened—not to be disappointed because what was expected to happen didn ' t occur. You should do the same thing in the laboratory: first observe what happens and then try to understand and explain i t .

The content o f lecture and laboratory sessions wi l l probably not be correlated completely. Several ideas already discussed in lectures are usually applied to a single laboratory problem. A few experiments may involve material that has not yet been covered in a lecture. To some people, this might appear to be a case of put t ing "the cart before the horse." However, you should be able to learn new chemical concepts in the laboratory just as you do in a lecture.

_ _ _

YOUR L A B O R A T O R Y R E S P O N S I B I L I T I E S

1. It is impossible to overemphasize the importance of preparation prior to the laboratory period. Vir tual ly every bad t r ip in the laboratory is due to an ''underdose" of preparation! If you are going to complete all of the necessary laboratory work wi th in the time allotted, you must be prepared when you come to class. This means that you should study the experiment in advance and have a plan for carrying out your work efficiently. The laboratory instructor wi l l answer any questions you may have, but this wi l l not help you i f you are not familiar enough w i t h the material to know what questions to ask.

2. You wi l l be expected to keep a laboratory notebook. Reserve a notebook exclusively for chemistry laboratory, and bring it to each laboratory session or to any other class meeting in which laboratory work is to be discussed.

You should make several kinds of appropriate entries in your notebook. First, you may need to prepare a work outline or set up a data table before you come to the laboratory. Or you may need to calculate the weight or the volume of a particular reagent that you wi l l need. Any preparatory material of this kind should be recorded in your notebook.

Second, you w i l l want to note your qualitative observations describing what happens and how things look while you carry out an experiment. For example, the colors of starting materials and products, the texture of a precipitate, or whether or not a reaction evolves heat cottlcl be recorded. Y o u should also note any mistakes or difficulties that might cause you to question the results in any part of an experi-

3 Safety in the Labo ra to r y

ment. For example', using the wrong concentration of a solution, spilling some of a carefully measured substance, or using too much of a particular reagent may affect ' the accuracy of your results. I f you record an error when you make i t , you w i l l remember it and this may help you to explain why your results are incorrect.

Th i rd , all numerical data should be carefully and completely recorded in your\ notebook. Be certain to label all numbers w i t h the appropriate units. A good way organize your data would be to make a data table in your notebook similar to the one on the Report Sheet for each experiment and then to enter numerical data in the appropriate places as you obtain them. The purpose of keeping a laboratory notebook is not to provide a work of art for your instructor to admire. Do not record data on scraps of paper and copy them in your notebook later—and perhaps more neatly. Have your notebook wi th you at all times and record data as you obtain them. If, for any reason, you decide later that you are not going to use a piece of data, you can simply draw a line through i t . But don' t run the risk of losing important data just to keep your records neat. A l l types of practicing technicians and scientists record data and observations this way. Learn ami practice proper recording of data from the start.

3. The experiments in this manual require you to submit wr i t t en reports. Unless your instructor's procedures differ, each report wi l l be due at the beginning o f the next laboratory meeting after the experiment has been performed.

Almost every experiment in this manual contains a Report Sheet to guide you in preparing your report. The "Results and Discussion" section at the end of each experiment tells you how to treat your data when fill ing out the Report Sheet. Questions which should serve to stimulate your th inking about experimental results are also included.

Your instructor may ask you to submit a report that includes additional items not covered on the Report Sheet. Such a report might include a statement of the procedures you used (especially if any of your procedures differed from the instructions you were originally given) and a discussion of the l imits placed on the accuracy of your results. A sample of such a report is presented in Appendix One (page 355). This type of report does not have to be more than two or three pages long.

In performing your laboratory experiments, supplying the " r igh t " answer is sometimes less important than your abili ty to observe accurately and explain your observations. If you obtain an unexpected result, try to explain why.

Some of the experiments in this manual consist of identifying an " u n k n o w n , " and the reports for these experiments wi l l simply be a recording o f your results. Your mastery o f the material involved in such an experiment w i l l be reflected in your abili ty to obtain the correct experimental results, and these results w i l l be o f prime importance in determining your grade for that experiment.

Safety in the Laboratory

One way in which a chemistry laboratory differs f rom, say, a philosophy or a calculus course is that you must take certain precautions so that you wi l l not be


physically injured. Because a laboratory accident cannot be predicted, you must constantly adhere to safety precautions while you are in the laboratory. The fol lowing list of procedures indicates areas where caution is essential.

1. Wear goggles or eyeglasses at all times. Even if you are careful not to splash chemicals into your eyes, someone close to you may accidentally do so. Goggles pushed up onto your forehead (or left in your desk) offer your eyes no protect ion. If your laboratory is equipped w i t h an eye shower, learn where it is located and how to use it .

2. Wear suitable clothing and shoes. Shorts or bare feet leave large areas of skin exposed to potential chemical burns. Long, full sleeves may drag in your chemicals or catch on your equipment; both the sleeves and the experiment could be ruined! Long hair should not be worn loose, especially when you are using a burner or centrifuge.

3. Take precautions to avoid cut t ing yourself. Discard cracked or chipped glassware. ( In the next section, you wi l l be told how to insert glass tubing in to a stopper and how to fire-polish tubing to smooth sharp ends.)

4. Work under a hood when so instructed. A few chemical experiments involve unpleasant or harmful substances. Their vapors w i l l be suctioned out o f the laboratory if you set up your equipment under a hood.

5. Dispose of waste chemicals as directed. Do not return any substances to reagent bottles. Usually, your instructor w i l l ask you to place waste chemicals in waste jars or to pour them down a sink w i t h plenty of running water. When in doubt, ask!

6. Always pour concentrated acid into water (not water into acid). Otherwise, some acid may spatter from the solution and burn you.

7. Report any i n j u r y - n o matter how minor it may seem-to your instructor.

8. Never attempt an unauthorized experiment. The procedures in this manual have been designed wi th your safety in mind. Don' t risk injury to yourself or to others by mix ing chemicals randomly or by setting up unauthorized reactions or equipment.

Some Laboratory Protocol

Fol lowing a few general rules of procedure in the laboratory can make your work and the work of the other students around you much more orderly and pleasant.

1. Clean up! Y o u are responsible for cleaning up any chemicals you spill in the reagent or balance area. Also, be sure to clean your own work area before you leave the laboratory; plan your work so that you have adequate time to clean up.

5 L abo ra to r y E q u i p m e n t and Techn ique

2. Never take reagent bottles to your desk. Another student's work may be needlessly delayed by searching for a reagent. The proper way to obtain the reagents you need is to use a clean beaker or a test tube for liquids or a beaker for solids to obtain substances from the reagent shelf.

3. Don ' t waste chemicals. Large quantities of scarce energy resources are consumed to make and to purify the chemicals you use in the laboratory. Take whatever you need, but do not waste resources or unnecessarily increase your school's expenses by taking more of a chemical than you need. Also, remember that any excess chemical you must throw away is a potential air or water pollutant.

4. Don ' t return leftover chemicals to reagent bottles. Sometimes mistakes are made: chemicals are returned to the wrong bottles or contaminated chemicals are returned to reagent bottles. You may have put the material into a d i r ty beaker; water, d i r t , or some other chemical may have splashed into a substance while it was sitting on your desk; you may have diluted a solution of a specific concentration by put t ing it into a wet container. As a result, the entire contents of a reagent bottle could be contaminated. This whole problem can be avoided if you dispose of any excess material as your instructor directs.

5. Don ' t put anything into reagent bottles. The obvious reason for this precaution is to eliminate the possibility of contaminat ion. Always pour a l iquid from a bott le and do not place the bottle l id or stopper on a desk or other laboratory surface. To pour a solid from a bott le , t ip the bottle slightly while rotating i t ; do not use your spatula.

Laboratory Equipment and Technique

Many operations are performed so often in the laboratory that a special section describing them is provided here rather than in individual experiments. Read this section now and then reread the relevant parts as you conduct the experiments.

EQUIPMENT

Several common pieces of laboratory equipment are shown and discussed on the fol lowing pages.

L A B O R A T O R Y BURNER

A mixture of natural gas and air is burned in the Bunsen-type burner to provide a source of heat. Figure i-1 shows burners. A Tirrill burner has a needle valve under the barrel to contro l gas How. A i r is admitted by unscrewing the barrel of the burner or by changing the position of the slots around the lower por t ion of the barrel. When the air intake is adjusted properly, the flame should look like the one depicted in Figure i -2 ; it w i l l have a bright blue inner cone and a darker blue outer cone. The

7 Laboratory Equipment and Technique

Introduction

Evaporating dish

Screw clamp

Pinchcock clamp

U n i v e r s a l c l a m p


F IGURE i-1

Bunsen-type burners: {left) a Bunsen burner; (right) a T i r r i l l burner.

hottest part of the name is the t ip of the inner cone. If insufficient air is being mixed w i t h the gas, the flame wi l l be long and yel low. I f this happens, adjust the air intake un t i l you have the proper blue flame. Too much air w i l l cause the flame to "flash back" ( to burn inside the barrel, making it very ho t ) ; the flame may even shoot out of the bo t tom of the barrel. If this happens, turn o f f the gas, adjust the air supply, and relight the burner.

C U T T I N G A N D B E N D I N G G L A S S

Soft glass rod or tubing can be cut easily if you place the tubing on the desk top and make a single scratch across it w i t h a triangular file or a glass scorer. Grasp the glass f i rmly w i t h your thumbs on the side of the glass away from the scratch; break the tubing away from you, as shown in Figure i-3. Glass broken in this way w i l l have a sharp and perhaps rough edge on which you could cut yourself. Always take the

FIGURE i-2

Flame of a properly adjusted Bunsen burner.

10 Introduction

Breaking glass tubing.

extra few minutes required to fire-polish the cut ends of glass tubing. First, remove any jagged edges by rubbing the end of the tubing on wire gauze held over a waste can. Then, hold the end of the tubing in the tip of a burner flame (as shown in Figure i-4) and turn the tubing so that it is heated evenly. After the glass has become partially melted and smoothed, set the tubing down on wire gauze or asbestos to cool.

To bend glass tubing, you w i l l need to heat a fairly long section o f the glass to softness. A flame spreader, often called a wing-tip, w i l l provide a wide flame. Rotate the tubing in this flame unt i l it is soft. Then remove the tubing from the flame and hold it for a second or two to permit even heat dis tr ibut ion. Bend the tubing to the desired shape and hold it unt i l it hardens (see Figure i-5). Place the

12 Introduction

FIGURE i-6

Inserting glass tubing into a stopper.

glass on wire gauze or asbestos to cool. Only soft glass can be worked in this manner. Borosilicate glass, such as Pyrex, has too high a melting point to be softened by a Bunsen burner flame.

I N S E R T I N G G L A S S T U B I N G INTO A S T O P P E R

Some of the worst injuries possible in the first-year chemistry laboratory result from t ry ing to force a thermometer, the stem of a glass funnel, or a piece of glass tubing into a stopper. If the glass breaks, it may be shoved so forcefully in to your hand that it causes a long, deep gash. The safest way to insert tubing is to wet it first wi th soapy water or glycerin. Then grasp the tubing just above the area to be inserted into the stopper. Holding both the tubing and the stopper w i th a towel , twist the stopper onto the tube (see Figure i-6). Don ' t hesitate to use a lot of soap or glycerin; it can always be washed o f f w i th water when you are finished.


P O U R I N G C H E M I C A L S

Many l iquid reagents are contained in stoppered bottles. To pour from a stoppered bott le , first remove the stopper and grasp the top of it between your index and third fingers w i t h your hand turned palm up. Then pick up the bottle w i th the same hand (see Figure i-7). The two advantages to this pouring method are that the stopper is never placed where it can become contaminated and that your other hand is free to hold the container into which you are pouring the l iquid .

A solid should be poured from a bottle by slightly t ipping and then rotating the bottle (see Figure i-8). If you must put the lid down while you pour the solid, be sure to place the top of the lid down on your desk or working area.

F I L T R A T I O N

A solid can be separated from a supernatant l iquid by f i l t ra t ion. One of two methods-gravi ty f i l t ra t ion or vacuum filtration—may be used. Both methods leave the solid on a piece of filter paper and the l iquid (the filtrate) in a flask or a beaker underneath a funnel.

In gravity filtration, gravitational force draws the l iquid through the funnel and the filter paper; in vacuum filtration, the l iquid is sucked through the filter paper by an aspirator vacuum.

Gravity f i l t ra t ion is accomplished wi th a conical funnel. The funnel stem touches the wall of the beaker to permit smooth l iquid How; the weight of the water column in the funnel stem speeds the f i l t ra t ion. Fold a piece of filter paper in half; then fold one half over the other half, keeping an angle of about 10° between the edges.

F IGURE i-8

Pouring a solid.

14 Introduction

FIGURE ¡-9

Folding filter paper into a cone.

Tear a small wedge o f f the smaller por t ion of the filter paper (see Figure i-9), open the paper into a cone, and place it in the tunnel. Wet the paper and press it f i rmly against the sides o f the funnel; having torn out the wedge wi l l enable you to form a seal at the fold of the filter cone to prevent air leakage behind the paper. The filter paper should be slightly below the r im of the funnel.

Decant the supernatant l iquid from the solid into the funnel by pouring it down a stirring rod (see Figure i-10). When you have filtered as much of the l iquid as you can using this method, transfer the solid using a stream of water from a wash bottle

FIGURE i-10

Gravity filtration.


and a stirring rod. Adding the solid last wi l l save you time in that most of the solution wi l l be filtered before the paper can become clogged by the solid. During the course of the f i l t ra t ion, the funnel should never be more than about three-quarters ful l .

Suction f i l t ra t ion requires a thick-walled flask wi th a sidearm, a Blichner funnel, thick-wallcd rubber tubing, and a trap (all shown in Figure i-1 1). The filter flask must be clamped, so that it wi l l not t ip over, i f the heavy vacuum tubing Hops around. Af ter clamping the flask, insert the Biichner funnel. Place a piece of filter paper in the bo t tom of the funnel; use a size that lies flat in the bot tom of the funnel wi thou t extending up the sides of the funnel. Connect suction tubing between the flask and the trap and between the trap and an aspirator. The trap wi l l prevent water from the aspirator from backing up into the filter flask if the water pressure drops. Wet the filter paper wi th a l i t t le solvent and press it t ightly against the bo t tom of the funnel. Turn on the aspirator and suck the solvent through the paper. Decant as much of the solution as you can into the funnel and draw it through the paper before adding the solid. After you add the solid, suction it unt i l it is as dry as possible. Always break the vacuum by disconnecting the tubing from the flask or from the trap; then turn o f f the aspirator. If you turn o f f the aspirator first, water may be drawn into the flask, where it w i l l dilute or contaminate the filtrate, and may even pass into the funnel.

C L E A N I N G G L A S S W A R E

For most laboratory purposes, it is sufficient to wash glassware w i t h detergent and tap water and to rinse wi th tap water and then w i t h distilled water. If a more

16 Introduction

FIGURE ¡-12

Clean and dirty glassware.

thorough cleaning is necessary, your instructor wi l l give you directions. The glassware is completely clean when water Hows smoothly down the sides of the vessel w i thou t collecting in droplets (see Figure i-12). If droplets collect, reclean the glass. High precision volumetric glassware (pipets, burets, volumetric flasks) should be thoroughly rinsed w i th tap water and wi th distilled water after each use and should always be clean when put away.

Scientific Measurement

Scientists are interested not only in knowing what happens during experiments but also in making quantitative measurements of natural events. For example, scientists were interested to learn that moon rocks contain both silicon and oxygen, but they also wanted to measure how much of each of these elements is usually present in the rocks. Do moon rocks contain eqtial amounts of silicon and oxygen, or is there far more of one element than of the other? When this kind of quantitative informat ion is available, comparisons between rocks on the moon and rocks on earth become possible. When copper and sulfur are heated together, it is obvious that they react to form a black compound. However, the chemist wants to know how many atoms of copper react w i th how many atoms of sulfur; that is, does the compound produced have the formula C u 3 S 2 , or CuS, or some other composition? This informat ion can be obtained by measuring the weights of copper and sulfur that combine to form the compound. Almost all experiments in the chemistry laboratory include quantitative measurements (weights, volumes, lengths, and so on) , and this section w i l l show you how to deal w i t h these numerical measurements.

A C C U R A C Y AND P R E C I S I O N

Accuracy and precision are important terms in discussing measurement. Accuracy refers to how well a measured value agrees w i th the true value. Precision

17 Scientific Measurement

refers to how well several determinations of one quant i ty agree wi th each other. Naturally, a scientist's goal is to make measurements that are both accurate and precise. However, it is possible to be quite precise wi thou t being at all accurate. For example, if the length of a table were measured several times, all of the values obtained might«*ie almost identical (let's say 3.13, 3.1 I, and 3.12 m) . But if the length of the measuring stick used was only 0.91 m ( 1 yd) when it was thought to be 1.00 m, these precise values would not accurately represent the true length of the table.

Several factors that lessen the accuracy of measurements are said to introduce systematic ox determinate errors (errors wi th in the system that can be determined and eliminated). One source of these errors—inaccurate equipment—has already been mentioned; two other sources are impure chemicals and inexact chemical procedures. Pure chemicals and tested procedures are provided in this laboratory. When necessary, equipment is calibrated to compensate for possible inaccuracies and minimize systematic errors. Therefore, the measurements you obtain should be reasonably accurate.

The precision of a measurement is affected by random or indeterminate errors (errors that are beyond the experimenter's control ) . Indeterminate errors include the effects of factors such as small temperature variations during the course of an experiment, the absorption of water by objects while they are being weighed, differences in judgment about the color change of an indicator, and the loss of extremely small amounts of material in pouring, fil tering, or other manipulations. Indeterminate errors may be just as apt to affect a measurement in a positive direction as in a negative direction, so a particular measurement may be either slightly too large or slightly too small. Hence, two or more measurements are usually made and their average is used in the hope that positive and negative random errors wil l cancel one .another. But random errors can never be completely eliminated, and perfect precision or reproducibi l i ty is never expected.

It must be noted here that, for the purposes of this discussion, we have assumed that your laboratory technique is good enough that you wi l l be able to work carefully, wi thout making such blunders as spilling half of a substance before you weigh it . In a first-year laboratory course, the process of developing a good experimental technique may make it more di f f icul t for you to obtain accurate and precise data, but this d i f f icul ty can be overcome by care and practice. However, no matter how good your laboratory technique is or how thoroughly systematic errors are eliminated, the existence o f random errors wi l l place l imits on the accuracy o f your measurements. Y o u should record the l imits placed on the accuracy of measurements whenever you report numbers.

Evaluation of Precision

As mentioned above, two or more measurements of a quanti ty are usually made. The average value of these measurements is then taken as the best estimate of the quantity being measured, since small random errors wi l l tend to cancel one another. Random errors are likely to be small, so we can have much greater confidence in our results if successive determinations of the same quant i ty deviate only slightly from the average. For example, three separate measurements of one side of a cubical box might be 10.32, 10.34, and 10.31 centimeters. The average of these three measurements is therefore

18 Introduction

10.32 10.34 10.31

3 )30.97 10.32 cm

But three other people might measure the box and obtain lengths of 10.39, 10.29, and 10.35 cm. The average value in this case would be 10.34 cm -a value that is very close to the first result but that is less precise. A quantitative indication of the precision of a set of data is the average amount by which individual measurements deviate from the average value of the set of measurements; this is called the average deviation. The calculation of the average deviation (a.d.) for each of these two sets of data is illustrated below:

Set I : Set II: a.d. a.d.

10.32 0.00 10.39 0.05 10.34 0.02 10.29 0.05 10.31 0.01 10.35 0.01

3 )30.97 3 )0 .03 3 ) 3 1.03 3 ) 0.1 1 10.32 cm 0 . 0 1 = a.d. 10.34 cm 0.04 = a.d.

The more precise data set (Set I) would be considered more reliable; in other words, it is probably more accurate.

Uncertainty and Significant Figures

Next, consider the actual process by which measurements are made, again using the example of measuring the length of a table wi th a meter stick. The stick is divided into 100 major divisions (centimeters), each of which is subdivided into 10 units (mill imeters). If the stick is placed along the table edge, the corner may be positioned between the 96- and 97-cm marks (see Figure i-13). An even closer estimate of the table's length can be obtained by noting that the corner falls between 96.7 and 96.8 cm; that is, the table is longer than 96.7 cm but is not as long as 96.8 cm. The length appears to be closer to 96.8 cm. A further refinement of the measurement can be made by mentally dividing the space between the 96.7- and 96.8-cm marks into 10 divisions and estimating where the corner falls. It falls about

Table

1 l l 1 1 i I I I I l l I I I I L

96 97

FIGURE i-13

Measuring the length of a table wi th a meter stick.

19 Sc ient i f i c Measurement

0.7 of the way between the two marks, so 96.77 em is the best measurement we can make of the table's length.

It is true that the last digit in the measurement 96.77 cm is somewhat uncertain. Someone else might estimate 6 rather than 7. Nevertheless, reporting this figure gives more information than simply saying that the length is 96.8 cm because the table's actual length is only close to (not exactly) 96.8 cm. We therefore report the table length as 96.77 cm. To be as honest as possible, the l imi t on the accuracy of this measurement should be indicated in reporting this figure. The distance between 96.7 and 96.8 cm can probably be estimated to about ±0.02 cm. This means that if the measurement were repeated under the same conditions, the last digit might be estimated to be 5, 6, 8, or 9 rather than 7. (This l imi ta t ion on reproducibil i ty exists because it is simply not possible to estimate the measurement any more closely wi th the unaided eye and a meter stick. If we wanted to measure the length any more accurately, we would have to use a caliper or a finer measuring instrument than a meter stick.)

Now we know that the length of the table is 96.77 ± 0.02 cm. We say that this measurement has four significant figures, four figures that tell tts how accurately the measurement was made; three of these figures are known exactly and the fourth is slightly uncertain because it was estimated. Generally, in any measured result that is reported, the last figure can be assumed to be estimated. Just how uncertain the final digit is depends on the measuring process and instruments employed. The closest estimate possible for the last digit of the table's length was ±2 . In the absence of information to the contrary, you can assume that the last digit is known to ±1 in figures reported in your text and in other scientific books and journals. Whenever you make any measurement in the laboratory (or anywhere else!), always try to place l imits on your own accuracy of estimation.

Scientific notat ion. Let's extend our consideration of the measured table length of 96.77 cm. We might also express this measurement as 0.9677 meter, as 967.7 millimeters, or as 9,677,000,000 Angstroms. T w o of these numbers contain more than four digits, but all three numbers are simply different ways of expressing the information obtained from one measurement—the length of a table-and not all of the digits are necessarily significant. Each of these three numbers contains only the three significant figures that were obtained exactly and the fourth significant figure that was estimated in the original measurement; the additional zeros are not significant. To write a number in a way that shows immediately how many figures are significant, we express the number in scientific notation; that is, we write it as a number between 1 and 10 mult ipl ied by some power of 10, and we retain only the significant figures. Thus, 0.9677 m would be wri t ten 9.677 X 10"' m; 967.7 mm = 9.677 X 10 2 m m ; and 9,677,000,000 A = 9.677 X 10 9 A. Even though three different sets of units are used here, all three numbers are the result of the same measurement and contain only the original four significant figures.

Now consider the number 0.000207. The zeros at the beginning of this number serve merely to locate the decimal point . The only three significant figures are 2, 0, and 7 (207). When this number is wri t ten in scientific notat ion as 2.07 X 1 0 " 4 , it is immediately obvious that it contains only three significant figures. Making the usual assumption that the uncertainty in the last digit is ± 1, this number has an accuracy of 1 part in 207.

20 Introduction

The number 1 200 g is a somewhat more confusing example. Does 1 200 g contain two, three, or four significant figures? The answer depends on the accuracy of the original measurement. Was the balance that was used accurate to ± l g ? If so, all four figures are significant: three digits are known exactly and the fourth digit is an estimate. This number is wr i t ten 1.200 X 10 3 g. But if the measurement is accurate to ± 10 g, then the number is 1.20 X 1 0 3 . And if the measurement is accurate to only ± 1 0 0 g, then the number is 1.2 X 1 0 3 . Scientific notat ion removes the ambiguity. You should use it in all your work to communicate informat ion about the accuracy of your measurements, just as all scientists do.

Calculations using significant figures. The results of many measurements are used in formulas to calculate other quantities. The use of significant figures is especially important in these calculated quantities, so that the proper degree of uncertainty in the result is shown. Suppose, for example, that a square table has a side s that is 96.77 cm long. Stated another way, the table length is known to 2 parts in 9677 or to roughly 1 part in 4800, if we assume ± 0.02 cm uncertainty in the measurement. If we wish to calculate the area of the table, we use the formula A = s 2 = (96.77 c m ) 2 = 9364.4329 c m 2 . On closer examination, however, this result does not make sense. Using all of the figures obtained in the mul t ip l ica t ion implies that we know the area to an accuracy of 1 part in 93,644,329 (roughly, 1 part in 94,000,000). But our knowledge of the length of the side of the table was accurate to only 1 part in about 4800. It is clearly impossible for us to know the calculated quant i ty more accurately than we know the measured quantities used in the calculation. Simply performing a mul t ip l ica t ion cannot increase the accuracy of an original measurement. So we cannot determine the area of the table to an accuracy greater than 1 part in 4800, or 9366 ±2 c m 2 . Reporting the area as 9366 c m 2

indicates the approximate accuracy to which this number is known on the basis of the original measurement.

This example illustrates the fact that we cannot simply perform arithmetic operations bl indly, wi thou t paying attention to the meaning of the numbers we use. In scientific calculations, we are not using mere abstract numbers but numbers that reflect our knowledge of a physical situation. Hence, in the above example, it was necessary to report the area using a specific number of significant figures ( four) , which made clear the accuracy to which we actually knew the measurement. Similar care must be exercised when any measurement is used to calculate other quantities. To maintain honesty about the uncertainty of calculated numbers, they must be reported to the proper number of significant figures that reflect the degree of accuracy of the original measurements. It is important to recognize that no calculated quantity can be known with more certainty than the least certain number used in the calculation. The fol lowing are guidelines for using significant figures in calculations.

1. For addition and subtraction: The term that is in error by the greatest por t ion of a unit l imits the accuracy of the answer. For example:

102.7 g 0.1 14 g 9.77 g

1 12.584 g= 1 12.6 g


The first quanti ty is known to only ±0.1 g, so the sum of the three quantities can be known to only this degree of accuracy.

2. For multiplication and division: When the units in which the answer is expressed and the units of the quantities used in the calculation are not the same, the number of significant figures in the answer is the same as the number of significant figures in the quanti ty having the fewest significant figures. For example:

1 1 7 ? - : m = 18.13 m/sec = 18 m/sec 65 sec

Counting. The kinds of measurements discussed thus far have all been of properties that can vary cont inuously-propert ies that can have any value. Measurements of this kind are subject to uncertainty, but there are some numbers that can be known exactly. The counting of discrete objects is a kind of measurement that can be made wi th complete accuracy and precision. Thus, in using such numbers in calculations, the uncertainty of the final result is determined by other measurements. For example, three people weigh 160.2, 175.0 and 180.9 lbs, respectively. What is their average weight?

160.2 lbs 175.0 lbs 1 80.9 lbs

3 ) 516.1 lbs \22-0 lbs = average weight

The average weight has four significant figures and is accurate to ±0.1 lb, as is each weight used in the calculation. The counted number 3 is exact and does not l imi t the accuracy of the result.

P R A C T I C A L H I N T S A B O U T L A B O R A T O R Y M E A S U R E M E N T

In making measurements in the laboratory, you wi l l often have more than one instrument at your disposal. For example, you could measure the volume of a solution using a buret or a graduated cylinder; the choice of which piece of equipment to use depends on the accuracy required of the measurement. A butcher doesn't weigh a steak to the nearest 0.0001 g on an analytical balance; steak is weighed to the nearest 0.01 lb, because-even at S2.99/lb—the l i t t le bit of extra money the butcher might collect from a more accurate weight would not justify the extra time required or the extra expense of a more accurate balance. You should be just as economical of your time and effort in the laboratory. Do not waste time making measurements w i th instruments that are far more accurate than required.

Two important kinds of measurement are volume and weight. The following table indicates the degree of accuracy that it is possible to attain w i th various instruments for measuring volume and weight.

22 Introduction

Measurement Available Instruments Instrumental Precision

Volume 5-ml pipet ±0.01 ml Buret ±0 .02 ml 10-ml graduated cylinder ±0.1 ml 100-ml graduated cylinder ±1 ml

Weight Analyt ical balance ± 0 . 0 0 1 , or better* Triple-beam balance ±0.01 g

•Depends on type of balance.

In first-year chemistry laboratory work, it is important to achieve the greatest possible accuracy in measuring your results, but not every measurement in a particular experiment has to be made wi th maximum accuracy. As an example, consider the reaction for the synthesis of zinc sulfide

A common laboratory experiment is to determine the formula of this sulfide - tha t is, to determine the numerical values of the subscripts x and y—by reacting a weighed quanti ty of zinc wi th sulfur and determining how much zinc sulfide is produced. The experiment consists of heating an accurately weighed amount of zinc wi th an excess of sulfur. No specific amount of sulfur must be used; the amount must simply be known to exceed the amount required for the reaction. Using more sulfur than is needed to react wi th the zinc ensures that all of the zinc wi l l react; the remaining sulfur wi l l be converted to gaseous sulfur dioxide and wi l l escape from the reaction vessel. The amount of sulfur in the resulting compound can be found as follows: 1.571 g (weight of zinc sulfide) - 1.052 g (weight of zinc) = 0.5 19 g (weight of sulfur). Thus, the amount of sulfur actually present in the compound can be accurately calculated. For this reason, it would be pointless to weigh the reactant sulfur to the nearest 0.001 g. Since we only need to be sure that an excess—say, 2 g—of sulfur is present, weighing the reactant on a triple-beam balance to the nearest 0.1 g is the only measurement necessary.

Before making any measurement, always ask yourself how accurate it must be to give the result of the required accuracy in the least amount of time using the least amount of effort . When in doubt , consult your instructor.

V O L U M E M E A S U R E M E N T S

You wi l l often need to measure the volumes of liquids in your laboratory work . Several vessels may be available, and their use wi l l depend on the accuracy o f the measurement required.

Graduated Cylinders

Rough estimates of volume can be made using beakers wi th graduation marks, but these measurements have a very low accuracy. If accuracy is important , a graduated cylinder should be used.

xZn(s) + vS(s)


F IGURE ¡-14

Measuring volume wi th a graduated cylinder.

A 10-ml graduated cylinder has graduation marks corresponding to one mi l l i l i te r . Each large division of 1 ml is divided into five smaller divisions of 0.2 ml each. The position of the bot tom of the l iquid meniscus gives the volume reading. In Figure i-14, the meniscus lies between 7.6 ml and 7.8 m l . Obviously, the volume is more than 7.6 ml but less than 7.8 ml . It is customary to divide the distance between the marks mentally into smaller divisions and then to estimate where the meniscus lies along this scale. But because the 0.2-ml divisions on the 10-ml graduated cylinder are already so close together, you wi l l probably find i t possible only to divide the intervening distance into halves rather than in to fifths or tenths; that is, your eye wi l l probably be able to distinguish a distance that corresponds to only half the length between each of the 0.2-ml divisions. Any volume from about 7.65 ml to 7.75 ml would be indistinguishable to your eyes, and the meniscus would look like it was about halfway between 7.6 ml and 7.8 m l . Your reading would therefore be 7.7 m l ; the second digit is estimated. Since any " t rue" volume from 7.65 ml to 7.75 ml would look the same to you , there is ah uncertainty of ±0.1 ml in the volume reading of 7.7 m l . It is customary to report the volume as 7.7 ± 0.1 ml , or simply as 7.7 m l , w i t h the understanding that an uncertainty of 1 exists in the second digit . Thus, the volume of the l iquid is know to 1 part in 77, or to about 1.3%. This measurement has two significant figures; one that is known exactly and one that is estimated. Perhaps this seems like a lo t of at tention to devote to such a simple measurement, but the l imi ta t ion on your abi l i ty to estimate the second digit is a fundamentally important concept in making any laboratory measurement. This random error is impossible to eliminate even w i t h practice, because it arises from the l imitat ions of your eyes and the equipment.

Volumetr ic Glassware

It is possible to make much more accurate volume measurements in the laboratory by using volumetric glassware—pipets, burets, and volumetric flasks. The construction of this type of glassware removes some of the l imitat ions on accuracy that graduated cylinders present. Volumetr ic vessels are more time-consuming to use

24 Introduction

and clean than graduated cylinders. As an example of the greater accuracy that is possible when you use volumetric glassware, a 5-ml pipet wi l l deliver 5.00 ± 0.01 m l of l iqu id . The volume is known to three significant figures. The uncertainty of ± 1 in the third figure means that the volume measurement is accurate to 1 part in 500, or to 0.2%—a measurement that is almost 10 times more accurate than the same measurement made w i t h a graduated cylinder. Pipets and burets are used to measure the volumes of liquids that are transferred into receiving vessels and are stamped " T D " (" to deliver"). Volumetr ic flasks are used to prepare solutions of precisely known volumes and are stamped " T C " (" to conta in") .

Pipets. A pipet (see Figure i-15) w i l l deliver a very accurately known quant i ty of l iquid . Liquid is drawn into the tube by suction unt i l its level is above the marking around the stem. Y o u should use a bulb for suction. Then the l iquid is

Control liquid level with index

25 Sc ien t i f i c Measurement

allowed to drain out of the pipet unt i l the bot tom of the meniscus just touches the calibration line; the flow is controlled by placing your index finger [not your thumb) on the top of the pipet. When the meniscus level has been adjusted, the tip of the pipet is touched to the side of a waste beaker to remove any drop of clinging l iquid and then to the wall of the vessel into which the l iquid is to be transferred. Your finger is then removed from the top of the pipet to permit the contents to flow out the t ip. Best results are achieved when the pipet is held vertically and the receiving vessel is tipped. When the l iquid has stopped f lowing, wait another 30 seconds for any remaining l iquid to drain down the sides o f the pipet. Then roll the pipet against the side of the receiving vessel so that an additional drop flows out. Some l iquid wi l l remain in the pipet. This is all right, because the pipet is calibrated to deliver the proper volume of l iquid when these few drops are retained.

To be certain that the pipet delivers an accurate volume, the l iquid must flow smoothly and completely down the walls of the pipet and not collect and adhere to the walls in droplets. Therefore, the pipet must be thoroughly cleaned. After the pipet has been cleaned and completely rinsed wi th distilled water, it should be rinsed three times w i th small portions of the solution to be transferred to make sure that no water remains in the pipet to dilute the solution. After you have finished using a pipet, rinse it several times wi th distilled water so that it can be put away clean.

Burets. Cleaning and filling the buret. Before using a buret, it should be thoroughly washed wi th a buret brush and detergent, rinsed wi th tap water, and then rinsed w i th three portions (about 10 ml each) of distilled water. Like the pipet, the buret should be cleaned unti l l iquid drains smoothly down its sides and does not collect in droplets on its walls. Next, grease the stopcock by placing a small ring of grease around it-on either side of the hole (make sure that no grease clogs the hole); burets w i t h Teflon stopcocks do not need to be greased.

Then rinse the buret w i t h a small amount of the solution that is to be put into it to remove any remaining water so that the solution wi l l not be di luted. Close the stopcock and pour about 5 ml of the solution into the buret. Hold the buret horizontally and roll it so that the solution washes over the entire inside wal l . Then drain the solution out of the stopcock, discard i t , and repeat this procedure two more times. Clamp the buret in place and fi l l it w i t h the solution. Open the stopcock and drain out solution unt i l the t ip of the buret is filled and does not contain an air bubble. If necessary, add more solution so that the buret is filled to just below the 0.00 mark.

Reading the buret. In any volume measurement, you should read the position of the l iquid meniscus—not the point at which the solution comes into contact w i t h the glass. Each graduated mark on the buret corresponds to 0.1 ml . As usual, you are to estimate visually the fractional distance between two of these marks to obtain a reading that is accurate to ±0 .02 m l . The position of the meniscus in Figure i-16 is 18.36 m l ; four significant figures are obtained.

It is important to eliminate the uncertainties in your readings that result if you view the meniscus from various angles. Note that the meniscus seems to change position, depending on whether you look at it from above or below (again, see Figure i-16). To make an accurate measurement, the eye must be level w i th the meniscus. To ensure this, look at the buret so that the back of the ml graduation

Introduction


mark nearest the meniscus is superimposed on the front of this same mark, and your eyes see only a single line. Read the meniscus level from this position.

Operating the buret. When you begin to use a buret, you may find it d i f f icul t or awkward to master the proper technique. However, i t w i l l pay you to exert the effort required to use a buret properly. To achieve the best results, operate the stopcock w i t h your left hand (see Figure i-17). Grasp the tip l ightly between your third and fourth fingers and control the flow of l iquid by turning the stopcock wi th your thumb and index finger. The flask in to which the solution flows should be

F IGURE i 18

Making up a standard solution using a volumetric flask.


swirled w i t h your right hand. If you are left-handed, you wi l l probably prefer to operate the stopcock wi th your right hand and to swirl the tlask w i th your left hand.

Before beginning any t i t ra t ion , touch the buret tip against the side of a waste beaker to remove the drop of l iquid that may be clinging to i t .

After running solution out of a buret, wait a minute before making a volume reading, so that the l iquid remaining on the upper walls has time to drain down.

Volumetr ic flasks. Volumetr ic flasks (see Figure i-18) are used to prepare solutions of precisely known molar i ty . For example, 0.100 mole of NaCl may be put in to a 1.00-liter volumetric flask and the flask filled wi th water to the calibration mark giving 1.00 liter of solution so that the concentration is known to be 0.100/V/. The solid that is to be dissolved must be very carefully weighed, but it is not necessary to measure the amount of solvent that is used; instead, the final volume of the solution is known precisely. After you put the solid into the tlask, use a l i t t le of the solvent to wash out the container in which the solid was weighed and add the washings to the contents of the tlask (be sure that none of the carefully weighed solid is lost). Add a l i t t le more water and swirl the flask to dissolve as much of the solid as possible before the tlask is completely filled wi th water. Then add more water unt i l the meniscus just touches the calibration line ( i t is helpful to use a medicine dropper to add the last few mil l i l i ters) . Stopper the tlask and mix the solution by repeatedly inverting and uprighting the flask (while holding your hand t ight ly over the stopper!).

Like pipets and burets, volumetric flasks must be scrupulously clean, especially in the area of the neck above the calibration line. If the neck of the tlask is d i r ty enough to allow water to collect in droplets, an unknown quantity o f water wi l l remain on the neck when the solution is being prepared, and the volume of solution wi l l not be the calibrated capacity o f the tlask. Consequently, the molari ty o f the solution wi l l not be accurately known. Al though the tlask must be clean, it does not have to be dry if you are using water as the solvent; just remember to rinse the flask wi th distilled water. When you are finished wi th the tlask, pour out the solution and rinse the tlask w i th tap water and then several times with a l i t t le distilled water. The tlask should be stored w i th the stopper in place so that dirt cannot get into i t ; the tlask does not have to be stored dry if there is only distilled water in i t .

MAKING MASS M E A S U R E M E N T S

You wi l l need to make many weighings to determine the masses o f the materials you wil l use in laboratory work. Because the instruments used for weight measurements require more instruction and more practice to master than the instruments used for volume measurement, the first experiment in this manual is devoted to weighing.

Experiment 1 The Use of Balances: Hydrated Compounds

Introduction

According to the Law of Constant Composit ion, any sample of a pure compound contains the same elements in the same proportions. For example, every sample of cobalt chloride (CoCl 2 ) , consists of 45.39% cobalt and 54.61% chlorine on a weight basis. However, this compound, like many others, also exists in a form known as a hydrate; the crystalline form characteristic of the solid contains water as well as cobalt chloride. The amount of water in the crystal may vary for some hydrated compounds. Some solids absorb water when they are in contact w i t h moist air; in contrast, water may evaporate from some hydrated crystals when they are in contact w i t h dry air. However, many compounds contain a definite proport ion of water. For example, the composition of hydrated cobalt chloride is represented by the formula CoCL - 6 H 2 0 , which means that the crystal contains six molecules of water for each C o C l 2 unit . On a weight basis, the crystal consists of 24.77% cobalt, 29.80% chlorine and 45.43% water. The water can be removed from this hydrate (and from many others) by heating it to a temperature of 100°C or higher.

The composit ion of a compound can be determined by carefully weighing each of the elements-that combine to form it or each of the substances into which it can be decomposed. In this experiment, you wi l l use a form o f the latter procedure to determine the percentage o f water in a hydrate. First you wi l l weigh the hydrated compound, and then you wi l l weigh the anhydrous compound that remains after the water has been removed from the solid by heating i t .

Both a triple-beam balance and an analytical balance w i l l be used in this experiment. Each of these balances has its own advantages. Much more accurate measurements ( to ±0 .0001 g) can be made on the analytical balance. The smallest graduation on a triple-beam balance is 0.1 g or 0.01 g, and it is d i f f icul t to estimate weights any more closely than this. However, the analytical balance has a more l imited total capacity-usually 160 g in comparison to 610 g for most triple-beam balances.

The triple beam balance is more convenient to use when you want to weigh out a particular amount of a substance. You can weigh a container, adjust the weights for the proper total weight of the container and the substance, and then pour the substance into the container unt i l you have the desired total weight.

A triple-beam balance is a rugged instrument. (Some are practically indestruc-table!) In contrast, an analytical balance can be easily damaged by misuse. If damage occurs, the top of the balance may have to be removed to enable your instructor or a technician to make an adjustment; even if the damage is not extensive, this wastes valuable laboratory time. More serious damage to the balance can be quite expensive to repair and w i l l deprive other students of the use o f the balance unt i l i t can be repaired. Even when damage is not immediately apparent, general misuse can

29

30 E x p e r i m e n t 1

gradually lessen the sensitivity of an analytical balance. Consequently* un t i l you are completely proficient in the proper use of the analytical balance, it is advisable to make preliminary weighings on the triple-beam balance so that you wi l l be less l ikely to misuse the analytical balance.

B R I E F B A L A N C E INSTRUCTIONS

Analytical Balance

Your instructor wi l l provide detailed instructions for the particular type of balance that you are using. A few general comments are given here.

1. The balance must be level when it is in use. If the leveling bubble indicates that the balance is not level, ask your instructor to make the necessary adjustment. Do not move or bump the balance. Do not lean on, bump, or move the table on which the balance is resting.

2. Always check and set the zero point of the balance each time you use it. This ensures that the balance records a weight of zero when there is no object on the balance pan.

3. Be sure tliat the balance is arrested, both when you place the object on the pan and later when you remove the object from the pan.

4. The object to be weighed should be clean and dry, so that its weight w i l l not change as a result of loss of di r t or moisture. This also protects the balance from damage that might be caused by moisture. If a solid chemical is to be weighed, it should be placed in a bott le, a beaker, or a metal weighing pan, and not on a piece of paper. If a l iquid is to be weighed, it should be in a closed container so that vapor cannot escape and possibly corrode parts of the balance. If a very accurate measurement of the weight of the object is required, handle the object w i th tongs or a test-tube holder or by a piece of paper wrapped around i t . Otherwise, your hands might deposit moisture, d i r t , o i l , salts from your skin, and so on, on the object, changing its weight.

5. Be sure that the arrest-release control of the balance is in the proper position before you dial the weights.

After recording your weighing, arrest the balance, return all weight settings to zero, remove the object, and close the doors of the balance case.

Triple-Beam Balance (Figure 1-1)

Place the object on the balance pan. Be sure that all sliding weights are set at their zero marks. First, locate the beam that has 100-g graduations and move the weight alone this beam to the right one notch at a time unt i l it causes the beam to

31 The Uses of Balances: Hydrated Compounds

F IGURE 1-1

Triple-beam balance.

drop. Then move the weight back to the left one notch; the beam wi l l rise. Second, move the weight along the beam that has 1 0 - g graduations unt i l the beam drops; move the weight back to the left one notch. Finally, move the weight along the beam that hasT-g graduations unt i l the scale is in balance. The pointer should move just as far upward as it does downward from zero; you do not need to wait for the pointer to come to rest. Read the weight to the nearest 0.1 g. Return the weights to zero, remove the object, and clean up any material spilled on or around the balance.

Your first step in weighing an object on an analytical balance is to check and set the zero point . On a triple-beam balance, this procedure can be fairly tedious and is usually omit ted unless your weighing must be as accurate as possible. You w i l l often weigh "by difference" on the triple-beam balance; that is, a container is first weighed empty, then it is weighed wi th a chemical in i t , and the weight of the chemical is the difference between these two weights. As w i th the analytical balance, the triple-beam balance must register a weight of zero when there is no object on the pan or the weighing wi l l be incorrect. However, each weighing wi l l be in error by the same amount, and the difference between the weights of an empty container and the container w i th the chemical w i l l give the correct weight of the chemical.

When you compare results obtained by weighing the same object on each type of balance, remember that the triple-beam balance may be in error by as much as several tenths of a gram if the zero point was not set.

E Q U I P M E N T N E E D E D

analytical balance Bunsen burner asbestos square clay triangle beaker tongs porcelain crucibles and lids

32 Exper iment 1

ring triple-beam balance ring stand

C H E M I C A L S N E E D E D

unknown hydrated compound

P R O C E D U R E

Wash and dry two porcelain crucibles and their lids. Record the numbers that are on the crucibles and lids; identify one set for Tr ia l 1 and the other for Trial 2 (see Report Sheet 1 on page 35). Support one of the crucibles and its l id on a clay triangle (see Figure 1-2), and heat them wi th the full heat o f a Bunsen burner for at least 10 minutes to remove any absorbed volatile or combustible material. If a soot deposit forms, burn it o f f completely. A l l o w the crucible and the l id to cool for about one minute on the triangle. Then use tongs to transfer them to a clean piece of asbestos and allow them to cool to room temperature. Cooling should take about 15 minutes, but it w i l l be more rapid i f the l id is removed from the crucible. Heat the second crucible and l id while the first crucible and lid are cooling.

Weigh the cooled crucible and l id on the triple-beam balance and then on the analytical balance. Remember to handle them wi th tongs. If they appear to be changing in weight during the measurement on the analytical balance, they are still

33 The Uses of Balances: Hydrated Compounds

warmer than room temperature; immediately remove them from the balance and wait 5 additional minutes before repeating the measurement.

Return to the triple-beam balance and weigh about 1-1.5 g of the hydrated compound into the crucible. Then use the analytical balance to weigh the crucible, l id , and hydrated compound as accurately as possible. Weight a similar sample of the hydrate into the second crucible (after it has been heated, cooled, and weighed). Be sure that you correctly identify each sample using the numbers on each crucible and l id .

Support the first crucible on a clay triangle wi th the l id very slightly ajar to permit water vapor to escape. Heat the crucible using the lowest possible flame; hold the burner in your hand so that you can remove it quickly if necessary. A puddle of water may form around the solid; gentle heating is required to ensure that the solid does not splash out of the crucible as this water boils away. Even if there is no apparent water loss, the hydrate may crackle and bounce around (like popcorn); again, gentle heating and a nearly closed lid are necessary to prevent escape of solid material from the crucible. You may notice that the compound changes color or changes from a crystal to a powder; on the other hand, there may not be any obvious change in its physical properties. When there is no longer any splashing or spattering, place the burner on the desk top and continue to heat the crucible w i th a low flame for a total of about 10 minutes. If the compound is heated too strongly, some decomposition other than just the loss of water may occur or the compound may react wi th oxygen that is in the air. A l l o w the crucible to cool to room temperature and then weigh it on both balances. During the cooling and weighing, the crucible must be covered to prevent the anhydrous compound from reabsorbing water from the air.-Repeat this procedure w i th the second crucible.

R E S U L T S AND D I S C U S S I O N

Review the introductory section on accuracy and precision in scientific measurement (page 16).

The difference between the weight of the crucible, l id , and hydrate and the weight of the crucible, l id , and anhydrous compound is the weight of water that was lost during heating. Express this weight as a percentage of the total weight of the compound that was originally present. Record all these data on the Report Sheet for this experiment. Be sure to retain the proper number of significant figures in your answer; express your result in scientific notation if the degree of accuracy would be ambiguous wi thout i t .

Q U E S T I O N S

1. How many significant figures are there in the result you obtained for the percentage of water in the hydrate, using weights measured on the triple-beam balance? Using weights measured on the analytical balance?

2. Imagine that your sample of hydrated compound weighed 137.2306 g. From the result you obtained using the analytical balance, calculate the weight of water in this sample. Be sure to retain the proper number of significant figures. If you

34 Experiment 1

had measured the sample weight and water loss for this sample using the triple-beam balance, what would these weights have been 9 What would your calculated value be for the percentage of water in the hydrate, using these triple-beam balance weights?

3. How would your experimental result for the percentage of water in the hydrate compare to the true value if you made each of the following errors? (a) Some of the hydrated compound is spilled after it is weighed. (b) Solid spatters out of the crucible during the heating. (c) The anhydrous compound absorbs moisture from the air while it cools. (d) Some of the anhydrous compound is spilled before it is weighed.

R E F E R E N C E

Yoder, C.H., Suydam F.H. , and Snavely, F .A. Chemistry. New York : Harcourt Brace Jovanovich, Inc., 1975. Chapter 2.

R E P O R T S H E E T

E X P E R I M E N T 1

Name Date

Section

T B B * A B * *

Note: A l l numbers must be reported with proper units attached.

Trial 1:

Weight of crucible and l id

Weight of crucible, l i d , and hydrate

Weight of hydrate

Weight of crucible, l i d , and anhydrous compound

Weight of anhydrous compound

Weight of water lost

Percentage of-water in the hydrate

Tria l 2:

Weight of crucible and lid

Weight of crucible, l i d , and hydrate

Weight of hydrate

Weight of crucible, l i d , and anhydrous compound

Weight of anhydrous compound

Weight of water lost

Percentage of water in the hydrate

Average of the two trials:

Percentage of water in the hydrate

* T B B = triple-beam balance. * * A B = analytical balance.

(Answer the questions from pages 33-34 on the back of the Report Sheet.)

E X E R C I S E S

1. How many significant figures are there in each of the fol lowing numbers? If the accuracy is ambiguous, state the min imum and the maximum number of significant figures that may be present.

152.36 0.0032 20.000

127.210 62000 0.0870

2. Express each of the above numbers in scientific notat ion.

3. Carry out each of the fol lowing calculations, retaining the proper number of significant figures. Assume that all numbers represent measured quantities. Express each answer in " n o r m a l " form and in scientific notat ion.

(a) 628.2 + 0.761 + 2.05

(b) 215.07 - 19.7623

(c) (2.18 X . • l ' 0 " s ) + (7.41 X 10" 6 ) + (8.54 X 10~ 4 )

(d) 2476 + 0.1207 + 0.0214 + 0.0081

(e) 417 X 2.1

( 0 6 2 8 . 7 1 8 - 8 . 7 2

(g) 31 - 0.0671

(h) (5.16 X 10 4 ) X (2.1 X 10~ 2) X (1.07 X 10~ 3 )

(i) 4 . 0 ^ ( 6 . 0 2 3 X 1 0 2 3 )

(j) 0.5761 X 0.0821 X 373 * 0.976 * 0.2824

(k) 12.01 1 15 + (1.66 X 1 0 " 2 4 ) - (6.023 X 1 0 2 3 )

( Use the back of this page for your calculations, if necessary.)

37

Experiment 2 Properties of Substances

Introduction

Chemists study chemical reactions; that is, they seek to learn about the nature of processes that convert one substance into another (for example, iron to rust, or lysergic acid to LSD). Several facets of chemical reactions can be investigated; how fast they occur; whether they evolve heat; whether they produce a large or a small amount of the desired product; and so on.

The first requirement for studying reactions is to recognize when one has occurred. Sometimes this is easy because the appearance of the product is different from the appearance of the reactants. For example:

copper metal + sulfur • copper sulfide (copper color) (yellow) (black)

But there are many reactions, especially in organic chemistry, in which the reactants and the products have the same appearance (for example, they may all be white powders). The chemist is then faced wi th the problem of how to tell whether mixing the reactants has actually produced a new substance. In other words, the chemist must be able to identify pure substances.

The physie-al properties of a substance are usually measured to identify i t . Some of the properties normally considered are;

1. Color. 2. Melting point: The temperature at which a substance changes from a solid to

a l iqu id . 3. Boiling point: The temperature at which the vapor pressure equals

atmospheric pressure. 4. Density: Mass per unit of volume. 5. Solubility: How much dissolves in a given quant i ty of a particular l iquid .

Several of these physical properties are measured to identify a substance. For example, suppose that a compound is a colorless solid w i th a melting point of 9 5 ° C and a boil ing point of 3 4 7 ° C . This information permits us to determine that the substance is benzil. Of course, many compounds are colorless crystals. Several of these colorless crystal compounds probably have melting points around 9 5 ° C , and more than one of these compounds may boil at 3 4 7 ° C . But it is quite unlikely that two compounds that are colorless crystals and that melt at 9 5 ° C and that boil at 3 4 7 ° C exist. The combination of these three properties leaves no doubt that the compound is benzil. (Properties other than melt ing and boil ing points can be used to identify substances. Y o u w i l l learn about some o f these in later experiments.)

It is most meaningful to measure physical properties that have numerical values, because there can be no doubt about the result (assuming that the l imits on the

39

40 Experiment 2

accuracy of the measurement are clearly stated). In contrast, the description of a property like color can be quite ambiguous; what one person describes as " 'p ink," another might call "rose," or " l ight red," or even "cora l . " Nonetheless, such a qualitative and subjective description sti l l has value; at least we know that the " p i n k " substance has color and it is not black or green!

Thus far, a very important aspect of the study of chemical reactions has been neglected: a reaction can produce several products; it does not always produce just one. When more than one product is produced, the chemist must separate these products from one another to identify them. This is generally accomplished by subjecting the mixture to dis t i l la t ion, crystallization, sublimation, or chromatography. After the components have been separated from one another by one of these processes (some o f which you w i l l use in later experiments), the substances can be identified by their physical properties.

In this experiment, yot i wi l l measure some o f the physical properties o f pure substances and mixtures o f substances. You wi l l then use these measured values to identify an unknown solid and an unknown l iquid . This experiment is designed to give you experience in (1) making accurate numerical measurements of the physical properties of substances and (2) ident ifying pure substances by their physical properties.


beakers cork borer rubber bands Bunsen burner paring knife spatula capillary tubes 5-ml pipet test tube (25 X 200 mm) copper wire ring thermometer cork ring stand wire gauze


l-chloro-4-nitrobenzene naphthalene methanol unknown l iquid (from list on page 45) mineral oil unknown solid (from list on page 45)

P R O C E D U R E

A. Liquid Density

Density is defined as mass per unit of volume. Measurements of density can be quite simple to make. You merely take an accurately known volume of a substance and determine its mass by weighing i t .

A pipet or a buret is usually used to obtain accurate volume measurements of liquids. In this experiment, you wi l l use a pipet that delivers 5.00 ml of l iqu id . (Before proceeding, review the discussion of volume and weight measurements on page 21 and the instructions for cleaning and using pipets on page 24.)

41 Properties of Substances

Cover a 50-ml beaker wi th a watch glass and weigh it on the analytical balance. Return to your desk w i th the beaker, and use the pipet to transfer 5.00 ml of the unknown l iquid into the beaker. Reweigh the filled beaker and watch glass. The weight of the l iquid is the difference between the weights of the filled beaker and the empty beaker wi th watch glass.

Dispose of the l iquid by pouring it into a waste jar; dry the beaker and repeat this density determination. The weight of the second sample should agree almost exactly w i th the weight of the first. If it does not , you can conclude that you are pipeting or weighing improperly, and you should ask your instructor for advice.

B. Boiling Point

Cautions: Work under the hood. The vapors of your unknown liquid may be toxic. Most of the unknowns are flammable; never pour them near a flame. Use the lowest flame possible to avoid the possibility of igniting the vapors as they leave the test tube. Do not leave your burner on after you have heated the hath.

1. K n o w n compound. The purpose of working wi th a pure l iquid of known boiling point is to enable you to assure yourself that your technique is correct. When you are confident that you can make this measurement accurately, proceed to Part 2 and collect data on your unknown l iquid .

Set up a ring and place a 250-ml beaker half-filled wi th water on wire gauze so that you can heat it wi th a burner. Clamp a test tube so that it extends into the water bath. A d d about 5 ml of methanol ( C H 3 O H ; boil ing point 6 5 ° C ) to the test tube; be sure that the water level in the bath is above the l iquid level in the test tube. Using a cork borer, bore a hole in a no. 10 cork into which you can fit a thermometer. Cut a wedge out of the cork w i th a paring knife (see Figure 2-1). At tach a capillary tube (wi th about half the length broken off ) to the thermometer, open end down and closed end up. Place the vented cork containing the thermometer in the tube, so that the thermometer bulb is immersed in the l iquid . (Your apparatus should look like that in Figure 2-2.)

As you begin to heat the water bath, watch what happens in the capillary tube. It is ini t ial ly filled wi th air at atmospheric pressure. As the bath is heated, the air (like any gas) expands and begins to escape from the open end of the tube as tiny bubbles. Vapor from the sample also rushes into the tube, expelling even more air. When the l iquid starts to boi l , the capillary tube provides a surface on which bubbles of vapor can fo rm; it acts as a boiling chip.

fESK"v'"'

FIGURE 2-1

Cork with vent.

HI Experiment 2

Remember that the boil ing point is defined as the temperature at which the pressure exerted by the vapor of the l iquid equals the atmospheric pressure on the surface of the l iquid . For the vapor bubbles to rise to the l iquid surface—against the downward force of atmospheric pressure and the pressure exerted by the l iquid that is above the bubbles-they must exert a pressure that is slightly greater than atmospheric pressure. Consequently, the temperature at which the l iquid boils may be slightly higher than its defined boil ing point .

Since the vapor pressure of the l iquid when it starts to boil is somewhat higher than atmospheric pressure, the vapor present in the capillary tube wi l l force l iquid out o f the tube and the l iquid level wi l l be higher in the test tube than it is in the capillary. At this point, turn o f f the burner and stop heating the bath. The temperature wi l l begin to fall approaching the boiling point of the l iquid from above. When the liquid's boil ing point is reached, the pressure due to the vapor in the capillary wil l be equal to the pressure o f the atmosphere; the l iquid levels wi l l be the same in the capillary and in the test tube. Record this temperature as the boiling point of the l iquid .

From the defini t ion of boil ing point , it can be seen that the temperature at which a l iquid boils depends on the pressure of the atmosphere. In tables, the boi l ing points of substances are given for a pressure of 760 torr; this is called the normal boiling point. If a boil ing point (b.p.) is measured at some pressure other than 760 torr, the pressure at which the boil ing point was determined is reported (for example, b.p. l 0 2 ° C / 6 0 0 tor r ) . You are measuring boiling points at a pressure that is probably not exactly 760 torr ; read the correct atmospheric pressure from a barometer.

2. Unknown compound. Use the procedure described in Part 1 to determine the boiling point of your unknown l iquid . Repeat the procedure using a fresh capillary tube. This measurement combined wi th your density measurement should enable you to identify the unknown l iquid from the list on page 45. Be sure to record the identification number for the unknown.

C. Melting Point

I. Pure, known compound. First, you are to determine the melting point of a known compound to assure yourself that your technique is accurate enough to reproduce a known result. As before, this should make you confident about the accuracy of your measurement on the unknown.

Pour about 1 5 ml of mineral oil into a 25 X 200 mm test tube that is clean and dry. Clamp the tube so that you can heat it wi th a burner.

Make a ring stirrer from a long piece of copper wire. Loop one end and bend the wire so that the plane of the loop is perpendicular to the rest of the wire. Dr i l l a small hole into your vented no. 10 cork and insert the stirrer.

Place a small sample of naphthalene ( C , 0 H S ; melting point 7 9 - 8 0 ° C ) on a piece of paper and crush it to a fine powder using a spatula. Heap the powder and scoop up some of it by plunging the open end of a capillary tube into the heap. Hold the tube upright and tap it on the desk top unt i l the powder is shaken to the bo t tom of


a

I I i i^-^—Thermometer

5 s i ^ r r:., •a——1__ \̂ _s ——"Rubber band

ci i j on T!j Inverted

- I capillary tube

F IGURE 2-2

Determining the boiling point of methanol.

the tube, where it should form a t ightly packed layer. Repeat this process unt i l the sample layer is about 5 mm deep.

Attach the capillary tube to a thermometer wi th a rubber band (see Figure 2-3). Carefully position the capillary tube end alongside the thermometer bulb. Be sure that the rubber band is well above the l iquid level.

Begin heating the mineral oil gently; for accurate results, the temperature should increase about 2° per minute. When you know the expected melting point for a substance, you can heat it fairly rapidly to a temperature about 20° below the melting point and then reduce the heat. During the heating, continually stir the o i l bath w i t h the ring stirrer. Stir gently so that you do not bump the thermometer or the capillary tube. Your sample should melt w i th in ±2° of 7 9 - 8 0 ° . If i t does not, repeat the determination and heat the bath more slowly. You should notice that a pure substance melts over a small temperature range of 0 . 5 - 1 . 0 ° C . The sharpness of its melting point (m.p.) is one way of telling that a substance is pure.

2. Mixtures. Crush well and thoroughly mix some l-chloro-4-nitrobenzene ( C 6 H 4 N O , C I ; m.p. 8 3 - 8 4 ° C ) w i th your naphthalene and determine a melting point for this mixture . Note that although the separate pure substances melt at very similar temperatures, the mixture melts at a lower temperature and over a much wider temperature range. The lowering and spreading of the melting point can also be used in compound identif ication. A material that is thought to be substance A can be mixed w i t h a . k n o w n sample of substance A. I f t h e melting point of the mixture is the same as the melting point of pure A, the unknown material is indeed substance A, for only pure A would have the correct, sharp melting point. If the unknown material is not substance A , the mixture wi l l melt over a fairly wide temperature range and at a temperature different from that of pure A (usually lower, but occasionally higher). This proceudre is known as taking a mixture melting point.

3. Unknown compound. Obtain an unknown solid from your instructor. Determine its melting point and identify the compound from the list of substances on page 45. (Addi t ional data of this type may be found in the Handbook of

44 Experiment 2

FIGURE 2-3

Determining the melting point of a solid.

Chemistry and Physics, published annually by the Chemical Rubber Company.) Samples of various solids are available in the laboratory, so that you can take mixture melting points to assist you in identifying unknowns. In general, you can expect your melt ing point data to agree wi th in ± 2°C of the literature. Be sure to record the identif ication number of your unknown. Dispose of mineral oi l by pouring it back into the original container or into the waste jar as your instructor directs.

D A T A A N D R E S U L T S

The density of a substance is defined as weight per unit of volume. Use the weights you measured for 5.00-ml volumes of your unknown l iquid to calculate its density for each of the two trials. Be sure to report your results to the correct number of significant figures. (First , review the discussion of the meaning and the use of significant figures in the in t roductory section on scientific measurement on pages 16-2 1.)

Use your density and boil ing point data to identify your l iquid unknown. Identify yotir unknown solid on the basis of its melting point and its mixture

melting points. Record your data and conclusions on the Report Sheet for this experiment.


Substance in.p.(°C) b.p.CO Density(g/ml)*

acenaphthene 95 278 1.024 acetanilide 114 305 1.21 acetoacetanilide 85 decomposes 1.103 benzoic acid 122.4 250 1.316 biphenyl 69-72 254-255 1.6 /;-bromochlorobenzene 67 196 1.576 14>romo-2,2-difluoroethane -74.5 57 1.817 bu ty l chloride -123 77-78 0.884 l-chloro-4-nitrobenzene 83-84 242/761 torr 1.520 cyclohexane 6.5 80.7 0.774 /;-dichlorobenzene 53 174 1.458 dimethylterephthalate 140-142 > 3 0 0 N,N-diphenylacetamide 103 sublimes heptane - 9 1 98 0.684 methanol - 9 8 65 0.792 methyl iodide - 6 6 42 2.279 naphthalene 80 218 1.145 phthalic anhydride 131-132 285 1.527 1- propanol -127 97 0.807 2- propanol -89 82 0.785

*At room temperature. Source: Data ate from N.A. Lange, Handbook of Chemistry (New York: McGraw-Hil l , 1967.); and Handbook of Chemistry and Physics (Cleveland: Chemical Rubber Publishing Co., published annually).

Q U E S T I O N S

1. Why is a pipet used instead of a graduated cylinder to make the volume measurement for the density determination?

2. Which measurement-weight or volume—limits the number of significant figures in your calculated density?

3. Why must the cork be vented in the boil ing point determination?

R E F E R E N C E

Yoder, C.H., Suydam, F .H. , and Snavely, F.A. Chemistry. New York : Harcourt Brace Jovanovich, Inc., 1975. Chapter 1.



Name Date

Section

(Note: All numbers must be reported with proper units attached.)

A. L iqu id density: Trial 1 Trial 2

Weight of beaker and watch glass

Weight of beaker, watch glass, and l iquid

Weight of l iquid

Density of unknown l iquid

Average of the two trials

B. Boiling poin t :

Methanol

Unknown l iquid

C. Melt ing poin t :

Naphthalene

Mixture of naphthalene and l-chloro-4-nitrobenzene

Unknown solid

Mixture melting points:

Compound added to unknown —_ m.p.

Compound added to unknown m.p.

Compound added to unknown m.p.

Results:

Number of unknown l iquid

Identi ty of unknown l iquid

Number of unknown solid

Identi ty of unknown solid -

(Answer the questions from page 45 on the back of the Report Sheet.)

Experiment 3 Compound Stoichiometry

Introduction

It is of great importance to chemists to know the formulas of compounds that form when elements react w i th one another. Stated another way, the chemist must determine exactly how many atoms of one element combine wi th how many atoms of another element. For example, we can allow magnesium to react wi th oxygen by heating the magnesium in 0 : to form magnesium oxide. One gram of magnesium wi l l produce 1.66 grams o f magnesium oxide. The weight of the oxygen in the oxide can be found quite simply by subtraction:

1.66 g (weight of magnesium oxide) -1 .00 g (weight of magnesium)

0.66 g (weight of oxygen)

Thus, 0.66 g o f oxygen wi l l react wi th 1.00 g o f magnesium to form an oxide. If.the gram-atomic weights of magnesium and oxygen are known, it is possible to calculate how many moles of each element reacted to form the oxide and the relative numbers of atoms involved in this reaction. With a table of atomic weights at our disposal, this is a straightforward process:

1 00 " moles Mg = L ^ ^ _ = 0.041 1 mole

24.3 g/mole moles O = —?-^Li—= 0.041 mole

16.0 g/mole

From these calculations, we see that the simplest or the empirical formula of the compound is

Mgo.4iOo.4i = MgO

It is easy to appreciate the difficulties which arose in the early history of chemistry before the atomic weights for the elements were known. Even then, it was simple to determine by the weighing operation described above that 1.00 g of magnesium reacted w i th 0.66 g of oxygen to form magnesium oxide. But w i thou t knowing the relative weights of magnesium and oxygen atoms, it was impossible for early chemists to discover how many moles of each element were involved in the reaction. They could not have calculated the formula wi thout knowing atomic weights.

On the other hand, / / t h e formula had been known, the relative atomic weights of magnesium and oxygen could have been calculated. Since MgO is the formula for magnesium oxide, an equal number of atoms must be contained in 1.00 g of magnesium as in 0.66 g of oxygen. This must mean that the relative weights of Mg

49

http://Mgo.4iOo.4i

50 Experiment 3

and 0 atoms arc 1.00/0.66 = 3/2. We now know that these atomic weights are 24 and 16, respectively; these weights are in a 24/16 or a 3/2 ratio.

The problems of determining the formulas of compounds and the atomic weights of elements seemed inseparably linked: if either one were known, the other could be found. But neither piece of information was available to early chemists. Obviously, the development of an accurate table of atomic weights was a real necessity for the progress of chemistry.

In this experiment you w i l l determine the formula o f the compound that forms when a metal reacts w i th sulfur. An accurately weighed sample of the metal is heated wi th an excess of sulfur to form a metal sulfide. Whatever sulfur is not needed to react w i th the metal combines wi th oxygen in the air to produce gaseous sulfur dioxide.

The weight of sulfur that reacts w i th the metal can be determined as the weight of the oxygen was in the MgO example. You can determine the empirical formula of the sulfide using the known atomic weights of the metal and the sulfur.


Bunsen burner crucible tongs clay triangle ring crucibles and lids wire gauze or asbestos square


copper wire or metal shavings powdered sulfur

P R O C E D U R E

Set up a clean crucible and l id as shown in Figure 1-2 (page 32). Apply full heat to the crucible and the lid for about 10 minutes to burn o f f any possible contaminants or volatile material. Then use crucible tongs to place the crucible and the lid on a piece of asbestos to cool. When they are at room temperature (after 15-20 minutes), weigh them on the analytical balance. Handle the crucible and lid w i th tongs after the heating.

While the crucible is cooling, obtain a piece of copper wire or metal shavings weighing about 1 g. If you use copper wire, coil it around a pencil and compress the coil so that it wi l l fit in the bo t tom o f the crucible. After you have weighed the crucible and l id , place the wire or the metal shavings in the crucible and weigh the crucible, l id , and metal.

Now add about 1.3 g of sulfur to the crucible; since an excess of sulfur is used, it need not be weighed any more accurately than to the nearest 0.1 g (see page 32 and Experiment 1). Cover the crucible, place it on the triangle, and heat it gently w i t h a low tlame un t i l the sulfur ceases to burn at the edge o f the crucible l id . Caution: This heating must be done in the hood. Sulfur vapor and sulfur dioxide are extremely i r r i ta t ing and can be toxic.

51 C o m p o u n d S t o i c h i o m e t r y

When you no longer see the blue flame of burning sulfur, increase the flame and heat the crucible unt i l the bot tom is dull red. Then continue to heat the crucible for an additional five minutes. {Be sure (he crucible remains covered during this heating, so that the metal or the metal sulfide cannot react with oxygen in the air.) Remove the covered crucible from the triangle, cool it to room temperature, and weigh the covered crucible and the metal sulfide that it contains. (Remember to handle the crucible w i th tongs.)

Since some o f the excess sulfur may still remain in the crucible at this point , you should return the covered crucible to the triangle and heat it strongly for about five more minutes. Then cool and weigh the crucible again. This weighing should agree with the first weighing to ±0 .002 g. If it does not, add another 0.2 g of sulfur to the crucible and repeat the heating, cooling, and weighing procedures unt i l you obtain two successive weighings that exhibi t this desired precision. Use the last weight you obtain in your calculations.

A second weight determination should be made using 2.0 g of sulfur. This wi l l allow you to determine whether or not the amount of sulfur used in the reaction affects the formula of the compound that forms. While the first crucible is cooling, heat the second crucible. Then heat the reaction mixture of metal and sulfur in the first crucible, while the second crucible is cooling. By dovetailing your operations for the two crucibles, you wi l l not waste t ime. One crucible can be heating while you are cooling and weighing the other crucible.

NOTEBOOK

In your notebook, prepare a data table like the one on the Report Sheet for this experiment and record your data there as you collect them. Also include any pertinent observations you make in the course of the experiment. For example, you wil l want to record the color and the appearance o f reactants and products. Be sure to record any difficulties you have or any mistakes you make that might affect your results.

R E S U L T S

Use the atomic weights of the metal and of sulfur to calculate (to the correct number of significant figures) how many moles of each element reacted in each of the two trials.

Calculate (to the correct number of significant figures) a value for the ratio of moles of metal to moles of sulfur present in the metal sulfide for each of the two trials.

Change the ratio of moles of metal to moles of sulfur to the simplest whole number ratio. Write the empirical formula for the metal sulfide.

QUESTIONS

1. What evidence is there that a reaction actually occurred between the metal and sulfur'.'

52 Experiment 3

2. Write an equation for the reaction that occurred between the metal and sulfur. 3. The uncertainty in each weighing you performed is probably about ±1 in the last

digit. This means that the figure farthest to the right of the decimal point in each weighing could have been estimated 1 higher or 1 lower than the figure you chose. Calculate the largest possible value that the weight of metal taken in Trial I could possibly have (assuming that you made an error in the last figure of each weighing). Now calculate the smallest possible value of the weight of the metal in Trial 1 for each weighing. Also calculate the extreme possible weights of metal sulfide produced in Tr ia l 1. Use these largest and smallest possible weights of metal and metal sulfide to calculate the min imum and the maximum values of the number of moles of metal for each 1.00 mole of sulfur in the metal sulfide. What effect, if any, would these weighing uncertainties have on the formula that you determined?

4. What difference, if any. did you observe between Trial 1 and Tria l 2, in which different amounts of sulfur were used'' What is the reason for this behavior?

R E F E R E N C E

Yoder, C M . , Suydam, F .H. , and Suavely, F .A. Chemistry. New Y o r k : Harcourt Brace Jovanovich, Inc., 1975. Chapter 3.



Name Date

Section


Data: Trial 1 Trial 2

Weight of metal sulfide, crucible, and lid

Weight of metal, crucible, and lid

Weight of crucible and lid

Weight of metal sulfide formed

Weight of metal reacted

Ident i ty of metal

Weight of sulfur in product

Calculation of formula:

Moles of metal reacted

Sample calculation:

Moles of sulfur reacted

. . .. moles metal Actual ratio

moles S Simplest whole number ratio

Empirical formula of the metal sulfide

(Answer the questions from pages 51-52 on the back of the Report Sheet.)

Experiment 4 Stoichiometry of a Reaction

Introduction

The net result of a chemical reaction is described by a chemical equation. For example: H 2 + C l 2 "+ 2HC1 states that the reaction of 1 molecule of hydrogen wi th 1 molecule of chlorine produces 2 molecules of hydrogen chloride. Of course,/V (Avogadro's number) hydrogen molecules w i l l require /V chlorine molecules to produce 2/V molecules of hydrogen chloride.

H 2 + C l 2 • 2HC1

1 molecule + I molecule • 2 molecules

1 mole + 1 mole • 2 moles

In order to write a chemical equation, a chemist must first determine experimentally what the reactants are, in what ratio they react (the stoichiometry of the reaction), and what quantities and types of products result from the reaction. This experiment is designed to determine the stoichiometry of the reaction between sodium hypochlori te (NaOCl) and sodium thiosulfate ( N a 2 S 2 0 3 ) , or potassium iodide ( K I ) , or-sodium sulfite ( N a 2 S 0 3 ) , or potassium thiocyanate (KSCN).

When two or more substances are mixed, it is possible to determine if a reaction has taken place by noting whether or not the properties of the mixture differ from the properties of the original components of the mixture . Further, by not ing how an observed property varies wi th the proportions in which the reactants are mixed, the stoichiometry of the reaction may be determined.

Consider the fol lowing example in which solutions of A and B are mixed to form a precipitate P. In a series of different experiments, solutions of A and B are mixed so that the sum of the moles of A and B remains constant and the amount of precipitate P that forms is determined. If A is in excess of the amount required to react w i t h B, then an increase in the amount o f B w i l l result in the formation of more precipitate and vice versa. The max imum amount of precipitate would be formed if A and B were mixed in the correct stoichiometric amounts. This stoichiometric point can be determined by p lo t t ing the amounts of precipitate that form against the amounts of A used in various reactions (see Figure 4-1). In this example, the maximum precipitate occurs at a ratio of 1 A : 1 B.

The format ion of a precipitate is only one of many properties that can be used to determine the stoichiometry o f a reaction. In this experiment, you w i l l base your decision about the proper ratio between reactants on the amount of heat that is evolved during the reaction. For example, w i t h sodium thiosulfate ( N a 2 S 2 0 3 )

x N a 2 S 2 0 3 + y NaOCl -* products + Q

55

J U txperiment 4

A + B = fixed value = .10 mole

FIGURE 4-1

Quantity of precipitate as a function of relative amounts of reactants.

where Q represents the heat that is evolved. This heat causes the temperature of the mixture to rise by an amount

AT - ( 7 f i n a l — ^ i n i t i a l )

As the amount of heat that is released in the reaction increases, AT increases. Consequently, you can use the temperature change AT to moni tor the amount of heat that is released in the reaction.

How does the quant i ty of heat released relate to the stoichiometry of a reaction? Suppose we know the values of x and y in the preceding equation. If we mix exactly x moles of N a 2 S 2 0 3 w i t h y moles of NaOCl, a certain number of calories of heat Q w i l l be produced. On the other hand, i f we mix fewer than .v moles o f N a 2 S 2 0 3

w i t h more than y moles o f NaOCl, not enough N a 2 S 2 0 3 w i l l be present to react w i t h all o f the NaOCl. Thus, a smaller amount o f heat w i l l be evolved because fewer units o f N a 2 S 2 0 3 w i l l be present to react, resulting in a smaller AT. This argument can also be applied when N a 2 S 2 0 3 is in excess.

It is important to note that the total number of moles (moles N a 2 S 2 0 3 + moles NaOCl) remains constant in any of these reactions. If the number of moles of N a 2 S 2 0 3 is increased, then a corresponding number of moles of NaOCl must be decreased and vice versa. We therefore conclude that the maximum amount of heat is released resulting in a maximum AT when the reagents are mixed in the correct stoichiometric ratio. In this experiment, you w i l l be asked to apply the above reasoning as you mix reagents in various ratios and observe the temperature increases. You can conclude that the ratio that produces the largest AT is the correct stoichiometric ratio.

In carrying out this experiment, you wi l l use a basic solution o f 0.50A/ N a 2 S 2 0 3 , or 0.50/Vf K I , or 0.50/W N a 2 S 0 3 , or 0.50A/ KSCN, as your instructor directs. The expression 0.50A/ means that the solution contains 0.50 mole of the solute in 1.0 liter of solution. Using solutions of known concentrations allows you

57 Stoichiometry of a Reaction

to obtain a certain number of moles of a reactant simply by measuring a specific volume o f solution. For instance, i f you measure 1 .0 liter o f a 0 . 5 0 Y W solution of NaOCl. you know that this solution contains 0 . 5 0 mole o f NaOCl; 5 0 0 ml o f this solution contains 0 . 2 5 mole of NaOCl; and so on.


1 0 0 - m l graduated cylinder Styrofoam cup thermometer graduated in 0 . 1 ° intervals

NOTF: Bring a piece of graph paper and a ruler w i t h you to the laboratory, so that you can plot the AT vs. moles of NaOCl data.


0 . 5 0 / W NaOCl

0 . 5 0 A / solution o f N a 2 S 2 0 3 , or K I , or N a 2 S 0 3 or KSCN (in 0AM NaOH)

P R O C E D U R E

Measure the temperature o f 5 0 ml o f 0 . 5 0 / V / NaOCl in a Styrofoam cup. Measure the temperature.of 5 0 ml o f 0 . 5 0 ; W N a 2 S 2 0 3 (or other 0 . 5 0 J W reactant, as directed by your instructor) in a graduated cylinder. In i t i a l ly , these temperatures should be identical. Mix the reagents thoroughly in a Styrofoam cup and record the maximum temperature. The mixture w i l l warm up at first and then gradually cool unt i l i t returns to room temperature. The Styrofoam acts as an insulator, so the mixture w i l l cool slowly and you should have no trouble observing the maximum temperature.

Now repeat the above procedure, using different amounts of each reagent and a total volume of 1 0 0 ml . You should use the fol lowing five volume combinations and any other combinations you feel would b e desirable: 5 0 : 5 0 ; 4 0 : 6 0 ; 6 0 : 4 0 ; 20 : 8 0 ; and 80 : 2 0 . After you have measured and recorded the temperature changes for at least these five different mixtures of the two reactants, construct a graph of the temperature change, AT. vs. the moles of NaOCl used. (Refer to Appendix I I , page 3 5 7 , for information on graphing.) From this plot , determine the stoichiometry of the reactants by applying the method used in the precipitation experiment described in the In t roduct ion . If you cannot draw a definitive conclusion from your graph, continue to work wi th different mixtures of the reactants and to plot additional points unt i l you can interpret your data conclusively.

R E S U L T S

Plot AT vs. moles of NaOCl on an 82" X 11 in . piece of graph paper. Then use this plot to determine the stoichiometry of the reaction. Write the equation for the

58 Experiment 4

reaction x X + v NaOCl -* products + Q, using the correct coefficients x and y determined from your data plot . (X refers to the material w i th which you allowed the NaOCl to react. You are not expected to know the identi ty of the products.) Explain how you obtained the values for.v and y from your plot of AT vs. moles of NaOCl.

Q U E S T I O N S

1. Why must the total volume of the reacting solutions remain the same when you work wi th different mixtures of the reactants to determine their stoichiometric ratio?

2. Imagine that a reaction occurs so that the absorption of heat causes the temperature of the mixture to drop as the reaction progresses. Sketch a graph that shows AT vs. number of moles of a reactant for a reaction of this type. Is AT a negative or a positive quanti ty?

R E F E R E N C E S

1. Bigelow, M.J. Journal of Chemical Education 46, 378 (1969) . 2. Yoder, C.H., Suydam, F .H. , and Snavely, F.A. Chemistry. New Y o r k : Harcourt

Brace Jovanovich, Inc., 1974. Chapter 3.



Name Date

Section


Data:

0.50yW NaOCl 0.50/v/ reactant X (ml ) (ml ) 7 W U , 7Vinai A7

Attach your graph of AT vs. number of moles of NaOCl. Explain how you determined the coefficients, and wri te the equation for the reaction.


Experiment 5 Synthesis of Co(NH3)6CI3

Introduction

One activity that occupies chemists is the synthesis or prepration of chemical compounds. A goal of synthetic chemistry is to make substances that have never been previously prepared. Some chemists work in the laboratory to prepare compounds that are known to occur in nature; other chemists strive to make total ly new, unknown compounds. In order to perfect the techniques involved in making compounds, chemists must spend a good deal of time duplicating syntheses of known compounds in the course of their academic training.

The purpose of this experiment is to gain experience in using the laboratory techniques involved in synthetic chemistry. The compound to be prepared is called a coordination compound. In C o ( N H 3 ) 6 C l 3 , the cation has six molecules of ammonia bonded to C o ( I I I ) in a regular geometric fashion (see Figure 5-1); the CP ions balance the charge of this cation. The compound wi l l be prepared by the reaction o f cobalt chloride, ammonia, and hydrogen peroxide, as in Equation (5-1):

2 C o C l 2 - 6 H 2 0 + 2 N H 4 C l + 1 0 N H 3 + H 2 0 2 • 2 C o ( N H 3 ) 6 C l 3 + 1 4 H 2 0 (5-1)

Many synthetic reactions are carried out in solution by dissolving the reactants and allowing them to form the desired product. After a compound forms in solution, the experimenter must isolate it (get it out of solution) and purify it ( r id it of all traces of the unused reactants and any side products, reaction products other than the one of interest).

Since the solubil i ty of most compounds decreases w i t h temperature, a common technique for isolating a product is to cool the solution un t i l the amount of compound dissolved exceeds its solubil i ty at that temperature. The excess then crystallizes out of the solution. Unfortunately, as the temperature drops, com-

i

i

i

H 3 N

FIGURE 5-1

Hexaamniinecobalt(lII) ion.

61

62 Experiment 5

pounds other than the desired one, such as unreacted starting material or other reaction products, often crystallize also. These contaminants can be removed by recrystallization; that is, by dissolving the impure product in a min imum amount of hot solvent, fi l tering the solution to remove insoluble impurit ies, and then cooling the solution to crystallize the desired compound while unwanted materials (hopeful ly) remain in solut ion. Sometimes the recrystallization procedure must be repeated many times to achieve puri ty . In this experiment, however, you can stop after only one recrystallization, since C o ( N H 3 ) 6 C l 3 can be purified fairly easily by this method.

It is necessary to plan your work carefully in order to use your laboratory time most efficiently. Bring w i t h you to the laboratory an outline of how you intend to proceed wi th the experiment.


beakers pneumatic trough Büchne r funnel ring stand Bimsen burner spatula 125-ml Erlenmeyer flask stirring rods 125-ml filter flask test tube w i th cork filter paper thermometer 100-ml graduated cylinder vacuum tubing medicine dropper wire gauze


C o C l 2 - 6 H 2 0 NH 4 C1 cone. HCl acetone 0.5A/ HCl activated charcoal 30% H 2 0 2 ice cone. N H 3

B R I E F O U T L I N E O F S Y N T H E T I C P R O C E D U R E

NH 4 C1 i n H 2 0 ; I A. boi l .

Dissolve C o C l 2 - 6 H 2 0 . i

Add to flask containing charcoal; i cool under running water.

Add cone. N H 3 ; I cool in ice bath.

Add H 2 0 2 dropwise; i A, destroy excess H 2 0 2 .

63 Synthesis o f C o ( N H 3 ) 6 C I 3

Cool un t i l product containing charcoal crystallizes. i

Collect product and charcoal by f i l t ra t ion. I

Recrystallize to remove charcoal.

Caution: All work must be performed under the hood!

P R O C E D U R E

A. Making the Compound

Weigh out 9 . 0 ± 0.1 g of CoCl 2 • 6 H 2 0 . Record the weight you actually take. This weight must be known to this degree of accuracy; the exact weights of other materials used in the synthesis are less important .

Weigh about 1 g of activated charcoal into a clean 125-ml Erlenmeyer flask. This charcoal is not a reactant in the synthesis; it simply acts as a catalyst, causing the reaction to occur rapidly.

Weigh about 6 g of NH 4 C1 (ammonium chloride) in to a clean 100-ml beaker. Add 15 ml of distilled water to the N H 4 C 1 . Place the beaker on wire gauze on a ring over your burner. Use a low flame to bring the solution to a bo i l ; stir it during this heating.

When the solution is boil ing, add the C o C l 2 ' 6 H 2 0 to it gradually. Continue stirring during-this addi t ion; be sure to stir gently so that drops o f hot l iquid wi l l not spatter out of the beaker. After all the C o C l 2 - 6 H 2 0 has dissolved, turn o f f the burner. Use beaker tongs or a towel to pick up the hot beaker and pour the hot solution in to the Erlenmeyer flask containing the charcoal. Cool the flask under running water. To this cooled mixture , add 23 ml of concentrated N H 3 solution (ammonium hydroxide) . Cool the flask in a pneumatic trough full of ice unt i l the temperature of the mixture is 10°C or lower. (Two students should share a trough of ice.)

While the mixture is cooling, use a graduated cylinder to measure 13 ml of 30% H 2 0 2 (hydrogen peroxide) and pour it into a small beaker. Caution: Hydrogen peroxide can cause severe burns. If you splash even a drop of it onto your skin, wash it off immediately w i t h lots of running water. If you spill any of it onto the desk top, wipe it up w i t h a sponge and then wash out the sponge repeatedly under running water. Hydrogen peroxide can also damage your clothing.

When the temperature of the reaction mixture in the flask is 10°C or lower, add the H 2 0 2 to the mixture a l i t t le at a t ime; use a medicine dropper so that you can add only a few drops at a time. During this addit ion, leave the flask in the ice bath; swirl the flask gently to mix the H 2 0 2 w i t h the other reactants.

After the H 2 0 2 has been added, return the flask to the wire gauze and heat the mixture gently to 5 0 - 6 0 ° C . Any temperature w i th in this range is satisfactory, but the temperature must not be lower than 50°C or higher than 6 0 ° C . Swirl the flask occasionally to keep the contents mixed. Continue heating and swirling for about 20 minutes, or un t i l the original pink color characteristic of C o C l 2 * 6 H 2 0 turns to a

64 Experiment 5

golden yel low. A good way to test the color o f the solution (which wi l l look black because of the charcoal) is to dip a stirring rod into the flask and transfer a drop of solution from the t ip o f the rod onto a piece o f white filter paper. The solut ion w i l l spread out enabling you to see its color.

After the proper color change has occurred, the flask and its contents may be stored in your desk un t i l the next period if you must interrupt your work . Otherwise, cool the solution in an ice bath. The product w i l l precipitate. Continue the cooling un t i l no more crystals seem to be forming. Filter the solution using a Büchner funnel containing two pieces of fi l ter paper (see page 15 for instructions on the use o f suction f i l t ra t ion) . It does not matter i f a l i t t le charcoal passes in to the filtrate, but be sure you don' t lose any of the crystals of the product. Wash the precipitate in the funnel wi th two 5-ml portions of ice water. To perform the washing operation, disconnect the suction, add the wash solvent, allow the mixture to stand about 10 seconds, and then suck the solvent through the filter paper. If you must interrupt the synthesis, place the filter paper and product in an evaporating dish and store it in your desk un t i l the next laboratory period.

B. Recrystal l ization of the Impure Product

In a 250-ml beaker, bring 75 ml of 0.5A/ HCl to a boi l . Transfer the impure product (f i l ter paper and all) to this beaker. Stir the mixture and continue heating it unt i l nothing else seems to be dissolving. Filter the hot solution through a B ü c h n e r funnel to remove the charcoal. In this f i l t ra t ion , you should be sure that the filtrate is free of charcoal. If any charcoal passes in to the fil trate, you must pour the mixture back into the beaker, reheat i t , and refilter it to remove the remaining charcoal.

Pour the golden ye l low, charcoal-free solution into a clean 125-ml flask; if any crystals form in the fil ter flask, scrape them all out. Add 21 ml of concentrated HCl to the solution. Cool the solut ion in an ice bath to precipitate the product. When no more crystals seem to be forming, filter the mixture through a Büchne r funnel containing two pieces of filter paper. Wash the crystals in the funnel w i t h two 5-ml portions of ice water to remove any HCl that may be on the crystals. Then wash them wi th 10 ml of acetone to wash away the water; acetone is much more volatile than water, so the crystals w i l l dry faster than they would i f they were wet w i th water. Dry the precipitate as thoroughly as possible by suction. Then scrape the crystals onto a large piece of dry fil ter paper and spread them out so that any solvent that is s t i l l on them can evaporate.

Weigh the dry compound by difference into a clean, dry test tube and stopper the tube. Label the tube w i t h your name and wi th the identi ty and weight of the compound. Turn i t in to your instructor.

R E S U L T S

The equation for the preparation of C o ( N H 3 ) 6 C l 3 is given as Equation (5-1) in this experiment. Perform the fol lowing calculations involving this preparation.

C5 Synthesis of C o ( N H 3 ) 6 C I 3

1. Calculate the number of moles of C o C l 2 - 6 H 2 0 that you used in the synthesis. 2. Using Equation (5-1), calculate the number of moles of C o ( N H 3 ) 6 C l 3 that your

synthesis could have produced. 3. Use the expected number of moles of C o ( N H 3 ) 6 C l 3 to calculate the expected

number of grams of product. This is the theoretical yield of the product. 4. Use the theoretical yield and your actual yield to calculate your percentage

yield.

Q U E S T I O N S

1. If the product had not been recrystallized, would your percentage yield have been larger or smaller? Explain.

2. Calculate the approximate number of moles of NH 4 C1 that you used in the synthesis. Would all of this react? Explain.

R E F E R E N C E S

1. Yoder, C.H., Suydam, F .H. , and Snavely, F .A. Chemistry. New York : Harcourt Brace Jovanovieh, Inc., 1975. Chapter 3.

2. Femelius, W.C. (ed.). Inorganic Synthesis, Volume I I . New York : McGraw-Hil l , 1946, p. 217.



Name Date

Section

Data:

Weight of CoCl 2 -6H 2 0 used

Weight of C o ( N H 3 ) 6 C l 3 isolated

Results:

Moles C o C l 2

- 6 H 2 0 used

Show calculation:

Moles C o ( N H 3 ) 6 C l 3 possible

Weight of C o ( N H 3 ) 6 C l 3 possible

Show calculation:

Percentage yield of C o ( N H 3 ) 6 C l 3

Show calculation:


Turn in your product in a labeled test tube.

Experiment 6 Line Spectra of Elements

Introduction

Early experimental results on the combining weights of the elements could be accounted for by Dalton's atomic theory. An element was believed to consist of a collection of identical atoms, each having the same weight. The observation that two elements always combine in the same weight ratio to form a particular compound was simply a reflection of the fact that a certain number of atoms of each k ind , each one having a characteristic weight, combined. As long as these were the only kinds of experimental results to be accounted for, atoms could be thought of as small bil l iard balls; if one were cut open, it would look the same throughout , just as a potato does.

In the nineteenth century, some experiments showed that the structure of atoms must be far more complicated than had been previously believed. These experiments investigated the behavior of atoms excited ( that is, provided w i t h excess energy) in an electric discharge. Experimenters found that light of characteristic colors was emitted by these excited atoms. This emitted light is called the atomic spectrum (plural , spectra). Passing the emit ted light through a prism separates it in to beams, each of a single wavelength and color. Prisms separate light by refracting i t ; when light of different wavelengths passes through a prism, it is bent through different angles (see Figure 6-1). The fact that excited atoms emit l ight of only certain discrete wavelengths is evidence that their internal structure must be more complex than a potato-like structure; otherwise, there would be no reason why light of every wavelength is not emit ted.

In this experiment, you w i l l study the light emitted by excited atoms and draw some conclusions about their inner structure. The apparatus you w i l l use to observe spectra is a Bunsen spectroscope (shown schematically in Figure 6-2). The function of the spectroscope prism is to disperse ( to separate according to wavelength) the light from the element by refraction. At the same t ime, the prism superimposes a scale on this refracted l ight by reflecting an image from the scale arm. The sharpness of the spectral lines you see through the viewing arm w i l l depend on the wid th o f the slit opening: the narrower the slit, the sharper the lines. However, a very narrow

Prism

59

70 Experiment 6

FIGURE 6-2

Schematic drawing of a Bunsen spectroscope.

slit w id th w i l l result in a d im image, because very l i t t le light w i l l enter the spectroscope. The best slit w i d t h w i l l obviously be a compromise between the sharpest lines possible and sufficient light intensity to see the lines easily. It should be apparent from Figure 6-2 that the alignment of the prism is cri t ical . Your spectroscope prism w i l l be adjusted properly by your instructor before you begin; any further prism adjustments should be made only by your instructor.

The purposes of this experiment are:

1. To observe qualitatively the light emitted by excited atoms of hydrogen, hel ium, neon, and argon, and to resolve this light in to lines of particular wavelengths w i t h the spectroscope (this is called observing the line spectrum of an element).

2. To make quantitative conclusions about the relationship of the line spectrum of the simplest a tom, H, to the internal (electronic) structure of the atom.

In order to accomplish the second objective, you wi l l have to (1) calibrate the spectroscope and then (2) take quantitative measurements of the lines that appear in the hydrogen spectrum. Alternat ively, your instructor may project a slide of a camera's view through the slit of a Bunsen spectroscope so that you can obtain data.

A. Calibration of the Spectroscope

Many scientific instruments are equipped wi th a numerical scale of some k ind . In order to make quantitative measurements, some correlation must obviously exist

71 Line Spectra of Elements

between the scale reading and the property being measured. A common way to establish such a correlation is to observe some property whose numerical value is already known and then to see what scale reading it produces. By making several such observations, a correlation can be established between the scale readings and values of the property. This correlation can then be used to measure the values of the same property on other unknown materials. This procedure is called calibrating the instrument.

The Bunsen spectroscope has a numerical scale across the top of the viewing field; these numbers are not wavelengths. Y o u must relate the numbers on the scale to actually known wavelengths. This is done by viewing the line spectrum of helium (the wavelengths of helium are given on the Report Sheet for this experiment) and not ing the scale reading for each line. A plot is then made of wavelength versus scale reading; this plot , called the calibration curve, w i l l enable you to read the wavelength of any line in a spectrum if you know the scale reading for that line. It is important to note that, because of the way the spectroscope prism is adjusted, different spectroscopes wi l l give different scale readings for the same line in a spectrum. Once a calibration curve is determined, the prism must not be moved; the hydrogen and helium data must be taken on the same spectroscope.

B. Energy Levels and the Hydrogen Line Spectrum

Light behaves according to the laws of wave mechanics; it is characterized by frequency and wavelength. The velocity of any traveling wave is equal to the product of its frequency and wavelength. For light waves, the equation describing this relationship- is

c = Xt> where c = speed of light X = wavelength v = frequency (6-1)

Light can be viewed as consisting of small bundles of energy called photons. The energy of a photon of light E is directly proport ional to its frequency

E = hv (6-2)

The h in Equation (6-2) is a proport ional i ty constant called Planck's constant; its value is 6.62 X 1CT 2 7 erg-sec. Solving Equation (6-1) for v and substituting the result in to Equation (6-2) gives

£ = — (6-3) X

Equation (6-3) relates the photon energy to the wavelength of l ight. The emission of light of only certain discrete wavelengths by excited atoms corresponds to the emission of certain discrete amounts of energy. (See Figure 6-3.)

According to the Bohr theory of the hydrogen atom, the single electron in the hydrogen atom can possess only certain energies. If this electron is in an excited (high) energy level En and drops to a lower energy level E\, then the atom wi l l emit a photon of light of an energy E that is exactly equal to the difference in energy

72 Experiment 6

between these levels:

E = E, Ex he (6-4)

Long before the Bohr theory of the hydrogen atom was developed, Johannes Robert Rydberg, a nineteenth-century Swedish physicist, discovered that the experimentally measured wavelengths in the hydrogen spectrum could be reproduced by Equation (6-5)

1 (6-5)

where R is a constant having the value 109,737 c m - 1 , and n { and «n are integers. This was a completely empirical result; that is, it reproduced the experimental results for hydrogen, but no one understood why it did so.

Later, Niels Bohr postulated that only certain energy levels characterized by integral quantum numbers were possible for the electron in a hydrogen a tom. Bohr also showed that the energy of each level was proport ional to - 1/n 2 , where n is the

73 Liny Spectra of Elements

principal quantum number. If Bohr's postulates are valid, then the difference between the energy levels characterized by quantum numbers n t and nn is

Eu-El=RH(-i--LA (6-6) \ " i " 1 1 7

where RH is the proport ional i ty constant. If this quanti ty of energy is emitted in the form of a light photon as the electron loses energy from level « n to level n l t then

E = E l l - E l = R H l - L - - ^ = i £ (6-7)

Dividing Equation (6-7) by he gives

A hc\ni~ flu*/

Equation (6-8) is the same as Equation (6-5) if R - RH/he. Or, stated another way, RH is simply the Rydberg constant expressed in ergs. The point of this argument is that assuming the Bohr postulates to be true, we wou ld expect to see a line-emission spectrum wi th wavelengths given by Equation (6-8) rather than a continuous spectrum. Moreover, the form of Equation (6-8) is the form of the empirically obtained Equation (6-5). This result lent credibi l i ty to the Bohr postulates and pointed out the physical significance of n t and n I ( as quantum numbers in Equation (6-5).

As an example of the use of Equations (6-7) and (6-8), consider an electron having nn = 5 falling to the level nx = 1. The energy of light emitted in this process is

E = RH(±-^j = R H ( \ - ± ) = RH (0.960) = (6-9)

The wavelength w i l l be

X = ^ £ ( _ L ) = i ( 1 . 0 4 ) = 9.48 X 10~ 6 cm (6-10) RH\0.96J R


aluminum foil Bunsen spectroscopes flashlights gas discharge tubes graph paper power supplies (high-voltage transformers) ring stands and clamps

t

74 Experiment 6

P R O C E D U R E

You should work in pairs for this experiment. Measurements arc made using equipment that has already been set up in the laboratory.

Your instructor wi l l show you how to connect the gas discharge tubes to the power supply. When the tubes are connected, observe and record the characteristic color of the emission of each gas that you use.

In your laboratory notebook, prepare a data table similar to the data section on the Report Sheet for this experiment, so that you can record data in an orderly fashion throughout the experiment.

A. Calibration of the Spectroscope

The gas discharge source must be properly aligned wi th the spectroscope slit, and the slit must be adjusted to its narrowest wid th so that you can view the sharpest spectrum. Your instructor w i l l show you how to do this.

The spectrum of helium is used to calibrate the scale. The Report Sheet gives the positions of the lines in the helium spectrum and their relative intensities on a comparison scale that assigns a value of 1000 to the most intense (brightest) l ine.* You wil l probably not be able to see the lines that have an intensity 20 or less; be sure to correlate the more intense, visible lines wi th the proper wavelength. Position the helium discharge tube, and wrap the tube and the slit area in a luminum foil to block out the laboratory light (overhead fluorescent lights emit a very strong green line). Clamp the flashlight at the end of the scale arm, so that the scale is visible in the eyepiece when the flashlight is turned on.

One student observes the spectrum (being careful not to introduce parallax error by moving the head while taking the readings) and calls out the colors and the scale values for the helium lines ( to at least three significant figures) to the other student, who records the data.

B. Position of H Lines

Remove the helium tube from the spectroscope slit and replace it w i th the hydrogen tube. Record the scale reading for each line in the hydrogen spectrum as you did for the helium spectrum. There are four visible lines in the hydrogen spectrum. The line farthest into the violet region of the spectrum is sometimes d i f f i cu l t -even imposs ib le- to see, because the human eye is not highly sensitive in this region. The two students should then reverse roles to see if they agree on the assignments. The readings obtained wi l l probably not be exactly the same; each student's data wi l l be internally consistent, however

C. Spectra of Neon and Argon

Replace the hydrogen tube w i th a neon tube and observe the line spectrum. Describe qualitatively what you observe (you do not need to record the scale

These data are from the Handbook of Chemistry and Physics (Cleveland: Chemical Rubber Publishing Co., published annually).

•

75 Line Spectra of Elements

readings of the lines). Note the colors of prominent lines; describe bundles of lines of a particular color; and so on. Your description should be clear and detailed enough that others can read it and know what they would see if they looked at the spectrum themselves. Repeat this procedure for argon.

R E S U L T S

Using your data for the helium spectrum, plot a calibration curve of wavelength vs. scale reading on graph paper. Remember to spread the graph over as much of the page as possible. A smooth curve should be drawn through the points. If your points do not lie along a smooth curve, you have probably incorrectly assigned one or more of the helium lines. For example, you may have assigned the red line at 7065 A to the scale reading that should be correlated w i th the red line at 6678 A. Try different assignments of lines unt i l your calibration curve is fairly smooth.

From the calibration curve and your scale readings for the hydrogen spectrum, determine the wavelength for each hydrogen line. Indicate on the calibration curve how you did this.

Transitions occurring between energy levels give rise to the hydrogen spectrum. We have not yet identified these energy levels, but it is possible to assign values to n{

and nn for the transition corresponding to each line in a simple way, by using Equation (6-5):

~ = R ( — 2 - - ^ ) = 109,737 cm" 1 (6-5) \ y i i «ii / \ " i « 1 1 /

This equation contains only one term that you do not know for each line, the term in parentheses containing the principal quantum numbers n. By using the known value of the Rydberg constant and by t rying various combinations of ti\ and n u , you can find a combinat ion for these numbers that reproduces your experimental 1/X values.

In deciding which n values to t ry , keep in mind the fol lowing considerations:

1. A l l the n values wi l l be fairly small. Since the H energy levels are proportional to \/n2, they wi l l be closer together as n becomes larger. Very large values o f n w i l l produce lines of very great X ( i f both nt and nn are large) or of very short X ( i f only n u is large). The lines produced by these transitions are of greater or shorter wavelength than we can perceive as visible l ight; hence, they cannot be a part of the spectrum that you see.

2. The «i and nu values o f lines toward the red end o f the spectrum wi l l differ less than those toward the violet end of the spectrum.

3. Begin w i th the simplest combinations: nx = 1 and nu = 1\ri\ = 1 and nn = 3 ; nx = 1 and nn - 4. If these values do not reproduce a value of 1/X that corresponds to the value you calculated from your experimentally determined wavelength, t ry values of rt\ - 2 and n\\ - 3 ; / i i = 2 and tin = 4; and so on.

Use the determined values of wavelengths of light for the lines in the hydrogen spectrum to calculate the energy of each transition. Construct an energy level

76 Experiment 6

diagram for the H atom. Use graph paper so that you wi l l have an accurate scale for the energy levels. As the base line for the diagram, use the lowest energy level involved in the transitions that produced lines in the spectrum. Along the energy scale, show the amount by which the energy of each higher level is greater than the energy of the lowest level. Label each level wi th the proper quantum number. Draw arrows indicating the transitions between the energy levels that produce lines in the spectrum. Label each arrow wi th the wavelength of the light that is produced by the transition.

Q U E S T I O N S

1. Show that a transition between two energy levels wi th large n values results in the emission of light of very long wavelength. (Hint: Choose similar values of 20 or higher for n { and n u to calculate the wavelength of the emit ted l ight.)

2. Show that a transition between energy levels of very large nu and very small n{

emits light of very short wavelength.

R E F E R E N C E S

1. Yoder. C.H., Suydam, F .H. , and Snavely, F.A. Chemistry. New Y o r k : Harcourt Brace Jovanovich, Inc., 1975. Chapter 5.

2. Handbook of Chemistry unci Physics. Cleveland: Chemical Rubber Publishing t Co., published annually.

Experiment 7 A Simple Covalent Compound: Synthesis and Proper ties

Introduction

Many compounds can be synthesized by the interaction of elements wi th one another. This is especially true of binary compounds—compounds that contain only two elements. The formula of the compound that results and the type of bonding wi th in the compound depend on the natures of the atoms that comprise the elements. A fairly great difference in the electronegativities of the atoms results in a transfer of electrons from one atom to the other so that ions are formed. Atoms of similar electronegativities share electrons; this covalent bonding produces a molecule. In this experiment, you wi l l allow t in and iodine to react w i th one another. The electronegativities of 1.8 for t in and 2.5 for iodine correctly suggest that the bonding should be covalent. The iodine atom, w i t h seven electrons in its valence level, attains a noble gas configuration by the formation of one covalent bond. The tin atom, w i th only four electrons in its valence level, can form four covalent bonds. Consequently, the formula of the resulting compound is S n l 4 ( t in tetraiodide or stannic iodide).

Ionic and covalent compounds can often be distinguished by their physical properties. The.attractive forces between molecules that hold them together in the solid or the l iquid state are generally much weaker than the strong attractive forces between the oppositely charged ions of an ionic compound. Consequently, compounds composed of molecules have much lower melting points and boil ing points than ionic compounds. They are also much more volatile. Another significant difference is in the solubilities of ionic and molecular compounds. Ionic compounds are more l ikely to be soluble in water than in other l iquids; however, some ionic compounds are not soluble even in water. Molecular compounds are soluble in many liquids that are composed of polar or nonpolar molecules, but are less l ikely to be soluble in water. In this experiment, you wi l l compare these properties for some selected molecular and ionic compounds.


beaker spatula Bunsen burner 4-ml test tubes 125-ml Erlenmeyer flask triple-beam balance graduated cylinder watch glass rubber stopper


benzene calcium sulfate

79

80 Experiment 7

carbon tetrachloride iodine dichlorome thane naphthalene granular tin sodium chloride

P R O C E D U R E

A. Synthesis of S n l 4

Into a clean and dry 125-ml Erlenmeyer flask, weigh 1.0 ± 0.1 g of granular t in using the triple-beam balance; record the weight as accurately as possible. To this t i n , add 30 ml of C H 2 C 1 2 (dichloromethane); stopper the flask loosely. Because C H 2 C 1 2 is quite volatile and is poisonous if large amounts are inhaled, you should uncover it only when necessary and then for as brief a time as possible. When you pour C H 2 C 1 2 , place the m o u t h of the bott le as close as you can to the graduated cylinder to minimize vaporization. Weigh 3.0 ± 0 . 1 g of iodine into a beaker and record the weight that you actually take. If aluminum weighing dishes are available, do not use one here; the a luminum and iodine may react on contact. Cover the beaker w i t h a watch glass, so that iodine vapor wi l l not pass into the air in the laboratory. Add a few crystals of iodine to the mixture of t in and C I I 2 C 1 2 ; immediately restopper the flask. Note the color that appears as the iodine dissolves in the C H 2 C 1 2 . Gently swirl the flask un t i l the solution has the red-brown color that is characteristic of S n l 4 instead of the purple-red color of a mixture of iodine and S n l 4 . Be patient; i t w i l l take at least five minutes and may take 15 minutes or more for this color change to occur. Af ter the color of the solution changes, add a few more crystals of iodine to the flask. Continue to swirl the flask and to add a few crystals of iodine at a t ime; after each addit ion, wait for the proper color change to occur before adding more iodine. Remember to keep the watch glass over the beaker as much of the time as you can and to keep the flask stoppered except when you are actually adding iodine to the solut ion. Have a beaker of cold water nearby as you add the iodine; if bubbling occurs that shows the C H 2 C 1 2 is starting to bo i l , immediately cool the flask in the water. After all the iodine has been added continue to mix the contents of the flask for at least ten minutes after the color suggests that all the iodine has reacted. A l l o w a min imum of one hour for the total reaction time from the first addition of iodine.

In the hood, decant the C H 2 C 1 2 solution of S n l 4 into a 50-ml beaker leaving the unreacted tin in the flask. Use a few ml of C H 2 C 1 2 to wash out the flask.

You wi l l now evaporate the C H ; C 1 2 to obtain the solid crystalline product. Several ways to do this are described here. Your instructor w i l l tell you which method to use.

Method I: Lower the hood window so that just enough space remains to place the beaker under the window. Position the beaker so that about half of it extends in front of the window and half of it extends behind the window. As air is drawn into the hood, it w i l l pass over the solution and speed the evaporation of the C H 2 C 1 2 . This is the simplest and most convenient method to use if a hood that draws well is available.

31 A Simple Covalent Compound: Synthesis and Properties

Method 2: If compressed air is available, direct a stream of it across the surface of the solution (as shown in Figure 7-1). This must be done in the hood.

Method 3: Use a water aspirator to pull a stream of air across the surface of the solution (see Figure 7-2). This also should be done in the hood.

Method 4: Instead of decanting the C H 2 C 1 2 solution into a beaker, decant it in to a f i l ter flask. Stopper the flask and connect it to a trap bottle that is connected to a water aspirator. Turn on the aspirator. The suction of the aspirator wi l l cause the C H 2 C I 2 to boil away. Again, this should be done in the hood.

When the S n l 4 is dry, scrape it out of the beaker onto a piece of filter paper. Break up large pieces of the solid to allow C H 2 C 1 2 trapped inside to evaporate. Do not continue your investigation of the properties of S n l 4 un t i l you are sure that it is completely dry. (Sniff it cautiously; if you can detect the odor of C H 2 C 1 2 , the compound is not yet dry . ) Weigh the S n l 4 so that you can calculate your percentage yield.

FIGURE 7-1

Using compressed air to evaporate solvent.

82 Experiment 7

B. Solubil ity of Covalent and Ionic Compounds

Place about 0.1 g of S n l 4 in each of three 4-ml test tubes. To one tube, add about 2 ml of water; to another tube, add about 2 ml of benzene ( C 6 H 6 ) ; and to the third tube, add about 2 ml of CC1 4 (carbon tetrachloride). In each case, record whether or not the S n l 4 is soluble. The compound does not have to dissolve completely to be classified as soluble; simply look for the characteristic S n l 4 color in the solvent. If the S n l 4 does not seem to dissolve, allow the mixture to stand and observe it again after five minutes and again after half an hour. Has there been any change in the appearance of the S n l 4 to suggest that it has reacted w i th the solvent? It wi l l react wi th one of these solvents. Repeat this procedure w i t h samples o f iodine, C 1 0 H 8 (naphthalene), NaCl (sodium chloride), and CaS0 4 (calcium sulfate). Categorize each of these compounds as soluble or insoluble in water, in benzene, in C H 2 C 1 2 , and CC1 4 . Dispose of the solutions as your instructor directs.

C. The Melting Point and Volat i l i ty of Covalent and Ionic Compounds

This port ion of the experiment must be done in the hood. Be sure that no one is evaporating C H 2 C 1 2 in the same hood. Your instructor may designate specific hoods for each procedure.

Hold one end of a spatula (preferably a broad, white porcelain one) in the flame

33 A Simple Covalent Compound: Synthesis and Properties

of a Bunsen burner for about one minute. Remove the spatula from the flame and immediately place a very small amount of S n l 4 on i t . Observe the behavior of the S n l 4 . Does it melt? Does it vaporize? If so, what is the color of the l iquid or of the vapor? Repeat this procedure wi th samples of iodine, naphthalene, sodium chloride, and calcium sulfate.

R E S U L T S AND DISCUSSION

On the Report Sheet for this experiment, briefly explain the reason for each step in the synthesis. Include answers to the fol lowing questions: Why was C H 2 C 1 2

chosen as the solvent in which to carry out the reaction? Why is excess t in used instead of excess iodine? Why is the iodine added in small amounts instead of all at once?

Provide the informat ion about yield and properties requested on the Report Sheet.

QUESTIONS

1. When the five materials were heated as described in the experiment, what compounds behaved similarly? Account for these similarities in terms of the type of particle of which each compound is composed. Account for differences in behavior.

2. Draw Lewis electron dot formulas for S n l 4 , naphthalene, benzene, and C H 2 C 1 2 . Is each of these molecules polar or nonpolar?

3. What similarities in solubilities did you observe for the five materials? Account for these similarities. Account for differences in behavior.

R E F E R E N C E S

1. Brauer, G. Handbook of Preparative Inorganic Chemistry. New York : Academic Press, 1963, p. 735.

2. Yoder, C.H., Suydam, F .H. , and Snavely, F.A. Chemistry. New Y o r k : Harcourt Brace Jovanovich, Inc., 1975. Chapters 6, 7, and 9.

3. Describe the physical appearance of S n l 4 .

Solubilities:

Substance In water In benzene In C H 2 C 1 2 In CC1 4

S n l 4

I 2

NaCl

CaS0 4

4. Describe the change in appearance of S n l 4 in any solvent w i th which it seemed to react.

Behavior when heated:

S n l 4

I :

C i 0 H 8

NaCl

(Answer the questions from page 83 on a separate sheet of paper and attach it to the Report Sheet.)

Experiment 8 Salt Identification Using Visible and Infrared Spectra

Introduction

Experiment 6, "Line Spectra of Elements," showed you (among other things) that gaseous atoms excited in an electric discharge emit light of characteristic color (for example, red for Ne). By using the Bunsen spectroscope, this colored light can be resolved into lines of a single wavelength. The behavior of excited atoms reveals interesting facts about the arrangement of electrons in atoms.

Characteristic light colors can also be put to a simpler, more practical use: they can be used to determine the presence of particular elements. Given the elements which may possibly be present in an unknown mixture , you could observe the light emission from each one separately and compare it w i th the light emission from the unknown. For example, if Ne and Ar are present in a mixture , then the emission spectrum w i l l show lines from both elements; i f only Ne is present, then only lines from Ne w i l l be observed.

During the nineteenth century, the Bunsen spectroscope was used extensively to make determinations of this k ind . Even before the reasons for the existence of line emission spectra were k n o w n , chemists relied on observations of the wavelengths of emitted light as evidence for the presence or absence of particular elements. If the light emit ted is in the visible region of the electromagnetic spectrum, an observation of its color is often sufficient to identify the emit t ing element. The color is, of course, determined by the wavelengths of emission.

Determining which elements are present in a sample is called qualitative analysis. The usual method for testing the presence or absence of elements is to observe a property that is unique to the element being tested for. Some ions when excited by the heat of a flame emit characteristic colors of light in the visible region (4000— 7000 A) of the electromagnetic spectrum as their electrons drop back into lower energy levels. Among the ions which can be identified by the characteristic colors of their electronic emission spectra are Na + , K + , C a 2 + , S r 2 + , and B a 2 + .

For the most part, the electronic spectra of simple anions do not exhibi t emission or absorption bands in the visible region. Many polyatomic anions do exhibi t vibrational spectra resulting from the absorption of electromagnetic radiation in the 1-100 p. (1 ju = 1 micron = 10,000 A) region as they stretch and bend.

The exact number and intensity of such absorptions depend on the number of atoms in the polyatomic species and how they are arranged w i t h respect to one another. As an example, the N 0 2 ~ ion has a bent planar structure and may undergo two kinds of stretches (known as the two stretching modes); the stretching modes and also a binding mode of this ion are depicted in Figure 8-1.

In the symmetric mode in Figure 8- l (a) ; 'both N—O bonds stretch and compress

37

88 Experiment 8

in unison; energy absorbed in the 1325-1380 cm" 1 region (~7.5 y ) w i l l cause the ion to vibrate in this fashion. In the antisymmetric mode in Figure 8-1(b), one bond stretches as the other compresses; this vibration is excited by energy in the 1230-1275 c m - 1 region (~8 .0 u ) . In addit ion, a bending mot ion in which the O - N — O angle changes is depicted in Figure 8-1 (c); this vibrat ion is excited by energy in the 810—850 c m - 1 region (~12 y ) . The infrared spectrum of N a N 0 2

(Figure 8-2) shows how strongly the salt absorbs infrared radiation at each wavelength. Y o u can see that there is a strong absorption at 830 c m - 1 corresponding to the O—N—O bend. The stretching absorptions are quite close energetically, so that only one rather broad band including both stretching absorptions appears in the spectrum o f N a N 0 2 .

You are not expected to know exactly how many absorptions any particular anion should have. The purpose of this discussion is simply to acquaint you wi th the analysis of one particular spectrum. (Note that it is more convenient to measure and discuss absorptions than emissions in the infrared region.)

Wavelength (microns)

2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5 8 9 10 11 12 14

4000 3000 2000 1800 1600 1400 1200 1000 800

Frequency ( C M - 1 )

FIGURE 8-2

Infrared spectrum of N a N 0 2 in KBr pellet. (Recorded on Perkin-Elmer 700.)

89 Salt I d e n t i f i c a t i o n Using Vis ible and In f ra red Spectra

Table 8-1

Salts Used as Unknowns

Dromates: K B r 0 3 , N a B r 0 3

Carbonates: ( N H 4 ) 2 C 0 3 , B a C 0 3 , L i 2 C 0 3 , K 2 C 0 3 , N a 2 C 0 3 , S r C 0 3

Nitrates: N H 4 N 0 3 , B a ( N 0 3 ) 2 , L i N 0 3 , K N 0 3 , N a N 0 3 , S r ( N 0 3 ) 2

Nitrites: K N 0 2 ) N a N 0 2

Phosphates: ( N H 4 ) H 2 P 0 4 , K 2 H P 0 4 - 3 H 2 0 Sulfates: ( N H 4 ) 2 S 0 4 , BaS0 4 , K 2 S 0 4 , N a S 0 4 , S r S 0 4

Sulfites: B a S 0 3 , K 2 S 0 3 • 2 H 2 0 , ~Na 2S0 3 , S r S 0 3

Thiocyanates: N H 4 N C S , B a ( N C S ) 2 - 2 H 2 0 , KNCS, NaNCS

Even when an experimenter does not know the particular atomic motions causing each absorption band, the infrared spectrum of a polyatomic anion serves as a property by which the anion can be identified. Several spectra of inorganic salts are depicted at the end of this experiment. You may notice that the appearance of some spectra depends not only on the anion but also on the cation—at least to some extent. This arises because the solid state packing of anions and cations introduces some complications that we wi l l not pursue here. Nevertheless, it should be obvious that infrared spectra are characteristic properties by which a particular salt can be identified.

Your ins t ructor ,wil l provide you wi th an unknown inorganic salt from the group whose infrared-spectra are given following page 9 1 . (These salts are listed in Table 8-1). Y o u wi l l then identify the cation and the anion present, using flame tests (for the cation) and infrared spectroscopy (for the anion).


BaCl 2 NH 4 C1 CaCl 2 SrCl 2

cone. HC1 acetone KBr (solid) unknown salt (from group KG whose infrared spectra LiCl are provided), dried NaCl overnight at 70°C


attenuator burner cobalt glass cork hexagonal wrenches (2)

infrared spectrophotometer mortar and pestle pellet press platinum or nichrome wire watch glass

90 Experiment 8

P R O C E D U R E

A. Flame Tests

Obtain a piece of plat inum or nichrome wire and make a small loop in one end by wrapping the wire around a pen or a pencil. Insert the other end of the wire in to a small cork, which wi l l serve as a handle. This wire w i l l be used to introduce samples into the burner flame. Obviously you wi l l want to be sure that the wire itself is free of any material that could color the flame. You can clean the wire by dipping the loop into some concentrated HC1 in a small test tube and then holding it in the hottest part of the burner flame. Repeat this process unt i l the wire no longer gives a color to the flame. It wi l l be necessary to clean the wire each time you change salts. (Why?) The HC1 wi l l convert any compounds on the wire to their chlorides, which can then be easily volatil ized. The HC1 wi l l eventually become contaminated and should be changed before working wi th your unknown salt.

Obtain a small sample of a known salt of each cation (L iC l , NaCl, K G , N H 4 C 1 , BaCl 2 , S rCl 2 ) . For each sample, dip the wet loop of the clean wire into the solid and thrust it into the (lame. Observe the characteristic flame color of each sample and note any pertinent features of the color (Does the color appear immediately when the sample is inserted into the flame? Does it last for a long time or a short time?) Record these observations in your notebook. You should observe each flame test several times. Then identify your unknown cation by its flame behavior.

The yellow color produced by Na + emission wi l l appear in all flame tests. This is because Na + impurities are impossible to remove (they are introduced from glass storage bottles). To determine whether an unknown is a sodium salt, you wi l l have to compare the intensity of the observed yel low color w i th a known sodium salt. In some cases, the yellow color from Na + impurities masks emissions from other cations that are present. This is particularly true for K + . So it is advisable to view the flame test for any unknown suspected of containing K + through a double thickness of blue cobalt glass. This glass wi l l filter out the yel low light from Na + .

B. Infrared Spectra

A convenient way to record the infrared spectrum of a solid is to use a solid solution in KBr. (Water absorbs strongly in the infrared region and therefore cannot be used as a solvent.) Potassium bromide does not absorb in the infrared region above 650 c m " 1 . This solid solution is pressed in to a translucent pellet.

To make a pellet, you should take only your solid sample and a spatula to the spectrophotometer area. A small mortar and pestle wi l l be provided there. Remove the KBr from the desiccator, where it is kept to exclude water. Place about 70 mg of KBr on the mortar and return the KBr to the desiccator. With your spatula, add 3 or 4 mg of your unknown solid. Grind the mixture to a very fine powder w i t h the pestle. This grinding w i l l take about two or three minutes. You should stop occasionally to scrape the powder into a pile in the mortar and then begin grinding again.

When the sample is a fine powder, obtain the pellet press, which consists of a metal block w i t h a threaded hole and two polished end bolts (see Figure 8-3(a)).

91 Salt Identification Using Visible and Infrared Spectra

<a) (b)

FIGURE 8-3

(a) Pellet press block and bolt . (b) Making a KBr pellet.

Screw one of these bolts about halfway into the block. With your spatula, sprinkle a thin coating o f the powder over the bolt surface un t i l i t is evenly covered. This w i l l probably not require all the powder you prepared. Now insert the other bolt and tighten it as much as you can wi th your fingers. Fi t the two hexagonal wrenches around the bolts, and tighten both bolts against each other as much as you can to create pressure on the powder inside (see Figure 8-3(b)). Maintain the pressure for a few seconds, and then let the block sit for about three minutes. At the end of this t ime, loosen both 'bol ts . Y o u should have a translucent pellet inside the threaded cavity of the block. If the pellet broke or is not translucent, repeat the procedure.

The block is placed in the sample beam of the infrared spectrophotometer. (See Appendix I I I concerning the construction of a spectrophotometer.) Record the spectrum, fol lowing the instructions supplied by your instructor. If your sample has absorbed water from the air, you w i l l see additional bands around 2.9 and 6.25 n.

Wash the mortar, pestle, block, and bolts w i t h distilled water; fol low wi th an acetone rinse.

R E S U L T S

On the Report Sheet for this experiment, record your observations regarding the flame tests you conducted.

Identify your unknown salt from the flame test and the infrared results.

R E F E R E N C E S

1. Yoder, C.H., Suydam, F .H. , and Snaveley, F .A. Chemistry. New York : Harcourt Brace Jovanovich, Inc., 1975. Chapter 8.

2. Ackermann, M . N . Journal of Chemical Education 47, 69 (1970) .

F IGURE 8-4 Infrared spectra of unknown salts in KBr pellets. (Recorded on Perkin-EImer 700.)

2.5


4 4.5 5 5.5 6 6.5 7 7.5 8 9 10 I I 12 14 - T -

4- I 1 1

* / •

r •—• A L u

i I, \

/ J f

/ / / /

a. KBr0 3 a. KBr0 3

400(1 3000

3.5

2000 1800 1600

Frequency (CM" ' )


4.5 5 5.5 6 6.

1400 1200 1000 800

10 11 12

4000 3000 2000 1800 1600 1400

Frequency ( C M ' 1 ) 1200 1000 800

H)00 3000 2000 1800 1600 1400 1200

Frequency (CM" 1 )

1000 800

92

•I —

Experiment 9 Synthesis of Coordination Compounds and Their Identification Using Spectrophotometry

Introduction

Several features of synthetic chemistry have already been discussed in the in t roduct ion to Experiment 5. You should review this material in conjunction w i t h Experiment 9 .

Here, you wi l l gain experience in synthetic techniques by preparing a synthetic intermediate, isolating i t , and using it as starting material to synthesize another complex. This intermediate is related to the cobalt complex of Experiment 5; the difference is that one of the neutral N H 3 molecules coordinated to C o 3 + has been replaced by the dinegative carbonate ion C 0 3

2 ~ (see Figure 9-1). This decreases the charge on the complex ion to + 1 , so that only one nitrate is needed to balance the charge. The formula for this compound is often wr i t ten [ C o ( N H 3 ) s ( C 0 3 ) ] ( N 0 3 ) to show that the molecules and ions w i t h the square brackets are attached to C o 3 + . The name of this compound is pentaamminecarbonatocobalt(III) nitrate.

The method o f preparation is given in Equation ( 9 - 1 ) :

4 C o ( N 0 3 ) 2 - 6 H 2 0 + 4 ( N H 4 ) 2 C 0 3 + 1 6 N H 3 + 2 0 2 ->

4 [ C o ( N H 3 ) 5 ( C 0 3 ) ] ( N 0 3 ) + 4 N H 4 N 0 3 + 2 6 H 2 0 (9-1)

The purpose of the oxygen is to oxidize the C o 2 + in the starting material to C o 3 + .

105

106 Experiment 9

This avoids using H 2 0 2 as in Experiment 5, which must be disposed of by heating:

2 H 2 0 2 >°~b0\ o2 + 2 H 2 0 (9-2)

After you have prepared [ C o ( N H 3 ) s ( C 0 3 ) l ( N 0 3 ) , your instructor wi l l give you an unknown substance to react w i th i t . This is to replace the carbonate ion w i t h some group X to produce the complex ion [ C o ( N H 3 ) 5 X ] " + . The value o f n wi l l depend on the charge ( i f any) on the group X. If X has no charge, then n - 3; the complex ion charge is determined only by the +3 charge on the Co, since all the groups bonded to Co are uncharged. The [ C o ( N H 3 ) 5 X ] 3 + ion wi l l pick up some o f the N 0 3 " ions in solution to form solid [ C o ( N H 3 ) 5 X | ( N 0 3 ) 3 . If the group X has one negative charge, it w i l l neutralize one o f the postive charges on the cobalt to produce a [ C o ( N H 3 ) 5 X | 2 + c a t i o n . In this case, the cation can utilize either N 0 3 ~ or X" anions to neutralize its charge, and [ C o ( N H 3 ) 5 X ] ( N 0 3 ) 2 o r [ C o ( N H 3 ) 5 X ] X 2

(whichever is less soluble) precipitates.

The reaction in which An X group replaces carbonate is carried out in H N 0 3

solution which destroys the C 0 3

2 ~ ion and causes C 0 2 gas to bubble off :

C 0 3

2 ~ + 2 H + ^ H 2 C 0 3 -> H 2 0 + C 0 2 ( g ) (9-3)

Once you have isolated a solid containing the [ C o ( N H 3 ) 5 A ' ] " + cation, you wi l l identify i t . Since these cations are brightly colored, they can be identified by their characteristic colors. You may remember that in Experiment 2 the property color was not used as a possible means of identification because there was no precise .numerical way to measure i t . You wil l now learn to measure the color of a compound quantitatively so that you can identify the complex you prepared.

The technique, called spectrophotometry, makes use of the fact that substances are colored because they absorb visible light of certain colors and transmit the remaining colors. As you probably remember from high school science courses, ordinary white light is composed of light of all colors visible to the naked eye. The color of light depends on its wavelength X. Figure 9-2 indicates the approximate wavelength ranges for various colors. Ordinary visible white light is a mixture of light having all wavelengths between 4000—7000 A. Light having wavelengths shorter than 4000 A is called ultraviolet light and cannot be seen by the human eye. Infrared light, which has wavelengths longer than 7000 A, is also invisible to the human eye.

107 Synthesis of Coordination Compounds and Their Identification Using Spectrophotometry

FIGURE 9-3

Light passing through a cell.

Solutions of various compounds are co'ored, then, because they absorb light of some wavelengths and transmit light of ov wavelengths. The color you see corresponds to the color of the transmitted l ight . For example, a solution of plant food appears green because it absorbs the light in the red, orange, yel low, blue, and violet wavelength regions and transmits green light. Light in the visible and ultraviolet regions of the spectrum causes electrons to j u m p to higher energy levels when it is absorbed.

Figure 9-3 diagrams a quantitative way of measuring light absorption by a solution. Some of the sample is placed in a small vessel called a cell or cuvette. Light of a single wavelength X having an intensity I 0 is shined into the cell. The solution may absorb some light of this wavelength and transmit the rest w i t h reduced intensity /. The quant i ty ( 7 / / 0 ) is obviously related to how strongly the solution absorbs the light. I f most o f the light is absorbed, / w i l l be quite small and ( / / / 0 ) w i l l be quite small. Conversely, if most of the light is transmitted, / w i l l be large and ( / / / 0 ) w i l l also be large. The quant i ty ( / / / 0 ) X 100 is called the percent transmit-tance T:

The defini t ion of A indicates that, at any wavelength where the sample absorbs strongly (where I 0 / I is large), /! w i l l also be large. When l i t t le light is absorbed,/! w i l l be small. A plot of the absorbance of a solution at various wavelengths, called an absorption spectrum, shows what colors are strongly absorbed and what colors are transmitted. The spectrum of [ C o ( N H 3 ) 5 ( C 0 3 ) ] ( N 0 3 ) 2 is depicted in Figure 9-4. The strong absorption of light in the orange, yel low, green, blue, and violet regions of the spectrum gives the complex the red color of the transmitted l ight. The color of the solution is complementary to the colors that are absorbed.

108 Experiment 9

Tire spectrum of a compound is a characteristic property and may be used to identify the compound. Rather than reproducing the entire spectrum, it is customary to report the wavelengths of maximum absorbance. These are indicated on the spectrum of [ C o ( N I l 3 ) 5 ( C 0 3 ) ] ( N 0 3 ) 2 in Figure 9-4. You can identify your compound by measuring its absorption spectrum and comparing it to the spectra of the possible products given at the end of this experiment. Note that this type of color comparison is similar to the flame tests for color that you made in Experiment 8. However, the more precise technique used here allows you to distinguish easily between different shades o f red, for example, because they w i l l have slightly different wavelengths of maximum absorption.

Locating the exact position of the maximum absorption X m a x can be d i f f icu l t , since the absorption curves are somewhat flat. A good technique is shown in Figure 9-5. Lines are drawn tangent to both sides of the curve and extrapolated to the point where they intersect. This point is taken as the point of maximum absorption.

The measured value of the absorbance A is related to several experimental factors by the Beer-Lambert Law

A = eel (9-6)

where e is the ext inct ion coefficient, c the concentration in moles/liter, and / the thickness of the cell.

This law simply states that the absorbance A at any wavelength X depends on (1) how many absorbing molecules a light beam encounters in passing through the cell (this number w i l l obviously increase as the cell thickness / and the concentration c increase), and ( 2 ) an intrinsic property of the molecules for absorbing at wavelength X (e).

Appendix I I I describes how measurements of absorbance at various wavelengths are made experimentally.



50-ml beakers ring Buchner funnel ring stand burner spectrophotometer 125-ml and 250-ml Erlenmeyer flasks thermometer filter flask universal clamp filter paper vacuum tubing graduated cylinder volumetric flask medicine dropper wire gauze pneumatic trough


CH3OH (methanol) 30 g o f ( N H 4 ) 2 C 0 3

20 g of C o ( N 0 3 ) 2 - 6 H 2 0 unknown solution (provided by ice instructor) cone. N H 3

P R O C E D U R E

A. Synthesis of Pentaamminecarbonatocobalt ( l l l )ni trate [Co(NH 3 ) s (C0 j ) ] (NO3K H 2 0

Weigh a clean, empty 50-ml beaker and record its weight. Now add about 20 g of C o ( N 0 3 ) 2 * 6 H 2 0 and record the weight of the beaker and salt. The weight of C o ( N 0 3 ) 2 * 6 H 2 0 is the one on which your yield calculation w i l l be based, and it is the only weight that must be known accurately.

To a 250-ml Erlenmeyer flask, add about 30 g of ( N H 4 ) 2 C 0 3 (ammonium carbonate) and dissolve this in about 30 ml of distilled water. Add about 50 ml of

110 Experiment 9

concentrated N H 3 solut ion. Dissolve the C o ( N O _ 1 ) 2

, 6 H , 0 in 10 ml of water and pour this solution into the Erlenmeyer flask; be careful to rinse the beaker to remove all of the cobalt salt. Continue to swirl the flask gently unt i l the contents of the flask are thoroughly mixed.

The solution is now oxidized by bubbling air into the flask. The most convenient method for doing this is to attach a piece of rubber tubing to the air jet and to place a medicine dropper in the other end of the tubing. The medicine dropper is placed in the flask, and the air jet turned on to provide gentle bubbling. The bubbling should be continued overnight or for at least 10 hours. You may need to add a l i t t le dist i l led water to replace some of the water lost by evaporation during the oxidat ion step. Try to be sure that all the material is sti l l in solution at the end of the oxidat ion . However, do not heat the solution to dissolve any precipitate.

Let the flask stand while you obtain a pneumatic trough full of ice. (This ice bath can be shared by two students.) Set the flask in the ice bath to cool , and when the temperature is near 0 ° , crystals should begin to form. Crystallization occurs slowly and may require scratching the beaker w i th a stirring rod. Do not expect all the product to precipitate immediately. Isolation of a crystalline product requires considerable patience.

Cool the flask in the ice bath un t i l no more crystals of the red product seem to be forming. If you have trouble crystallizing the product, you may add some methanol to the solution. (The solution absolutely cannot be allowed to sit in your drawer unt i l the next laboratory period and then cooled.) Af ter this cooling, collect the crude product by f i l t ra t ion on a Buclmer funnel; wash the product wi th a l i t t le ice water and then wi th cold methanol.

Place the compound in an evaporating dish in your desk; allow air to dry the product for several days. Then weigh the dry product.

B. Synthesis of [ C o ( N H 3 ) s X ] ( N 0 3 ) 2

Your instructor wi l l provide you wi th enough of a compound that is a source o f an X group to react w i t h 3.0 g of f C o ( N H 3 ) s ( C 0 3 ) ] ( N 0 3 ) . Depending on the nature o f your unknown solut ion, your instructor wi l l direct you to fol low Procedure A or Procedure B below in synthesizing your final product.

Procedure A. Place 3.0 g of [ C o ( N H 3 ) 5 ( C 0 3 ) ] ( N 0 3 ) in a 250-mI Erlenmeyer flask Add about 15 ml of distil led water. Then, working under a hood, add the unknown solution provided by your instructor. C A U T I O N : This solution contains concentrated H N 0 3 ! After stirring occasionally for about 30 minutes, add 150 ml of methanol to precipitate your product. Collect the product by suction f i l t ra t ion on a Biichner funnel (see In t roduc t ion , page 15). Wash the compound w i t h cold water and then w i th methanol. A i r dry the solid and then weigh i t .

Procedure B. Place 3.0 g of [ C o ( N H 3 ) s ( C 0 3 ) ] ( N 0 3 ) in a 125-ml Erlenmeyer flask. Add about 15 ml of distilled water. Then, working under a hood, add the unknown solution provided by your instructor. C A U T I O N : This solution contains concentrated H N 0 3 ! Place the flask on a ring stand w i t h wire gauze and heat the flask flask w i t h a burner. The solution should be heated at a temperature range of 6 5 - 7 5 ° C for one hour, either using a burner or a steam bath. Cool the solution in an


ice bath and collect the precipitate by Filtration on a Buchner funnel (see Introduct ion, page 15). Wash the precipitate w i th cold water and then wi th methanol. A i r dry the solid and then weigh i t .

C. Product Identification

You wi l l identify the product you have prepared by measuring its absorption spectrum. As precisely as possible, weight out 0.5 g of your dry unknown compound. Transfer the compound quantitatively to a clean, 250-ml volumetric tlask. After completely dissolving the compound in a few ml of distilled water, dilute the solution so that the bo t tom of the meniscus matches the mark on the neck of the tlask. Then invert the tlask several times to ensure complete mixing.

Place some of the solution in the cuvette of the spectrophotometer and read the percent of transmittan.ee at intervals of 250 A over the visible range. Your instructor wi l l provide directions for using the spectrophotometer. (A common type o f visible-UV spectrophotometer is discussed in Appendix 111.)

( I f the 0.5 g quanti ty of your product wi l l not dissolve completely in 250 ml of distilled water, discard the solution and rinse the volumetric tlask wi th distilled water. Then, as precisely as you can, weigh out about 0.2 g of your dry unknown and proceed as before.)

http://transmittan.ee

113 Synthesis o : Coordination Compounds and Their Identification Using Spectrophotometry

R E S U L T S

1. Using the weight of C o ( N 0 3 ) 2 - 6 H 2 0 wi th which you began the experiment, the weight of your product, and Equation (9-1), calculate the percentage yield of ( C o ( N H 3 ) ' 5 ( C 0 3 ) ] ( N 0 3 ) .

2. Plot absorbance vs. wavelength on a piece of graph paper. Draw a smooth curve through your experimental points to obtain the spectrum of your unknown compound in the visible region.

3. Locate the positions of maximum absorbance. 4. Compare your spectrum wi th the spectra for known cobalt complexes, and

identify the final product you prepared. (Consult Figures 9-6 through 9-10.) 5. (a) Knowing the identi ty of your product, compute its formula weight by using

a table of atomic weights. (b) From the formula weight and the amount of the product that was weighed

. out , calculate the concentration c of the solution you prepared. (c) From your plot, read the absorbance values at the points of maximum

absorbance. Use this information and the solution concentration to compute the ext inct ion coefficient e at each maximum from Equation (9-6).

6. Knowing the identi ty of the final product you synthesized, write an equation for its product ion from [ C o ( N H 3 ) 5 ( C 0 3 ) ] ( N 0 3 ) . Assume that anions are provided as their sodium salt or come from H N 0 3 . Compute the percentage yield of this compound.

7. Turn in your products in test tubes labeled w i th your name, section number, and instructor's name.

Q U E S T I O N

What are possible sources of random or systematic error in the values of e you calculated?

R E F E R E N C E S

1. Yoder, C.H., Suydam, F .H. , and Suavely, F.A. Chemistry. New Y o r k : Harcourt Brace Jovanovich, Inc., 1975. Chapter 8.

2. Basolo, F., and Murman, R.K. Inorganic Syntheses 4, 1 7 1 (1948) . 3. Olson, G.L. Journal of Chemical Education 46, 508 (1969) .

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Name Date

Section

6. Equation for product synthesis:

Percentage yield of product

Show calculation:

(Answer the question from page 113 in the space remaining below; use the back of the Report Sheet, if necessary.)

Experiment 10 Vaporization of a Liquid: Vapor Pressure and Heat of Vaporization

Introduction

Any l iquid contained in a closed vessel evaporates to at least a small extent. In other words, some molecules leave the surface of the l iqu id and become vapor. As these molecules in the vapor phase move about randomly, some may strike the surface of the l iquid and again become part of the l iquid phase; that is, they may condense. Eventually, a state o f equil ibr ium wi l l be reached in which molecules vaporize and condense at exactly the same rate and there is no change in the number of molecules in either the l iquid or the vapor phase. This vapor, like any gas, exerts pressure. When the l iquid and vapor phases of a substance are in equi l ibr ium, this pressure is called the vapor pressure of the l iquid . The vapor pressure is the maximum pressure that can be exerted by the vapor phase of a substance at a particular temperature. It is possible to tell when this equi l ibr ium between phases has been reached, because at this point the volume of l iquid (which can be measured) no longer decreases and the pressure exerted by the vapor (which can be measured) no longer increases. In this experiment, you w i l l determine when equil ibr ium has been established between the l iquid and vapor phases of a substance by measuring the pressure exerted by the vapor.

The size of a l iquid's vapor pressure is a measure of the number of molecules in the vapor phase at equi l ibr ium. It therefore reflects the tendency of molecules to escape from the surface of the l iqu id . In the l iquid phase, molecules exert attractive forces on one another. A certain amount of energy is required for a particle to overcome these forces—to leave the l iquid and to become part of the gas phase. Since temperature is a measure of average molecular kinetic energy, it is reasonable to expect that as the temperature increases, the probabil i ty increases that the molecules wi l l possess the energy necessary for escape. Consequently, vapor pressure should increase as temperature increases. In part of this experiment, you wi l l measure the vapor pressure o f one l iquid at several temperatures; you wi l l see that these two measurable properties of the l iquid do indeed decrease or increase together.

You w i l l take measurements that show that the vapor pressure o f a l iquid is related to temperature and then use these measurements to calculate the energy^ needed to overcome intermolecular attractions in the l i qu id . The quanti ty you w i l l calculate is called the molar heat of vaporization AHV. This is the heat necessary to convert one mole of l iquid in to one mole of vapor at a constant temperature. For example, CC1 4(1) + AIIV -*• CCl 4 (g ) . The mathematical relationship between vapor pressure and the molar heat of vaporization is given by the Clausius-Clapeyron equation:

119

120 Experiment 10

log P v = - A H v (-) + C (10-1) 2 . 3 0 3 R\TJ

This equation was derived in the branch of chemistry and physics known as thermodynamics; you must use this equation here, but you do not need to be concerned wi th its derivation. Equation (10-1) is in the form of the equation for a straight line y - mx + b. Here ,^ = log Pv and x - \/T. If the logarithm (to the base 10) of the vapor pressure is plot ted against the reciprocal of the Kelvin temperature, a straight line is obtained. The y intercept b corresponds to the constant C in the Clausius—Clapeyron equation; you wi l l not be concerned w i t h this constant. The slope of the line equals -AHv/2.303 R, where R is the gas constant 1.987 ca l / °K mole. After determining the slope of the line from your p lot , a value for AHV in cal/mole can easily be calculated. This determination of the heat of vaporization is a good example of how scientists measure a quanti ty which is not easily accessible (A/ /„) by using measurements that are easy to make (Pv and T) and by applying an equation (the Clausius—Clapeyron equation) that relates these quantities.

EQUIPMENT N E E D E D

400-ml beaker ring stand - v » ^ v f *

capillary tubes rubber band - kr— A*- ^ ^a**.*^ clamp serum cap copper wire stirrer syringe cork ( w i t h vent) test tube

W;!-*^^-^ 0 1 2 5 - m l filter flask thermometer manometer Tygon tubing ring wire gauze

CHEMICALS N E E D E D

ice organic liquids

MEASURING P R E S S U R E WITH A MANOMETER

The pressure of a confined gas sample can be measured w i t h a manometer. This simple instrument consists of a U-shaped glass tube partially t i l led w i t h mercury. When both arms of the U are open to the atmosphere, the pressure on the mercury in the two arms is the same; thus, the mercury levels wi l l be the same in both arms. Attach one arm of the manometer to a closed flask containing air at atmospheric pressure; the mercury levels w i l l remain the same in both arms, since the pressure on the mercury in the two arms is still the same (atmospheric pressure). Then a l iquid w i l l be put in to the flask. As the l iquid vaporizes, the pressure inside the flask w i l l rise above atmospheric pressure and the mercury w i l l be pushed down in the arm that is connected to the flask (see Figure 10-1). The difference in the mercury levels

121 Vaporization of a Liquid: Vapor Pressure and Heat of Vaporization

FIGURE 10-1

Using a manometer to measure vapor pressure.

Ah is the amount by which the pressure in the flask exceeds atmospheric pressure:

Pfiask ^ . t m + A/; (10-2)

The total pressure in the flask after the l iquid vaporizes is the air pressure in the flask plus the vapor pressure of the l iqu id :

. v / > f l « . k ^ . i r + / > , (10-3)

Since Pa„ = Patm , the total pressure is the same as Patm + Pv, and the difference Ah is simply the vapor pressure of the l iqu id . You can measure Ah in either mm or cm of mercury; just be sure that you know which unit you are reading and that you record the proper unit .

If the vapor pressure of the l iquid were extremely high, the mercury might be pushed down past the bend of the U, so that the height of the mercury column could not be measured; mercury might even be pushed out of the open arm. To measure the pressure of such a l iqu id , the flask should be partially evacuated before the l iquid is put in to the flask. To do this, the needle from a syringe is inserted through the cap closing the flask un t i l the needle extends into the flask. Vacuum tubing is then used to attach the needle to a water aspirator. The aspirator is turned on and air is drawn out of the flask un t i l the mercury in the arm connected to the flask is w i th in about a centimeter of the bend (see Figure 10-2). The air pressure in the flask is now less than atmospheric pressure by the amount Ah':

f a i r ^ t m - A V (10-4)

After the l iquid is put into the flask and has vaporized, the mercury w i l l be pushed down in the arm connected to the flask, as First shown in Figure 10-1. Once again, the difference Ah is the amount by which the pressure in the flask exceeds atmospheric pressure (see Equation 10-2). As before, the total pressure inside the flask is the sum of the air pressure and the vapor pressure (see Equation 10-3).

However, now the air pressure does not equal the atmospheric pressure. You must use Equation (10-4) to calculate the air pressure, Equation (10-2) to calculate the total pressure inside the tlask, and, f inally, Equation (10-3) to calculate the vapor pressure. Obviously, this procedure and the accompanying calculations here are more troublesome than they were in the first example._If your attempt to measure vapor pressure at room temperature indicates that the second procedure is necessary, consult your instructor before you begin i t .

P R O C E D U R E

The filter tlask must be clean and dry. Stopper it wi th a serum cap. The center port ion of the cap should be placed inside the neck of the tlask and the por t ion that extends upward should be pulled down around the neck on the outside of the tlask; pull the outside port ion of the cap down as low as you can so that it fits as t ight ly as possible. Clamp the flask to a ring stand. Use a short piece of Tygon tubing to attach the flask to the manometer; do this after you have positioned the serum cap. When you put the tubing onto the manometer, grasp the glass tubing as close to the end as possible; push the Tygon tubing on only far enough to be sure that it cannot fall off. The mercury levels should be the same in both arms of the manometer; if they are not, remove and reconnect the tubing.

Draw about 1 ml of your l iquid into a syringe. Turn the syringe upside down and push the plunger far enough to expel any air bubbles. Then plunge the needle through the center of the serum cap and inject the l iquid into the tlask. As you do this, keep a firm grip on the plunger of the syringe so that it cannot be pushed upward by pressure from the vaporizing l iqu id . Remove the needle from the cap. As you do this, be sure that the needle and the syringe do not separate from one another; if they do, air and vapor can escape through the needle. After approximately three minutes, read the mercury levels in the two arms of the manometer. Repeat this reading at one-minute intervals unt i l the levels longer change. When

123 Vaporization of a Liquid: Vapor Pressure and heat of Vaporization

the levels have stabilized, the difference All is the vapor pressure of the l iquid . If you find that the difference Ah is decreasing after several minutes, there is a leak in the system which must be found and eliminated; you may need your instructor's assistance. After completing this measurement, disconnect the tubing from the flask, pour the l iquid in the flask into a waste container, and thoroughly dry the flask (preferably w i th a strong stream of compressed air). It is absolutely essential to dry the flask. I f vapor remains in the flask, it w i l l l imi t the amount o f additional l iquid that can vaporize; this wi l l cause your measurements to be low.

Next, measure the liquid's vapor pressure at 0 °C . Immerse the stoppered flask as deeply as possible in a beaker of water and ice; let it stand for at least 10 minutes. Be sure to use enough ice to maintain a temperature of 0°C down to the bot tom of the flask. Then attach the flask to the manometer. Wait five minutes; if the mercury levels remain the same in both amis, inject the l iquid and measure its vapor pressure. If the levels have changed during this time, the air in the flask has not yet cooled to 0°C and its pressure is still dropping as it cools further; disconnect the flask and reattach it before injecting the l iquid. After you are finished, pour out the l iquid and dry the flask.

Now measure the vapor pressure of your l iquid at a temperature of about 10—15°C. Immerse the stoppered flask in a beaker of cold tap water or in water that has been cooled by the addition of ice. Add ice or cold water as needed to maintain a constant temperature. Be sure to record the actual temperature of this bath. After you are sure that the flask and the air have cooled sufficiently, add the l iquid and measure its vapor pressure.

A fourth set of temperature/vapor pressure data can be obtained by recalling the definit ion of boiling point : the temperature at which the vapor pressure of a l iquid equals the pressure of the atmosphere above the l iquid . Use the technique you learned in Experiment 2 to measure the boiling point . Read the barometer to determine the atmospheric pressure; that is, the vapor pressure of the l iquid at the measured boil ing point.

R E S U L T S A N D D I S C U S S I O N

Review Appendix II (page 357) on graphing experimental data. Prepare a graph plo t t ing the vapor pressure of your l iquid on the ordinate and

the temperature on the abscissa. On the back of the graph, compile a data table showing the values of P v and of T that you used. The line through the experimental points should be a continuous one; discuss what the general shape of this plot tells you about the relationship between vapor pressure and temperature.

Now prepare a graph plot t ing log P v along the ordinate and 1 /T( in ° K ) on the abscissa. Y o u must convert each \/T to a decimal; do not use fractions. Draw the best straight line through the points. On the back of the graph, compile a data table showing Pv, log Pv, T, and 1/7 for each measurement.

Determine the slope of the straight line for this second graph. From the slope and the Clausius-Clapeyron equation, calculate the value of AHV for your l iquid . Be sure to include the proper units.

124 Experiment 10

Q U E S T I O N S

1. How w i l l each o f the fo l lowing affect the measured vapor pressure in comparison to the true vapor pressure for your liquid? (a) A long length of slightly porous rubber tubing is used instead of a short

length of Tygon tubing. (b) The flask is immersed only halfway in the 0°C bath. (c) A large air bubble is injected in to the flask along w i t h the l iquid. (d) A 250-ml flask is used instead of a 125-ml flask.

2. Liquid A has a vapor pressure of 10.2 cm of Hg at room temperature and l iquid B has a vapor pressure of 26.8 cm of Hg at the same temperature. In which l iquid are there stronger intermolecular attractions? Which l iquid would be expected to have the higher normal boil ing point? A higher heat of vaporization?

R E F E R E N C E

Yoder, C.H., Suydam, F .H. , and Snavely, F .A. Chemistry. New York : Harcourt Brace Jovanovich, Inc., 1975. Chapters 9 and 10.

Experiment 11 Molar Weight by Vapor Density

Introduction

The ideal gas law describes the quantitative relationship among the pressure exerted on or by a gas sample, the volume that the gas occupies, its temperature, and the number of moles of gas n in the sample:

PV=nRT (11-1)

An important aspect of this relationship is that it applies to all gases: it does not depend on the size, shape, mass, or composition of the particles of the gas. Consequently, it is possible to use measured pressure, volume, and temperature data to " coun t " the number of moles (or number of molecules) of gas in any gas sample wi thou t knowing the identi ty of the gas. If one more piece of data is known, it is also possible to learn one important physical property of the specific gas in the sample: if the mass of the gas sample is known, the weight of one mole of the gas can be calculated. The number of moles of a substance in any weighed sample can be calculated by dividing the mass of the sample by the weight of one mole of the substance:

• ' i grams g M l « moles = — - — or / ; =— (11-2)

grams/mole M

This expression for n can be put in to the ideal gas law:

PV=(l-)RT (11-3)

The equation now contains the four measurable variables of a gas sample—pressure, volume, temperature, and mass—that can be used to calculate the molar weight of the gas. Rearranging the terms in Equation (11-3) to solve for molar weight gives

M = g R T (11-4) PV

In this experiment, you w i l l work w i t h a compound that is a l iquid at room temperature but that has a boiling point below the boil ing point of water. Y o u w i l l measure the mass of gas that occupies a particular volume at 100°C and at atmospheric pressure. Then, you wi l l use your measured values for these properties to calculate the molar weight of the compound.

Remember that the equation you are using to calculate molar weight is called the ideal gas law. It is an exact description of the behavior of a gas only if the gas is ideal; that is, if there are no attractive forces between the particles of the gas and if the particles of the gas are volumeless points. Of course, neither of these characteris-

127

128 Experiment 11

tics is true for a real gas. However, the behavior of a real gas is nearly described by the ideal gas law if the total volume occupied by the gas is much greater than the volume that is actually taken up by the gas particles and if the temperature is high enough that the gas particles have enough energy to resist intermolecular attractions. Because the compounds that you w i l l use do not behave ideally, your calculated molar weight might not be exactly correct even if you could eliminate all random errors. However, you should be able to obtain a molar weight that is wi th in 3-10% of the true value.

After you have calculated an approximate molar weight for your unknown, your instructor wi l l give you the elemental percentage composition o f the compound. From these data, you can calculate the empirical formula of the compound. Use your approximate molar weight and the empirical formula to determine the molecular formula of the compound and then to determine its true molar weight. Chemists actually use this procedure to determine the ident i ty of new, unknown substances. The main difference between your work here and the work of a chemist ident ifying unknown substances is in the sophistication of the equipment that is used-not in the procedure itself.

The equation PV = nRT applies to ideal gases. It has been modified to describe real gases that behave nonideally. This modif icat ion is called the van der Waals equation:

Two terms here account for the ways in which real gases differ from ideal gases. The van der Waals constant a is a measure of the attractive forces between particles of the gas; the greater the attractive forces are, the greater the numerical value of this constant is for the gas. These attractions cause the measured pressure P to be lower than predicted by the ideal gas law for a particular volume of gas. The measured pressure plus the correction factor containing a equals the ideal pressure. The van der Waals constant b is the volume occupied by the molecules themselves in one mole of a gas ( l i ters/mole); the compressible volume (that is, the free space that is available and in to which the particles can be pushed) is reduced by this amount. In Equation (11-5), V is the measured volume (including the actual molecular volume) and (V- nb) is the compressible or ideal volume. After you have calculated the approximate molar weight o f your unknown, your instructor w i l l also give you the density of the l iquid ( in g /ml) . In a l iqu id , particles are about as close together as they can possibly be, so the l iquid density can be used to calculate a good approximation of b (the volume taken up by the molecules). Then, you wi l l use this calculated value of b, the correct molar weight for the unknown, and your measured values of pressure, volume, temperature, and mass to calculate a numerical value for a (the measure of intermolecular attractions w i th in the gas).


analytical balance 800-ml beaker barometer Bunsen burner 50-ml beaker 250-ml Florence flask

129 ' lolar Weight by Vapor Density

ring triple-beam balance ring stand universal clamp thermometer wire gauze #4 one-hole Neoprene stopper

wi th glass tubing

C H E M I C A L S W E E D E D

Unknown l iquid w i th boil ing point below 9 0 ° C (elemental analysis and density at room temperature to be provided by your instructor) .

T E C H N I Q U E

A small amount of the l iquid is poured into a previously weighed tlask. The flask is closed wi th a Neoprene stopper. A short piece of glass tubing extends through the stopper to allow vapor to escape from the tlask. The tlask, w i t h the stopper in place, is heated in a water bath to about 100°C. As the l iquid boils, its vapor and air rush out o f the flask. Vapor w i l l escape from the flask un t i l only enough remains to exert a pressure that exactly balances the pressure being exerted on the vapor by the atmosphere. The flask is then cooled to condense the vapor to a l iquid . As the flask cools and the vapor condenses, the pressure inside the flask drops; air rushes back into the flask, unt i l the pressure inside the flask again equals atmospheric pressure. During this cooling, the stopper is replaced by a beaker that is inverted over the neck of the flask and rest's on the l ip of the flask. Of course, the beaker is. not an airtight seal and air cari easily return to the flask, but the beaker does decrease the possibility of the loss of a small amount of vapor by diffusion into the atmosphere. The flask containing the l iquid is then weighed (on the analytical balance) to determine the mass of the compound that was present as a vapor when the flask was in the water bath.

The volume of the flask is determined by weighing the flask when it is empty and then weighing the flask filled wi th water. Since the density of water is 1.00 g/ml, the measured mass of water equals the volume of water, which, in turn, equals the volume of the flask. A triple-beam balance can be used for this measurement to obtain a volume wi th four significant figures. Since the mass o f the vapor wi l l contain only four figures (or three, depending on the type o f balance used), this w i l l not noticeably lessen the accuracy of the calculated molar weight.

P R O C E D U R E

Weigh the clean, dry flask (w i th the beaker over its neck) on the analytical balance. Put 3 ml of the unknown liquid into the flask. Insert the stopper very loosely; it should f i t snugly enough that it w i l l not fall out but loosely enough that it can be removed easily and quickly .

Caution! Your unknown is probably flammable. Do not pour it near a flame and do not use more of the unknown than instructed. If a fire does start, immediately turn o f f your burner and use a towel to smother the flames.

130 Experiment 11

FIGURE 11-1

Flask during heating.

Caution! Avoid inhaling the vapor from your unknown. Do not stand over the flask while the sample vaporizes.

Immerse the flask as completely as possible in a water bath clamping it in place as shown in Figure 1 1-1. Heat the water to boiling, and then lower the burner flame to maintain a gentle boi l . You wi l l probably be able to see the l iquid boil and to see a stream of vapor (and perhaps droplets of l iquid) rush from the flask. After the water is boiling and after you can no longer tell that there is l iquid in the flask or that vapor is being emitted from the flask, leave the flask in the bath for at least another five minutes to ensure that all the vapor is at the same temperature. Note that you are to leave the flask in the bath for five minutes after both of these conditions exist. Measure the temperature of the bath.

Now remove the flask from the bath. I t wi l l probably be easiest to remove the clamp, sti l l holding the hot flask, from the ring stand. Cool the flask under cool tap water or in a beaker of cool water; be sure that no water enters the flask through the glass tubing in the stopper. Remove the clamp when the tlask is cool enough that you can handle it easily. Dry the outside of the flask. If water has splashed into the lip of the flask, absorb it w i th a tissue. Remove the stopper and immediately replace it wi th an inverted 50-ml beaker, so that the bo t tom of the beaker rests against the lip of the flask, as shown in Figure 1 1-2. A l low the flask to continue to cool un t i l it is at room temperature (at least five minutes, but probably not more than ten minutes). Weigh the cooled flask and the beaker on the analytical balance.

Add another 2 ml of your unknown to the flask and repeat the experiment. It is not necessary to remove the remainder of the first sample of the l iqu id ; no matter how much l iquid is present originally, vaporization wi l l occur unt i l only enough vapor remains to exert a pressure that is equal to atmospheric pressure.

Carry out a third determination.

131 Mol?r Weight by Vapor Density

F IGURE 11-2

Flask during cooling.

After you have completed three determinations of the mass of vapor in the flask, measure the volume of this vapor. Pour the remaining l iquid out of the flask ( into the appropriate waste container, not down the drain). Dry the flask and weigh it on the triple-beam balance. Then f i l l the flask wi th water and insert the stopper loosely in the flask to push out a l i t t le water. The volume of water remaining in the flask w i l l equal the volume occupied by the vapor when the stopper was in place. Remove the stopper, dry the outside of the flask, and weigh the water-filled flask. Then pour out the water and fi l l and weigh the flask again. As usual, use an average o f your results.

Read the laboratory barometer to determine the atmospheric pressure.


Calculate the molar weight of your unknown l iquid . Use Equation (1 1-4), inserting your measured values of volume, mass, temperature, and the atmospheric pressure. Report the average of the three trials. Briefly describe any difficulties you encountered while performing the experiment that might affect this result. List what you th ink your main sources of error were and how they affected your results.

Determine the empirical formula of your unknown from the elemental analysis provided by your instructor. Then use this formula in combination w i th your calculated value for the molar weight of your unknown to determine the molecular formula and the true molar weight of your unknown compound.

Use the l iquid density of your unknown (given to you in g/ml) and its true molar weight to calculate an approximate value for the van der Waals constant /; in liters/mole.

Next , consider Equation (1 1-5), the van der Waals equation for real gases. You have a numerical value for each variable in this equation. You have just calculated a value for the constant You can now calculate a value for the constant a for your unknown. You can rearrange the equation to solve for a, or you can insert your numerical values in the equation and then complete the arithmetic to obtain a value for a. The latter procedure wi l l probably be easier. Use your experimental measurements of volume, temperature, and atmospheric pressure. The number of moles of

132 Experiment 11

gas should be based on an a.erage of the measured weights of your gas and the true molar weight of your gas. As you work through this problem, retain and pay close attention to all units.

QUESTIONS

1. In this experiment, there are many avoidable sources of error. Explain how the calculated molar weight of your unknown compound would compare to its true molar weight as a result of each of the following errors: (a) The tlask is not completely submerged in the boiling water bath. (b) The flask is removed from the bath before all the vapor is at the boil ing

point of water. (c) Vapor is still being emit ted from the flask when it is removed from the bath (d) The temperature of the vapor is still well above room temperature when the

flask and the vapor are weighed. (e) The flask is open to the atmosphere for several minutes, w i t h neither the

stopper nor the inverted beaker in place. (f) Several air bubbles are trapped in the tlask when you f i l l i t wi th water to

determine the volume of the flask. 2. The most important unavoidable source of error in this experiment is the

nonideality of the vapor, especially the existence of intermolecular attractive forces. Will this nonideality make the calculated molar weight of your unknowr compound higher or lower than its true molar weight? Explain your answer.

3. Consider your answers to Questions 1 and 2 and other possible sources of error in this experiment. Does it seem more likely that your calculated molar weight wi l l be higher or lower than the true value? Does the direction in which your calculated molar weight is in error agree w i th your answer to this question? If it does not, try to account for the difference.

4. What does a large value for the van der Waals constant a tell you about a substance? What would a value of zero for this constant mean?

R E F E R E N C E

Yoder, C.H., Suydam, F .H. , and Suavely, F.A. Chemistry. New Y o r k : Harcourt Brace Jovanovich, Inc., 1975. Chapter 10.

Experiment 12 Molecular Interactions in Solution

Introduction

There are several kinds of solutions: solid—solid, so l id - l iqu id , liquid—liquid, gas—liquid, and gas—gas. In first-year chemistry, you wi l l work mostly wi th solutions of solids and liquids. This experiment, however, is concerned wi th how the properties of l i q u i d - l i q u i d solutions are reflected by their vapor pressures.

The properties of two-component solutions are related to the properties of the solvent, the component that is present in greater amount. In fact, for solutions in l iquid solvents, several properties (vapor pressure, freezing point, boiling point , and osmotic pressure) depend only on the relative numbers of molecules of each component.

How the presence of a solute affects the vapor pressure of the solvent depends on the interactions that occur in the solution. First, consider the simplest solution of two l iquid components A and B. In this case, the interactions between A molecules and B molecules in the solution are no different from the interactions between A molecules in component A's pure state and between B molecules in component B's pure state. Another way of thinking about this is to realize that the potential energy of an A molecule or a B molecule in the solution is the same as the potential energy of the molecule in the liquid's pure state. Because the energy state of each molecule is the same in solution and in the pure l iqu id , no energy in the form of heat must be given out or absorbed when A and B are mixed. For this solution, the enthalpy change A / / s o i u t i o n is zero. A solution for which A / / S O | u t i o n = 0 is called an ideal solution.

Now consider the vapor pressure of such a solut ion. You should remember from Experiment 10 that the vapor pressure of a l iquid reflects the tendency of molecules to escape from the surface of a l iquid into the vapor phase. As long as there are no differences between A - A and A - B interactions in solution (that is, as long as the solution is ideal), the only factor affecting the escape tendency of A molecules in solution wi l l be the fact that there are fewer o f them per unit of surface area, since some space is occupied by B molecules (see Figure 12-1). Statistically, there are fewer chances per second for an A molecule to leave the surface of the solution. Obviously, the oppor tuni ty for vaporization of A in solution is related to the fraction of the molecules that are A. This behavior is formulated mathematically as Raoult's Law:

PA=PA°XA

The law states that the vapor pressure due to A in solution is the vapor pressure of pure A ( / 7 A ° ) mul t ip l ied by the mole fraction of A .

As long as B is volatile (as long as B has an appreciable vapor pressure), the same argument can be made for B:

1>U=PB° Xa=pB°{\ ~X,0 (12-2)

135

136 Experiment 12

The second equality follows from the fact that the sum of the mole fractions of all (here, onl> two) components of the solution must be 1.

The total vapor pressure above the solution is simply the sum of the individual vapor pressures of the components of the solution (Dalton's Law):

Ptotal = PA +PB= PA° *A + PB° (1 " * A ) (12-3)

A plot of the total vapor pressure and the partial pressures of A and B as a function of the composition of the solution is shown in Figure 12-2.

It is obvious that a simple cri ter ion can be used to determine whether liquids form an ideal solution. Solutions of various compositions can be prepared and their vapor pressures measured. I f the solutions are ideal, a plot o f Ptot:ii vs. composit ion wi l l be a straight line, as in Figure 12-2.

Y o u can readily imagine one factor on a molecular level that tends to produce interactions in solution that are similar to interactions in pure components: a similarity in the shape and the polari ty of component molecules.

137 Molecular Interactions in Solution

Ideal solutions that obey Raoult's Law are a particularly simple case. Addi t iona l complications leading to deviations from Raoult's Law can occur when the molecular interactions in a solution are different f rom those in a pure l iqu id . Suppose, for example, that two liquids C and D are mixed and that D molecules attract C molecules more strongly than C molecules attract one another. This k ind of interaction is called solvation. Such a solution would have the structure represented in Figure 12-3. The additional attraction that C and D molecules have for one another means that they are in a more stable condit ion (they have lower energy) in solution than in their pure states. Consequently, energy in the form of heat w i l l be given out during thedissolving process, or A / / s u k l t j o n < 0. This solvation can result when C and D are polar molecules of suitable shape and dipole moment to be attracted toward one another.

In a solution displaying solvation effects, the vapor pressure of C is influenced not only by the statistical factors present in ideal solutions but also by the fact that solvation by D molecules keeps C molecules from escaping into the vapor phase. A similar argument can be made for the vapor pressure of D. Figure 12-4 shows the vapor pressure curve for a two-component solution as a function of composit ion

138 Experiment 1 2

when solvation is impor tant . The dotted lines indicate what the vapor pressures would be if" the solution were ideal. The real solution shows negative deviation from Raoult's Law; that is, the vapor pressures are lower than they are for an ideal solution. Notice that very dilute solutions (Xc — 0.0 or = 1.0) obey Raoult's Law fairly closely.

The final possible solution behavior is that molecules of each of the two components wi l l have stronger attractions for their own kind o f molecule than for a different kind of molecule. In this case, the solution structure (shown in Figure 1 2-5) would consist of regions of like molecules. The vapor pressures of the components would be enhanced in comparison to the pure l iqu id , because each k ind of molecule would tend to "squeeze out'" the other type of molecule from the solution in order to interact wi th its own type. Figure 12-6 is a plot of vapor pressure vs. composit ion for a solution displaying interactions of this type, known as association. Again, the dotted lines show the expected behavior if the solution were ideal. Dilute solutions behave approximately according to Raoult 's Law. More

139 Molecular Interactions in Solution

concentrated solutions show positive deviations; that is, their vapor pressures are higher than Raoult's Law would predict.

A solution exhibi t ing association w i l l be in a higher energy state than its pure components, so heat energy wil l be absorbed when the components are mixed. In this case, AH> 0.

You should be able to determine whether a solution exhibits ideal behavior, solvation, or association by preparing solutions of" several different compositions, measuring their vapor pressures, and constructing plots o f p v a p o r vs. mole fraction. Of course, you can use the procedure given here to measure only the total vapor pressure of the solution and not the individual pressure of each component. After obtaining these results, you wi l l be asked to account for the observed solution behavior on the bases of molecular shapes and polar i ty .

Y o u or your instructor should select one of the pairs of liquids listed below for investigation:

acetone-hexane acetone—ethanol acetone-isopropylamine acetone-propionic acid ethanol-isopropylamine hexane-heptane


buret serum cap clamp syringe 125-ml filter flask thermometer manometer Tygon tubing


Liquids from the list above.

P R O C E D U R E

Prepare nine different solutions: 9 ml of one l iquid and 1 ml of the other, 8 ml of one and 2 ml of the other, and so on. The volumes should be measured to ±0 .02 ml w i t h a buret (see Int roduct ion, page 25). You need not measure exactly 9.00 m l ; just be sure to record the volume you actually take. Mix each solution thoroughly. Measure the vapor pressure of each solution and the vapor pressure of each pure component l iqu id , using the technique outl ined in Experiment 10. For some solutions, the vapor pressure wi l l be so great that i t w i l l be necessary to partially evacuate the flask before the measurement can be made. Consult your instructor before you do this. Use only 1 ml of solution for each measurement. Be sure that

140 Experiment 12

the flask is dried thoroughly between each tr ial . Waste liquids should be poured into the proper waste container; they should not be poured into the sink. You may work in pairs; each student should make half of the measurements.

DATA AND R E S U L T S

Obtain the density of each l iquid from the Handbook of Chemistry and Physics. Use the volumes measured for each l iquid to compute the number of moles of each l iquid and the mole fraction of the l iquid you designate as A for each solution.

Calculate the vapor pressure of each solution, as you did in Experiment 10. Plot vapor pressure for each solution, vs. mole fraction of A on 8r X 1 1 in.

graph paper. Draw the best possible smooth curve through your data points.

QUESTIONS

1. Explain why the mole fraction of acetone in a solution of 9.00 ml of acetone and 1.00 ml of chloroform is different from the mole fraction of acetone in a solution of 9.00 ml of acetone and 1.00 ml of carbon disulfide.

2. How would the plot that you prepared have differed if you had graphed P t o t a | vs. XB instead of Ptotai vs. /VA?

3. The instructions in Experiment 10 tell you to inject about 1 ml of l iquid into the flask for each vapor pressure measurement. Why is it important for you to use only a small volume of the solution instead of the entire 10 ml of solution that you prepared'.'

R E F E R E N C E

Yoder, C.H., Suydam, F.H. , and Suavely. F.A. Chemistry. New York: Harcourt Brace Jovanovich, Inc., 1975. Chapter I 1.

-

Vapor pressure measurements:

Solution

After evacuation * A fter injection

"vapor Solution Flask arm Open arm Flask arm Open arm "vapor

Pure A

1

2

j>

4

5

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Pure B

Pure A

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j>

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Pure B

Pure A

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j>

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Pure B

Pure A

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j>

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Pure A

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j>

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Pure B

Pure A

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j>

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Pure B

Pure A

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j>

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Pure B

Pure A

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j>

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Pure B

Pure A

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j>

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Pure B

Pure A

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j>

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Pure B

Pure A

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j>

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Pure B

Pure A

1

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j>

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9

Pure B

*A partial evacuation should be performed only if necessary and only wi th your instructor's approval. If a partial evacuation is not necessary, leave these data columns blank.

Results: At tach your plot of .Ptotai v s - %A - D° your solutions obey Raoult's Law or do

they show positive or negative deviation from Raoidt's Law? Do the solutions behave ideally or does solvation or association occur? Draw electron dot structures for the component molecules. Discuss the predominant type of intermolecular interactions in the system you studied. Suggest how molecular shape, dipole moments, and other characteristics might be related to the vapor pressure behavior you observed.

(Answer the above questions and the questions from page 140 on a separate sheet of paper and attach it to the Report Slicet.)

Experiment 13 Freezing Point Lowering

Introduction

In this experiment, you wi l l study the nature of solutions by making use o f one of their colligative properties—freezing point lowering. You have already been briefly exposed to this phenomenon in Experiment 2. There you observed that a mixture of comparable amounts of two solids has a melting point lower than the melting point of either pure solid. Since melting and freezing points are really the same phenomenon considered from opposite directions, you already dealt wi th freezing point lowering when you took a mixture melting point .

The subject of this experiment is also related to Experiment 12, except that one of the two solution components here is a solid and has a negligible vapor pressure. The /^vapor o f the solution is determined by the properties o f the solvent, which is more volatile than the solute. The reason for the lowering of the freezing point of a solvent by the addition of a nonvolatile solute is discernible from Figure 13-1. The solid curves represent vapor pressure vs. temperature plots for both the pure solid (frozen) solvent and the pure l iquid solvent. At point A, the curves for the solid and the liquid intersect; that is, both the solid and the l iquid have the same vapor pressure at point A This means that any l iquid which vaporizes at this temperature and pressure could just as l ikely recondense as a solid, and viee versa. This is just another way of.saying that solid, l iqu id , and vapor are in equil ibr ium at point A The temperature at p o i n t / I is the freezing temperature of the pure solvent in the presence of air.

Now consider the dashed curve in Figure 13-1, which represents the vapor pressure of a solution of given composition made by dissolving some nonvolatile solute in the solvent. The vapor pressure of this solution is always lower than the

143

144 E x p e r i m e n t 13

vapor pressure of the pure solvent. The situation is related to the case of solvation discussed in Experiment 1 2. wi th the qualification that the solid component has a negligible vapor pressure. This means that the point of intersection of the dashed curve wi th the vapor pressure curve of the solid solvent wi l l occur at a lower temperature than before and therefore that the solid, solution, and vapor wi l l be in equil ibrium at a lower temperature (see point B in Figure 13-1). This temperature is what we call the freezing point of the solution. When the temperature of the solution is lowered to rs, the solution wi l l be in equil ibr ium wi th the pure solid solvent; crystals o f solvent wi l l appear. As the solvent begins to crystallize, the composition o f the solution wi l l change becoming more concentrated in the solute. O f course, this w i l l change the position o f the solution vapor pressure curve and w i l l necessitate further cooling to cause more freezing. For our purposes, this complication can be ignored by defining the freezing point of the solution as the temperature at which the first crystals of solvent appear.

The lowering of the freezing point of any solvent by a solute in moderately dilute solutions depends on two factors, as shown in Equation (13-1):

Atf=tp-ts = Kfm (13-1)

fhe two factors are the identi ty of the solvent (through the value of Kf) and the concentration of solute particles (as measured by the molali ty m).

Table 13-1 lists the freezing points and the molal freezing point constants Kf for four common solvents.

By combining the information in Table 13-1 and Equation (13-1), you can see that dissolving a solute which supplies a mole o f particles in a kg of water w i l l result in a freezing point of - 1.86° for the solution, compared to 0 ° C for pure water.

Freezing point lowering data can be used to find the molecular weight of a dissolved solid. From Equation ( 13-1), you can see that once a particular solvent is chosen, K f is fixed and the freezing point lowering Atf depends only on the molal i ty of the solution. By measuring the Atf, you can easily evaluate the molal i ty of the solution: in = Atf/Kf. Using weighed quantities of an unknown solute and of the solvent benzene (C '„I I 6 ) , you can set up and solve an expression for the molal i ty of the solution, having the molecular weight of the solute as the unknown. The solvent benzene can be measured wi th a buret and the volume can be converted to weight by knowing that the density is 0.874 g/ml .

T A B L E 13-1 -\

Freezing Point Kf '

Solvent . CO (°C kg/mole)

Acetic acid 16.6 3.90 Benzene 5.5 5.12 Naphthalene s o . : 6.80 Water 0.0 1.86

145 Freezing Point Lowering


800-ml beaker clock wi th second hand freezing point apparatus graph paper thermometer (from your desk) thermometer wi th 0.1° graduations (from stockroom)


benzene (reagent grade) ice unknown solids

T E C H N I Q U E

Freezing points can be measured using the apparatus shown in Figure 13-2. The 800-ml beaker contains an ice-water mixture for cooling the solution and a stirrer. It is covered w i t h a metal or an asbestos l id , through which a test tube is inserted. The test tube contains another tube, which, in turn, contains the solution whose freezing point is to be determined. An air jacket between the tubes permits slow and uniform

146 Experiment 13

cooling in the inner tube. The inner tube is equipped w i t h a stirrer; the solution should be continually stirred during the cooling. Stirring should be gentle to avoid disturbing'the thermometer or breaking the bo t tom ot" the tube. A thermometer that is graduated to 0 .1°C wi l l be provided to measure the temperature o f the solution as it is cooled.

P R O C E D U R E

Obtain a freezing point apparatus and a thermometer (graduated to 0 .1°C) from the stockroom. Also obtain some ice, and prepare an ice and water bath in the 800-ml beaker. Throughout the experiment, this bath must always contain several pieces of ice to keep the temperature as close to 0°C as possible. Put the large stirrer, the l id , and the outer test tube in place. Place your own thermometer {not the special thermometer from the s tockroom) in the bath to determine its temperature.

A. Determining the Freezing Point of Benzene

From a buret, measure about 20 ml of benzene into the clean, dry inner test tube. You do not have to take exactly 20 ml ; just be sure to record to the nearest 0.02 ml the volume that you actually take. Place this inner tube inside the test tube in the freezing point apparatus; stopper it w i th the cork fit ted w i t h a stirrer and the thermometer graduated to 0 . 1 ° . Be sure that the thermometer bulb extends into the l iquid .

Record the temperature of the benzene every 20 seconds. (You can estimate the temperature to the nearest 0 .05°C . ) Tap the thermometer l ightly wi th your finger before you take each reading to ensure that the mercury in the thermometer does not stick to the walls. It is possible that the mercury column may separate; examine the mercury very closely before you begin to use the thermometer and again after you conduct each part of the experiment to see if this has happened. If the mercury does separate, ask your instructor for advice or assistance; you cannot make accurate temperature readings when the thermometer is in this condi t ion.

The benzene may supercool; that is, it may drop to a temperature below its freezing point before it starts to freeze. I f this happens, the temperature wi l l rise to the freezing point t p as freezing occurs because of the release of the heat of fusion. For this reason, you should keep recording the temperature unt i l it remains the same for at least four successive readings as the l iquid is converted into crystals. To determine tp, the freezing point of pure benzene, plot the temperature vs. t ime. Take the horizontal part of the curve, corresponding to a constant temperature, as the freezing point . Your plot should look like the one in Figure 13-3.

Remove the tube containing the benzene from the apparatus and melt the benzene by warming the tube in your hand. Then repeat the determination of the freezing point in the same manner that you conducted the first t r ial . The two trials should agree wi th in 0 . 1 ° . If they do not, determine the value a third time and use the average of the three trials as tp. If you compare your results w i t h those of other students, you may find that your values for the freezing point of pure benzene differ by a degree or more. Do not let this upset you . The thermometer that you are using

147 Freezing Point Lowering

FIGURE 13-3

Cooling curve of a pure l iquid.

may not be perfectly accurate. However, since you are concerned in this experiment wi th the difference between two temperatures-the temperature of a pure l iquid and the temperature of a solution in which the l iquid is solvent—both of your measurements w i l l contain the same error and you can accurately determine a difference despite this error. You should recognize that, because of this possible inaccuracy in your thermometer, it is extremely important that you use the same thermometer to take measurements for benzene and for the solution.

B. Molecular Weight of an Unknown Solid

Obtain an unknown solid from your instructor. The molecular weight w i l l be roughly between 100 and 200. On the analytical balance, weigh a 0.8-g sample of the unknown; record the weight you actually use. Add the unknown to the benzene in the sample tube. Be sure that all of the solid dissolves.

Now measure the freezing point of the solution ts, as you did for pure benzene A plot of temperature vs. time for the solution wil l look somewhat different from your plot for benzene. In Figure 13-4, you can see that the solution temperature

148 Experiment 13

continues to fall as crystallization occurs, because o f the changing solution concentration as the solvent crystallizes. The rate of this decline is slower than the temperature drop was before the solvent began to freeze; the freezing point ts can be located at the break in the curve.

You may find it easier to locate the freezing point of the solution as you warm it and watch the crystals of the solvent disappear. Begin by placing the solution in the freezing point apparatus (in the ice bath) and stir it un t i l nearly all the solution has frozen. Remove the outer tube and the inner tube from the apparatus, but keep the inner tube inside the outer tube. The air jacket wi l l permit slow and uniform warming of the inner tube. If you held the inner tube in your hand to warm i t , some spots would probably be warmed more than other spots and the temperature would not be un i fo rm. A l l o w the solution to warm up as you stir it gently. Determine the temperature at which the last solvent crystals just disappear. This is the freezing point of the solut ion.

Repeat this freeze-thaw cycle unt i l you are satisfied that you have a reproducible figure for the freezing point . Then discard the benzene solution into the designated waste container; do not pour the solution down the drain.


Use the volume of the benzene that you took and its density to calculate the weight of benzene.

Set up an expression for calculating m in terms of the weights of solvent and solute and the molecular weight of the unknown. Equate this expression for m to the value of m that you found from the freezing point lowering. Solve the equation to determine a numerical value for the molecular weight of the unknown solute.

QUESTION

How would the molecular weight of the unknown solute that you calculated be affected by each of the fol lowing blunders?

(a) Some benzene is accidently spilled before the unknown solid is added. (b) Some of the solution is spilled. (c) Not all of the solid is dissolved in the benzene.

R E F E R E N C E 1

Yoder, C.H., Suydam, F .H. , and Snavely, F.A. Chemistry. New York ; Harcourt Brace Jovanovich, Inc., 1975. Chapter 1 1.

Experiment 14 Chemical Kinetics: Oxidation of Ethanol by Chromium(VI)

Introduction

Chemical kinetics is the study of the rates of chemical reactions and of the mechanisms by which they occur. The rate of a reaction is affected by (1) what the reactants are, (2) the temperature at which the reaction occurs, and (3) the concentrations of the reactants. The rate of a particular reaction at a constant temperature wi l l depend only on reactant concentrations. For a general reaction

aA + bB • cC + dD

this dependence can be expressed mathematically by a rate law of the form

rate = k[A\m [B]n

The exponents m and n are usually the integers 0, 1,2, or possibly 3; these exponents may occasionally be fractions or negative numbers. The sum (m + n) is known as the overall reaction order. The A: is a constant called the rate constant; its value is characteristic of a particular reaction at a particular temperature.

The reaction you w i l l study in this experiment is the oxidat ion of ethanol or ethyl alcohol ( C H 3 C H 2 O H , or C 2 H 6 0 ) to acetic acid ( C H 3 C 0 2 H , or C 2 H 4 0 2 ) . The oxidizing agent is prepared by dissolving potassium dichromate ( K 2 C r 2 0 7 ) in hydrochloric acid; during this dissolution, the acid and the dichromate ion react to form C r 0 3 C l ~ :

C r 2 0 7

2 " + 2 H 3 0 + + 2CF • 2 C r 0 3 C F + 3 H 2 0 (14-1)

As the alcohol is oxidized to acetic acid, the chromium atom in C r 0 3 C l ~ i s reduced from C r ( V I ) to the ion C r 3 + :

12H 3CT + 3 C 2 H 6 0 + 4 C r 0 3 C F • 3 C 2 H 4 0 2 + 4 C r J + + 4CF + 21 H 2 0 (14-2)

To emphasize the impor tant feature of the oxidizing agent- the fact that it contains a chromium atom that is reduced from the oxidat ion state + V I to the oxidat ion state + I I I—i t wi l l be referred to in the fol lowing discussion as simply C r ( V I ) .

The rate law for Equation (14-2) might be expected to have the form

rate = / t [ H 3 0 + 1 x [ C 2 H 6 0 ] v [ C r ( V I ) ] 2 (14-3)

However, if the concentrations of hydrochloric acid and of alcohol are far greater than the concentration o f C r ( V I ) , they wi l l undergo only a slight relative change

151

152 Experiment 14

during the reaction. When these two reactant concentrations remain nearly constant, the rate law reduces to

rate = i f c ' [C r (VI ) ] z (14-4)

where k' = k[H30+] x [ C 2 H 6 0 ] - v . This means that if the concentrations of acid and alcohol are relatively constant, it is possible to determine the effect that a change in the concentration of C r ( V I ) has on the rate of the reaction wi thout being concerned w i t h the simultaneous effects of changes in the concentrations of all three reactants.

The rate of a chemical reaction can be measured in either of two general ways: (1) by observing the rate of appearance of the products, or (2) by observing the rate of disappearance of the reactants. In this experiment, one could determine how rapidly C r 3 + or acetic acid is produced, or one could determine how rapidly alcohol or C r ( V I ) is consumed. Since a fairly convenient way of determining [ C r ( V I ) ] exists, you wi l l use the latter method to fol low the reaction in this experiment.

The rate law states that the rate of a reaction depends on the concentrations of the reactants, each raised to some power. Of course this means that the reaction rate w i l l be constantly changing as the reaction progresses, because the reactant concentrations keep changing as the reactants are consumed. If the concentration of C r ( V I ) is measured at two different times after the reaction begins, the average rate of reaction over the time interval can easily be calculated. For example:

At r, = 24 sec, [ C r ( V I ) ] = 0 .00545M At t2 = 35 sec, [ C r ( V I ) ] = 0.00500A/

rate = ~ A [ C r ( V 1 ) ' = - ( [ C r ( V I ) ] f , - [ C r ( V I ) ] f \ At \ t 2 - t { )

_ - (5 .00 X 10" 3 - 5.45 X 1 0 " 3 ) M _ 0.45 X 10" 3A/ (35 - 24) sec 11 sec

= 4.4 X 10"5/v//sec

The rate calculation above gives an average rate over the time interval At = t2 ~ t,. In reality, there is a different rate for every instant of t ime, since the rate of disappearance of C r ( V I ) changes continuously as this reactant is consumed. The instantaneous rate can be found by making At smaller and smaller. The rate at any instant t is defined as

v r a t e = : l im - A [ C r ( , V I ) l = - r f [ C r ( V I ) ] 4 _ 5 )

A r - * 0 At dt

where the times 11 and t2 are taken as an arbitrarily small interval on either side of t. This l im i t as At -> 0 is the derivative of concentration w i t h respect to time—a form that may be familiar to you from calculus. However, you wi l l not need to use calculus here.

In this experiment, you w i l l be asked to determine values for z, the reaction order w i t h respect to C r ( V I ) , and for k1, the rate constant in Equation (14-4). To do

153 Chemical Kinetics: Oxidation of Ethanol by Chromium (V I )

this, yon wi l l measure the concentration o f C r ( V l ) that has not reacted wi th the alcohol at /arious time intervals and then determine z and k' graphically.

To determine the rate of the reaction at various times, a graph is prepared plot t ing [ C r ( V I ) ] (on the y axis) as a function of time (on the x axis). This plot w i l l be a curve showing how [ C r ( V I ) ] decreases as time passes. The average rate over any time interval can easily be determined from the graph by picking two points on the curve and dividing the change in ( C r ( V I ) ] by the change in t ime, as described above. The rate at a particular instant t is not as easy to determine. The instantaneous rate is

l im - A [ C r ( V I ) ] st -o " A/

This l imi t is equal to the slope of a straight line that is tangent to the curve at the point (t, [ C r ( V I ) ] ) . Thus, to determine the reaction rate at various times, you must draw tangents to the curve at various points, as shown in Figure 14-1. The slope of the tangent describes the instantaneous rate of change of [ C r ( V I ) ] ; this is a negative quant i ty , since (Cr (VI ) ) decreases as the reaction progresses. The reaction rate, expressed as a positive quant i ty , is equal to this slope mul t ip l ied by - 1 . At t\, for example, when [ C r ( V I ) ] = c { , the reaction rate is equal to -1 times the slope of tangent A. At t2, when [ C r ( V I ) ] = c2, the reaction rate is equal to -1 times the slope of tangent B. Note that the slope of tangent B is less than the slope of tangent A, showing that the reaction rate is lower for a lower [ C r ( V I ) ] .

After determining the reaction rate at several points on the curve, you are to determine the order of the reaction w i t h respect to [ C r ( V l ) ] . First, suppose that z = 0. Then

rate = / c ' [Cr (VI) ] 0 = k' (14-6)

This equation states that the rate of the reaction is constant; the reaction rate does not change as [ C r ( V I ) ] changes. If the rate at which the reaction occurs does not depend on the concentration o f this particular reactant, the rate w i l l be the same at every [ C r ( V I ) ] ; all the tangents would have the same value, and the graph o f

154 Experiment 14

[Cr(VT)] vs. t would be a straight line. However, for the reaction you are investigating in this experiment, the order is not zero; you wi l l find that the reaction rate does depend on [ C r ( V l ) ] .

Suppose the reaction is first order w i th respect to [ C r ( V I ) ] ; that is

rate = fc'[Cr(VI)] (14-7)

According to this equation, the reaction rate is directly proport ional to [ C r ( V I ) | . When this is the case, a graph of rate vs. [ C r ( V I ) ] is a straight line wi th a slope equal to k'.

Now suppose that the reaction is second order; that is

rate = A ' [ C r ( V I ) | 2 (14-8)

Here, the reaction rate is directly proportional to [ C r ( V I ) J 2 , and a graph of rate vs. [ C r ( V I ) ] 2 is a straight line wi th a slope equal to k'.

To determine the order of the reaction wi th respect to [ C r ( V I ) ] , you must prepare the graphs described here: rate vs. [ C r ( V I ) ] and rate vs. [ C r ( V I ) ] 2 . Once you decide which graph is linear, you w i l l know the reaction order. Then use the linear plot to determine the value of k'.

T E C H N I Q U E

First, a solution is prepared containing C r ( V I ) by dissolving potassium dichromate in hydrochlor ic acid. To a por t ion of this solution, ethanol is added at time t = 0. The redox reaction'. Equation (14-2), then begins to occur.

At various time intervals, the concentration of C r ( V I ) remaining unreacted in solution wi l l be determined. This is done by withdrawing an accurately measured volume of the reaction mixture and adding it to some potassium iodide solution. The potassium iodide immediately reacts wi th all the remaining C r ( V I ) :

2 C r 0 3 C l " + 6 I " + 1 2 H 3 0 + • 3 I 2 + 2 C r J + + 1 8 H 2 0 + 2CI" (14-9] (orange) (brown) (green)

The fairly intense iodine color wi l l mask the pale green color o f C r 3 + . The addit ion of a starch solution to this mixture causes an even more intensely colored starch-iodine complex to form. Sodium thiosulfate is used to reduce the iodine back to colorless iodide: '

L + 2 S 2 0 3

2 " > 2 r + S 4 0 6

2 - (14-10)

When all the iodine is reduced, the pale green color o f Cr 3 1" wi l l be apparent. From the amount of sodium thiosulfate required for the reaction of Equation (14-10), it is possible to calculate the amount of iodine present in the solution. This iodine is formed by the reaction to Equation (14-9); consequently, it is possible to calculate the amount of C r ( V I ) that was still present in the solution and that reacted wi th iodide to form this iodine.

155 Chemical Kinetics: Oxidation of Ethanol by Chromium (VI)

Before you use the sodium thiosulfate solution as described above, you must determine its concentration. It is diff icul t to prepare a solution of known concentration of this salt, because it exists as a hydrate in which the amount of water may vary. Also, the thiosulfate ion slowly decomposes in solution so that its concentration gradually decreases. Consequently, the only way to be sure of its concentration is to measure it shortly before you use the solution. To do this, you w i l l add sodium thiosulfate to a por t ion o f your C r ( V l ) solution that has been treated in the same way as the solution you w i l l use for the kinetics experiment, except that no alcohol has been added. You know the concentration of C r (VI ) in this solution. You wi l l add potassium iodide solution to i t , so that all of the Cr (VI ) reacts as shown in Equation (14-9). The iodine that forms is t i trated w i th the sodium thiosulfate solution to the pale green end point , Equation (14-10). This determinat ion also gives you the oppor tuni ty to practice t i trations for the kinetics experiment. Remember that when you begin the kinetics experiment, each t i t ra t ion is unique: it measures [ C r ( V I ) ] at a specific time. That time cannot be recaptured; the t i t ra t ion cannot be repeated if you do it sloppily.


analytical balance glass stirring rod burets (2) 10-ml graduated cylinders (2) clock w i th minute markings 100-ml graduated cylinder

and a second hand 2-ml pipet Erlenmeyer flasks 250-ml volumetric flask


ethanol 3% KI solution 12/V/HC1 N a 2 S 2 0 3 solution K 2 C r 2 0 7 0.2% starch solution

P R O C E D U R E

A. Preparation of Cr(VI) Solution

Before you begin this experiment, review the procedure for cleaning and using volumetric glassware in the In t roduct ion (pages 23 -28) . Prepare 300 ml of 3.6/W HC1 from concentrated (11.6/V/) HC1. Make the necessary calculations before you come to the laboratory.

Prepare a 0.0035-0.0040/W solution of potassium dichromate in a 250-ml volumetric flask using your 3.6A/ HC1 as the solvent; any concentration in this range w i l l be satisfactory. (Again, you should make the necessary calculations before you come to the laboratory.) Weigh the potassium dichromate on the analytical balance for the greatest possible accuracy. Use Equation (14-1) to calculate the concentra-

156 Experiment 14

t ion o f C r 0 3 C l (referred to simply as C r ( V l ) here) that wi l l form in this solution. Rinse and fill one buret wi th this solution.

B. Standardization of N a : S 2 0 3 Solution

Rinse and fi l l a second buret wi th sodium thiosulfate solution provided by your instructor. You wi l l now determine the concentation o f the sodium thiosulfate solution. From the buret, measure about 10 ml of C r ( V I ) solution into a clean Erlenmeyer flask; record the volume to the nearest 0.02 ml . To this solution add 4 ml of 3% potassium iodide solution; the color of the solution wi l l immediately change from orange to brown as iodide is oxidied to iodine by C r ( V l ) ; see Equation (14-9). To this mixture add 2 m l o f starch solut ion; a dark brownish-black color w i l l result from the starch-iodine complex that forms. Titrate this solution w i t h sodium thiosulfate solution to a pale green endpoint; the color wi l l gradually change from dark brown to dark blue and then to light blue just before the endpoint. Repeat this procedure un t i l you have two results that agree wi th in 2% for the concentration of the sodium thiosulfate solution.

C. Reaction of Ethanol and Cr(VI)

Use a graduated cylinder to measure 1 50 ml of the C r ( V I ) solution into a clean, dry 250-ml Erlenmeyer flask. Be sure to rinse the cylinder wi th this solution first to remove any distilled water which would lower the C r ( V l ) concentration. Use a pipet to add 2.00 ml of ethanol to this C r ( V l ) solut ion; then use a glass rod to mix the solution well . Note the exact time (to the nearest second) that this addition is made; it w i l l be "zero t ime" for the remainder o f the experiment.

Drain the buret containing the C r ( V l ) solution. Rinse and fi l l the buret w i t h the a l c o h o l - C r ( V l ) mixture . Into a clean Erlenmeyer tlask, measure about 10 ml of this solution to the nearest 0.01 m l . Five to ten minutes after "zero t ime , " add 4 ml of 3% potassium iodide solution to this por t ion of the alcohol—C'r(VI) mix ture ; record the exact time of this addit ion. A l l of the C r ( V I ) that has not yet reacted w i th the alcohol wi l l react immediately w i th iodide, converting it to iodine. Add 2 ml o f starch solution and titrate the solution w i th sodium thiosulfate to the pale green endpoint. Repeat this procedure wi th a fresh 10-ml sample of the a l coho l -Cr (VI ) mixture every five to ten minutes. Make a total of eight to ten titrations spread over a 50—70 minute time period.

The flask into which you put the alcohol-Cr(VT) mixture should £e clean and as dry as possible. If a few drops of distilled water remain, do not allow the solution to stand in the tlask for more than a few seconds before adding the potassium iodide; extra water wi l l dilute the solution and lower the reaction rate.

R E S U L T S AND C A L C U L A T I O N S

To determine the sodium thiosulfate concentration, use the volume of C r ( V I ) solution required in a t i t ra t ion and the molari ty of the C r ( V I ) solution to calculate the number of moles of C r ( V I ) used in the t i t ra t ion . Use Equation (14-9) to calcu-

157 Chemical Kinetics: Oxidation of Ethanol by Chromium (VI)

late the number of moles of iodine that formed when potassium iodide reacted w i t h this C r ( V I ) solution. Use Equation (14-10) to calculate the number of moles of S 2 0 3

2 ~. tha t would react w i th this amount of iodine. Then use this calculated number of moles of S 2 0 3

2 - and your measured volume of sodium thiosulfate solution to calculate the molari ty of the sodium thiosulfate solution. Record the average value for your best trials.

f o r the kinetics experiment, use the molar i ty and the volume of the sodium thiosulfate solution to calculate the number of moles of S 2 0 3

2 ~ required for each t i t ra t ion. Use Equation (14-10) to calculate the number of moles of iodine that reacted w i th S 2 0 3

2 ~\ Use this amount of iodine to calculate the number of moles of C r ( V l ) that reacted according to Equation (14-9). Use this number of moles of C r ( V I ) and your measured volume of C r ( V I ) solution to calculate [ C r ( V I ) ] in the solution at each t ime.

Prepare a graph plo t t ing [Cr(VT)] on the y axis vs. time on the x axis. Include a point for t = 0 at which [ C r ( V l ) ] is the value you calculate on the basis of the weight of the potassium dichromate used to prepare the solution. Be sure to spread your graph over as much of the page as possible. Use a sharp pencil to draw a fine line through the points to produce a smooth curve. The [ C r ( V I ) ] changes continuously w i th t ime; therefore, any deviation from a smooth curve wi l l be the result of experimental error. Then, using a ruler and sharp pencil, draw tangents to the curve at a min imum of four points spread about evenly over the total t ime; the curve should bend away from a tanget line by an equal amount on each side of the point of tangency. Determine the slope of each tangent. Mul t ip ly the value of each slope by -1 to determine the reaction rate at each point of tangency.

Now, on one graph, plot reaction rate on the y axis vs. [ C r ( V I ) ] on the x axis and also vs. [ C r ( V I ) ] 2 on the x axis. You wi l l need two different scales for the two quantities plotted on the x axis. Position one across the top of the graph and the other across the bo t tom of the graph; be sure that each axis is clearly labeled and that each plot is properly identified. In addition to the points for reaction rate that you determined experimentally, each plotted line should also include the point (0,0), since the reaction rate wi l l be zero i f no C r ( V I ) is present to react. Use the plot that is a straight line to determine the order of the reaction w i t h respect to [ C r ( V I ) ] . Use this straight-line plot to determine the value of the rate constant k' in equation (14-4) to two significant figures.

QUESTIONS

1. If the concentration of alcohol is far greater than the concentration of C r ( V I ) , the alcohol concentration wi l l undergo a very small relative change during the reaction and wi l l not affect the rate at which the reaction occurs. Calculate the molar concentration of alcohol in the alcohol—Cr(VI) solution to show that it is far greater than the concentration of C r ( V I ) . The density of e thyl alcohol is 0.789 g/ml .

2. How w i l l the calculated [ C r ( V I ) ] compare to the true [ C r ( V I ) ] for a particular t ime, if : (a) The a l coho l -Cr (VI ) buret is accidentally filled w i th the original C r ( V I )

solution that did not have alcohol added to it? (b) A measured port ion of the alcohol—Cr(VI) solution is allowed to stand w i t h

158 Experiment 14

several ml of distil led water for several minutes before the potassium iodide is added?

(c) Some of the a l c o h o l - C r ( V I ) solution is spilled during the titration? (d) Four ml rather than 2 ml of starch solution is added?

3. Use your value of the rate constant to calculate the reaction rate when [ C r ( V I ) ] = 0.002CW and when [ C r ( V I ) ] = 0.001 OA/.

R E F E R E N C E S

1. Yoder, C.H., Suydam, F .H. , and Suavely, F.A. Chemistry. New Y o r k : Harcourt Brace Jovanovich, Inc., 1975. Chapter 12.

2. Finlayson, M.E. , and Lee, D.G. Journal of Chemical Education 48, 473 (1971) .

Experiment 15 A Complete Rate Law

Introduction

The rate at which any chemical reaction proceeds depends on several factors. The most obvious factor is what the reacting species are. Other factors are the temperature, the concentrations of the reactants, and whether or not a catalyst is present. In this experiment, you wi l l study in some quantitative detail how reactant concentrations influence a reaction rate. Instead of moni tor ing reaction speed by following the concentration of one reactant over time (as you did in Experiment 14), you wi l l use a simpler technique here a measurement o f how long it takes for a particular amount of reaction to be completed. Obviously, the longer the time required for completion, the slower the reaction and the smaller the rate. Similarly, you would conclude that a track runner who took 14 seconds to complete the 100-yard dash traveled at a slower rate than a runner who took 10 seconds. Mathematically, the relationship is an inverse one:

rate a 1 (15-1) completion time

The reaction to be studied here has been selected because it can be visually monitored by a color Change.

The reaction you wi l l investigate i s the oxidat ion of I" to I 3 ~by H 2 0 2 (hydrogen peroxide):

I I 2 0 2 + 31"+ 2 H 3 0 + - > I 3 " + 4 1 I 2 0 (15-2)

The rate law for this reaction can be wri t ten

rate = /c 1 H 2 0 2 1 x [ I " V [H30+\z (15-3)

Your task wi l l be to determine the exponents x, y, and z. In other words, you wi l l be asked to find the order of the reaction in H 2 0 2 , I " , and H 3 0 + , respectively.

An obvious way to begin would be to vary the [ H 2 0 2 ] in several trials, while keeping [ I ~ ] and [ H 3 0 + ] the same, and to measure the relative rates. For example, if the [ I I 2 0 2 ] is twice as large in one trial as it is in another and the reaction rate is four times as large (all other concentrations being equal for the two trials), then we could conclude that x = 2 (that is, that the rate depends on [ H 2 0 2 ] 2 ) . A similar procedure could be used to determine y and z.

The rate of the reaction in Equation (15-2) can be measured easily by visual means. Y o u can determine how long it takes for the same [ I 3 ~ ] to be produced under varying sets of concentration conditions. A small concentration of N a S 2 0 3

and some starch indicator are added to each tr ial . I 3 ~ reacts very rapidly wi th thio-sulfate:

I 3 " + 2 S 2 0 3

2 ~ ^ S 4 0 6

2 ' + 31" (15-4)

163

164 Experiment 1 5

Thiosulfate converts I 3

_ back to iodide and prevents any change in [ 1 ~ | as long as it is present. When the small amount o f [ S 2 0 3

2 ~ ] is used up in the reaction. I 3 ~ w i l l no longer be able to react w i t h anything and w i l l begin to accumulate in solution. The I 3 ~ wi l l form a deep blue complex with the starch indicator, and the solution wi l l suddenly change color. Thus, the length of time that it takes for the blue color to appear is a measure of how long it takes the reaction in Equation (15-2) to produce enough I 3 ~ to consume a given concentration of S 2 0 3

2 ~. Obviously, this time interval is related to the reaction rate.

For this method of fol lowing the reaction rate to work, the concentrations of the reactants should not change very much during the time interval required for the blue color to appear. If this condit ion is fulf i l led, the concentrations of all starting reactants w i l l be effectively constant over the measurement time. As pointed out above, this w i l l definitely be true for I " ; [ H 3 0 + ] w i l l also remain the same, because buffer solutions designed to keep a constant [ H 3 0 + | are used. The concentration of S 2 0 3

2 ~ in all solutions wi l l be about .003/V/. As long as a sufficiently large [ H 2 0 2 1 is chosen, the amount that reacts to produce the necessary 1 3~ to consume the S 2 0 3

2 ~ w i l l be negligible compared to the original amount o f H 2 0 2 present in the reaction mixture .


100-ml and 400-ml beakers ring 50-ml burets (3) ring stand burner stirring rods 125-ml Erlenmeyer flasks watch wi th second hand or 10-ml and 100-ml graduated cylinders stopwatch (to be provided 10-ml pipet by student) pneumatic trough wire gauze


Buffer D: [ H 3 0 * ] = 1.0 X 10~S.V/ (contains 2 g/liter starch indicator) Buffer E: [ H 3 0 + ] = 2.0 X 10~ SA/ (contains 2 g/liter starch indicator) 0.1 OA/ H 2 0 2

0 . 1 0 M K I 0.03A/ N a 2 S 2 0 3

ice

P R O C E D U R E

You w i l l need to measure out volumes o f the stock solutions o f reactants in such a way that: (1) the [ S 2 0 3

2 ~ ] is the same in every solution, and (2) the concentrations of the other reagents are varied systematically. One simple way of doing this is to mix reagents in proportions that always yield the same total volume and to add the same volume of sodium thiosulfate solution to each mixture.

155 A Comp le te Rate Law

Solution Volume (ml)

I 11 111 IV V

N a 2 S 2 0 3 10 10 10 10 10 H 2 0 2 15 20 10 10 20 Buffer D 50 45 30 — 20 Buffer E — — — 55 —

K I 25 25 50 25 50

Each of the five sample solution volumes suggested in the table above w i l l yield a total final volume of 100 m l . The decomposition of H 2 0 2 is catalyzed by many substances; therefore you must be sure that all your glassware is completely clean. Use a pipet to measure N a 2 S 2 0 3 . Be certain to use a pipet bulb. Measure H 2 0 2 , buffers D and E, and KI w i t h a buret. Mix all solutions except KI in the same Erlenmeyer flask. Measure the KI into a separate flask. When you are ready to begin your tr ial , pour the KI solution into the other flask as you start t iming. Pour the contents back and for th between the two flasks a couple of times to mix the solut ion completely. Set the flask on a piece of white paper and stop t iming when the first faint but definitely blue color appears. Record this as the completion t ime. If you have enough laboratory time, make two trials for each solution.

C A L C U L A T I O N S AND R E S U L T S

1. Prepare a table showing the concentration of each reactant in each of the five sample solutions. The [ H 3 0 + ] w i l l be the same as the H 3 0 + concentration in buffer D or E, because buffer solutions do not change pH on d i lu t ion .

2. Record the complet ion times you measured and their average for each solution. 3. Recalling the relationship between reaction rate and completion time shown in

Equation (15-1), determine the dependence of the rate on each reagent. Use average completion times in your calculations. Note that sometimes more than one reactant concentration is varied.

4. Write the rate law for the reaction you studied.

S U G G E S T I O N S F O R F U R T H E R W 0 R K

You may also study the influence of temperature on the reaction rate by cooling (in an ice bath) or heating (in a water bath) all reactants to the same temperature before the final mixing. Note that varying T changes the rate by changing k in Equation (15-3).

R E F E R E N C E

Yoder, C.H., Suydam, F.H. , and Snavely, F .A. Chemistry. New Y o r k : Harcourt Brace Jovanovich, Inc., 1975. Chapter 12.

I 11 I I I IV V

[ N a 2 S 2 0 3 ] [ H 2 0 2 ] [ H 3 0 + 1 [ K I ] Time (trial 1) Time (trial 2) Time ( t r ia l 3) Average time

1.0 X 10~5 1.0 X 10" 5 1.0 X 10^ 5 2.0 X 1 0 - 5 1.0 X 10" 5

Show sample calculations for determining the dependence of rate on [ H 2 0 2 ] , [ H 3 0 + ] , and [ I - ] .

Rate law:

Experiment 16 Chemical Equilibrium

Introduction

Reactants are not completely converted in to products in every chemical reaction. In fact, it is more common for a reaction to reach a state such that both reactants and products are present in a reaction mixture and their concentrations do not change. For example, consider the reaction of nitrogen wi th oxygen to produce nitric oxide:

N 2 ( g ) + 0 2 ( g ) >2NO(g) (16-1)

It is obvious that these two gaseous elements do not react extensively in this manner at what we consider normal temperatures, because our atmosphere is primarily a mixture of nitrogen and oxygen.

From a commercial point of view, it would often be advantageous if reactions did reach complet ion. Ammonia is an important industrial chemical and is also used as a fertilizer. Ammonia is produced commercially by the reaction of nitrogen and hydrogen:

N 2 ( g ) + 3 H 2 ( g ) - 2 N H 3 ( g ) (16-2)

At a temperature high enough for a convenient reaction rate and at a pressure of 1000 atm, only 57% of the nitrogen is converted to ammonia.

Both reactions (16-1) and (16-2) are examples of chemical systems that reach equi l ibr ium. When the reaction appears to "s top," bo th reactants and products are present; the rate at which products are forming equals the rate at which products are being reconverted to reactants, so there is no net change in the amount of reactants or products. The equations for such reactions are often wr i t ten w i t h two arrows to indicate that both the forward and the reverse directions of the reaction can occur.

N 2 ( g ) + 0 2 ( g ) ^ 2NO(g) N 2 ( g ) + 3 H 2 ( g ) * 2 N H 3 ( g )

The relative amounts of reactants and products present at equi l ibr ium can be expressed quantitatively by an equil ibrium constant K. This constant states the ratio of product to reactant concentrations, wi th each concentration raised to a power that is the coefficient from the balanced equation. For reaction (16-1)

K = [ N O ' 2 . i Q - 4 0 a t 2 5 ° c (16-3) [ N 2 ] [ 0 : ]

For reaction (16-2)

() K= | N H 3 J ~ = Q.Q8 at 5 0 0 ° C (16-4) [ N 2 ] [ H 2 ] 3

169

170 Experiment 16

The value of A" is characteristic of the particular reaction at a particular temperature. For both of these reactions, K < 1. This must mean that the numerator is

smaller than the denominator in the concentration ratio. For both reactions, the amount o f products present at equi l ibr ium w i l l be small compared to the amount o f reactants that is sti l l present. For these reactions, the position of equil ibr ium is said to lie to the left, meaning that when chemical equi l ibr ium is established, the materials on the left side of the equation (the reactants) predominate in the reaction mixture .

For many reactions, however, K > 1. One example is the reaction of hydrogen and chlorine to produce hydrogen chloride:

Kt 2(g) + CI 2 (g ) * 2HCl(g) (16-5)

K'mjlkr,0"M25°c - (l6-6)

This large value of K shows that when the reaction "stops," nearly all the hydrogen and chlorine has been converted to hydrogen chloride. The position of equi l ibr ium lies far to the right.

Changing the conditions of a chemical system in equil ibr ium wil l generally result in a change in the amounts of substances present. The change that occurs can be predicted using Le Chatelier's principle: If a stress is applied to a system in equilibrium, the system will adjust in such a way as to relieve the stress and establish a new equilibrium state. Possible ways of disturbing a system in equil ibrium include changing reactant or product concentrations by adding or removing some substance, changing the temperature, and changing the pressure (in reactions involving gases).

In this experiment, you wi l l work wi th reversible reactions, studying the extent of reaction and the shifting of equilibria and investigating some of the changes that disturb systems in equi l ibr ium. The differing colors of the reactants and the products that you wi l l work wi th in this experiment wi l l allow you to determine whether the position of equi l ibr ium lies to the left or to the right for a particular reaction.


beakers ring stand burner test-tube holder 10-ml graduated cylinder test-tube rack medicine droppers 20-ml test tubes ring wire gauze


2 M A g N 0 3 K 2 C r 0 4

. 0 . 5 A / C o ( N O 3 ) , 6M HC1 0 . 5 A / C r ( N O 3 ) 3 12M HC1 0.5M C u ( N 0 3 ) 2 6 / V / H N 0 3

171 Chemical Equilibrium

K 2 C r 2 0 , 6M NaOH KNCS ice 6M KNCS

T E C H N I Q U E AND P R O C E D U R E

A. A Reversible Reaction

A solution that is prepared by dissolving a salt containing the chromate ion C r 0 4

2 ~ or by dissolving a salt containing the dichromate ion C r 2 0 7

2 _ can actually have both of these ions present. If the solution is basic, C r 0 4

2 - predominates; if the solution is acidic, C r 2 0 7

2 ~ predominates. Because the characteristic colors of these two ions are different, it is possible to tell which ion is present in a solution, or at least to tell which ion is present in a greater amount when both ions may be present.

Reactions for the reversible conversion of one ion into the other are

C r 2 0 7

2 - + O H " ^ H C r C V + C r C V (16-7)

H C r C V + O H " ^ C r 0 4

2 ~ + H 2 0 (16-8)

These reactions can be summarized in one equation that overlooks the presence of H C r 0 4 " :

C r 2 0 7

2 ~ + 2 0 H - ^ 2 C r 0 4

2 ' + H 2 0 (16-9)

This simplification wi l l be used in this experiment. Equation (16-9) makes it clear that C r 2 0 7

2 ~ in solution can be converted to C r 0 4

2 ~ by the addition o f hydroxide ions (that is, by adding a base). You w i l l use NaOH (sodium hydroxide) and watch for a color change indicating that the forward reaction given in Equation (16-9) is occurring.

It is not as obvious from Equation (16-9) how C r 0 4

2 ~ in solution can be converted to C r 2 0 7

2 " . Keep in mind that a solution prepared from one o f the ions w i l l also contain at least a trace of the other ion because of the interconversion. If a solution is prepared from C r 0 4

2 ~ , the extent to which the reverse reaction of Equation (16-9) occurs wi l l be increased i f hydroxide ions are removed from the solut ion; this is the same as saying that the position o f equi l ibr ium w i l l be farther to the left. The easiest way to remove hydroxide ions is to add an acid (a source of hydronium ions, H 3 0 + ) to the solution, so that the fol lowing reaction wi l l occur:

H 3 0 * + O H - » 2 H 2 0 (16-10)

As O H " ions are removed, the extent to which the reverse reaction given in Equation (16-9) occurs w i l l increase, and the accompanying color change w i l l be your visual evidence of the formation of C r 2 0 7

2 ~ . Dissolve 0.3 g of K 2 C r 2 0 7 (potassium dichromate) in 20 ml of water. Dissolve

0.4 g of K 2 C r 0 4 (potassium chromate) in 20 ml of water. Record the color of each

172 Experiment 16

solid and of each solution. The two solutions contain different weights of materials and different concentrations of chromium-containing ions, but each solution has the same concentration of chromium atoms.

1. Pour about one-fourth of the K 2 C r 2 0 7 solution into a 20-ml test tube. Save the rest of the K 2 C r 2 0 7 as a standard for color comparisons. To the solution in the test tube, add 6M NaOH dropwise; after each drop shake the tube to mix the contents thoroughly. Add NaOH unt i l a definite color change occurs. Record the amount of base required. Account for the color changes in terms of the preceding equations. Now add 6M H N 0 3 (ni tr ic acid) dropwise to this basic solut ion; as before, mix the solution after you add each drop. Add the acid unt i l a definite color change occurs, and explain this color change in terms of the appropriate equations. Again, add base to the solution, and explain any color change. Remember that d i lu t ion wi l l make the colors slightly lighter than they were in the original solutions.

2. Pour about one-fourth of the K 2 C r 0 4 solution into a 20-ml test tube. To this solution, add bM nitric acid dropwise; mix the solution unt i l a definite color change occurs. Now, add NaOH unti l a further color change occurs. Then add ni t r ic acid, as before. Account for your observations in terms of the appropriate equations.

B. Relative " P o s i t i o n " of Equi l ibr ium

Transition metal ions exist in aqueous solution as complex ions in which water molecules are bound by coordinate covalent bonds to the metal. For cobal t ( I I ) , ch romium(HI) , and copper(II) , these complex ions are believed to have the formulas C o ( H 2 0 ) 6

2 + , C r ( H 2 0 ) 6

3 + , and C u ( H 2 0 ) 6

2 + , respectively. The water molecules can be replaced by other ions or molecules to form different complex ions. If two complex ions of a particular metal differ in color, then it is easy to tell which ion is present in a solution.

In this experiment, you w i l l at tempt to convert the aquo complex ions listed above to chloro complex ions, using concentrated HCI as a source of chloride ions. For C o 2 + , for example, the reaction is

C o ( H 2 0 ) 6

2 + + 4CP - C o C l 4

2 " + 6 H 2 0 , (16-11)

A general equation for various metal ions would be

M ( I f 2 O ) x

+" + yCI" - MC1} + ""y> + xH2O (16-12)

In working wi th the three metal ions chosen for this experiment, you wi l l be asked to decide whether or not there is any visual evidence of the formation of a complex ion wi th in a short period o f time. You wi l l then be asked to list the ions on the basis of how extensive the reaction of Equation (16-12) is for a particular amount of chloride ion. In other words, you wi l l be asked to order the ions on the basis of the size of the equi l ibr ium constant for the reaction of Equation (16-12).

After comparing the chloro complex ions of the three metal ions in this experiment, you wi l l then be asked to compare different complex ions o f one metal ion.


You wi l l investigate the reactions in Equations (16-1 1) and (16-13)

C o ( H 2 0 ) 6

2 + + 4C1" * C o C l 4

2 " + 6 H 2 0 (16-11)

C o ( H 2 0 ) 6

2 + + 4NCS" ^ Co(NCS) 4

2 " + 6 H 2 0 (16-13)

to determine which reaction is more extensive for a particular amount of a complex-ing agent (Cl~ or N C S - ) ; that is, which reaction has a larger equi l ibr ium constant.

1. Pour 5 ml of 0.5M C o ( N 0 3 ) 2 (cobalt nitrate) into each of two 20-ml test tubes. Pour 5 ml of 0.5/W C r ( N 0 3 ) 3 (chromium nitrate) in to each of a second set of two 20-ml test tubes. Pour 5 ml of 0.5A/ C u ( N 0 3 ) 2 (copper nitrate) in to each of a third set of 20-ml test tubes. Record the color of each solution. These solutions contain the aquo ions for which the formulas are given above. Set aside one test tube of each solution to use as a standard for color comparisons. To each of the three remaining test tubes, add 1 2M HC1. Add the HC1 wi th a medicine dropper from a 10-ml graduated cylinder containing 10 ml of the acid. Record, to the nearest 0.5 ml , the volume of HC1 that is required to produce a definite color change in each solution. After you think that a color change has occurred, add another drop or two of acid to each solution to decide if you are seeing a distinct color rather than a mixture of two colors. For example, if a solution changes color from red to orange, the orange may be the color of the product or it may be a mixture of the red-colored reactant and a yellow-colored product . Do not add more than 10 ml of acid to each solution. If you are not sure whether the solution has changed color, add to the color comparison tube a volume of water equal to the volume of acid that you used; a slight color change may occur just from the d i lu t ion . Account for the differences among the three solutions on the basis of the position of equil ibrium for the reaction of Equation (16-12). In other words, for which of the three metal ions does the position of equi l ibr ium appear to be farthest to the right for a particular amount of chloride? For which of the three metal ions does the position of equi l ibr ium appear to be farthest to the left for a particular amount of chloride?

2. Mix 3 ml of 0.5M C o ( N 0 3 ) 2 and 3 ml of 0 . 5 M C u ( N 0 3 ) 2 in a 20-ml test tube. Add 12A/ HC1 to this mixture dropwise un t i l a definite color change occurs. Record the volume of acid that you use. Does the chloro complex of cobalt or the chloro complex of copper form first? Does this agree wi th your previous conclusions about the positions of equi l ibr ium for the reactions of these two metal ions w i th chloride ions? The colors that you see in this mixture wi l l not be exactly like the colors you observed for the solutions containing only one metal ion. However, the colors of C o C l 4

2 " and CuCl 4

2 ~ are sufficiently intense and different enough from one another that you should be able to recognize either one despite the presence of another colored species.

3. Pour 5 ml of 0.5yV/ C o ( N 0 3 ) 2 into each of two 20-ml test tubes. To one test tube, add 6M KNCS (potassium thiocyanate) dropwise from a graduated cylinder unt i l a definite color change occurs. To the second test tube, add 6M HCI drop-wise, unt i l a definite change occurs. Do not add more than 10 ml of either solution to the C o ( N 0 3 ) 2 ; record the volume of solution that is required in each case. The

174 Experiment 16

colors o f the two complex ions C o C l 4

2 - a n d Co(NCS) 4

2 ~ wi l l appear to be the same. Account for any difference in the amounts of complexing agent required in terms of the relative positions of equi l ibr ium for the reactions of Equations (!(>-! I) and (16-13). You should also be able to account for the difference in your results here and your results when you added 12M HC1 to C o ( N 0 3 ) 2 .

C. "Shif t ing the Pos i t ion" of Equi l ibr ium

The colors of C o ( H 2 0 ) 6

2 + and Co(NCS) 4

2 " are quite distinct, so you can easily tell which of these complex ions is present in the greater amount in a solution. If comparable amounts o f these two ions are present, the solution wi l l have an intermediate color that can be described as reddish-purple. If anything is added to this solution that causes the forward reaction in Equation (16-13) to occur to a greater extent (that is, that shifts the position of equi l ibr ium to the r ight) , the color should become more blue. If a change in conditions causes the reverse reaction of Equation (16-13) to occur to a greater extent (that is, shifts the position of equil ibr ium to the left) , the color should become more pink. You wi l l investigate how the position of equi l ibr ium for this reaction is affected (1) by adding each of two materials involved in the reaction, (2) by removing one of the materials, and (3) by varying the temperature.

In a small beaker, add 7 ml of 6M KNCS to 3 ml of 0.5M C o ( N 0 3 ) 2 . The solution should have the intense blue color characteristic of Co(NCS) 4

2 ~. Dilute this solution with no more than 15 ml of distilled water (perhaps a l i t t le less), unt i l it has a purplish color intermediate between the colors of C o ( H 2 0 ) 6

2 + and Co(NCS) 4

2 ~ This intermediate color indicates that a moderate amount of each complex ion is present. Divide this solution as evenly as possible among seven test tubes. Set tube 1 aside to use as a standard for color comparisons.

1. Effect of concentration of a reactant. To tube 2, add 1 ml of distilled water. Record any color change, and account for this change in terms of the forward or the reverse direction of the reaction of Equation (16-13); that is, in terms of a shift in the position of equi l ibr ium when the material that is involved in the reaction is added to the reaction mixture .

To tube 3, add 5 ml of distilled water. Compare the colors-of the solutions in tube 2 and tube 3; account for your observations.

To tube 4, add about 0.5 g of solid KNCS. Account for your observations in terms of the position of equi l ibr ium for the reaction of Equation (16-13).

To tube 5, add 1 ml of 2M A g N 0 3 . The white precipitate that forms is insoluble AgSCN; the addition of A g + causes the removal of NCS" from the solut ion. Af ter the precipitate settles, compare the color of the solution in tube 5 to the color of the standard solution in tube 1. Has the position of equil ibr ium shifted? In which direction? Why? Compare the color of the solution in tube 5 to the solution in tube 2, to which 1 ml of distilled water was added. Is the color change that you observe in tube 5 simply the change that you would expect after adding the water in the A g N 0 3 solution, or has a greater change occurred? Explain your answer.


2. Effect of temperature. Heat a beaker of water to boil ing. Use a test-tube holder to place tube 6 in this boiling water. After several minutes, compare the color of the solution in tube 6 to the standard solution in tube 1. Explain what has happened.

Place tube 7 in a beaker of ice and water. After several minutes, compare the color of the solution in tube 7 to the color of the standard solution in tube 1, which remained at room temperature, and to the color of the solution in tube 6, which was heated. Determine whether the two color changes in tubes 6 and 7 are reversible.

What can you conclude about the effect of temperature changes on the position of equil ibr ium for the reaction of Equation (16-13)'.' Is the forward or the reverse reaction favored by an increased temperature? Is the forward or the reverse reaction favored by a decreased temperature? Which direction of the reaction is endo-thermic? Which direction of the reaction is exothermic?

R E S U L T S

Letters and numbers match those in the Procedure and on the Report Sheet.

A. A Reversible Reaction

On the Report Sheet for this experiment, provide the requested information about the colors of the original materials and about the color changes that provide visual evidence of the reaction that is occurring. Account for each color change in terms of the forward or the reverse direction of the reaction in Equation (16-9).

B. Relative "Position" of Equilibrium

1. Provide the requested information about colors that is the basis for comparing the extent of the reaction in Equation (16-12) for the three metal ions. In addition, calculate the concentration of chloride ion that has been added to each solution (that is, the total concentration of C I " that would be present if none reacted with a metal ion) when the color indicates that the aquo complex has been converted to the chloro complex. If no color change occurred, calculate the chloride ion concentration that is present after the addition of the entire 10 ml of concentrated HC1. As an example of such a calculation, assume that a color change occurs when 2.5 ml of \2M HC1 is added to 5.0 ml of the original solution; the total volume is now 7.5 ml . In the total solution

[ C r ] _ (12 moles CI"/1 liter HC1) X (0.0025 liter HC1) 0.0075 liter of solution

[CP] = 4.0M (16-14)

2. Because it is d i f f icul t to assign specific terms to the descriptions of colors, especially when they are mixtures, not all students w i l l have the same answers here. However, your conclusions about positions of equi l ibr ium should be the same, regardless of how you describe the colors of your solutions.

176 E x p e r i m e n t 16

3. In addi t ion to describing your observations and explaining them in terms of the relative positions of equil ibrium for the reactions in Equations (16-1 1 ) and (16-13), calculate the concentrations of chloride and thiocyanate ions that you put into the solutions.

C. "Shifting the Position" of Equilibrium

Describe the appearance of the solution in each tube. Account for each observation in terms of the effect that each change has on the position of equil ibr ium for the reaction of Equation (16-13).

Write an equation for the reaction of Equation (16-13) put t ing the word "heat" on the proper side to indicate its effect on the reaction.

QUESTION

Is it possible to reach any definite conclusion about whether or not C r 3 + reacts wi th C I " in aqueous solution to form a chloro complex ion?

(a) If the reaction is extensive but slow, how would you show that this is the case''

(b) If the reaction is extensive at some temperature other than room temperature, how wotdd you show that this is the case?

(c) If the reaction is extensive at room temperature but forms a chloro complex that is nearly the same color as the aquo complex, how would you show that this is the case'.' {Hint: Refer to Experiment 9.)

R E F E R E N C E

Yoder, C.H. , Suydam, F .H. , and Suavely, F .A. Chemistry. New York : Marcourt Brace Jovanovich, Inc., 1975. Chapter 13.

B. Relative "pos i t i on" of equ i l ib r ium:

M ( H 2 0 ) 6

+ " + > C 1 " ^ M C l / * " - ^ + 6 H 2 0 (16-12)

1 . C o : + C r 3 + C u 2 +

Color of aquo complex

Color of chloro complex

ml of 12M HC1 to form complex

[ C P ] to form complex

Show calculation of [CP] added to C o ( H 2 0 ) 6

2 + w h e n CoCl 4

2 ~ color appears:

Most extensive reaction w i th

Least extensive reaction wi th

Reaction w i t h largest equi l ibr ium constant

Reaction w i t h smallest equi l ibr ium constant

2 . Mixture containing C o ( H 2 0 ) 6

2 + and C u ( H 2 0 ) 6

2 + :

Original color of solution

Color after addition of ml of \2M HC1

Ion for which the chloro complex is present *

Conclusion about relative positions of equi l ibr ium for chloro complexes o f C o 2 + and C u 2 + :

Does this agree w i th your conclusions in Part 1?

) Experiment 17 Spectrophotometric Determination of an Equilibrium Constant

Introduction

The extent to which a chemical reaction occurs can be expressed in terms of the equil ibrium constant for the reaction. This number is characteristic of the reaction at a particular temperature and states the ratio of product concentrations to reactant concentrations (each raised to a power that is the coefficient from the balanced equation for the reaction) in the reaction mixture when chemical equil ibrium is attained (that is, when these concentrations no longer change). For example, for a general reaction

aA + bB ^ cC + dD (17-1)

j g ^ c i M D i ; ( 1 7 . 2 )

[ A ] a [ B ] b

The greater the numerical value of K, the more extensive the reaction wil l be and the ) greater the tendency for the reaction to occur w i l l be.

Another indicator of the tendency and the extent of a reaction is the free energy change AC that accompanies that reaction. This property takes into account the enthalpy change AH and the entropy change AS that occur during any chemical or physical process:

AG = AH - 7 A S (17-3)

A process is most l ikely to occur if AH is negative (energy is evolved) and AS is positive (disorder increases). This combination causes AG to be negative. Consequently, a negative value o f A G for a chemical change shows that this change w i l l occur spontaneously. The more negative the value of AG, the greater the tendency o f the reaction to occur, that is, the more extensive the reaction wi l l be before chemical equi l ibr ium is established. The relationship between AG° (the standard free energy change) and K is

A G ° = -RT(2.3 log A.') (17-4)

Calculations involving AG and K are easy when one of these variables is known. You have also performed calculations involving K and concentrations of reactants and products in a reaction mixture. Of course, this type of calculation requires that you know the value for an equil ibrium constant in the first place. The purpose of this experiment is to acquaint you w i t h one method that can be used to determine the numerical value of an equil ibrium constant. In problems in your chemistry text ,

181

182 Experiment 1 7

you have calculated equi l ibr ium constants from data stating the number of moles of each reactant and product in a reaction mixture . However, for many reactions it is d i f f icul t , or even impossible, to measure these concentrations directly. In this experiment, you wi l l be asked to use an indirect method that makes it unnecessary to measure the concentration of each component of an equil ibr ium mixture .

The reaction that you wi l l investigate is the formation of a complex ion in aqueous solution from reactants in aqueous solution:

F e ( a q ) 3 + + N C S ( a q ) " ^ FeNCS(aq) 2 + (17-5)

The concentration ratio that you w i l l determine is

^ J F e N C S * * ] [ F e 3 + | [ N C S - ]

You w i l l need to know the total concentration of F e 3 + and the total concentration of NCS" put into the solution. However, you wi l l not need to know how much o f either of these materials is in the form of the complex ion or how much remains uncomplexed. A n d you wi l l never need to know the actual concentration o f F e N C S 2 + i n the solution. Instead, the only piece of experimental data that you must determine is the absorbance A due to F e N C S 2 + i n a solution prepared from known amounts of F e 3 + and NCS".

Y o u should recall from Experiment 9 that absorbance is a measure of the amount of light absorbed by a colored material. Its numerical value is directly related to the concentration of the absorbing product in solution c, to the depth or thickness of the sample through which light passes /, and to an intrinsic physical property of the absorbing product called its molar ext inct ion coefficient e:

A=ecl (17-7)

In this experiment, you wi l l not need to know the values of any of these terms except A; however, the symbols e, c, and / wi l l be used in the fol lowing discussion.


beakers 10-ml pipet cuvettes or test tubes spectrophotometer 1-ml pipet 100-ml volumetric flask


0AM F e ( N 0 3 ) 3 in 0.5M I I N 0 3

0.5M H N 0 3

0.002M KNCS in 0.5M H N 0 3

133 Spectrophotometry Determination of an Equilibrium Constant

TECHNIQUE

You w i l l begin w i t h an accurately known volume o f a solution that contains an accurately known concentration of KNCS (potassium thiocyanate). To this solution, you wi l l add an accurately measured volume o f a solution that contains an accurately known concentration o f F e ( N 0 3 ) 3 (ferric nitrate). Then you wi l l measure the absorbance of the solution, using a wavelength of visible light at which FeNCS 2 +

absorbs strongly. Af ter adding more ferric nitrate solution, you w i l l measure the absorbance again. The increased amount of F e 3 + in the solution w i l l cause the position of equi l ibr ium for the reaction of Equation (17-5) to shift to the right; because more FeNCS 2 + i s present in the second solution, the absorbance characteristic of this species w i l l increase. This process is repeated for increasing amounts o f Fe 3 + . Before conducting this experiment, read Appendix IV for a discussion of spectrophotometers. Before you begin work, your instructor w i l l provide directions for the specific spectrophotometer that you w i l l be using.

T R E A T M E N T OF T H E DATA

You are t rying to determine a value for the ratio

K- [F

3fCS2+l (17-6)

[ F e 3 + ] [NCS-]

A way must be found to relate these three unknown concentrations to known quantities.

The concentration of FeNCS 2 + can be related to your one experimental measurement, absorbance A:

A = eel = e / [FeNCS 2 + ] (17-8)

Rearranging this equation gives

[FeNCS 2 + ] =A/el (17-9)

You w i l l know Tn, the total concentration of NCS"pu t in to the solution, but you wi l l not know how this concentration is divided between complexed and un-complexed (free) NCS":

T n = [ N C S - ] f r e e + [ F e N C S 2 + ] (17-10)


[ N C S - ] f r e e = r „ - [ F e N C S 2 + ] (17-11)

Substituting Equation (17-9) into Equation (17-1 1)

[ N C S - ] f r e e = T„-A/el (17-12)

184 Experiment 1 7

On the basis of the d i lu t ion that occurs, you can calculate 7)-, the total concentrat ion of F e 3 + , put into the solut ion, wi thou t knowing how much of this concentration is complexed or uncomplexed F e 3 + :

7>= [ F e 3 + ] f r e e + [FeNCS 2 + ] (17-13)


[ F e 3 + ] f r e e = 7 > - [FeNCS 2 + ] (17-14)

Substituting Equation (17-9) in to Equation (17-14)

[ F e 3 + ] f r e e = Tf-Alel (17-15)

When Equations (17-9), (17-11), and (17-15) are substituted into Equation (17-6), the new expression for the concentration ratio becomes

K = (17-16) (Tf - A/el) (Tn-A/el)

Mul t ip ly ing by the denominator gives

KTfTn- KTn A/el - KTfA/el + K(A/el)2 = AI el (17-17)

These terms can be rearranged and combined into the more familiar form

K(A/el)2 - (KTn + KTf + 1) (A/el) + KTfTn =0 (17-18)

which you should recognize as a quadratic equation of the general form

ax2 +bx +c = 0 (17-19)

In this case, the term A/el is the x in Equation (17-19). Quadratic equations w i t h numerical coefficients are simple although sometimes tedious to solve. Here, the complexity of the coefficients makes the equation quite dif f icul t to solve. An approximate solution is

A/el = Tfln (17-20) Tf+TH + UK

A stepwise rearrangement of this equation puts it in to the form that you w i l l use:

(17-21)

(17-22)

(17-23)

Tf + Tn + \/K_ TfTn

el A

Tf+Tn + \/K_ TfTn

el el A

Tf+Tn + 1 . TfTn

el elK A

185 Spectrophotometric Determination of an Equilibrium Constant

The final form of this equation is

T T 1112. = (1 /e/)( 7> + T„ ) + (1 lelK) (17-24)

A

This equation is now in the form of a general equation for a straight line

y = mx + b (17-25)

If (TfTnlA) is plot ted along the y axis and (7} + 7"„ ) is plotted along the x axis, a straight line results. Its slope is (1 /el), and its y intercept is ( \ / e l K ) . To obtain a value of K, the numerical value of the slope is divided by the numerical value of the intercept

slope =_Uel_=K ( n . 2 6 )

intercept \/e!K

Thus, the value of K has been determined by using two pieces of easily calculated data and one simple experimental measurement, wi thou t ever knowing the equil ibr ium concentrations of the components of the mixture .

P R O C E D U R E

Prepare a 2 X 10~ 4A/ solution of KNCS (potassium thiocyanate) in 0.5M H N 0 3

(nitr ic acid) in the fol lowing manner: Clean and rinse a 100-ml volumetric flask; then rinse the flask wi th three 5-ml portions of 0 .5M H N 0 3 . Clean and rinse a 10-ml pipet giving it a final rinse w i t h the 2 X 1 0 _ 3 A / KNCS solution that is provided. Pipet 10 ml of 2 X 1 0 _ 3 A / KNCS into the volumetric tlask, and then add Q.5M H N 0 3 to the 100-ml calibration mark on the neck of the tlask. Invert the stoppered flask several times to ensure homogeneity. Be sure to record the exact molar i ty of the KNCS solution that is provided.

In a small clean beaker obtain about 20 ml of the 0. \M solution of F e ( N 0 3 ) 3

(ferric nitrate) in 0.5A/ H N 0 3 that is provided. If your beaker is not completely dry , rinse it several times w i th a few ml of the F e 3 + solut ion, so that the 20 ml of F e 3 +

solution wi l l not be diluted by any water remaining in the beaker. Clean and rinse a 1-ml pipet, and then rinse it several times w i t h the F e ( N 0 3 ) 3 solution.

Pour the KNCS solution that you prepared into a clean, dry, 400-ml beaker. Assume that the volume of solution transferred is 0.100 liter. In to this solution, pipet 1.00 ml of 0.1A/ F e ( N 0 3 ) 3 . Gently swirl the beaker to ensure thorough mixing. Throughout the experiment, it is essential not to lose any of the solut ion, because each measurement and calculation in a series of ten is based on the assumption that all of the original solution has been retained.

Adjust the wavelength control of the spectrophotometer to 447 nm (4470 A ) . If the wavelengths on your instrument are marked every 5 nm, rather than every 1 n m , set the wavelength control at 445 to be sure of a reproducible setting; if wavelengths are marked only every 10 n m , set the control at 450. Adjust the zero contro l . Rinse a clean sample cuvette or test tube several times w i t h 0.5A/ H N 0 3 , and then f i l l i t

186 Experiment 17

with 0 .5M H N 0 3 . Place this tube in the sample holder and adjust the light control to " 0 " absorbance to ensure that the absorbance you measure is due only to reaction components. It is assumed that you w i l l share a spectrophotometer w i t h other students. Before you use the instrument each time, check the wavelength setting and the zero setting; then use the tube of 0.5M H N 0 3 to check the light control . The settings may dr i f t slightly while the instrument is in use, or you or another student may accidentally bump the knobs and change the settings.

Into a clean and dry matching sample tube, pour some of the K N C S - F e ( N 0 3 ) 3

solution, and then measure its absorbance. Pour this solution back in to the beaker and retain the sample tube.

Pipet another 1.00 ml of 0. \M F e ( N 0 3 ) 3 into the solution; mix thoroughly by swirling the beaker gently. Pour about 2 ml of this solution in to the sample tube, and then pour it back into the beaker. Repeat this process three more times washing the solut ion over the entire inside wall of the tube to ensure that no solution is lost and that all o f the solut ion has the same concentration of components. F i l l the tube and measure the absorbance. Pour the solution back into the beaker and retain the tube.

Repeat this procedure, adding 1.00 ml of F e ( N 0 3 ) 3 at a t ime un t i l a total of 10 ml has been added. It is important not to lose any solution. If any solution ever drips down the side of the tube during pouring, rinse only the outside of the tube wi th distilled water and dry it so that no solid residue can remain to affect the absorbance. Do not add the distilled water to the solution. Record the temperature in the laboratory.

C A L C U L A T I O N S

For each of the ten mixtures, calculate the total concentation of NCS~ on the basis of the d i lu t ion that has occurred

_ (M of NCS" in stock solut ion) x (V of stock solut ion) (17 27) V of total solut ion

For each mixture , calculate the total concentration of F e 3 + added to the solution

T - CM of F e 3 + in stock solution) X (V of stock solution) r\n~>Q\ If — ( l I-la) V of total solution

In each calculation of concentration, be sure to retain the proper number of significant figures given on the stock solutions of KNCS and F e ( N 0 3 ) 3 .

For each mix ture , calculate the terms TfTn/A and (7} + T„) that are to be graphed. Be careful in placing your decimal points!

Plot TfTn/A (on the y axis) vs. (7} + T„) (on the x axis). Use scales that enable you to plot data to three or four significant figures. Draw the best possible straight line through your plot ted points. Determine the slope and the y intercept of the line. Calculate K to three significant figures (K = slope/intercept).

137 Spectrophotometric Determination of an Equilibrium Constant

Q U E S T I O N S

1. Use your value of the concentration ratio to calculate the concentration of each component in the tenth mixture. Use a calculator, and retain as many figures as possible throughout the calculation; then round o f f each answer to the appropriate number of significant figures.

2. According to your answers to Question 1, what proportion of the total NCS"is in the complex ion? What proportion of the F e 3 + is in the complex ion?

3. Use your experimentally determined value of A' to calculate the standard free energy change for the reaction between F e 3 + a n d N C S ' t o form FeNCS 2 + .

4. What experimental conditions are used to define the standard free energy change? How do the conditions under which you prepared the first mixture differ from standard conditions'.'

5. It is assumed that the reaction between F e 3 + a n d N C S ' i s quite rapid, so that equil ibrium has been attained by the time you measure absorbance. If this is true, what is the free energy change at the time you measure absorbance?

6. Does it seem more reasonable to expect the entropy change for the reaction you arc investigating to be positive or negative? Explain your answer.

R E F E R E N C E S

1. Yoder, C.H., Suydam, F.H. , and Suavely, F .A. Chemistry. New York : Harcourt Brace Jovanovich, Inc., 1975. Chapter 13.

2. Ramette, R.W. Journal of Chemical Education 4 0 , 7 1 (1963) . 3. Frank, H.S.,.and Oswalt, R.L. Journal of the American Chemical Society 69,

1321 (1947) .



Name Date

Section

Data and calculated quantities for the ten mixtures:

Molarity of stock KNCS

Molarity o f stock F e ( N 0 3 ) 3

7} Tn A TfTn T/TJA 7} + T„

1.

2.

3.

4.

5.

6.

7.

8.

9. .

10.

Show the calculation of Tf for mixture 10:

Show the calculation of T n for mixture 10:

Attach the graph of TfTn/A vs. (7} + T„) to the Report Sheet. Show the calculation of the slope of the straight-line plot :

K- Show calculation of K:

(Answer the questions from page 187 on the back of the Report Sheet; attach extra sheets of paper, if necessary. )

Experiment 18 Enthalpy of Reaction

Introduction

Every chemical reaction is accompanied by an energy change in which energy is absorbed or evolved, usually in the form of heat. Energy must be supplied to break the chemical bonds in the reactants and energy is released when bonds form in the products. Whether the net result of these two steps requires energy to be supplied or released depends on the relative sizes of the energies associated w i t h bond breaking and bond formation.

The amount of heat associated w i t h a reaction depends not only on what the reaction is but also on the temperature at which the reaction occurs, and on whether the reaction occurs under conditions of constant pressure or constant volume. In the laboratory, reactions are conveniently carried out at the constant pressure of the atmosphere in beakers or flasks that are open to the air. The amount of heat absorbed or released under these conditions is called the enthalpy change AH for the reaction

AH — //products ^reactants

Enthalpy H can be thought of as the "heat content" of a substance; this heat is stored as potential energy in the form of bond energies, energy of solvation, and other energies. When atoms rearrange in the course of a reaction, the heat content of the products is usually different from the heat content of the reactants, because the bond energies and other energies in the products are not the same as in the reactants. The difference in heat content appears as heat absorbed or released and is generally expressed in calories per mole (cal/mole) of a particular material. For example, in this experiment you wi l l study an acid-base neutralization in aqueous solution

//*(aq) + OH-(aq) H 2 0 (18-1)

The enthalpy change that you wi l l be asked to determine could be given in cal/mole of I T , or in cal/mole of O H - , o r in cal/mole of H 2 0 .

Note that the enthalpy change AH = H p r o i i u c t s - H n e t C t a n t s is defined as positive if heat is absorbed ( i f # p r o d u c t s > / / r e a c t a n t s ) ; such a reaction is called endothermic. Release of heat is associated wi th a negative enthalpy change (// p r 0ducts *> Reactants); this kind of reaction is called exothermic.

The enthalpy change for a reaction helps to determine whether the reaction w i l l be spontaneous. Consider an exothermic process. Because heat is released in the course of an exothermic reaction, the system is changing from a high energy state to a low energy state. The formation of new bonds between atoms causes the system to be in a state of greater stabili ty, and the "excess" energy is released as heat. In tu i tively, it seems plausible that reactions would have a natural tendency to proceed spontaneously to a state o f min imum energy, as a ball rol l ing down a h i l l . Indeed, this is part of what causes some reactions to occur spontaneously.

191

192 EXPERIMENT 18

Whether a reaction w i l l be spontaneous depends not only on AH but also on the entropy change AS that accompanies the reaction. A n y process in which a system becomes more disordered (AS > 0) is more likely to occur than a process in which a system becomes more ordered (AS < 0) . During a chemical reaction, an entropy increase may result from an increase of material in a less ordered physical state (gas rather than l iquid or solid; l iquid rather than solid), from an increase in number of particles, or from a chemical mixing of one material w i t h another (as in the conversion of H 2 and C l 2 to HCl).

The Gibbs free energy change AG considers changes in both enthalpy and entropy

AG = AH -TAS (18-2)

If bo th factors are conducive to spontaneity (AH < 0 and AS> 0), then AG is negative; AG < 0 can be used as a criterion for spontaneity. If both factors oppose spontaneity (AH > 0 and AS < 0), then AG > 0; such a reaction is not spontaneous. Or one factor may oppose spontaneity and the other factor may favor i t ; in such cases, the temperature and the relative magnitudes of AS and AH determine whether the reaction w i l l occur spontaneously.


asbestos square 100-ml graduated cylinder clamp ring stand clock w i t h a sweep second hand 8-oz Styrofoam cups (2) cork 0. l° thermometer glass stirrer


0.503/ C u S 0 4 2.OA/ NaOH 2.0A/ HCl 6A/ N H 3

MEASURING AH: C A L 0 R I M E T R Y

Measuring the heat change for a reaction at constant pressure is easier if the reaction is exothermic than if it is endothermic. The heat that is produced in the reaction is trapped and allowed to cause some measurable physical change. This change is then compared w i t h the size of the change produced by adding some known amount of heat to the heat trap.

The device used to measure heat changes is called a calorimeter. In this experiment, your calorimeter wi l l be constructed using two nested Styrofoam cups and a square o f asbestos for a l i d ; a thermometer and stirrer w i l l be inserted through the lid in to the cups, as shown in Figure 18-1. A perfect calorimeter should trap all the heat from the reaction, allowing none to escape. The trapped heat causes the temperature inside the calorimeter to rise, and the amount of trapped heat can be

133 En tha lpy o f React ion

where C t o t a i is the heat capacity of the entire system, or the number of calories required to raise the temperature of the system 1 ° . The reactions you wi l l study in this experiment w i l l occur in aqueous solutions in the Styrofoam cups. The tota l heat capacity is the sum of the heat capacities of the individual parts of the system:

^total "^solution ^calorimeter ( 1 8 - 4 )

Heat capacity data for various kinds of solutions can often be found in chemical tables. However, because every type o f calorimeter is different, C c a | 0 r j m e t e r must be determined experimentally. This procedure w i l l be described later in the experiment. Basically, it consists of introducing a known amount of heat in to the calorimeter and measuring the temperature change that occurs; this wi l l enable you to calculate the heat capacity in cal /C°.

It might seem quite simple to measure AT. Ideally, if no heat escaped from the calorimeter, we would only need to measure r i n i t i a i , mix the reactants, and then measure r f i n a l . Unfortunately, some heat leakage that must be compensated for does occur when using any type of calorimeter. A graphical procedure is generally used to obtain AT. A typical plot of temperature 7 vs. t ime t for an exothermic reaction is shown in Figure 18-2. A reaction was begun by mix ing the reactants at time t - 0. A large increase in temperature is followed by a steady drop in T, caused by heat leakage from the calorimeter. The ini t ia l points plot ted on the graph may

194 Experiment 18

fluctuate wi thou t revealing any real trend if the mix ture was not thoroughly stirred immediately. Then the temperature drop becomes fairly linear w i th time. A straight line is drawn (perhaps omi t t ing the first point or two before temperature equil ibr ium was established) and extrapolated to time t = 0. The value of T at t = 0 in terms of your measurements is 7Y,nai l o r the reaction that occurs as rapidly as the reactant solutions can be stirred together. This is the temperature that the calorimeter and its contents would have reached if equi l ibr ium had been established instantaneously and if no heat leakage had occurred.

P R O C E D U R E

A. Determination of Calorimeter Heat Capacity

Heat about 75 ml of distilled water to 6 0 - 7 0 ° C . If the thermometer (graduated in 0 .1° ) you are using does not register this high a temperature, heat the water to about 5° above the highest temperature that your thermometer can record.

Use a graduated cylinder to measure 50.0 ml of distilled water, and then pour the water in to the calorimeter. Measure its temperature T c w i t h the 0.1° thermometer.

Now use the graduated cylinder to measure 50.0 ml of the heated water. A l low the water to stand in the cylinder for about 30 seconds. Then stir the water gently and briefly and record its temperature Th. Reinsert the 0.1° thermometer in the calorimeter l id . Pour the hot water into the calorimeter; make the transfer as rapid and as completely as possible. Note the time of the addi t ion; this is t = 0 for your temperature measurements. Quickly put the calorimeter l id , thermometer, and stirrer in place. Stir the water gently w i t h an up and down mot ion for a few seconds;

195 Enthalpy of Reaction

be careful not to bump the thermometer. Wait 30 seconds after the hot water is added and then record the temperature; continue to record the temperature at 30-second intervals for three minutes.

Dry the calorimeter thoroughly w i t h a clean towel. Repeat the determination. Dry the calorimeter again when you finish the second determination.

B. Heat of Neutralization

You are to combine 50.0 ml of 2.OA/ HC1 and 50.0 ml of 2.OA/ NaOH. Measure one of these solutions and pour it into the calorimeter. Measure the temperatures of both solutions. Be sure to rinse and dry the thermometer when transferring it from one solution to the other. If the temperatures of the two solutions are different, allow them to stand unt i l their temperatures are the same, or record the average of the temperatures of the two solutions as the init ial temperature.

Pour the second solution into the calorimeter containing the first solut ion; note the time. Quickly put the l id , thermometer, and stirrer in place, and.stir the mixture gently for a few seconds. Record the temperature at 30-second intervals for three minutes.

Rinse the calorimeter w i t h distilled water and dry it thoroughly. Repeat the determination. When you finish the second determination, rinse and dry the calorimeter again.

C. Heat of Formation of C u ( N H 3 ) 4

2 +

The reaction is

Cu 2 + ( aq ) + 4 N H 3 ( a q ) -» C u ( N H 3 ) 4

2 + (18-5)

An excess of ammonia is used to ensure that the position of equi l ibr ium w i l l be far enough to the right that vir tual ly all of the C u 2 + reacts as shown.

Measure 50.6 ml of 0.5A/ C u S 0 4 and 50.0 ml of 6M N H 3 . Pour one of these solutions into the clean, dry calorimeter. Record the ini t ial temperature of each solution or an average of the temperatures of both solutions. Note the time when you add the second solution to the calorimeter. Position the l i d , thermometer, and stirrer, stir the mixture gently and briefly, and record the temperature at 30-secorid intervals for three minutes.

Thoroughly rinse and dry the calorimeter, and repeat the determination. Clean and dry the calorimeter again before you return it to the s tockroom.


A. Determination of Calorimeter Heat Capacity

Ideally, the final temperature after mixing the cold and hot water should be exactly halfway between Tc and Th ; the heat lost by the hot water should exactly

196 Experiment 18

equal the heat gained by the cold water. In reality, the final temperature wi l l be lower than this, because the hot water w i l l lose more heat than the cold water w i l l gain. The difference between these two heat terms is the heat absorbed by the calorimeter.

Plot T vs. t on a piece of graph paper. Draw the best possible straight line through your data points and extrapolate this line to t = 0 to determine 77 f i n a l

in terms of the heat transfer process w i t h i n the calorimeter. Calculate the heat gained by the cold water Qc:

Qc = Vc X 1.00 g/ml X 1.00 cal/g C° X ( 7 r i m i l - Tc) (18-6)

In a similar fashion, calculate the heat lost by the hot water Qh. The difference between Qh and Qc was absorbed by the calorimeter. Divide this

heat term {Qh - Qc) by the temperature change that occurred w i t h i n the calorimeter (Tfinai ~ Tc) when the hot and cold water were mixed. Your result is the heat capacity of the calorimeter in ca l /C° .

B. Heat of Neutralization

Plot T vs. t on a piece of graph paper. Extrapolate the straight line to t = 0 to determine Tftna\ for the reaction.

The heat released during the reaction is absorbed by the solution and by the calorimeter, raising the temperatures of both:

A / / = (^calorimeter + Cisolution ( ' 8-7)

Each of these terms must be calculated separately using the heat capacity of the particular part of the system being considered and the temperature change that it undergoes. For the calorimeter

Gcalorimeter = C c a i o r i m e t e r X. AT (18-8)

After the neutralization, a solution of NaCl is present. Because heat capacities arc defined for one gram of a material, the volume must be converted to mass to calculate the heat absorbed by the solution:

Solution = V X density X C s o l u t i o n X AT (18-9)

The density of a sodium chloride solution of the concentration you are using here is 1.06 g/ml. The heat capacity of this solution is 0.931 cal/g C° .

After calculating the heat of the reaction for the amounts of the reactants used here, convert this heat term to cal/mole of a reaetant.

C. Heat of Formation of C u ( N H 3 ) 42 +

Fol low the procedure described in Part B for dealing w i th the data. Here, you wi l l also need to know that the resulting C u ( N H 3 ) 4 S 0 4 solution has a density o f 1.06 g/ml and a heat capacity of 1.00 cal/g C ° . Express AH as cal/mole of C u 2 + .

197 Enthalpy of Reaction

QUESTIONS

1. Assume that after determining the heat capacity of your calorimeter, y o u study an endothermic reaction for which 77 i n i t i a | equals 2 5 ° C . (a) Draw a small sketch, showing what a graph of T vs. t would look like for

such a reaction. On the T scale, mark 2 5 ° C to make it obvious how r f i n a | would compare to 7 " i n i t i a i .

(b) What would the direction of the heat "leakage" be? 2. Both of the reactions that you studied in this experiment are spontaneous; this

means that AG is negative. For each reaction, explain whether it seems more likely that AS is positive or negative. In each case, does the entropy change act w i t h or against the enthalpy change to determine spontaneity?

R E F E R E N C E

Yoder, C.H. , Suydam, F.H., and Snavely, F .A. Chemistry. New Y o r k : Harcourt Brace Jovanovich, Inc., 1975. Chapter 13.



Name

Section

Date

A. Determination of calorimeter heat capacity:

Tr ia l 1 Tr ia l 2

Tc

Th

^f i l i a l

AT for cold water

AT for hot water

Qn

^calorimeter

^calorimeter

Average C c a l o r i m e t e r

Show the calculation of Q c for one t r ia l :

Show the calculation of Q h for one t r ia l :

Show the calculation o f C c a | 0 r i m e t e r for one tr ia l :

(Attach your two graphs ofT vs. t. On the back of each graph, construct a table showing the time and the temperature data that you used to prepare the graph.)

B. Heat of neutralization:

* i n i t i a l

^ f i l i a l

AT

öcalorimeter

ösolution

AH

A/7/molc of IP

Average A/ / /mole o f H +

Trial 1 Trial 2

On a separate sheet of paper, show calculation of öcalorimeter. ösolution. A / / , and A/Y/inole:

C. Heat of formation of C u ( N H 3 ) 4

2 + :

Tr ia l 1 Trial 2

^initial

^t'inai

AT

Öcalorimeter

Oso, ution

AH

A// /mole o f C u 2 +

Average AH/mole o f C u 2 +

On a separate sheet of paper, show calculation of öcalorimeter. ösolution. AH, and A/// 'mole:

(Attach your graphs ofT vs. t with data. Answer the questions from page 197 on a separate sheet of paper and attach it to the Report Sheet.)

Experiment 19 Solubility of an Electrolyte

Introduction

Different substances have considerably different solubilities in water. Calcium carbonate ( C a C 0 3 ) , which occurs naturally as the mineral limestone, is practically insoluble; it dissolves to the extent of only about 1CT4 mole in 1 liter of water at 25°C. On the other hand, sodium chloride (NaCI), or common table salt, is fairly soluble; at 2 5 ° C , 1 liter of water w i l l dissolve about 1.1 moles o f NaCI. Solubil i ty depends not only on the identi ty of the substance but also on the temperature and the solvent. Thus, the solubilities of C a C 0 3 and NaCI in water would be different at a different temperature and might be very different in some other solvent.

In this experiment, you wi l l study the solubil i ty of C u ( I 0 3 ) 2 (copper iodate), a strong electrolyte, in water at room temperature. The purpose is to study a simple electrolyte system as a model for more complex systems, such as physiological solutions that contain several electrolytes and nonelectrolytes in aqueous solution. Water is often the best solvent for dissolving electrolytes such as copper iodate

C u ( I 0 3 ) 2 ( s ) ^ Cu 2 + (aq) + 210 3"(aq) (19-1)

for two reasons. First, water is a l iquid of high dielectric constant. This means that the electro

static attraction between cations and anions is considerably reduced by the intervening water molecules compared to the cation-anion attraction in the crystalline state. In other words, the water molecules help to break down the crystal lattice. Second, water molecules, which are polar, can surround ions to produce ion-dipole attractive interactions (see Figure 19-1). This further lowers the energy of ions in solution compared to ions in a crystal.

Copper iodate is practically insoluble in water; the equi l ibr ium in Equation (19-1) can be described by the value of A s p , the solubility product constant for this salt. In this experiment, you wi l l be asked to determine the value o f A s p at room temperature, using the technique described below.

If no substances are present in the solution except water and C u ( I 0 3 ) 2 ) each ion wi l l be in an environment like the one pictured schematically in Figure 19-1: each ion w i l l be completely separated from all other ions by the surrounding water molecules. Because C u ( I 0 3 ) 2 is not very soluble, there w i l l be many more water molecules than ions and each ion wi l l "see" only water molecules in solution.

However, if a reasonable concentration ( ~ 1 0 _ 1 A / ) of other ions, such as K + and N O J , is present in the solution, then the C u 2 + a n d l O j i o n s have some probabili ty of encountering and interacting wi th these ions as well as w i t h the water molecules. In effect, the presence of other ions changes the nature of the medium in which the copper iodate is being dissolved. In the second part of this experiment you wi l l investigate the effect of the added, chemically inert electrolyte K N 0 3 (potassium nitrate) on the solubil i ty of copper iodate.

201

202 Experiment 19


beakers 10-ml pipet Büchne r funnel ring buret ring stand burner rubber stoppers Erlenmeyer flasks stirring rods fi l ter flask vacuum tubing filter paper wire gauze long-stemmed funnel


C u S < V 5 H 2 0 , o r C a ( N 0 3 ) 2 - 4 H 2 0 , o r C d ( N 0 3 ) 2 - 4 H 2 0 K I 0 3

O.lAf, 0 . 2 M , 0 . 3 M , and 0 A M K N 0 3

0.0S7W N a 2 S 2 0 3

acetic acid 0.2% starch solution

T E C H N I Q U E

The experimental procedure to be followed in this experiment is straightforward. A sample o f copper iodate w i l l be prepared and then this sample wi l l be used to prepare saturated solutions of the salt in distilled water and in K N 0 3 solutions. (Alternatively, these solutions may be supplied by the instructor.) Carefully measured 10-ml portions o f these solutions wi l l be taken, and the amounts o f C u 2 + and I 0 3 " ions that each solution contains wi l l be determined. Using this informat ion , the solubil i ty of the salt in moles/liter and the solubil i ty product can be calculated.

203 S o l u b i l i t y of an E l e c t ro l y t e

An indirect procedure is employed in calculating the amounts of C u 2 + and I0 3 ~ ions in each 10-ml por t ion of saturated solution. Some solid KI (potassium iodide) is added to the solution causing the fol lowing reactions to occur:

C u 2 + + 2 I " ->CuI(s) + f [, (19-2)

I 0 3 - + 51" + 6 H 3 0 + • 3 I 2 + 9 H 2 0 (19-3)

For each mole of C u ( I 0 3 ) 2 that dissolves, 6.5 moles of F are produced: 0.5 mole from the reaction of one mole of C u 2 + ions and 6.0 moles from the reaction of the two moles o f I0 3 ~ ions. By determining the amount o f I 2 produced, it wi l l be easy to calculate the quanti ty of copper iodate originally present in the solution.

The amount of iodine produced in the reactions of Equations (19-2) and (19-3) can be determined by t i t ra t ion wi th a standardized N a 2 S 2 0 3 (sodium thiosulfate) solution, using a starch indicator. Iodine forms a deep-blue complex wi th starch; the color wi l l disappear when all the I 2 has been reconverted to colorless iodide ions by its reaction w i th the thiosulfate ion

2 S 2 0 3

2 - + I 2 • S 4 0 6

2 " + 2 r (19-4)

Equation (19-4) states that for every mole of I 2 in the solution, two moles of S 2 0 3

2 - w i l l be needed for t i t ra t ion . Since each mole o f C u ( I 0 3 ) 2 produces 6.5 moles o f I 2 , each mole o f C u ( I 0 3 ) 2 originally present in the solution wi l l require 13 moles of S 2 0 3

2 " . Knowing the molari ty and the volume of the S 2 0 3

2 " solution used in the t i t ra t ion permits the calculation of the number of moles of S 2 0 3

2 ~ used and consequently of the number of moles of C u ( I 0 3 ) 2 originally present in the solution.

I t w i l l be necessary to determine the exact molar i ty o f the N a 2 S 2 0 3 solut ion; this is called standardizing the solution. The solution that w i l l be supplied is ~0.08/V/, but its molarity must be measured to a greater number of significant figures. Weighed samples of K I 0 3 (potassium iodate) can be dissolved in distilled water and then treated w i th I" to produce I 2 via the reaction shown in Equation (19-3). The I 2 can then be t i trated w i t h N a 2 S 2 0 3 solution. Since the number of moles of K I 0 3 i s known from weighing, the number of moles of N a 2 S 2 0 3 required for the t i t ra t ion can be calculated from Equations (19-3) and (19-4). Because this number of moles must be contained in whatever volume of N a 2 S 2 0 3 solution was used in the t i t ra t ion, the molarity can then be calculated.

Each t i t ra t ion analysis should be made three times and the average of the results should be used. Before coming to class, reread the section on the use of a buret in the In t roduc t ion (see page 25).

»

P R O C E D U R E

The class should be divided in to teams of four students. Each student should prepare C u ( I 0 3 ) 2 and determine its Kip in water and in one of the four K N 0 3

solutions. Data on the solubil i ty in K N 0 3 solutions wi l l then be shared among team members.

204 Experiment 19

Preparation of Cu( IO , ) 2

You may be directed to omi t this part of the experiment if the copper iodate is to be supplied by your instructor.

Use the triple-beam balance to weigh out about 2 g of CuS0 4 ' 51120 and 3.5 g of K 1 0 3 separately. In a 250-ml beaker, dissolve the K I 0 3 in about 50 ml of distil led water. In another beaker, dissolve the C u S 0 4 in about I 0 ml of distilled water. Pour the C u S 0 4 solution into the K I 0 3 solut ion; stir constantly. Use a Buchner funnel and a filter flask to collect the precipitated C u ( I 0 3 ) 2 . You are not interested in saving the solution, so you may discard the contents of the filter flask. Wash the precipitate wi th three 5-ml portions of distilled water.

Preparation of Saturated C u ( I O , ) 2 Solution

Y o u may be directed to omi t this part of the experiment if saturated solutions are to be supplied by your instructor.

Place about 1 g of C u ( I 0 3 ) 2 in a 125-ml Erlenmeyer flask and add about 100 ml of distilled water. Swirl the flask repeatedly as you warm the mixture to 50—60°C wi th a burner. Then cool the tlask and its contents to room temperature. ( I f time permits, it is simplest to stopper the tlask and let it cool in your desk un t i l the next laboratory period.)

A. Standardization of N a 2 S 2 0 j Solution

Your instructor wi l l provide a sodium thiosulfate solution o f about 0.0SW concentration. You must determine this concentration to as many significant figures as possible. Record your results in your notebook. (Prepare a data table similar to the one on the Report Sheet for this experiment.)

Weigh 0.1 0-0.1 2 g of K I 0 3 as acctirately as possible on the analytical balance. Place this sample in an Erlenmeyer tlask and dissolve it in 25-30 ml of distilled water. Add about 2 g of solid KI to this solution and swirl the tlask unt i l all of it dissolves.

Obtain about 400 ml of N a 2 S 2 0 3 solution. This is all of this solution that you wil l use for the entire experiment. After the standardization, save the remaining N a 2 S 2 0 3 solution in a stoppered container. Clean a 50-ml buret and rinse it w i th three 5-ml portions o f the N a 2 S 2 0 3 solut ion; then f i l l the buret w i th this solution. Be sure to record the ini t ial buret reading.

Now add 2 ml of acetic acid to the mixture in the Erlenmeyer tlask and begin to titrate this mixture wi th the N a 2 S 2 0 3 solution. The mixture in the tlask w i l l originally be yel low-brown in color because I 2 is formed in the reaction of I03"~ and I " . As S 2 0 3

2 _ converts I 2 to colorless I ~ , the yel low-brown color wi l l begin to fade. When the solution is a very pale yel low, add 2 ml of starch solution. The remaining I 2 w i l l form a deep-blue complex wi th the starch. Titrate unt i l the deep-blue color just disappears signaling that all of the 1 2 has reacted. Record the final buret reading to the nearest 0.02 ml .

205 S o l u b i l i t y of an E l ec t ro l y t e

Make two more determinations, using two new portions of K I 0 3 . Obviously, you wi l l not need to rinse the buret before each t i t r a t ion , but you wi l l undoubtedly need to refill i t .

Place the rest of the N a 2 S 2 0 3 solution in a clean, dry, stoppered vessel. If necessary, your work may be interrupted here and continued during the next laboratory period.

B. Determination of Cu ( IO . , ) 2 Concentrations

First, remove any undissolved solid from the saturated C u ( I 0 3 ) 2 solution by f i l t rat ion. Before you begin, reread the technique for gravity f i l t ra t ion , described on p. 13 in the In t roduct ion . Be sure the funnel and all glassware that you use in this f i l t rat ion are clean and dry; because you want to know the concentration of C u ( I 0 3 ) 2 in a saturated solut ion, it is essential that you do not dilute the solution. When all the solution has been filtered, check to see whether the filtrate seems cloudy. If it does, repeat the f i l t ra t ion.

Pipet 10.00 ml of the clear filtrate [the saturated C u ( I 0 3 ) 2 solut ion] in to an Erlenmeyer flask. A d d about 1 5 ml of distilled water and then about 2 g of K I . Titrate this solution w i t h your standardized N a 2 S 2 0 3 fol lowing the same procedure you used in the standardization. Be sure to add 2 ml acetic acid just before beginning the t i t ra t ion and to add starch as the yel low-brown color fades. Remember to record the init ial and the final buret volumes to the nearest 0.02 m l . Make this determination a total of three times.

C. Solubil ity of C u ( l 0 3 ) 2 in K IM0, Solutions

To test the effect of an electrolyte on the solubil i ty of C u ( I 0 3 ) 2 , you should prepare a saturated solution of C u ( I 0 3 ) 2 in one of the fol lowing solutions: 0 . 1 , 0.2, 0.3, or 0.4/V/ K N 0 3 . A l l o f these K N 0 3 solutions wi l l be provided. Fol low the technique for preparing a saturated solution given earlier, but substitute one of the K N 0 3 solutions for the distilled water. (Alternatively, saturated solutions of C u ( I 0 3 ) 2 in K N 0 3 may be provided by your instructor.)

This saturated solution is then filtered, just as you filtered the aqueous solutions, and three different 10-ml portions of the saturated solution are t i trated with standardized N a 2 S 2 0 3 .

D. Solubil ity of C a ( l 0 3 ) : or C d ( l 0 , ) ;

Calcium iodate or cadmium iodate can be prepared by substituting 4.9 g of C d ( N 0 3 ) 2 - 4 H 2 0 or 4 g of C a ( N 0 3 ) 2 - 4 H 2 0 for the C u S 0 4 in the first procedure ("Preparation of C u ( I 0 3 ) 2 " ) in this experiment.

The solubilities of these salts in distilled water and in K N 0 3 solutions can be determined using the technique described for C u ( I 0 3 ) 2 . However, there is one important difference in the way that the calculations for these two electrolytes are

206 Experiment 19

made. The C a 2 + and C d 2 + ions do not react wi th I" as C u 2 + does. Consequently, the only source of the 1 2 that is t i trated w i t h S 2 0 3

2 ~ i s the reaction of the I 0 3 ~ wi th I" [see Equation (19-3)] . This means that for each mole of the electrolyte that dissolves, 6.0 moles of 1 2 w i l l be produced, instead of the 6.5 moles o f iodine produced from C u ( I 0 3 ) 2 .

C A L C U L A T I O N S AND R E S U L T S

A. Standardization of N a 2 S 2 0 j Solut ion

Calculate the number of moles of K 1 0 3 in each sample that you weighed. Equation (19-4) shows that 2 moles of S 2 0 3

2 _ a r e required to titrate each mole of I 2 , while Equation (19-3) indicates that 3 moles of I 2 are produced by each mole of I 0 3 ~ in your weighed sample of K I 0 3 . Thus, the number of moles of S 2 0 3

2 ~ used in a t i t ra t ion must be six times the number of moles of K I 0 3 you weighed out in the original samples. This is the number of moles contained in the number of liters of N a 2 S 2 0 3 solution used in the t i t ra t ion . The molar i ty can be calculated from the defini t ion

, , moles S 2 0 3

2 ~ 6 X moles KICK , . n , , M = — = ( l z - 5 )

liters of solution liters of N a 2 S 2 0 3 solution

Average your three trials, and use this value as the molari ty M of the N a 2 S 2 0 3

solution in the remaining calculations.

B. Determination of C u ( l 0 3 ) 2 Concentrat ions

From the volume of N a 2 S 2 0 3 solution required to titrate the 10-ml portions of saturated solution and from the molari ty of the N a 2 S 2 0 3 solut ion, calculate the number of moles of S 2 0 3

2 ~ used:

moles S 2 0 3

2 ~ = M X liters S 2 0 3

2 ~ solution used (19-6)

From the introduct ion to this experiment

molesCu(I0 3 ) 2 = m o l t f S ^Oa 2 " ( 1 9 . 7 )

However, this is the number of moles contained in 10.00 ml of solution. From this informat ion, calculate the concentration of C u ( I 0 3 ) 2 in the saturated solution in moles/liter. This is the solubi l i ty s of C u ( I 0 3 ) 2 .

Write the expression for / v s p for C u ( T 0 3 ) 2 in terms of [ C u 2 + ] and [ I 0 3 ~ ] and then in terms of s.

For each of the three trials you performed on the saturated solution, evaluate A ' s p using the value of s that you just determined. Take the average of these three trial values as the value of A" s n .

207 S o l u b i l i t y of an E l e c t ro l y t e

Q U E S T I O N S

1. From your team members, obtain the data for the three K N 0 3 solutions that yot i did not study. What trend ( i f any) do you see in the solubil i ty of this typical strong electrolyte as a function of the concentration of the chemically inert electrolyte K N 0 3 ? Can you account for this trend wi th a physical model? How?

2. Why is the saturated solution filtered to remove undissolved solid? Explain how your calculated values of solubil i ty ,y and of A ' s p would be affected if the undissolved solid were not removed.

3. Why is it important to filter the saturated solution into a dry vessel? Explain how your calculated values of 5 and A ' s p wotdd be affected if the saturated solution were collected in a wet vessel.

R E F E R E N C E

Yoder, C.H., Suydam, F .H. , and Suavely, E.A. Chemistry. New York : Harcourt Brace Jovanovich, Inc., 1975. Chapter 14.


E X P E R I M E N T 1 9 I

Name Date

Section

A. Standardization of N a 2 S 2 0 3 solution:

Trial 1 Trial 2 Tr ia l 3

Weight o f K I 0 3

Moles K I 0 3

Moles S 2 0 3

2 ~ needed

Volume S 2 0 3

2 " used

Molari ty o f N a 2 S 2 0 3

Average molari ty

Show a sample calculation of the molarity of the N a 2 S 2 0 3 solution. Be sure to include all units-and to retain the proper number of significant figures.

B. Determination o f C u ( I 0 3 ) 2 concentrations:

1. Solubil i ty of C u ( 1 0 3 ) 2 in water

Tria l 1 Tr ia l 2 Tr ia l 3

Volume S 2 0 3

2 ~ used

Moles S 2 0 3

2 ~ used

Moles C u ( I 0 3 ) 2 present

Solubil i ty o f C u ( I 0 3 ) 2

in moles/liter

KsP

Average A ' s p

Show a sample calculation of moles S 2 0 3

2 _ used:

I

Show a sample calculation of moles C u ( I 0 3 ) 2 present:

Show a sample calculation of solubil i ty of C u ( I 0 3 ) 2 in moles/liter:

Show a sample calculation of Ksp:



Name D a t e

Section

2 . Solubil i ty of C u ( I 0 3 ) 2 in M K N 0 3

Trial 1 Trial 2 Trial 3

Volume S 2 0 3

2 ~ used

Moles S 2 0 3

2 " used —

Moles C u ( I 0 3 ) 2 present

Solubil i ty o t " C u ( I 0 3 ) 2

in moles/liter

Average solubil i ty

Solubil i ty o f C u ( I 0 3 ) 2 i n M K N 0 3 :

(data from _^ )

Solubil i ty o f C u ( I 0 3 ) 2 in M K N 0 3 :

(data from )

Solubil i ty o f C u ( I 0 3 ) 2 i n M K N 0 3 :

(data from )

(Answer the questions from page 207 below and on the back of the Report Sheet.)

Experiment 20 Gravimetric Determination of Iron

Introduction

Chemistry severed its connections wi th the mystical practices of alchemy and became a modern science when the importance of precise quantitative methods of investigation was recognized. Perhaps the most fundamental quantitative measurement in chemistry is the measurement of the amounts of materials involved in chemical reactions. The nature of many chemical reactions and the composition of many compounds could not be understood unt i l accurate mass (or weight) determinations became possible. We stil l base much o f our knowledge about the composit ion of substances on measurements of the weights of their components. In this experiment, you wi l l deal wi th gravimetric analysis—analysis by measuring weights— an historical but still an important method of quantitative chemical analysis.

Step by step, the general procedure for the gravimetric determination of a particular ion is as follows:

1. Dissolve a known weight of the sample containing the ion of interest. 2. Add to the solution a substance that w i l l form an "insoluble" compound

wi th the ion of interest. The solubil i ty of this compound must be low enough that .the weight of the ion remaining in solution is too low to be measured w i t h the laboratory instruments available.

3. Collect the compound by f i l t ra t ion. 4. Wash the compound to remove soluble contaminants. 5. Dry the compound and weigh i t . If the compound has been collected in filter

paper, the paper must be burned o f f so that only the pure, dry compound remains.

6. Calculate the weight of the ion of interest contained in the precipitate. 7. Calculate the percentage by weight of the ion of interest in the original

sample.

For an analysis of this type, you must learn three new skills that have long been a part of the chemist's repertoire: quantitative precipitat ion, quantitative f i l t ra t ion , and the igni t ion of a precipitate to free it of f i l ter paper and water.

In this experiment, you wi l l be asked to perform a gravimetric analysis for i ron . This procedure can be used to determine the pur i ty of iron salts, to measure the amount of i ron that is present in alloys such as brass or bronze, and to determine the amount of iron in some ores. A precipitate forms when aqueous ammonia is added to a solution containing F e 3 + . From the simplest point of view, the equation for this reaction is

F e 3 + + 3 0 f P • Fe(OH) 3 (s ) (20-1)

Actual ly , the precipitate is hydrated w i th an unknown, variable amount of water. Removal of water by drying above 8 0 0 ° C converts the precipitate to F e 2 0 3 (ferric

213

214 Experiment 20

oxide), one of the common forms in which iron occurs in nature. Knowing the weight of F e 2 0 3 , the weight and the percentage of iron in the original sample can be calculated.


ashless filter paper 10-ml graduated cylinder 100-ml beaker 100-ml graduated cylinder 400-ml beakers (2) medicine dropper 800-ml beaker Meeker (or other high-temperature) burner Bunsen burner porcelain crucibles w i t h lids (2) clay triangle ring desiccator ring stand funnels (2) small test tubes glass stirring rods (2) wash bottle


0.1 A/ A g N 0 3

6M HC1 cone. H N 0 3

cone. N H 3

N H 4 N 0 3

unknown

T E C H N I Q U E S

A. Preparing the Samples

The sample is a pure or an impure salt that contains i ron as F e 2 + or F e 3 + . The sample is dissolved in an acidic solution, because many iron salts are insoluble at pH = 4 or greater. The analysis is based on the presence of Fe 3 + ; consequently, F e 2 +

must be oxidized to this state. Nitr ic acid is used to accomplish this. The reaction is

3 F e 2 + + N O J + 4 H 3 0 + • 3 F e 3 + + NO(g) + 6 H 2 0 (20-2)

The solution may darken, depending on what other ions are present in solut ion. It is boiled to hasten the oxidat ion reaction and then to remove the NO. A clear, yel low color characteristic of F e 3 + should be present.

B. Precipitation

The precipitate is not crystalline and does not even appear to be composed of particles. Instead, it has a gelatinous appearance because its particles are very small,

215 Gravimetric Determination of Iron

below the l imits of ordinary microscopic vis ibi l i ty . The smaller the particles are that comprise a precipitate, the more diff icul t it is to collect the precipitate by filtration; extremely small particles may clog the pores of the filter paper or may simply pass through the paper. However, when the precipitate of hydrated F e ( O H ) 3 forms at a high temperature, its particles are somewhat larger (although st i l l invisible) thaji when it forms at room temperature; consequently, the solution is maintained at a near boi l during the precipitat ion. After the precipitat ion is complete, boiling is continued for one minute to allow the particles to coagulate and to settle out of suspension as a gelatinous mass. Longer heating must he avoided, for it produces the opposite effect: aggregates break up and the precipitate becomes slimy and more diff icul t to collect by f i l t ra t ion.

C. Collecting and Washing the Precipitate

Reread the technique for gravity f i l t rat ion described on page 13 in the Introduct ion . Folding the fil ter paper properly, filtration by decantation, and the complete transfer of the precipitate to the filter paper are very impor tant techniques here.

The supernatant l iquid is decanted through the filter paper keeping as much of the precipitate as possible in the beaker; a large amount of solid in the funnel would slow the f i l t ra t ion . A stirring rod is used to direct the solut ion, to prevent splashing, and to prevent dr ipping from the beaker spout.

The precipitate w i l l be contaminated w i t h other ions that were in the solution: CI" from the hydrochlor ic acid used to dissolve the salt, the anion that was originally combined w i t h the i ron, and any other impur i ty that was present in the sample. To ensure that only..ferric oxide remains after the igni t ion , these adsorbed ions must be removed. The ions are simply washed away. Pure water could be used, but the aggregated precipitate has a tendency to disperse unless it is in contact wi th an electrolyte solut ion. Therefore, a dilute solution of ammonium nitrate is used to wash away the unwanted ions. Any ammonium nitrate that is retained by the precipitate w i l l volatilize during ignit ion.

Al though not all of the contaminating ions that might be present are known, chloride is known to be among them. A test for this ion w i l l determine whether the precipitate has been washed sufficiently. A few drops of the filtrate are collected in a small test tube containing a silver nitrate solution. If Cl~ is sti l l present, a white precipitate o f AgCl w i l l form. I f this happens, the test solut ion should be discarded and the precipitate should be washed unt i l the filtrate shows no precipitate or only a very faint tu rb id i ty in silver nitrate solution.

After the precipitate in the beaker has been washed, it is transferred into the funnel w i th the aid of a stream of water from a wash bott le and allowed to drain as thoroughly as possible.

D. Ignition of the Precipitate

The crucible is placed in a slanted position on a clay triangle on a ring; the lid is slightly ajar. Gentle heat is applied first to dry the fil ter paper. Strong heat could cause spattering and some precipitate could be lost.

216 Experiment 20

When the filter paper is d ry , it is charred at as low a temperature as possible. The crucible must be partially uncovered to allow contact wi th air. In the absence of air, F e 2 0 3 could be reduced to F e 3 0 4 . The filter paper must not be allowed to burst into flame, because gas currents might carry particles of the precipitate out of the crucible. If flames or even sparks appear, the crucible should be covered immediately to smother them.

When the filter paper is entirely charred, the heat is increased to complete the combustion of the carbon that remains and to drive all water out of the precipitate. When the igni t ion is complete, the crucible is cooled in a desiccator to keep it clean and to prevent the crucible and the precipitate from absorbing moisture.

P R O C E D U R E

In an 800-ml beaker, prepare 600 ml of a 1% (by weight) solution of N H 4 N 0 3 , which you w i l l use to wash the precipitate. Begin heating this solution while you prepare the samples.

Use the analytical balance to weigh out , by difference from the sample bot t le , two 0.7-0.9 g samples of your unknown in to two clean 400-ml beakers. Be sure to mark the beakers, so that you w i l l know which beaker contains which sample.

Add 50 ml of distilled water and 10 ml of 6M HC1 to each beaker. Heat one of the solutions to a near boi l and add 2 ml (40 drops from a medicine dropper) of concentrated H N 0 3 to the solution dropwise. Boi l the solution gently un t i l it has a clear, ye l low color (this takes at least 3-5 minutes).

In a small beaker, add 10 ml of concentrated N H 3 to 10 ml of distilled water. Stir constantly w i t h a glass rod as you add about 15 ml of this solution in a slow stream to the F e 3 + solution. A l l o w the precipitate to settle slightly and then run a few drops of the N H 3 solution down the inside wall of the beaker; watch closely to see if any more precipitate forms. If it does, add N H 3 dropwise unt i l there is no evidence of precipitate format ion; then add about a 2 ml excess of N H 3 . If precipitat ion appears to be complete, simply add the excess 2 m l o f N H 3 . There wi l l be a noticeable odor of N H 3 in the steam above the mixture after an excess of N H 3 has been added.

Now gently boi l the mixture for one minute. Then remove the burner and allow the precipitate to settle. Decant as much of the supernatant l iquid as you can through fil ter paper in a funnel: transfer as l i t t l e as possible of the precipitate. Then add about 30 ml of hot N H 4 N 0 3 solution to the precipitate in the beaker. Stir the mixture thoroughly and allow the precipitate to settle. Decant as much of the l iquid as you can in to the funnel. Repeat this washing two more times. Test a few drops of the filtrate for CP; if CP is present, repeat the washing and the test for CP as often as necessary. When washing is complete, transfer the precipitate to the funnel. Use a stream of distilled water from a wash bott le to remove particles from the beaker wall and stirring rod.

During the f i l t ra t ion of the first sample, carry out the oxidat ion and precipitat ion of the second sample. Be careful not to interchange the two samples after you begin fil tering the second sample.

Once you start a f i l t r a t ion , you must complete i t . If the precipitate is allowed to dry , its surface w i l l crack and form channels through which the wash solution w i l l run w i t h o u t effectively washing the precipitate.

217 Gravimetric Determination of Iron

After washing, drain each precipitate as thoroughly as possible in the funnel. If yotir schedule permits, allow the precipitates and filter paper to dry in your desk (where dir t cannot fall in to them) unt i l the next laboratory period.

Wash a crucible and l i d ; support them on a clay triangle on a ring, and apply the full heat of a Meeker (or other high-temperature) burner to them for at least 15 minutes. A l l o w the crucible and l id to cool on the triangle for a minute and then to cool to room temperature in a desiccator (this should take about 15-20 minutes). While the first crucible is cooling, wash and heat a second crucible and l i d . Weigh each crucible (wi th l id) on the analytical balance. When you record the weights, also record the ident ifying numbers and letters that are on each crucible and l i d , so that you w i l l know which sample you put in to which crucible. Heat, cool , and weigh the crucibles as many times as you need to unt i l you obtain two consecutive weighings that differ by no more than 0.0006 g. Use the final weight in your calculations.

Fold the open edge of the fil ter paper over to enclose the precipitate, and place the paper, open-end down, in a crucible that has been heated to constant weight. Heat the nearly covered crucible slowly to dry the paper; hold the burner in your hand, so that you can remove the heat immediately if the paper begins to flame or spark. Char the paper w i thou t inflaming, and then burn o f f the carbon wi th free access to the air. Be sure that the l i d , as well as the crucible, is freed of carbon. Then continue heating the open crucible at the highest possible temperature for at least 10 minutes to remove the last traces of water from the oxide. Be careful to keep the gases of the flame from entering the interior of the crucible, for they could react wi th the F e 2 0 3 . Cool the crucible for a minute on the triangle; then cover the crucible and cool it to room temperature in a desiccator. Repeat this process w i t h the second crucible. Weigh each crucible, l i d , and contents on the analytical balance. Repeat the h igh»tempera ture heating, cooling, and weighing un t i l two consecutive weighings differ by no more than 0.0006 g. This wi l l indicate that all volatile and combustible materials (water, N H 4 N 0 3 from the wash solution, and filter paper) have been removed, leaving pure F e 2 0 3 . Use the final weight in your calculations.


For each sample, determine the weight of F e 2 0 3 , the weight of iron in the oxide, and the percentage of the original sample weight that was i ron. Report an average for the two samples. Be sure to retain the proper number of significant figures.

Describe any departures that you made from the procedure given here and any problems you encountered or any blunders you made that could affect the accuracy of your results for either t r ia l .

Q U E S T I O N S

1. What effect ( too high, too low, no change) would each of the fol lowing have on your calculated percentage of iron in comparison to the correct percentage of iron for your unknown? (a) Some unknown sample is lost after it is weighed. (b) The F e 3 + is not completely precipitated.

218 Experiment 20

(c) The precipitate is not thoroughly washed. (d) The filter paper is not completely burned off. (e) Some F e 2 0 3 is reduced to F e 3 0 4 . (f) The original sample contained a mixture of F e 2 + and F e 3 + .

2. Using the procedure described in this experiment, a 0.8014 g sample of wet, hydrated F e ( N 0 3 ) 3 yields a residue of F e 2 0 3 that weighs 0.1483 g after ignit ion . Calculate the percentages of Fe, F e ( N 0 3 ) 3 , and H 2 0 in the sample.

3. If you use the quantities of reagents described in this experiment, the concentrat ion of O H " remaining in the solution should be not less than \0~SM. (a) What concentration of F e 3 + can remain in solution w i t h this concentration

of OH", rather than be precipitated as Fe(OH) 3 ? (b) What weight of F e 3 + would remain in the volume of solution that you had in

this experiment? (c) What weight of F e 2 0 3 would contain the weight of F e 3 + remaining in

solution? If this F e 3 + were precipitated entirely, could you detect the presence of the additional F e 2 0 3 ? Explain your answer. Can this be considered an essentially complete precipitat ion of Fe 3 + ?

R E F E R E N C E S

1. Yoder, C.H., Suydam, F .H. , and Snavely, F .A. Chemistry. New Y o r k : Harcourt Brace Jovanovich, 1975. Chapter 14.

2. Vogel, A . I . Quantitative Inorganic Analysis, Second Edi t ion . New Y o r k : John Wiley & Sons, Inc., 1961. Chapter V.


E X P E R I M E N T 2 0

Name Date

Section

Data:

Trial 1 Trial 2

Weight of unknown sample

Weight of crucible and l id

After first heating

After second heating

After thi rd heating

After four th heating

Weight of crucible, l i d , and F e 2 0 3

After first heating

After second heating

After thi rd heating

After fourth heating

Weight o f F e 2 0 3

Weight of Fe in sample

Percentage Fe in sample

Average percentage Fe

Average deviation

On the back of the Report Sheet, show the calculation of weight and percentage of Fe in the unknown for one tr ial .

f Answer the questions from pages 217-218 on the back of the Report Sheet.)

Experiment 21 Acid-Base Titration

Introduction

Much of chemistry and biology is concerned w i t h the behavior of acids and bases. Many chemical reactions can be enhanced ( i f they are desirable) or can be retarded ( i f they are undesirable) by adjusting the acidity of the reaction mixture . For example, the concentration of acid in the stomach must be quite high (in comparison to the rest of the body) to break down proteins in the digestive process.

This experiment is designed to serve as an in t roduc t ion to a quantitative determination of the amount of acid or base in an aqueous solut ion. Acid-base t i t ra t ion is a common laboratory procedure of substantial chemical importance and it has many applications in biochemical and medical laboratory work as well . The procedure you w i l l use in this experiment is based on the most fundamental property o f acids and bases: their neutralization reaction

HA + BOH > H 2 0 + BA

or

acid + base • water + salt

The fol lowing discussion wi l l show that from a quantitative point of view, if the amount of base (or acid) in one solution is accurately k n o w n , then the amount of acid (or base) in a second solution can be determined by measuring how much of it w i l l exactly neutralize the base (or acid) in the first solut ion.

Concentrations of acid and base solutions are often given in terms of normali ty N:

..^ number of equivalent weights N = normali ty = -—

solution volume (in liters)

The equivalent weight of an acid or a base is defined as

molecular weight equiv. w t . = ; —— number of reactive H or OH / tormula unit

The equivalent weight of an acid or a base is therefore the number of grams that contains one mole of reactive H + or O H " . The def in i t ion of equivalent weight makes it possible to state a simple relationship between acids and bases: one equivalent weight of an acid will react with exactly one equivalent weight of a base (for example, 0.04 equivalent weight o f acid wi l l react w i t h 0.04 equivalent weight of base). An equivalent weight is often referred to as an equivalent.

Thus, in an acid—base neutralization reaction, a sample of acid (or base) of unknown equivalent weight w i l l be neutralized when it has combined wi th the same

221

222 Experiment 21

number of equivalent weights of the titrant base (or acid) of known concentration; at neutralization

no. e q u i v s a m p l e = no. e q u i v , i t r a n t (2 l - l )

In this experiment, you w i l l be asked to determine the equivalent weight o f an unknown acid by t i t ra t ing an accurately weighed sample of the acid wi th a base solution of known normal i ty A ^ N a O H ) - You wi l l use Equation (2 l - l ) and the definitions of the equivalent weight and the normali ty of a solution to accomplish this determination. The t i t rant is the sodium hydroxide solution; the sample is an unknown solid acid:

g acid 1 — = no. e q u i v a c i d = no. e q u i v N a O H

equiv wtacjd = ^ ( N a O H ) X K ( N a 0 H ) (21-2)

By neutralizing a carefully weighed sample of an acid of known equivalent weight w i th a base solution you prepare, you can calculate the normali ty o f the base solution by using the relationship

p arid A W > H ) = • f V T / (21-3)

equiv w t a c i d X K ( N a O H )

After the base has been standardized (that is, after its normali ty has been determined), you can use it to titrate a carefully weighed sample of unknown acid, and thus find the equivalent weight of the unknown acid:

g acid ,~ , equiv w t a c i U = - * L — . (21-4) ' V ( N a O H ) * ' ( N a O H )

This experiment consists of three parts:

1. Using the stock solution o f 6 A / N a O H , you wil l prepare a solution o f ~0.1/V NaOH by d i lu t ion w i th distil led water that has been boiled ( to expel C 0 2 , which dissolves in water to form carbonic acid) and cooled.

2. You wi l l standardize your NaOH solution by t i t rat ing it wi th an acid o f known equivalent weight.

3. You w i l l then use your standardized NaOH to determine the equivalent weight of an unknown solid acid.

This experiment emphasizes how necessary it is to work accurately in the laboratory at all times. If your standardization is not carefully done, your determination o f the equivalent weight o f the acid w i l l probably be inaccurate.

Before coming to class, reread the section concerning the use of burets in the In t roduct ion (pages 25 -28 ) .


beakers clamps 50-ml buret Erlenmeyer flasks

223 Acid-Base Titration

flat-bottomed flask rubber stopper graduated cylinder


K H C 8 H 4 0 4 (potassium hydrogen phthalate) 6N NaOH phenolphthalein indicator unknown solid acid

P R O C E D U R E

A. Preparation of NaOH Solution

Boil about 350 ml of distilled water in a 400-ml beaker for a few minutes to expel the dissolved C 0 2 . Pour about 295 ml of this water in to your flat-bottomed flask; cool the flask under tap water. Now, add about 5 ml of the 6N NaOH to the water. Swirl the flask to mix the contents thoroughly. You can prepare the NaOH solution the week preceding the titrations and store the solution in your desk if it is well stoppered. Be careful not to permit the solution to come in contact w i t h the stopper. The flask should be stoppered as much as possible throughout the remainder of the experiment.

B. Standardization of NaOH Solution

The NaOH solution is standardized by measuring the volume needed to titrate an accurately known weight of K H C 8 H 4 0 4 (potassium hydrogen phthalate), which has an equivalent weight of 204.2 g. One advantage of using a solid for a standardization is that you do not have to rely on the accuracy of anyone else in preparing solutions you w i l l use. The only sources of experimental error in the amount o f acid are your own weighings, and the errors in these should be minimal .

Clean and f i l l a 50-ml buret w i th the ~0.1yV NaOH solution that you prepared. Don' t forget to rinse the buret w i th the solution first. Before f i l l ing the buret, swirl the flask to be sure that the solution is homogeneous.

On the analytical balance, weigh about 0.6 g of K H C 8 H 4 0 4 into an Erlenmeyer flask. This weight must be known to the maximum precision of the balance. If small bottles of this standard acid are provided, it is advisable to weigh the bott le , take a sample of the appropriate size, reweigh the bott le and determine the weight of the sample by difference. If you determine the sample weight by weighing the flask and then weighing the flask and the sample, you must be sure that the flask is dry; otherwise, evaporation wi l l occur during the weighing and your sample weight w i l l be inaccurate. Dissolve the sample in about 30 ml of distilled water. Add no more than 2 or 3 drops of phenolphthalein indicator, and titrate to the endpoint—the palest possible pink color. Carefully note the volume of base that you used; estimate each reading to the nearest 0.02 ml .

224 Experiment 21

Make a total of three standardizations using samples of K H C s H 4 0 4 of about 0.6 g each t ime. If two of your calculated values for the concentration of your NaOH solution do not agree w i t h i n 2%, repeat the standardization procedure one more t ime.

C. Ti t rat ion of an Unknown Acid

Obtain a sample of an unknown solid acid from your instructor. You must determine its equivalent weight by t i t ra t ing accurately weighed samples of this acid w i t h the standardized base and noting the exact volume of base used for each t i t r a t ion . First you must decide how large a sample of the acid to weigh out for a t i t ra t ion . You do not want to take so much acid that more than 50 ml o f base wi l l be required to neutralize i t . If this were to happen, you would have to refill the buret in the course of one t i t r a t ion ; not only would this be inconvenient, but it would also force you to make four (instead of t w o ) volume readings to determine the volume of the base that you used. Because each reading is subject to an uncertainty of ±0 .02 m l , your total uncertainty could be ±0 .08 m l , or twice as great as when only two volume readings are required. On the other hand, it is advantageous to use as large a volume o f base as possible (as long as you do not have to refill the buret and you do not deplete your supply of base solution). For any volume determinat ion , you must make an ini t ial and a final reading for a total uncertainty of ± 0 . 0 4 m l . This amount o f uncertainty w i l l make less percentage difference in a volume of, say, 40 ml than it w i l l in a volume of, say, 10 ml . To obtain the greatest accuracy, you must choose an amount o f acid that wi l l consume as large a volume o f base as possible wi thou t exceeding 50 m l .

The unknown acids for this experiment have equivalent weights ranging from ~ 5 0 g to ~ 2 0 0 g. Since you do not know the exact equivalent weight of your acid, you should make a rough t i t ra t ion to determine its approximate equivalent weight. Weigh out about 0.2 g of the unknown acid and titrate i t . The results of this t i trat ion w i l l help you choose an appropriate sample weight for the next two trials. Weigh out two more samples o f the unknown in amounts that wi l l require about 3 0 - 4 0 ml o f base. Keep in mind that i f you run out o f base, you w i l l have to prepare and standardize a new solut ion. Therefore, you wi l l probably want to take small enough samples that some base solution remains for a fourth t i t r a t ion , in case two of your three trials do not agree w i t h i n 2%. Titrate these two samples of the unknown.


Calculate the normal i ty of the base [Equat ion (2 1-3)] for each of your three standardization trials. Report the average of these three trials or of your best two trials as the normali ty of the base.

Using the normali ty you just calculated for the NaOH solution, determine the average equivalent weight of the unknown acid from Equation (21-4). Note that you could calculate the molecular weight of this acid only when you know how many acid hydrogens it has. T u r n in your calculations and your identif ication code of the acid. Indicate the uncertainties in your results on the Report Sheet.

225 Acid-Base Titration

Q U E S T I O N S

1. How would your calculated normali ty of the sodium hydroxide solution be affected ( too high, too low, no change) by each of the fol lowing errors? (a) The buret is not rinsed w i t h the solution before it is fi l led. (b) Some K H C 8 H 4 0 4 is spilled out of the flask after it is weighed. (c) Some solution splashes out of the flask during the t i t ra t ion . (d) You go past the endpoint in the t i t ra t ion.

2. How would your calculated value for the equivalent weight of the acid be affected ( too high, too low, no change) by each of the fol lowing errors? (a) Some acid is spilled out of the flask after it is weighed. (b) Some solution splashes out of the flask during the t i t ra t ion . (c) You go past the endpoint in the t i t ra t ion .

R E F E R E N C E

Yoder, C.H., Suydam, F .H. , and Snavely, F.A. Chemistry. New York : Harcourt Brace Jovanovich, Inc., 1975. Chapter 15.

Experiment 22 Antacids

Introduction

An antacid is a substance that prevents or neutralizes acidity. We most commonly use the term to refer to a medication (almost always nonprescriptive) used to neutralize acid in the stomach. Part of the digestion process takes place in the stom-stomach, where food is acted on by gastric juices that contain digestive enzymes and hydrochloric acid. Pure gastric juice is quite acidic, wi th a pH of ~1 ( [ H 3 0 + ] = lCT 'Af but stomach contents consisting of a mixture of gastric juice and food are less acidic, wi th a pH of ~3 ( [ H 3 0 + ] = \0~3M). Hence, an "acid s tomach" is perfectly normal, because the acidity is necessary for proper digestion to occur. However, under some circumstances such as emotional stress, a person may have an unusually large quanti ty of acid in the stomach. Highly acidic or spicy foods may also cause a feeling of abnormal acidity. The terms "excess stomach acid," "acid indigestion," "hyperacidi ty ," "sour stomach," and "hear tburn" are all used to describe these conditions.

Antacids contain basic compounds that neutralize acid; these compounds either contain hydroxide ions or react w i t h water to produce hydroxide ions. Usually these are compounds that have a low solubil i ty in water; they dissolve more readily in acid and neutralize the acid as they dissolve in i t . Antacids may be purchased either as tablets consisting of acid-neutralizing compounds, nonreactive " f i l l e r s , " and perhaps flavoring agents pressed together or as a water suspension of the ingredients. Common active ingredients are M g ( O H ) 2 (magnesium hydroxide) ; A l ( O H ) 3 (a luminum hydroxide) ; C a C 0 3 (calcium carbonate); M g C 0 3 (magnesium carbonate); M g 2 S i 3 0 8

(magnesium trisilicatc); and N H 2 C H 2 C 0 2 H (glycine). Just one or several of these ingredients may be present in a particular antacid.


50-ml burets (2) Erlenmeyer flasks clamps ring stand


\2M HC1, or standardized \M HC1 K H C 8 H 4 0 4 (potassium hydrogen phthalate) 6M NaOH, or standardized 0.5A/ NaOH antacids methyl orange-xylene cyanole indicator phenolphthalein indicator

229

230 Experiment 22

T E C H N I Q U E

In this experiment, the antacids wi l l be in tablet form. A l l o f these antacids are commercial products that can be purchased at almost any drugstore wi thou t a prescription. The brand name o f the antacid w i l l not be known (unless it has recognizable marks on i t ) , and the acid-neutralizing compound or compounds contained in the tablet w i l l not be identified. First, the weight of hydroxide that is present in or that can be produced by a particular tablet w i l l be determined. A l l the antacids used for this experiment contain compounds that are only slightly soluble in water; these compounds are combined wi th fillers that are insoluble in both water and in dilute acid solutions.

Because the stomach pH is normally ~ 3 , it w i l l be necessary to find how much acid a tablet w i l l neutralize while maintaining a pH of ~ 3 . To accomplish this, a tablet wi l l be treated wi th more than enough HC1 (hydrochloric acid) to dissolve the tablet and to react w i th the acid-neutralizing compounds it contains. Then the remaining acid w i l l be titrated w i t h a standardized NaOH (sodium hydroxide) solution. The acid-base indicator to be used is a mixture of methyl orange and xylene cyanole. Its color change from violet through neutral gray occurs at pH = ~3 .6 . By knowing the number of moles of hydroxide used in the t i t ra t ion and the number of moles of hydronium ion in HC1, the number of moles of hydroxide ion provided by the antacid can be determined.

P R O C E D U R E

Standardized solutions of HCI and NaOH may be provided or you may be instructed to prepare and standardize one or both of these solutions.

The standardization of sodium hydroxide is described in detail in Experiment 21 . Prepare 0.5M NaOH by di lu t ing about 25 ml of 6M NaOH to 300 ml of solution wi th distilled water that has been boiled and cooled. This standardization w i l l require three samples of K H C 8 H 4 0 4 , each weighing about 2 g. Be sure to make the weighings on the analytical balance.

After you know the concentration of the NaOH solution, you can use this solution to standardize the HCI solution. A 1/V/ HCI solution can be made by d i lu t ing about 17 ml o f \2JVI HCI to 200 ml o f solution. Three 10-ml portions o f this solution (each measured to the nearest 0.02 ml from a second buret) can be t i trated to the phenolphthalein endpoint w i t h the NaOH solution.

You wi l l use three one-tablet samples to determine the hydroxide content o f the antacid. Weigh a tablet on the analytical balance (preferably by difference from the bottle containing the tablets), and put it into an F.rlenmeyer flask. Weigh all three tablets before you begin the experiment instead of completing the analysis of one tablet before you weigh another. Immediately put a few milli ters of distilled water into the flask; the tablet w i l l begin to crumble in the water and w i l l dissolve more readily later. Be sure to mark each flask clearly, so that you know the weight of each tablet.

Use a buret to measure about 15 ml of HCI into the flask. Remember to take the initial buret reading, but do not take a final buret reading yet. Swirl the flask to speed the dissolving. It may help to crush the tablet w i t h a glass rod (do not use a

231 Antacids

metal utensil, because it might react w i t h the acid); be sure to rinse all solution and solid particles o f f the rod and into the flask when you remove the rod. Ten to fifteen minutes may be required for complete dissolution of the active compound. However, at this point , undissolved solid w i l l st i l l be present; the inert filler w i l l remain as a powder. Be certain that no obvious chunks or chips of the tablet remain; just a uniformly fine powder should be present in the flask. It is essential to dissolve the antacid compound completely, because any hydroxide-producing compound that does not dissolve cannot be detected in the analysis. This may be the most time-consuming part of the experiment.

I f the tablets you are analyzing contain carbonate ions, they wi l l react w i th HCI to produce C 0 2

2 H 3 0 + + C 0 3

2 - <• H 2 C 0 3 • C 0 2 ( g ) + 3 I I 2 0 (22-1 )

As a tablet dissolves, i t w i l l effervesce. After the effervescence subsides, the C 0 2 in solution must be removed before you begin the t i t ra t ion ; i f any C 0 2 remains, it wi l l react wi th N a O l l . To remove the C 0 2 from the solution, heat the mixture to a near boi l ; be careful not to let any solution spatter out of the flask. Continue to heat the flask gently for about five minutes; then cool the flask to room temperature under cold water.

When you are sure that dissolution is as complete as possible, add 4 drops of methyl orange-xylene cyanole indicator to the mixture . From a buret, rapidly add a stream of NaOH solution unt i l the solution is definitely basic (that is, unt i l you intentionally go past the endpoint) . Then rapidly add HCI dropwise, unt i l you are sure the solution is acidic. Now, carefully titrate the solution w i th NaOH to the endpoint, which is reached when a neutral gray or brown-gray color persists in the turbid solution. If you think you have passed the endpoint, add another drop or two of HCI and titrate the solution again wi th NaOH unt i l you reach the endpoint. This kind of t i t ra t ion is often faster than the ti trations you have performed in earlier experiments; it can also be more accurate. Remember to record the final readings for both burets. If the antacid and the acid produced C 0 2 that was not completely removed from the solution, the endpoint color of the solution may fade to gray when the endpoint is reached; if this happens, be more thorough in removing C 0 2

from the next mixtures. In t i t rat ing the remaining samples, you should use a greater amount of HCI if the amount of base solution that you used for the first sample was less than 10 ml .


1. Use the volume of the NaOH solution and the molecular weight of K H C 8 H 4 0 4

to calculate the concentration of the base solution for each of your trials. For the rest of the experiment, use the average value from your three trials.

2. Use the concentration of the base solution and the volumes of the acid and base solutions to calculate the concentration of the HCI solution for each of the three trials. For the rest of the experiment, use the average of these three values.

3. For each t i t ra t ion performed on an antacid tablet: (a) Use the concentration and the volume of the acid solution to calculate the

number of moles of H 3 0 + used.

232 Experiment 22

(b) Use the concentration and the volume of the base solution to calculate the number of moles of O H " used.

(c) How many moles of H 3 0 + were neutralized by the NaOH solution? (d) The remainder of the H 3 0 + was neutralized by the antacid. Calculate the

number of moles of O H - provided by the antacid. (e) Convert moles of O H " to weight of O H " . (f) Determine the percentage by weight of the antacid tablet that is O H "

4. Report an average percentage of O H " for the three tablets.

QUESTION

How w i l l the calculated percentage o f hydroxide be affected (too high, too low, or not at all) by each of the fo l lowing factors? (a) Some of the acid-neutralizing compound remains undissolved. f (b) The C 0 2 produced by the reaction of acid w i t h a carbonate compound is not

fully removed. (c) Three acid-neutralizing compounds (rather than just one) are present in the

antacid. (d) You go past the endpoint in the t i t ra t ion w i t h the antacid. (e) You incorrectly record your init ial buret reading for the acid in the t i t ra t ion of

an antacid tablet as 0.03 ml rather than 0.30 m l .

FOLLOW-UP

After the Report Sheets have been submitted for this experiment, your instructor may compile your results and then issue a report identifying the tablets and giving the percentage by weight of O H " for each antacid. The costs of the tablets might also be compared.

R E F E R E N C E

Yoder, C.H., Suydam, F .H. , and Snavely, F .A. Chemistry. New York : Harcourt Brace Jovanovich, Inc., 1975. Chapter 15.

(Attach a separate sheet of paper with the answers to the question on page 232.)

) Experiment 23 Acidic and Basic Salts

)

Introduction

Water makes up about two thirds of the weight of a human body. Other animals contain comparable proportions of water, and some plants are almost 95% water. Water covers much of the earth's surface; it is in the soil, and it is present as vapor in the atmosphere. Consequently, reactions which occur in aqueous solutions and reactions in which water participates are important to us; many are essential to the life and growth of all organisms. Many of these reactions affect the pH of water by producing hydronium or hydroxide ions.

It is obvious from their formulas that some compounds wi l l produce H 3 0 + or O H " in aqueous solution. A l l Arrhenius acids contain hydrogen, which is usually wri t ten first in the formula to emphasize its abil i ty to react w i t h water. Ionic Arrhenius bases contain hydroxide ions that become separated from the positive ions when these compounds dissolve in water.

Salts such as N a 2 S 0 4 are ionic compounds that contain a cation from a base (in this case, Na + , which is present in the ionic Arrhenius base NaOH) and an anion from an acid (here, S0 4

2 ~, which is present in the Arrhenius acid H 2 S 0 4 ) . Al though not all of these compounds contain H or O H , many of them do react w i t h water to alter its p H . Such a reaction is called the hydrolysis of a salt—the reaction of an ion of a dissolved salt wi th water that produces H 3 0 + or OH" .

The hydrolysis of a negative ion produces the negative ion of water. This react ion occurs if the anion is the conjugate base of a weak acid. For example, HCN is a weak acid; Ka = 4.9 X 1 0 " 1 0 :

HCN + H 2 0 H 3 0 * + C N " (23-1)

The acid molecule HCN has l i t t le tendency to donate protons; the conjugate base C N " accepts protons readily. Consequently, when a salt containing C N " , such as NaCN, is dissolved in water, a Br0nsted—Lowry acid-base reaction occurs between H 2 0 (acid) and C N " (base); that is, a proton is transferred from H 2 0 to C N " :

H 2 0 + C N " =^ HCN + O H " (23-2)

The resulting solution is basic, because hydroxide ions in excess of those produced by the autoionization of water are present.

The hydrolysis of a positive ion produces the positive ion of water. This reaction occurs if the cation is the conjugate acid of a weak base. Molecular bases such as N H 3 are weak bases:

N H 3 + H 2 0 N H 4

+ + O H " (23-3)

For this reaction, Kb = \ .8 X 10" 5 . The forward reaction shows the proton-accepting behavior of the base N H 3 . The reverse reaction indicates the proton-donating abil i ty

235

236 Experiment 23

of N H V , the conjugate acid of the base. The ammonium ion can donate H + to any substance that is capable of acting as a base. Consequently, when a salt containing N H 4

+ is put into water, N H 4

+ and H 2 0 react in a Brcmsted-Lowry acid-base fashion:

N H 4

+ + H 2 0 - N H 3 + H 3 0 + (23-4)

Some hydrated metal ions also hydrolyze to produce acidic solutions. The strong attraction between a small, perhaps highly charged metal ion and the oxygen atom of a watei" molecule weakens an oxygen—hydrogen bond in the water molecule. As a result, the hydrate acts as a proton donor:

[ M ( H 2 0 ) J + " + H 2 0 ^ [ M ( H 2 0 ) ( , ^ u ( O H ) ] * " - 1 + H 3 0 + (23-5)

Reactions of this type often are responsible for the acidity of soi ls .^ The extent to which a hydrolysis reaction occurs is related inversely to the

extent to which the parent acid or base reacts w i th water; that is, the weaker the acid or base, the stronger the conjugate base or acid. The constant describing the equil ibrium concentration ratio is called the hydrolysis constant Kh. For the hydrolysis of an anion A -

[ H A ] [ O H - ] ^ [ A " ] Ka

where Kw is the water ionization constant. For the hydrolysis of a cation BH*

[ H 3 Q - | [ B l = / ^ [ BH ] Kb

A buffer solution can neutralize an acid or a base; the solution utilizes the abil i ty of an ion from a salt to behave as an acid or as a base. A buffer solution may consist of a weak base and a salt o f the base, such as a mixture of N H 3 and N H 4 C I . The N H 3 is a base and can neutralize added acid by the reaction

N H 3 + H 3 0 + - N H 4

+ + H 2 0 (23-8)

The N H 4+ is a proton donor and can neutralize added base by the reaction

N H 4

+ + O i l " - N H 3 + H 2 0 (23-9)

Or, a buffer solution may consist of a weak acid and a salt of the acid, such as a mixture of acetic and sodium acetate: H C 2 H 3 0 2 neutralizes added base, and C 2 H 3 0 2 ~ neutralizes added acid.

In the third part o f this experiment, you wi l l rely on the acidic nature o f a cation to determine the molecular weight o f a salt in which it is present. You wi l l dissolve a 1:1 electrolyte containing N H 4

+ in a carefully measured volume of standardized NaOH (sodium hydroxide) . The reaction in Equation (23-9) w i l l occur, and the ammonia that forms can be boiled out of the solution. The remaining O H " w i l l

237 Acidic and Basic Salts

be titrated w i t h a hydrochloric solution. You wi l l know the number of moles of O H " put into the solution from NaOH and the number of moles of O H " that remained to react w i t h HC1 after the reaction in Equation (23-9) occurred. The difference between these two quantities is the number of moles of O H " that reacted w i t h N H 4

+

according to Equation (23-9); it equals the number of moles of N H 4

+ in the salt sample. Knowing both the weight of the salt sample and the number of moles that it contains, you can calculate the molecular weight of the salt.


beaker ring Bunsen burner ring stand buret stirring rod Erlenmeyer flasks test tubes p l l meter or pH paper watch glass 20-ml pipet wire gauze


bromothymol blue standardized 0.5A/ NaOH 1AÍ F e ( N 0 3 ) 3 IM NaOH \M H C 2 H 3 0 2 \M N a N 0 3

standardized 0,57V/ HC1 lAf N H 3

I/V/HNO3 " ' 1 / V / N H 4 C 2 H 3 0 2

l A í N a C 2 H 3 0 2 LMNH4CI \M N a 2 C 0 3 thymol blue \M N a 2 H P 0 4 unknown ammonium salts \M N a H 2 P 0 4 \M solutions of unknown hydrolyzable salts

P R O C E D U R E

A. Hydrolysis and pH

1. Known acids, bases, and salts. Use pH paper or a pH meter to measure the pH o f \M solutions o f H C 2 H 3 0 2 (acetic acid), N H 3 (ammonia), N H 4 C 2 H 3 0 2

(ammonium acetate), NH 4 C1 (ammonium chloride), F e ( N 0 3 ) 3 (ferric nitrate), H N 0 3 (ni t r ic acid), N a C 2 H 3 0 2 (sodium acetate), N a 2 C 0 3 (sodium carbonate), N a 2 H P 0 4 (disodium hydrogen phosphate), N a H 2 P 0 4 (sodium dihydrogen phosphate), NaOH (sodium hydroxide) , and N a N 0 3 (sodium nitrate). Also measure the pH of distilled water; do not assume that it is exactly 7.0. If the pH of the water is not 7.0, then the pH of a solution may be different from 7.0, even if no acid—base reactions occur between water and the solute.

If you use pH paper, put a small amount of a solution into a small test tube. Place a piece of wide-range pH paper on a clean watch glass, dip a clean glass stirring

238 Experiment 23

rod into the solution, and transfer a drop of the solution onto the paper. After determining the pH of the solution wi th in one or two units, select the proper narrow-range pH paper and determine the pH of the solution wi th in a few tenths of one pH unit . Use an educated imagination about color to make approximations when necessary between the colors on the comparison charts.

I f you use a pH meter, your instructor w i l l provide the necessary instructions. Report the pH of each solution to the nearest 0.1 pi I uni t .

2. An unknown salt. Measure the pi I of a 1.1/ solution of an unknown salt in which one o f the ions hydrolyzes. The salt w i l l contain only one hydrolyzable ion per formula unit ; that is, the formula wi l l be of the type B ( X,. if B x + hydrolyzes and of the type X V A 1 i f A * - hydrolyzes. The salt does not contain any hydrolyzable ion that you worked wi th in the first part of this section of the experknent.

B. A Buffer Solution

T h y m o l blue is an acid-base indicator that changes from red to yellow over the p l l range 1.2-2.8 and from yellow to blue over the p l l range 8.0—9.6. You may be instructed to use either this indicator or a p l l meter for this section of the experiment. To see the colors characteristic of each form of thymol blue indicator, put 5 ml of distilled water in to each of three 20-ml test tubes and add 2 drops of t hymol blue to each test tube. Then to one tube, add \M I I N 0 3 dropwise unt i l the color of the solution is the light red characteristic of pi I < 1.2. To a second tube, add 1/17 NaOH dropwise, unt i l the blue color characteristic of pH > 9.6 appears. Save these three tubes to make color comparisons later in the experiment.

In a small beaker, mix 10-ml of 1/V/ H C 2 H 3 0 2 and 10-ml of \M N a C 2 H 3 0 2 . This produces a buffer solution that contains 0.5/W of the acid and 0.5/1/ of the salt. Divide this mixture eveidy among four 20-ml test tubes.

To each 5-ml sample of the buffer solution, add 2 drops of t hymol blue. The color of the solution should be yel low, showing that the pH of the buffer solution is between 2.8 and 8.0. Calculate the p l l of this solution before you come to the laboratory.

To one 5-ml sample of the buffer solution, add 1A7 NaOH dropwise unt i l the color indicates that the solution has become strongly basic; count the drops as you add them. If you are using a pH meter, add NaOH unt i l the solution is neutral (pH - 7.0).

To a second 5-ml sample of the buffer solution, add IM H N 0 3 dropwise unt i l the color indicates that the solution has become strongly acidic. If you are using a p l l meter, add H N 0 3 unt i l the solution is neutral (pH = 7.0). Does it seem surprising that an acidic solution can have an acid-neutralizing capacity comparable to its base-neutralizing capacity? f low does the amount of \M H N 0 3 required to change the buffer's pH (already well below 7) to about 1 compare to the amount of 1/V/ H N 0 3 required to change water's pH to about 1? Does it surprise you that a solut ion that is already acidic has a greater resistance than water to being made more acidic?

Combine 5 ml of \M H C 2 H 3 0 2 and 5 ml of distilled water to produce 0.5.1/ H C 2 H 3 0 2 . Divide this solution evenly between two test tubes and add 2 drops of t h y m o l blue to each test tube. Calculate the pH of this solution before you come to

239 Acidic and Basic Salts

the laboratory; note the color of the indicator in this solution and account for this color in terms of the pH. To show that the base-neutralizing capacity of the buffer is comparable to the base-neutralizing capacity of a solution containing only the same concentration of acid, add IM NaOH dropwise to 5 ml of 0.5M H C 2 H 3 0 2 un t i l the color of the solution turns blue.

To show that the acid-neutralizing capacity of the buffer results from the presence of C 2 H 3 0 2 ~ , a substance that can act as a proton acceptor, prepare 10 ml of 0.5M N a C 2 H 3 0 2 , divide this solution evenly between two test tubes, and add 2 drops of thymol blue to each test tube. Calculate the pH of this solution before you come to the laboratory; note the color of the indicator in this solution and account for this color in terms of the pH. Then add IM H N 0 3 dropwise to one test tube unti l the color of the solution changes to light red; count the drops as you add them. Are the acid-neutralizing capacities of the buffer and of the salt comparable?

To a third test tube containing 0.5,1/ 11C 2 11 3 0 2 and 0.5.1/ N a C 2 H 3 0 2 , add \M H N 0 3 unt i l the solution is strongly acidic. Then, add a few crystals of solid N a C 2 H 3 0 2 . What happens'.' Why' '

C. Molecular Weight of an Ammon ium Salt

On the analytical balance, weigh out three 0.1-0.3 g samples of the unknown ammonium salt into Erlenmeyer flasks. Into each flask, pipet 20.00 ml of standardized 0.54/ NaOH.

Heat the solutions to boi l ing and maintain a gentle boi l (so that no solution spatters out of the flask) for at least 10 minutes to remove the ammonia that forms according to the reaction of Equation (23-9). To determine whether ammonia is sti l l present after the boiling, hold a piece of moist litmus paper over the neck of the llask. When the vapors no longer test basic, you may discontinue the heating. Cool each flask under cold water.

To each flask, add 4 drops of b romothymol blue. This acid-base indicator changes from yellow to blue over the pH range 6.0-7.6; it is more convenient to use b romothymol blue indicator than to try to determine the point at which the more common indicator phenolphthalein changes from pink to colorless as you titrate base with acid. Titrate each solution wi th standardized 0.5.17 HC1 to the yellow endpoint.

R E S U L T S

A. Hydrolysis and pH

1. Known acids, bases, and salts. List the solutions from most acidic to most basic; give the pH of each solution. State whether each substance is an acid, a base, or a salt. For each solution, write an equation for any reaction that occurs and affects pH. For a particular salt, more than one proton-transfer reaction may occur. If this is the case for any of the salts, show all possible reactions and state which reaction is more extensive as indicated by the pH. (Space is not provided on the Report Sheet for this experiment; attach a separate sheet of paper w i th these results and equations.)

240 Experiment 23

2. An unknown salt. Use the pH of the \M solution to calculate the [ H 3 0 + | or the [ O H - ] of the solution as a result of hydrolysis. Calculate K,, for the hydrolyz-able ion.

B. A Buffer Solution

Describe the color of the indicator in each solution: buffer solut ion, acid solut ion, and salt solution. Account for the color of each solution in terms of its pH and the pi I at which the indicator changes color.

For each solution, state the quanti ty of acid or base that is required to produce the requested change in pH. Compare the acid- and the base-neutralizing capacities of the solutions.

C. Molecular Weight of an Ammonium Salt

Calculate the number of moles of HC1 used in the t i t ra t ion and the number of moles of NaOH that the HCI neutralized. Calculate the number of moles of NaOH that you pipetted into the tlask. The difference between these two quantities is the number of moles of NaOH that reacted wi th N H 4

+ according to Equation (23-9); i t equals the number of moles of N H 4

+ in the salt sample. Use the weight of the salt sample and the number of moles of N H 4

+ that it contains to determine the molecular weight of the salt N H 4 X . Report an average of your results for the three trials. Round the figure o f f to three significant figures, even if you are entitled to four.

QUESTIONS

1. Some compounds that are used as fertilizers are N H 3 , N U 4 N 0 3 , and ( N H 4 ) 3 P 0 4 . What effect would each of these compounds have on the pH of ground water?

2. Some compounds that are tised for the specific purpose of altering the soil's pH are C a C 0 3 , FeS0 4 , and A 1 2 ( S 0 4 ) . What effect would each of these compounds have on soil pH?

3. Equimolar amounts of NaOH and H C 2 H 3 0 2 are combined in aqueous solution. Will the resulting solution be neutral? Explain yotir answer.

4. Describe a way of preparing an acetic acid/sodium acetate buffer solution that has an acid-neutralizing capacity twice as great as its base-neutralizing capacity.

5. Describe a way of preparing an acetic acid/sodium acetate buffer solution that has a base-neutralizing capacity three times as great as its acid-neutralizing capacity.

R E F E R E N C E

Yoder, C.H., Suydam, F .H. , and Suavely, F.A. Chemistry. New York : Harcourt Brace .lovanovich. Inc.. 1975. Chapter 15.

C. Molecular weight of an ammonium salt:

Tr ia l 1 Trial 2 Trial 3

Weight of salt

Molar i ty of NaOH

Volume NaOH used

Moles NaOH

Molar i ty of HCI f

Volume HCI used

Moles HCI

Moles NaOH neutralized by HCI

Moles NaOH

neutralized by N H 4

+

Moles N H 4

+

Moles N H 4 X

Molecular weight o f N H 4 X _ _ _

Average molecular weight

(Answer the questions from page 240 below; if necessary, attach a separate sheet of paper to the Report Sheet.)

Experiment 24 Electrochemistry: Voltaic Cells

Introduction

This experiment deals wi th cells in which spontaneous oxidat ion-reduct ion reactions can be used to produce electricity. The reactants in the o x i d a t i o n -reduction reaction are separated physically, so there cannot be a direct electron transfer from one reactant to the other as there would be if the reactants were in physical contact. Instead, the two reactants are joined by a wire through which electrons travel from one reactant to the other. This How of electrons through a wire constitutes an electric current. You should recognize one important difference between voltaic and electrolytic cells: in the voltaic cell, electrons flow unassisted from one reactant to the other, because the reaction is thermodynamically spontaneous; in an electrolytic cell, a nonspontaneous reaction is forced to occur by the application of an outside power supply (an external battery) that "pumps" electrons through the electrolytic cell.

Figure 24-1 shows a voltaic cell in which C u 2 + ions are reduced by zinc by the cell reaction

Zn + C u 2 + • Cu + Z n 2 + (24-1)

244 Experiment 24

The voltaic cell consists of:

1. A Zn strip immersed in a 0.54/ Z n S 0 4 solution. 2. A Cu strip immersed in a 0.54/ C u S 0 4 solution. 3. A salt bridge that is constructed by f i l l ing a U-shaped tube wi th 0.5M

K 2 S 0 4 , plugging the ends of the tube wi th cot ton, inverting the tube, and immersing one arm of the tube in each beaker.

4. A conducting wire through which electrons flow from the Zn electrode to the Cu electrode.

A voltmeter is inserted in the circuit in Figure 24-1 to measure the cell voltage. The value of the voltage, or the electromotive force (emf), can be regVrded as a quantitative measure of the tendency of electrons to How in the cell, that is, the tendency of electrons to be transferred from one rcactant to the other.

The overall reaction in a voltaic cell (or in an electrolytic cell) can be divided into two half-cell reactions, an oxida t ion half and a reduction half, each occurring at one electrode:

Oxidation: Zn • Z n 2 + + 2e (at the anode) (24-2)

Reduction: C u 2 + + 2e~ • Cu (at the cathode) (24-3)

Overall reaction: C u 2 + + Zn • Cu + Z n 2 + (24-1)

Separating the total reaction into two half reactions, each occurring in a half cell, implies that we can construct several different voltaic cells simply by replacing one of the half cells, say, Z n / Z n 2 + , w i th different half cells.

The concept that voltaic cells consist of two half cells also suggests that the measured cell voltage is the sum of the contributions from both half cells. In mathematical language

A total — ^-oxidation A reduction (24-4)

In this experiment, you are to construct several voltaic cells, measure their voltages, and then investigate the effect of solution concentrations on cell voltage.

M E A S U R I N G C E L L V O L T A G E

The instrument that measures emf is called a voltmeter. It consists of a scale w i th a deflecting needle mounted in a case; two lead wires attached to the voltmeter are connected to the electrodes of the cell (see Figure 24-2). When the lead wires are connected to the proper electrodes, the needle wi l l move from zero and the voltage generated by the cell can then be read. If the lead wires are connected in reverse, the needle w i l l not move. Before you begin the experiment, you must first determine how to connect the voltmeter to a cell. In the reaction between C u 2 + and Zn, you know that the Zn electrode is the anode and that the Cu electrode is the cathode. Determine and record which lead wire on the voltmeter you must connect to which electrode before a voltage wi l l register. For the remaining cells, use this information

245 Electrochemistry: Voltaic Cells

to determine which electrode is the anode and which electrode is the cathode, and, consequently, in which direction the electrons How through the external cell circuit .


alligator clips sandpaper 250-ml beaker voltmeter (1000 f i / v o l t resistance) porous cup


Cu strip - ' 0.5M F e S 0 4

Fe strip 0.5yV/ M g S 0 4

Mg strip cone. N H 3 solution Pb strip 0.5M P b ( N 0 3 ) 2

Sn strip 0.5A/ SnCl 2

Zn strip 0.5A-/ Z n S 0 4

0.5M C u S 0 4

P R O C E D U R E

To avoid'the time-consuming process of making U-tube salt bridges, you can construct a cell as illustrated in Figure 24-3. The porous cup allows ions to flow back and for th through its walls to equalize the electrical charges in the two halves of the cell. The cup performs the same function as a salt bridge but is easier to handle.

Cell Voltages

First, pour about 150 ml of 0.5M C u S 0 4 in to a 250-ml beaker. Bend a sandpapered Cu strip and hang it over the side of the beaker so that it is partly immersed

246 Experiment 24

in the solution. Place the porous cup in the beaker and hang a cleaned strip of Zn over the side. Add 0.5A/ Z n S 0 4 to the porous cup unt i l the l iquid level is the same as it is in the beaker.

Use alligator clips to connect the metal electrodes to the terminals of the voltmeter. Record the ini t ia l voltage of the cell.

Now disconnect the voltmeter and remove the porous cup from the beaker. Discard the Z n S 0 4 solution and clean and dry the cup.

Repeat this procedure, replacing Zn and 0.5/1/ Z n S 0 4 w i th the fol lowing metal strips and solutions of their salts:

he and 0.5.4/ FeS0 4

Mg and 0.5/1/ M g S 0 4

Pb and 0.5.1/ P b ( N 0 3 ) 2

Sn and 0.5/1/ SnCl 2

Identify the anode and the cathode for each cell.

Concentration Effects on Cell Voltage

A l l the solutions that you are using in this experiment are 0.5A/. Next you w i l l design experiments that determine (at least quali tat ively) what effect changing the concentration of a salt solution has on cell voltage. As you design your experiments, keep the following points in mind :

1. Use a Z n - C u cell , so that the results of different students can be easily compared.

2. If your time is l imi ted , the effect may be investigated for oidy the cathode compartment solution instead of for both solutions.

3. Vary the concentration of the solution slightly from O.SM to see what effect this has. Then try larger variations, such as 0.05A7, 0.005A7, and so on.

4. Try to determine if the cell voltage varies in some regular way wi th concen-

247 Electrochemistry: Voltaic Cells

t ra t ion. For example, is the variation linear 0 That is, does a tenfold decrease in concentration lower the voltage to one tenth of its former value? and so on.

Complexation Effects on Cell Voltage

The blue color of the 0.5,1/ C u S 0 4 solution is actually due to the presence of a complex ion of copper C u ( I I 2 0 ) 6

2 + . Design and carry out an experiment that wi l l show what effect ( i f any) is produced by the presence of copper in the form of the ammonia complex C u ( N H 3 ) 4

2 + instead of in the form of the hydrated ion. The hydrated ion can be converted to the ammoniated ion by adding concentrated N H 3

solution to a C u S 0 4 solution unti l an intense deep-blue color results. For both of the above experiments that you design for yourself, yot i r report

should (1) describe the experimental procedures yoti use, (2) include the data you collect, and (3) discuss the conclusions you can draw from your data.


1. For each cell for which you measured voltage, write the anode half reaction and the cathode half reaction. In each case, this involves determining which electrode is the anode and which electrode is the cathode.

2. For each cell, add the two half reactions to obtain the overall spontaneous cell reaction.

3. The emf of each cell is the sum of the contributions from each half cell. The two half-cell contributions are d i f f icul t to separate; however, if we define that the Cu 2 + /Cu couple has E° - 0.000 volts, we can assign the entire measured cell potential emf to the other couple. The potential emf wi l l then represent in quantitative fashion the tendency of each couple to reduce C u 2 + to Cu. Write each half-cell reaction as a reduction. Then assign a standard cell potential to each reduction, as described above, attaching the correct algebraic sign.

4. Arrange the list of metals in order of their decreasing abi l i ty to donate electrons to C u 2 + ; that is, list the metal w i t h the greatest tendency to donate electrons first.

5. From your textbook or from some other data source, obtain the accepted values of standard half-cell potentials for the couples used in this experiment. Next to the appropriate half reaction wri t ten as a reduction, record this value of E°. For each half reaction, record the difference between this value and the value of E° that you assigned in this experiment.

6. What do the differences between the tabulated standard potentials and the values you assigned from your measurements tell you about the E° of the C u 2 + / C u couple? Explain your answer.

7. Discuss your data concerning cell voltage and its concentration dependence. 8. What effect does the complexat ion of C u 2 + by N H 3 to produce C u ( N H 3 ) 4

2 +

have on the half-cell potential of Cu 2 + /Cu? Which species is the better oxidizing agent, C u ( H 2 0 ) 6

2 + or C u ( N H 3 ) 4

2 + , >

248 Experiment 24

Q U E S T I O N S

1. In this experiment, you assigned zero contr ibut ion to cell potentials from C u 2 + / C u ; this enabled you to compare the tendencies of other couples to reduce C u 2 + to Cu. Is this a valid procedure? Explain your answer.

2. Using your measured data, calculate the free energy change AG° which would occur in the cell Zn|0.5A/ ZnSO 4 | |0 .5Af SnCl 2 |Sn, if 0.5A/ of S n 2 + were reduced while the concentrations of Z n 2 + and S n 2 + remained at 0.5M.

R E F E R E N C E V

Yoder, C.H., Suydam, F .H. , and Snavely, F.A. Chemistry. New Y o r k : Harcourt Brace Jovanovich, Inc., 1975. Chapter 17.

!

) Experiment 25 Electrochemistry: An Electrolytic Cell

Introduction

In an electrolytic cell, an oxidat ion-reduct ion reaction that does not proceed spontaneously can be forced to occur by using an external device to pull electrons from one electrode (the anode), causing oxidat ion to occur, and to pump them into another electrode (the cathode), causing reduction to occur. The process that occurs in the cell is called electrolysis. This experiment deals w i t h the relationship between the amount of oxidat ion and reduction that takes place during electrolysis and the quanti ty of electricity that passes through the cell.

Figure 25-1 is a diagram of a cell for the electrolysis of a concentrated C u C l 2

solution; here, a strip of copper is the cathode and a strip of platinum is the anode. Platinum is a fairly inert metal and undergoes no reaction in this cell; it serves only as an electron-transfer medium. The reactions at the electrodes in Fi mi re 25-1 are

Anode: 2C1 >-Cl2(g) + 2t' (oxidat ion)

Cathode: C u 2 + + 2e~ • Cu(s) (reduction)

(25-1)

(25-2)

If the current (in amperes) is known and the time of electrolysis ( in seconds) is recorded, then the quanti ty of electricity (in coulombs) passing through the cell can be calculated: -••

amp X sec = coul/sec X sec = coul (25-3)

Faraday's First law states that the quantity of a metal deposited at an electrode in an electrolytic cell is proport ional to the number of coulombs of charge passing

I

254 Experiment 25

through the electrical circuit . Thus, the greater the amount of charge that passes through the cell, the greater the weight of metal deposited onto the cathode.

Faraday's second law states that the weights of different substances produced by the passage of a given amount of charge are proport ional to the equivalent weights of those substances. For example, the passage of 96,500 coul of charge would deposit different weights of silver, copper, and lead.

Of course, the laws describing electrolysis can also be applied to cells in which reactions do not involve the plating out of a metal onto a cathode. It is simply that the use of such cells in the laboratory provides a convenient way to measure the amount of reduced material (in this case, the metal).

In this experiment, you w i l l electrolyze a solution of C u S 0 4 (copper sulfate), so that copper is deposited onto the cathode. In addi t ion, oxidat ion at the anode must occur. You wi l l use copper for both electrodes. Depending on the voltage supplied by the batteries, any of these three oxidat ion reactions might occur:

Sulfate ion might be oxidized:

2 S 0 4

2 " • S 2 0 ^ %Je (25-4)

The copper anode itself might be oxidized:

Cu • C u 2 + + 2c" (25-5)

Water might be oxidized:

2 H 2 0 • 0 2 + 41F + 4e (25-6)

You must determine which of these three reactions occurs by observing the anode during and after the electrolysis. Reaction (25-5) would cause the anode to lose weight; weighing the anode before and after the electrolysis wi l l determine whether this has occurred. Reaction (25-6) would result in bubbles of oxygen gas being given o f f around the anode. If there is no evidence of anode weight loss or oxygen gas emission, then reaction (25-4) must be occurring (assuming that there is no other unknown material present that can be oxidized). The increase in the mass of the cathode where copper wi l l plate out serves as a measure o f the quanti ty o f the reaction taking place at this electrode. This mass increase can he related to the number of moles of electrons f lowing through the cell (calculated from the total electric charge and the charge of 1 mole of electrons). The equivalent weight of copper (that is, the weight deposited b y . l mole of electrons) can then be calculated easily.

It is best to plate metals by electrolysis under conditions of long plating time and low current. The currents used in this experiment wi l l be substantially higher than op t imum to save t ime; consequently, some inaccuracies wi l l be introduced. But your results w i l l probably be wi th in 10-15% o f the theoretical results.


alligator clips direct-current power supply ammeter sandpaper beakers variable resistance

255 Electrochemistry: An Electrolytic Cell


acetone \M C u S 0 4

Cu strips

P R O C E D U R E

Clean two copper strips w i th medium or fine sandpaper. Bend each strip about 2 cm from the end, so that it can be hooked over the side of a beaker. Handle the cleaned strips only by the edges to keep them as free from contamination as possible. Weigh each strip as accurately as possible on the analytical balance. Be sure that you can distinguish between the two strips. One w i l l be the anode and one wi l l be the cathode. Y o u must know which is which, so that you can compare the weight of each strip after the electrolysis wi th its weight before the electrolysis.

Place the copper electrodes over opposite sides of a 100-ml beaker; be sure that the strips remain parallel to one another. Assemble the electrical equipment as shown in Figure 25-2. Be careful to connect the cathode to the negative terminal of the power source. Before proceeding further, ask your instructor to check your apparatus.

Pour 60 ml of C u S 0 4 solution into the beaker. Adjust the variable resistance to its maximum value. Turn on the power-supply switch; note the time. Immediately adjust the resistance so that a current of about 100 mil l iamp (read on the ammeter dial) flows in the cell.

256 Experiment 25

A l l o w the current to flow for 25 minutes to carry out the electrolysis. Check and record in your notebook the current reading once a minute. If necessary, readjust the resistance to maintain the current of 100 mil l iamp.

At the end of 25 minutes, tu rn o f f the power supply. Then disconnect the clips to the electrodes. L i f t each electrode by the edge and hang it over the lip o f a dry, 250-ml beaker. Wash each electrode carefully by dipping it into a beaker of distilled water or w i t h water from your wash bottle. Do not squirt the water directly onto the metal deposited during the electrolysis, because this might cause the metal to fall from the surface of the electrode. Instead, direct the water at the upper por t ion of each stip and allow it to run down the strip. Be sure to wash both sides of each strip. Then repeat this washing procedure using acetone.

A l l o w the electrodes to dry for about 15 minutes, and then weigh them. Be careful to handle the electrodes only by the edges and above the area where the metal has been deposited.

Your instructor may ask you to make a second tr ial .

D A T A AMD R E S U L T S

1. To obtain the average value of the current, add all the current readings ( in amp) and divide by the number of readings.

2. From Equation (25-3), calculate the amount of charge (in coul) that passed through the cell, using the average value of current (in amp) and the time (in sec) of your electrolysis.

3. Since the charge is carried by electrons, you can calculate the number of moles of electrons needed to carry the charge that passed through your cell. (The charge on one mole of electrons is 96,500 coul.)

4. From the cathode weight gain and the atomic weight of copper, determine the number of moles of copper that plated out onto the cathode.

5. The number of grams of copper deposited per mole of electrons passing through the cell is the equivalent weight of copper.

Q U E S T I O N S

1. Compare your experimental value for the equivalent weight of copper w i t h the theoretical value.

2. If you lost some of the copper deposit on the cathode before weighing i t , how would your value of the equivalent weight be affected?

R E F E R E N C E

Yoder, C.H., Suydam, F .H . , and Snavely, F.A. Chemistry. New York : Harcourt Brace Jovanovich, Inc., 1975. Chapter 17.

Write an equation for the reaction that occurred at the anode in the electrolytic cell.

Describe the physical evidence on which you based your conclusion about the anode reaction.

Answer the questions from page 256 in the space below. (If necessary, attach a separate sheet of paper to the Report Sheet.) ^»

Experiment 26 /{J An Oxidation-Reduction Titration U

Introduction

In one type of chemical reaction, electrons are transferred between the reactants. These reactions are called oxidation-reduction (or redox) reactions.

An electron transfer reaction between the enzyme cytochrome a 3 (which contains F e 2 + ) and oxygen enables living organisms to use atmospheric oxygen to support life. The energy released in fuel-cell redox reactions has provided power for Apol lo space flights. The energy of ordinary automobile and dry-cell batteries is released by spontaneous redox reactions. Experiment 24 was concerned wi th electrical energy supplied by these reactions.

In addit ion to being sources of energy, oxida t ion-reduct ion reactions are useful for analytical purposes. This experiment is designed to study a redox reaction used in chemical analysis. Al though you have used redox reactions before in this course-for example, in the synthesis of S n l 4 and in studying the solubil i ty product of C u ( I 0 3 ) 2 — t h e focus has often been only on the stoichiometric relationships involved. This experiment wi l l examine in detail some features o f ti trations in which the reactants are oxidizing and reducing agents. The goal o f this experiment w i l l be to use a redox t i t ra t ion to report the percentage of F e 2 + in an unknown sample (to be provided by your instructor) .

Whether any- particular species tends to be reduced is reflected in the value of the standard electrode potential E° for the reduction half-reaction. Equation (26-1) shows that the ferric ion F e 3 + has a tendency (in comparison to H 3 0 + ) to be reduced to the ferrous ion F e 2 + :

Fe 3 + (aq) + e • Fe 2 + (aq) E° = +0.77 volts (26-1)

Similarly, Equation (26-2) shows that the dichromate ion C r 2 0 7

2 ~ is easily reduced t o C r 3 + :

C r 2 0 7

2 ' ( a q ) + 1 4 H + + 6e > 2 C r 3 + + 7 H 2 0 E° = +1.33 volts (26-2)

Although the E° values in Equations (26-1) and (26-2) indicate a tendency for both F e 3 + and C r 2 0 7

2 " to be reduced (in comparison to the tendency of FT to be reduced to y H 2 ) , it is obvious that C r 2 0 7

2 ~ has a considerably greater tendency to be reduced than F e 3 + . The dichromate C r 2 0 7

2 ~ w i l l even abstract electrons from Fe 2 + (aq) and oxidize it to Fe 3 + (aq) , because C r 2 0 7

2 " has a very great tendency to be reduced. When C r 2 0 7

2 ~ and F e 2 + are allowed to react in an acidic aqueous solution, the reaction that occurs can be obtained by combining Equations (26-1) and (26-2) in such a way that the number of moles of electrons to be supplied to C r 2 0 7

2 _ is exactly provided by the F e 2 + :

6Fe 2 + (aq) + C r 2 0 7

2 " ( a q ) + 14H +(aq) • 6Fe 3 + (aq) + 2Cr 3 + ( aq ) + 7 H 2 0 (26-3)

259

260 Experiment 26

Equation (26-3) shows that 6 moles o f F e 2 + react w i t l i 1 mole of C r 2 0 7

2 " . Because of this simple relationship, it is convenient to use a dichromate solution of known molari ty to titrate a solution that contains iron as F e 2 + . The number of moles of iron present can easily be determined from Equation (26-3) and from the volume and molari ty of dichromate solution used in the t i t ra t ion .

T w o problems are encountered when this simple procedure is actually carried out. f i r s t , it is necessary to be sure that all the iron in the sample is actually present as Fe 2 + . Second, some way must be devised to detect the endpoint; that is, the point when exactly 6 moles of C r 2 0 7

2 ~ have been added for each mole of F e 2 + present. These problems and their solutions wi l l be discussed in turn .

Oxygen in the air can oxidize F e 2 + to F e 3 + and the 0 2 can be reduced to H 2 0 in aqueous solutions, as indicated by the fol lowing reduction potentials:

0 2 + 4 H + + 4<f • 2 H 2 0 E° = +1.23 volts (26-4)

F e 3 + + e • F e 2 + E° = +0.77 volts (26-1)

Thus, in any solution containing F e 2 + , some F e 2 + may be oxidizsd to F e 3 + by the oxygen present in the air; sometimes, the pale-yellow color characteristic of a small amount of F e 3 + can be seen. Because F e 3 + does not react wi th dichromate ions, whatever iron is present as F e 3 + instead of as F e 2 + cannot be detected in the dichromate t i t ra t ion , and low results for iron w i l l be obtained. To avoid this, any iron that is present as F e 3 + must be converted to F e 2 + before the t i t ra t ion begins; this is done by adding SnCl 2 to the hot solution:

S n 4 + + 2e > S n 2 + E° = +0.15 volts (26-5)

F e 3 + + e" > F e 2 + E° = +0.77 volts (26-1)

2 F e 3 + + S n 2 + -* 2 F e 2 + + S n 4 + E° = +0.62 volts (26-6)

However, this addi t ion, in turn, generates another d i f f i cu l ty ; any remaining S n 2 + in excess of the quanti ty needed to reduce the F e 3 + can be oxidized by the dichromate ion, as can be seen from the standard electrode potentials E° for reactions (26-2) and (26-5). Because it is not possible to distinguish the dichromate that reacts wi th F c 2 + from the dichromate that reacts w i th S n 2 + , the calculated amount of F e 2 +

would be high. Therefore, the next step is to remove the excess Sir2"1"; this is accomplished by adding a solution of H g C l 2 . Any remaining S n 2 + w i l l immediately be oxidized by H g 2 + to S n 4 + while the H g 2 + is reduced. The relevant equations are

2 H g 2 + + 2C1" + 2e • Hg 2 Cl 2 ( s ) E° = +1.44 volts (26-7)

2 H g 2 + + S n 2 + + 2CP • Hg 2 Cl 2 ( s ) + S n 4 + E° = +1.29 volts (26-8)

The insoluble H g 2 C l 2 precipitates from the solution as a small quanti ty of white solid. An examination of the standard electrode potentials E° in Equations (26-7) and (26-1) indicates that H g 2 C l 2 should react wi th the F e 3 + that is produced in the dichromate t i t ra t ion of F e 2 + . Fortunately, this reaction is quite slow; if the t i t ra t ion

251 An Oxidation—Reduction Titration

is carried out at a reasonably rapid pace, no interference from this source is encountered.

To summarize the procedure discussed in the last several paragraphs: Only i ron present as F e 2 + w i l l react w i t h the C r 2 0 7

2 ~ solution that is used as the t i t rant . Therefore, SnCl 2 solution is added to reduce any F e 3 + to F e 2 + . Excess SnCl 2 is removed by adding HgCl 2 solution to oxidize S n 2 + to Sn 4 + . It is theoretically possible for various mercury species to react w i t h other solution components, but these reactions are slow enough that the t i t ra t ion can be completed before they occur.

Using an indicator to detect the endpoint in a redox t i t ra t ion presents a different problem than the one you encountered in acid-base t i trations. There, the indicator was a species w i t h protonated and deprotonated forms that exhibited different colors:

HIn + H 2 O ^ H 3 0 + + In (26-9) Color I Color II

As [ H 3 0 + ] changes during a t i t ra t ion , the equi l ibr ium shifts to the left or to the right, and the solution displays color I or color I I , respectively.

An equi l ibr ium shift also occurred when you t i t rated iodine using thiosulfate as a reducing agent and starch as an indicator (Experiments 14, 15, 19). In the presence of iodine, starch forms a deep-blue complex if some iodide is also present:

I 2 + I" + starch «<* deep-blue complex (26-10)

The disappearance of the blue color, resulting from an extensive shift to the left, means that the iodine was just consumed in the t i t ra t ion .

For most r.edox reactions, no indicator substances are available that form different colored species w i th oxidized and reduced forms of a couple (as starch does w i th iodine). Consequently, some substance that can be oxidized is used:

In ^ In* +e (26-11) Color I Color II

The position of this equi l ibr ium (and therefore the color of the solution) depends on the potential of the solution at the endpoint.

A suitable indicator for the t i t ra t ion of F e 2 + by C r 2 0 7

2 ~ is diphenylamine sulfonic acid, wi th E° - 0.85 volts, which changes from colorless to purple at the endpoint. In other words, the indicator is oxidized to the purple I n + form when the endpoint is reached.

The essential point of the preceding discussion is that the electron-accepting abili ty of a t i t ra t ion solution at equil ibrium can be measured. An indicator w i th an E° close to this value is chosen, so that i t w i l l be oxidized at equi l ibr ium. I f In and I n + exhibi t different colors, the endpoint can be detected visually.

This in t roductory discussion has dealt pr imari ly w i t h two important problems normally encountered in redox titrations in general: (1) how to be sure that all the titratable material is in the same oxidat ion state, and (2) how to select a visual indicator. These two points have been applied to the i ron-dichromate t i t ra t ion . The simple conclusion is that it is possible to carry out such a t i t ra t ion to determine the percentage of iron in a sample, if the amount of dichromate required for the ti trat ion is known , by using the stoichiometry of Equation (26-3).

262 Experiment 26


50-ml buret 250-ml volumetric flask burner medicine dropper 10-ml graduated cylinder ring stand Erlenmeyer flasks wire gauze


diphenylamine sulfonate indicator H 3 P 0 4

6A/HC1 K 2 C r 2 0 7

HgCl 2 solution SnCT2 solution

P R O C E D U R E

In conducting this experiment, you w i l l prepare a potassium dTchromate solution of accurately known molar i ty and then use this solution to titrate weighed samples of your unknown.

Preparation of K ; C r 2 0 7 Solution

On the analytical balance, weigh out about 1.3 g of K 2 C r 2 0 7 in to a clean, dry beaker or weighing pan; record the weight that you actually take as precisely as the balance permits. Transfer all the salt to a clean (see In t roduct ion , p. 28), 250-ml volumetric flask; you do not have to dry the flask if there is only distilled water in i t . Rinse the beaker several times w i t h a few ml of distilled water to be sure that all the K 2 C ' r 2 0 7 is washed into the flask. Add about 50 ml of distilled water to the flask anil swirl the flask unt i l the potassium tlichromate is completely dissolved. Then dilute the solution w i t h distilled water to the mark on the neck of the flask. Stopper the flask and invert it several times to mix the solution completely. If necessary, this solution may be stored in your desk unt i l the next laboratory period, provided it is t ightly stoppered.

Titration of Samples

Obtain an unknown and record its identif ication number. Clean a buret, rinse it w i th three 10-ml portions of your standard K 2 C r 2 0 7

solution, and then fi l l the buret w i th this solution. The buret should be filled early in the procedure, because the samples must be titrated rapidly once they are prepared.

On the analytical balance, weigh out about 1.4 g of your unknown; record the weight that you actually take. Dissolve the sample in 15 ml of 6M HC1 and then add 35 ml of distilled water. The solution w i l l probably have a yel low color because o f the presence of F e 3 + . Heat the solution almost to boiling and add SnCl 2 solution dropwise unt i l the yellow color just disappears; then add 1 or 2 extra drops. If very

263 An Oxidation— Reduction Titration

l i t t le F e 3 + is present, only 1 or 2 drops of SnCl 2 solution may dispel the yel low color. Even if no yellow color is apparent, you should still add 1 or 2 drops (no more) of SnCl 2 solution to be sure that any trace of F e 3 + is reduced to Fe 2 + .

Cool the solut ion under running water to room temperature. Then measure 10 ml of saturated HgCl 2 solution in a graduated cylinder and quickly add this to the flask as you swirl i t . The appearance of a silky, white I I g 2 C l 2 precipitate indicates that the reduction has proceeded satisfactorily. If the precipitate is gray, indicating the presence of metallic mercury, you have added too much SnCl 2 and must discard the sample, because the mercury wi l l react wi th C r 2 0 7

2 ~ . If the precipitate is white and silky, add 5 ml of 85% H 3 P 0 4 and 5 drops of

diphenylamine sulfonic acid indicator. Titrate at once wi th your standard K . 2 C r 2 0 7

to the purple endpoint; the solution may appear green just before the endpoint is reached because of the presence of intermediate oxidation products of the indicator.

Weigh out a second sample of your unknown, dissolve it and treat it as described above, and then titrate i t . You may want to adjust the sample size so that a more convenient volume of the dichromate solution is required.

Prepare and titrate a third sample. Because it takes a non-negligible amount of dichromate to react w i th the indi

cator and oxidize it to the purple form, you must determine what volume of dichromate solution is required to achieve this effect and then subtract this required volume from the volume used in each t i t ra t ion . You can do this by using K 2 C r 2 0 7

solution to titrate a solution containing all the components mentioned previously except your unknown; that is, you can mix 15 ml of (vVl HCl , 35 ml of distilled water, 2 drops of SnCl 2 solution, 10 ml of HgCl 2 solution, 5 ml of H 3 P 0 4 , and 5 drops of indicator. Then titrate this solution w i th dichromate. The volume required is the blank correction. In the absence of F e 2 + , the reaction of C r 2 0 7

2 ~ wi th d i phenylamine sulfonate is quite slow. Be sure you allow enough time for each drop to react before adding the next drop. The blank correction should be less than 1 ml .


Using the exact weight of K 2 C r 2 0 7 that you took, the formula weight of the salt, and the fact that the volumetric flask contains 250.0 ml , calculate the molari ty of your standard potassium dichromate solution to the proper number of significant figures.

For each t i t ra t ion , determine the volume of standard K 2 C r 2 0 7 solution that reacted wi th F e 2 + by subtracting the blank correction from the total volume used.

For each t i t ra t ion , calculate the number of moles of K 2 C r 2 0 7 used. From Equation (26-3), determine the number of moles of iron in the sample.

Calculate the number of grams of iron in each sample. Convert this figure to the percentage of iron in the sample. Be sure to use the correct number of significant figures.

QUESTION

How would your calculated percentage of iron compare to the true value (high, low, same) if each of the fol lowing blunders were made?

264 Experiment 26

(a) Not all of the K 2 C r 2 0 7 that was weighed out was transferred to the volumetric flask.

(b) Some of the iron-containing sample was spilled after it was weighed. (c) No SnCl 2 solution was added to the iron-containing sample solution, even

though the solution had a yellow color. (d) No HgCl 2 solution was added after the SnCl 2 solution was added to the yel low,

iron-containing sample solution. (e) After the addit ion of H g C l 2 , a gray precipitate indicating presence of metallic

mercury appears and the solution is titrated instead of discarded.

R E F E R E N C E S

1. Yoder, C.H., Suydam, F .H . , and Snavely, F .A. Chemistry. New York : Harcourt Brace Jovanovich, Inc., 1975. Chapter 17.

2. Kol thof f , I . M . , and Belcher, R. Volumetric Analysis, V o l . I I I . New York : Inter-science Publishers, John Wiley & Sons, Inc., 1957. Chapter V.

3. Quane, D. Journal of Chemical Education 48, A105 (197 1).

Show a sample calculation of moles of K . ,Cr . ,0 7 reacting w i th F e 2 + :

Show a sample calculation of grams of iron present in a sample:

Show a sample calculation of the percentage of iron in a sample:

(Answer the questions from pages 263-264 on a separate sheet of paper and attach it to the Report Sheet.) *

)

)

Experiment 27 Metal Chelate Complexes

Introduction

Even before studying transition metal complexes in your textbook, you encountered several such species in your laboratory work . The first of these was [ C o ( N H 3 ) 6 ] C l 3 , which you synthesized in Experiment 5. As you may recall, this compound contains a C o 3 + ion bound to six ammonia molecules that are arranged in octahedral fashion around the ion (see Figure 5-1, page 61). These surrounding groups are called ligands. Each ammonia ligand has only one atom (N) that can bond to a metal cation. In some possible ligands, more than one atom is capable of bonding to a metal cation. One example of such a ligand is N H 2 C H 2 C H 2 N H 2 (ethylene-diamine). Here, the two N atoms can both be bonded to the same metal ion simultaneously; the ligand is said to be bidentate. The resulting complex contains rings of atoms and is called a chelate. An ion such as C o 3 + can be surrounded in octahedral fashion w i th six coordinated N atoms by forming a complex wi th six N H 3 molecules (refer again to Figure 5-1) or by forming a complex w i t h three ethylenediamine molecules (see Figure 27-1).

Chelating ligands form complexes that are especially stable thermodynamically. In other words, the equi l ibr ium constants for the formation of these complexes are extremely large in comparison to the equil ibr ium constants for the formation of similar complexes that contain ligands wi th only one donor atom. For example, reaction (27-1) has an equi l ibr ium constant K = 1 0 3 4 :

268 Experiment 27

This difference is especially surprising when we consider that the C o 3 + is in substantially the same environment in both complexes; that is, the C o 3 + ion is surrounded octahedrally by six coordinated N atoms. The greater stability of complexes containing chelating ligands, compared to the stabili ty of similar complexes containing nonchclating ligands, is called the chelate effect.

The essential reason for the added stabili ty of chelate complexes is a statistical (ent ropy) one. Once the first N of ethylenediamine attaches itself to the metal cation, the second N atom is held nearby. The metal cation is far more l ikely to find this second N and coordinate wi th it than to encounter and react wi th an entirely different ligand molecule in the solution. This statistical effect shows up in the entropy of complex formation. Another way of viewing the particularly favorable entropy change in the formation of chelates is to note that seven particles are produced from four in Equation (27-2), while a monodentate ligand produces seven particles from seven [see Equation ( 2 7 - 1 ) ] . The chelate effect occurs wi th many ligands besides ethylenediamine. To produce this effect, the only requirement is that each ligand molecule must have more than one atom that can coordinate to one metal cation. v .

The purpose of this experiment is to study the consequences of chelate formation compared to the formation of complexes wi th monodentate ligands. The chelating ligands that you w i l l study in this experiment appear in Figure 27-2. The behavior o f these ligands wi l l be compared wi th the behavior o f the monodentate ligands water, ammonia, and chloride. Note that the ligands ethylenediamine and oxalate are bidentate; each ligand has two atoms that are capable of coordinating

269 Metal Chelate Complexes

simultaneously w i th one metal ion. Ethylenediaminetetraacetate ( E D T A ) is sexa-dentate; it has six atoms for coordinating (2 N and 4 O). Figure 27-3 shows how an ethylenediaminetetraacetate ligand can coil itself around a cation so that all six of the octahedral coordination positions are f i l led. Acetate may be either monodentate or bidentate; i t is probably bidentate w i th P b 2 + , which you wi l l use in this experiment.

One of the main practical uses of chelate formation is in the dissolution of nearly insoluble salts. This aspect o f chelate behavior wi l l be emphasized here.

Consider the equil ibr ium representing complex formation between a metal cation M 2 + and a monodentate or chelating ligand L, where n is the number of ligands coordinated t o M 2 +

M 2 + + « L ^ M L „ 2 + (27-3)

As w i th any equi l ibr ium, reaction (27-3) can be described by an equi l ibr ium constant; for this type of reaction, the equi l ibr ium constant is called a formation constant:

[ M L „ 2 + ] f [ M 2 + ] [ L ] ' !

By adding sufficient L, the equi l ibr ium of Equation (27-3) can be shifted to the right and essentially all the M 2 + can be converted to M L „ 2 + . If M L „ 2 + has a characteristic color, i t w i l l be easy to determine when this shift occurs. Because chelating ligands have especially large values o f K;-, a lower concentration o f L w i l l be required to shift the equi l ibr ium of Equation (27-3) to the right if L is chelating than if L is monodentate (assuming that the same k ind of atom is bonded to the metal ion in each case, so that the ligands are as much alike as possible).

Now consider the process of dissolving a salt MX in a solution of some ligand L. If MX is only slightly soluble, it w i l l dissolve according to Equation (27-5):

MX(s) ^ M 2 + ( a q ) + X 2~(aq) (27-5)

The salt wi l l stop dissolving when [ M 2 + ] and [ X 2 - ] are large enough that their

270 Experiment 27

product is equal to Ksp for the salt:

Ksp = [ M 2 + ( a q ) | [X 2 " (aq ) ] (27-6)

I f the solution also contains L, then the concentration o f M 2 + ( a q ) w i l l constantly be diminished by conversion to M L „ 2 + and the solubil i ty equi l ibr ium wi l l be shifted to the right by removal of M 2 + ( a q ) :

MX(s) * M 2 + ( a q ) + X 2 _ ( a q ) (27-7) +

nL It

M L „ 2 +

The more effective L is at complexing M 2 + ( a q ) (that is, the larger the formation constant for ML„ 2 + ) , the less L wi l l be required to shift the equi l ibr ium o f l iquat ion (27-7) to the right by removal of M 2 + ( a q ) to dissolve MX(s) completely.

In more mathematical terms, the solubil i ty of a nearly insoluble salt in the presence of a complexing ligand L can be determined as follows. The equilibria in solution are

MX(s) - M 2 + ( a q ) + X 2 "(aq) (27-5)

M 2 + ( a q ) + n L - M L „ 2 + (27-3) I

Assume that the solubil i ty of MX is s moles/liter. Then when equi l ibr ium is established for bo th Equations (27-5) and (27-3) and when MX stops dissolving (that is, when the solution is saturated), s = [ X 2 ~ ] and

[ M 2 + ( a q ) ] s = Ksp (27-8)

The concentration of M 2 + ( a q ) is determined by the equil ibr ium of Equation (27-3) as well as of Equation (27-5)

K _ [ML„ 2 + 1 _ |ML„ 2+1 f 2 ? 9 v f [ M 2 + ( a q ) l [ L I " (Ksp/s) [L]n

A rearrangement indicates how s, the equil ibr ium concentration of X 2 - , is determined:

s - ¥± ^sp C7-10) [ M L , 2 + |

Equation (27-10) shows that the solubil i ty of MX in the presence of a complexing ligand L depends on how effectively L complexes (depends on Kf and [ L ] ) as well as on how soluble MX is in pure water (depends on Ksp). In this experiment, you wi l l study how solubil i ty depends on each o f these factors.



medicine droppers test-tube rack test tubes


l . O A / C u ( N 0 3 ) 2 l .OA/Nal 0 . 1 A / E D T A 6.0 /1 /NH 3

3.OA/ H C 2 H 3 0 2 3.OA/ N H 2 C H 2 C H 2 N H 2

l.OA/ K 2 C 2 0 4 3.0M N H 4 C 2 l i 3 0 2

l .OA/NaBr 1 . O A / N i ( N 0 3 ) 2

l .OA/NaCl l . O A / P b ( N 0 3 ) 2

PLAN

In the first part of this experiment, you w i l l investigate the complexing abil i ty of monodentate and bidentate ligands wi th the N i 2 + and Cu 2 + ca t ions . Complexing ability wi l l be reflected by the ease wi th which solutions o f the ligands dissolve precipitates of N i C 2 0 4 (nickel oxalate) and C u C 2 0 4 (copper oxalate). I f you begin wi th several test tubes containing equal amounts of precipitate, only a small amount of the ligand that is the best complexing agent (that is, the ligand wi th the largest Kf for complex format ion) w i l l be required to dissolve the precipitate. If Kf is small, you wi l l have to'add a large amount o f L to shift the solubil i ty equi l ibr ium to the right. By measuring the volume of each L required to dissolve the precipitate, you can order the ligands according to the size of Kf for complexation wi th each metal.

In the second part o f this experiment, you w i l l work wi th a sexadentate ligand, E D T A , which is an excellent complexing agent, to determine how the A^p of a low-solubili ty salt influences the ability of a chelating ligand to dissolve the precipitate. The nearly insoluble compounds that you w i l l attempt to dissolve are PbCl 2 , PbBr 2 , and P b l 2 . You w i l l then compare these E D T A results wi th the results you obtain by dissolving the lead halides as acetate complexes. Acetate is most l ikely a bidentate ligand toward lead, rather than a monodentate. Whether acetate is mono-or bidentate, E D T A coordinates several more atoms to P b 2 + than acetate does.

Final ly, you w i l l make a few observations on the formation o f complexes in solution.

P R O C E D U R E

A. Varying the Ligands

In this part of the experiment, you w i l l use a single metal ion at a t ime. Af te r forming a precipitate of the low-solubili ty metal oxalate, you w i l l try to dissolve it in solutions of various ligands by complex format ion.

272 Experiment 27

In labeled test tubes, obtain about 10 ml each of the ligand solutions 6.0/1/ N H 3 . 3.OA/ N H 2 C H 2 C H 2 N H 2 , 1.0/1/ K 2 C 2 0 4 , 0.1A/ E D T A , and 1.0A/ NaCl. Also obtain about 10 ml of 1 .OA/ N i ( N 0 3 ) 2 .

Prepare six test tubes containing N i C 2 0 4 precipitate by adding to each test tube 0.5 ml ( 10 drops from a medicine dropper) of the N i ( N 0 3 ) 2 solution followed by 7 drops (no more) of 1 .OA/ K 2 C 2 0 4 solution. After the tubes stand for a few minutes, you should see a precipitate of nickel oxalate. If you do not, add one or t w o more drops of potassium oxalate and wait two or three more minutes for the precipitate to form. Do not add any more potassium oxalate than you need to obtain a precipitate; be sure to add the same amount of potassium oxalate to each of the tubes so that your comparisons wi l l be valid.

To each test tube, add one of the five ligand solutions dropwise. (Be sure to use a different dropper for each ligand solution.) Shake each tube to mix the contents thoroughly after the addition of each drop of the ligand solution. Add water to the sixth tube. For each ligand solution, count and record the number of drops required to dissolve the precipitate and form a clear solution. Note and record the color of each solution. Add several more drops of the particular ligand solution to each test tube. Note and record whether any further color change develops, as well as the number of drops required to produce any color change that does occur. A dissolution or a color change may result simply from the effect of the water in the solut ion, rather than from the effect of adding more drops of a particular ligand. To the sixth test tube, add distilled water dropwise to match the total volume in each of the other five tubes.

Some of the solubili ty equilibria are established slowly. If the first 30 or so drops of a ligand solution do not dissolve the precipitate, set the test tube aside and shake it occasionally, adding 10 drops of ligand solution each time. The ligands provided may not all dissolve the precipitate. I f an entire test tube full o f a mixture does not produce a clear solution, you can conclude that the ligand you are testing wi l l not dissolve the precipitate.

Next prepare six test tubes containing copper oxalate by combining 10 drops of 1.0A/ C u ( N 0 3 ) 2 and 1 1 drops of 1.0/1-/ K 2 C 2 0 4 in each tube. Investigate the abi l i ty of each ligand solution to dissolve this precipitate, fol lowing the same procedure you used in examining nickel oxalate.

B. Vary ing the Precipitate for One Metal Ion

Obtain 5 ml of P b ( N 0 3 ) 2 solution, 1 ml each of NaCl, NaBr, and Nal solutions, and 5 ml of E D T A solution. Prepare three test tubes containing PbCl 2 , PbBr 2 , and P b l 2 by adding 2 drops of lead nitrate solution and 4 drops of the sodium halide solution to a test tube. At tempt to dissolve each of the three precipitates in 3.OA/ N H 4 C 2 F I 3 0 2 solut ion; record the number of drops required to dissolve each precipitate. Then repeat this procedure using E D T A to dissolve the precipitates. When making comparisons between E D T A and ammonium acetate, it is important to recognize that the E D T A solution is only one-thirt ieth as concentrated as the ammonium acetate solution.

Prepare a precipitate of PbCl 2 as before. Dissolve this precipitate in 3.OA/ H C 2 H 3 0 2 and compare the number of drops required to dissolve PbCl 2 w i t h the


number of drops of ammonium acetate solution that were required to dissolve this precipitate. Account for the difference.

C. Shifting Equil ibria in Solut ion: Monodentate vs. Chelating Ligands

Prepare a set of test tubes for color comparison as follows. Add 2 drops of 1.04/ C u ( N 0 3 ) 2 to each of three test tubes. Then, to the first tube, add 10 drops of distilled water; to the second tube, add 10 drops of 6.0/17 N H 3 ; and to the third tube, add 10 drops o f E D T A solution. The tubes w i l l contain C u ( H 2 0 ) 6

2 + , C u ( N H 3 ) 4

2 + , and C u ( E D T A ) 2 - , respectively, and w i l l be used to observe the color characteristic of each complex.

Prepare a test tube containing the copper-ammine complex by adding 2 drops of C u ( N 0 3 ) 2 solution to a test tube followed by enough 6.047 N H 3 to produce the color of the C u ( N H 3 ) 4

2 + complex. Then add a 200% excess of N H 3 to be sure that essentially all the C u 2 + is present as C u ( N H 3 ) 4

2 + rather than as C u ( H 2 0 ) 6

2 + . To this solution, add E D T A dropwise un t i l the complex is converted to C u ( E D T A ) 2 " , as evidenced by the color change. Record the number of drops required.

Prepare a solution of C u ( E D T A ) 2 " by combining 2 drops of C u ( N 0 3 ) 2 solut i on and enough E D T A to produce the color of this complex. Add a 200% excess of E D T A . Then add 6.04/ N H 3 dropwise to t ry to convert the E D T A complex to the ammine complex. Record the number of drops required.


1. The ligand solutions that you used in this experiment were not all of the same concentration. The ligand concentrations were varied for reasons that include solubil i ty difficulties ( E D T A ) and the necessity to avoid the side reactions that would occur w i t h more dilute solutions of ammonia and ethylenediamine. Therefore, to obtain an effective comparison of the complexing abilities of the ligands you used here, you must make some concentration and statistical corrections in the volumes you measured. It is convenient to convert all the volumes to the comparison basis of 1.04/ N H 3 solution. Such a solut ion would contain 1.0 mole/l i ter of N atoms that could coordinate. Each drop of 6.047 N H 3 that you used would then be equivalent to 6 drops of 1.047 N H 3 .

Each molecule of ethylenediamine contains two N atoms that can coordinate. Thus, a 3.04/ N H 2 C H 2 C H 2 N H 2 solution contains 6.0 moles/liter of N atoms that are capable of metal coordinat ion; 1 drop of this solution is then equivalent to 6 drops of 1.04/ N H 3 .

The 1.04/ solution of oxalate contains bidentate ligands that have two O atoms for coordination, whereas each E D T A ion has six atoms that can coordinate. ( I t is l ikely that only five o f these six atoms w i l l coordinate w i t h N i 2 + and that only four o f these six atoms w i l l coordinate w i t h C u 2 + , but you can ignore this refinement for the purposes of this experiment.)

Use the reasoning outl ined here to convert the volumes of all the solutions you used in Part A of this experiment to the comparison basis of a 1.047 concentrat ion of coordinating atoms.

274 Experiment 27

2. Assume that all complexes of a particular metal ion have the same number of ligand atoms coordinated from the added ligands. You can then rank the ligands that coordinate w i t h N i 2 + and C u 2 + according to their format ion constants for metal complexes by not ing which requires the smallest corrected volume to dissolve precipitates. Do this separately for the ligands reacting w i t h nickel oxalate and w i t h copper oxalate. Write the series w i t h the best coordinating ligand first: A > B > C

Al though the assumption that all complexes of a particular metal ion have the same number of ligand atoms coordinated from the added ligands is not s tr ict ly true, departures from this assumption w i l l not significantly affect your results and you do not have to account for them here.

3. Explain how your observations for the N H 3 and NH2CH2CH2NH2 reactions w i t h N i 2 + and C u 2 + illustrate the chelate effect. Why is this a better comparison than the reactions o f N H 3 and C 2 0 4

2 " w i t h N i 2 + and Cu 2 + ? 4. Account for the relative volumes of ammonium acetate solut ion required to

dissolve PbCl 2 , PbBr 2 , and P b l 2 . Account for the relative volumes of E D T A solution required to dissolve the lead halide precipitates. Refer t6 Equation (27-10) in your discussion.

5. Compare the corrected volume of E D T A required to dissolve each lead halide precipitate w i t h the corrected volume of ammonium acetate required to dissolve each of these precipitates. What does this tell you about the formation constants of the E D T A and acetate complexes of Pb 2 + ?

6. Account for the different quantities of 3.OA/ ammonium acetate and 3.OA/ acetic acid solutions required to dissolve lead chloride as the acetate complexes.

7. Explain the results y o u obtained in Part C. Can the EDTA-complex ^ ammine-complex equi l ibr ium be shifted both f rom left to right and from right to left? Explain why this is the case.

R E F E R E N C E

Yoder, C.H., Suydam, F .H. , and Snavely, F .A. Chemistry. New Y o r k : Harcourt Brace Jovanovich, Inc., 1975. Chapter 22.

Experiment 28 Synthesis of an Organometallic Compound

Introduction

Organometallic compounds contain metal atoms bonded to carbon atoms of organic groups. Transition metals form a large number of such compounds. In this experiment, you wi l l use one organometallic compound as a starting material to prepare another. The starting compound that you wi l l use is shown in Figure 28-1. This compound is composed of two identical halves bonded together and is therefore called a dimer (di: two) . Each Fe atom is bonded to a C 5 H 5 " ring. The dashed circle inside each ring indicates that several Lewis structures can be wri t ten for the ring, as depicted in Figure 28-2. The delocalized 7r-orbital system of the ring is involved in bonding to the Fe atom. Each Fe atom is also bonded to CO groups, two of which bridge the Fe atoms. Finally, the two Fe atoms form a covalent bond. The oxidat ion state for Fe in this compound is + 1 .

The iron atoms in this molecule can be easily oxidized to the +2 state that is more common for i ron. Iodine, like the other halogens, is a good oxidizing agent and reacts readily w i th the dimer just described to produce (C 5H s)Fe(CO)2L depicted in Figure 28-3. In this product, there is a covalent bond between Fe and I. Iodine is reduced to its very stable -1 oxidat ion state, and iron is oxidized to the +2 state. The equation for the reaction is

[ ( C s H s ) F e ( C O ) 2 ] 2 + I 2 - 2 ( C 5 H 5 ) F e ( C O ) 2 I (28-1)

278 Experiment 28

279 Synthesis of an Organometailic Compound

iodine mineral oil

0.5A/ N a 2 S 2 0 3

T E C H N I Q U E

Both of the reactants and also the product are nearly insoluble in water, so the reaction cannot be carried out in this solvent. In this experiment, you wi l l use the polar organic solvent dichloromethane (CH 2 C1 2 ) in which all three materials are quite soluble.

An excess of iodine wi l l be used to make sure that the organometallic starting material reacts wi th in a fairly short t ime. Before you come to the laboratory, calculate the approximate amount o f iodine that you wi l l use in the experiment. Check your figures wi th your instructor before you begin.

After the reaction, the excess iodine must be removed so that only the product remains in the dichloromethane solution. This is done by shaking the dichloromethane solution w i t h an aqueous solution of sodium thiosulfate ( N a 2 S 2 0 3 ) . Iodine and sodium thiosulfate react to form sodium iodide and sodium tetrathionate, both of which are water soluble:

I 2 + 2 N a 2 S 2 0 3 • 2NaI + N a 2 S 4 0 6 (28-2)

Water and dichloromethane are immiscible; consequently, the two solutions can be easily separated from one another using a separatory funnel.

After treating the,dichloromethane solution w i th sodium thiosulfate, the solut ion should confain only the product ( C 5 H 5 ) F e ( C O ) 2 I . The solution wi l l be wet wi th water and must be dried before the product can be isolated. This is done by mixing a l i t t le anhydrous sodium sulfate ( N a 2 S 0 4 ) into the solution. This drying agent absorbs water by forming hydrated crystals that are insoluble in dichloromethane; these crystals can be removed from the dried solution by f i l t ra t ion .

To isolate the product, hexane ( C 6 H 1 4 ) is added to the solution, which is then concentrated by removing solvent under reduced pressure. Of these two organic solvents, dichloromethane is more volatile; as the mixture of the two solvents boils under reduced pressure, the vapor w i l l contain a higher propor t ion o f dichloromethane than it wi l l o f hexane. The solution wi l l become richer in hexane and the product, which is nearly insoluble in this nonpolar solvent, w i l l precipitate. The solution cools considerably during this concentration, so it may be helpful to place the tlask in a beaker of warm water. The precipitate can be collected by f i l t ra t ion .

Iodine, [ ( C 5 H 5 ) F e ( C O ) 2 ] 2 , and ( C 5 H 5 ) F e ( C O ) 2 I are all dark-colored crystalline materials. To convince yourself that the material you isolate is actually different from the two starting materials, you wi l l examine some physical properties o f the three materials. Al though the crystals of these materials look similar, they have noticeably different colors when they are powdered. Concentrated solutions of all three substances are quite dark and may be indistinguishable, but their dilute solutions have distinctively different colors. Their melting points are also different. A l l three substances melt above 100°C; you wi l l need to use a mineral-oil bath to measure their melting points. The melting point of the product , if it is pure, should be fairly sharp; melting should occur over not more than a 2° range. Make a prel imi-

280 Experiment 28

nary determination w i th one sample of the product by rapidly heating the bath. Then, wi th a second sample, heat the bath very slowly to make a precise determinat ion . The melting point of the organometallic starting material w i l l not be sharp; report a range for i t . To determine the melt ing point of iodine, the open end of the capillary tube must be sealed in a low burner flame after the iodine is put in to i t ; if this is not done the iodine can vaporize out of the tube wi thou t melting.

P R O C E D U R E

Caution! Both the starting material [ (C 5 H 5 )Fe(CO ) 2 ] 2 and the product ( C 5 H s ) F e ( C O ) 2 I can decompose and liberate carbon monoxide. In this experiment you w i l l work w i t h small quantities o f these materials, from which there should be no danger. However, to avoid any possibility of liberating carbon monoxide in to the laboratory atmosphere, handle these materials only as described in this experiment. Do not at tempt any other reaction or investigation of the properties of these compounds.

Caution! Hexane is extremely flammable. Be sure that you are not near a lighted burner when you pour i t .

A. The Reaction

Measure out 50 ml of dichloromethane. Place 20-30 ml of it in a 125-ml Erlen-meyer flask, and dissolve 1.0-1.5 g of [ ( C 5 H s ) F e ( C O ) 2 ] 2 in the dichloromethane. Record the weight of this starting material that you actually use. This weighing (and all others in the experiment) can be made on the triple-beam balance. To the solut ion , add a 200% excess of iodine. Stopper the flask, and swirl it gently for at least 1 5 minutes. Occasionally wash down the sides of the flask wi th a few ml of the remaining dichloromethane.

B. Removal of Unreacted Iodine

Pour the solution into a separatory funnel; to i t , add about 50 ml of 0.5A/ sodium thiosulfate solution. Stopper the funnel and shake the mixture vigorously; occasionally open the stopcock (while holding the stoppered funnel upside down) to release the pressure that builds up from the vaporization of dichloromethane. Support the funnel in a metal ring on a ring stand, and allow the funnel and its contents to stand for several minutes. The mixture wi l l separate in to two layers; the more dense dichloromethane solution w i l l settle to the bo t tom. Unstopper the funnel and drain o f f and collect the dichloromethane layer. Discard the water layer. Return the dichloromethane solution to the funnel, and repeat this sodium thiosulfate treatment to be sure that all the iodine has been removed. Drain the dichloromethane layer in to a flask.

C. Drying the Dichloromethane Solut ion

Add 3-5 g of anhydrous sodium sulfate to the dichloromethane solution and vigorously swirl the mixture to ensure contact between the sodium sulfate and any

281 Synthesis of an Organometallic Compound

water that may be present. Set up a funnel and a piece of filter paper for gravity filtration. In the bo t tom of the paper place a small amount of sodium sulfate; filter the dichloromethane solution through this in to a clean, dry suction f i l t ra t ion flask. Use a few ml of the remaining dichloromethane to wash out the flask and to wash as much solution as possible out of the sodium sulfate in the funnel.

D. Precipitating and Collecting the Product

To the dichloromethane solution, add 100 ml of hexane. Put a broken wooden stick into the flask to provide a rough surface on which small bubbles may form to prevent " b u m p i n g " of the solution when it boils. Stopper the flask and connect it to a trap bot t le ; connect the trap bottle to the water aspirator. Turn on the aspirator and continue the suction un t i l the solution is reduced to less than 50 ml . If not all of the [ ( C 5 H s ) F e ( C O ) 2 ] 2 has reacted, it w i l l precipitate first as a dark red f i lm on the sides o f the flask. Then the product wi l l precipitate as t iny crystals. Collect these crystals by gravity f i l t ra t ion; reuse the filtrate to wash crystals out of the flask; be careful not to scrape out any [ ( C s H 5 ) F e ( C O ) 2 ] 2 if it is present. Open the filter paper and let the crystals stand un t i l they are dry.

E. Determining Properties

Using a mineral oi l bath and a thermometer graduated to at least 2 0 0 ° C , determine the melting points of iodine, [ ( C 5 H 5 ) F e ( C O ) 2 1 2 , and ( C 5 H 5 ) F e ( C O ) 2 I .

Determine the color of each of these materials in its crystalline form and the color of each when a small crystal is crushed on white paper to a very fine powder.

Determine the color of a very dilute dichloromethane solution of each of the materials. Be sure that each solution is sufficiently dilute that you can describe it by a single color such as " y e l l o w , " rather than by a mixture of colors such as "yel low-brown . "

R E S U L T S A N D D I S C U S S I O N

Report gram and molar quantities of both starting materials and of the product. Calculate the percentage yield of the product.

Provide the requested information on the physical properties of iodine, [ ( C 5 H 5 ) F e ( C O ) 2 ] 2 , and ( C 5 H 5 ) F e ( C O ) 2 I .

Turn in your product in a labeled bottle.

Q U E S T I O N S

1. Explain, in terms of sound chemical principles, why it is desirable to make the dichloromethane solution of the reactants as concentrated as possible. Why is an excess of iodine used?

2. Write an equation to show what happens to sodium sulfate when it is used as a drying agent.

3. Look up the normal boil ing points of dichloromethane and hexane. Which

282 Experiment 28

should have the higher vapor pressure at room temperature? If the atmospheric pressure over these liquids drops, what wi l l happen to their boil ing points? Can these liquids boi l at 20°C? Why does the mixture of liquids cool as it boils?

R E F E R E N C E S

1. Yoder, C.H., Suydam, F .H. , and Snavely, F .A. Chemistry. New York: Harcourt Brace Jovanovich, Inc., 1975. Chapter 22.

2. King, R.B. Organumetallic Syntheses, Volume I. New York : Academic Press, 1 9 6 5 , p . 1 7 5 .

) Introduction II Inorganic Qualitative Analysis

A. Chemical Systems for Study

Qualitative analysis is the study of how to identify the components of a mixture. The particular kinds o f mixtures that w i l l concern you are aqueous solutions of inorganic salts. It would be impossible in this section to study all the known inorganic ions, so we shall confine our at tention to 28 cations and 7 anions.

Most of the cations you wi l l study here are metallic. This is not surprising i f you recall that the most characteristic feature of metals is their electropositivity—the ease wi th which they give up electrons.

It is also wor th noting that bo th cations and anions probably exist in aqueous solution as more complex species than the formulas used in this section indicate. For example, the C o 2 + ion is k n o w n to be closely surrounded by exactly six molecules of water, arranged in a regular fashion w i t h the negative end of the water dipole pointing toward the positively charged cobalt cation. Other cations are also surrounded by several water molecules, but the exact number is not always known . Anions are also surrounded by water molecules, but these molecules are arranged so that the positive end of the water dipole points toward the negatively charged anion. Again, the precise number of water molecules bound in this way is not always known . For this reason, it is more convenient to refer to the ions as C o 2 + or Br", for example, even when they are in solution. Sometimes anions and cations are labeled Co 2 + ( aq ) , as an example, if emphasis is to be placed on the surrounding water molecules.

B. Identifying Ions

The cations and anions you w i l l be asked to identify can be recognized by the characteristic reactions that they undergo. For instance, a solution containing silver ions wi l l react w i th chromate ions to form a dark maroon compound, silver chro-mate. This compound is not very soluble in water, and so it precipitates from solut ion . The reaction between silver and chromate is represented by the equation

2Ag + (aq) + C r 0 4

2 - ( a q ) • A g 2 C r 0 4 ( s ) (1)

The (s) indicates that the compound is a solid precipitate. The silver solution could not be composed merely of dissolved silver cations; enough anions must also be present to balance the positive charge. In fact, solutions containing A g + ions are usually made by dissolving silver nitrate

A g N 0 3 ( s ) • Ag +(aq) + N 0 3 - ( a q ) (2)

285

286 I n t r o d u c t i o n I I I ,

Because the nitrate ion remains in solution and does not participate in reaction ( I ) , it is of no interest. The nitrate ion is therefore omit ted from the equation showing the reaction of silver ion w i t h chromate ion. For the same reason, the cation in the chromate solution is omi t t ed from the reaction, and we speak of the solution as if it contained only chromate.

To summarize, one reaction characteristic of the silver ion is the formation of a dark maroon compound wi th chromate ion. Thus, if you have a solution that forms such a precipitate when you add chromate, you can conclude that the solution contains Ag*. This reaction can serve as a test for the presence of the ion.

Similarly, the barium ion forms a bright yel low precipitate of barium chromate when chromate ion is added, by the reaction

Ba 2 + ( aq ) + C r 0 4

2 _ ( a q ) • BaCr0 4 ( s ) (3)

This reaction can serve as a test for barium. The procedure explained here shows you the reasoning used in qualitative analysis: A solution that is known to contain some ion is found to give a characteristic reaction, such as the precipitation of B a C r 0 4 . If some other solution of an unknown composition behaves similarly, then this behavior is regarded as evidence that the B a 2 + ion is present. Obviously, if no reaction is observed wi th Cr0 4

2 ~ , B a 2 + must be absent. But suppose that a solution contains both B a 2 + and A g + ; if we try to confirm the

presence of either of these ions by adding chromate, both maroon silver chromate and yellow barium chromate wi l l precipitate at the same time

2Ag + (aq) + C r 0 4

2 " ( a q ) - + A g 2 C r 0 4 ( s ) (maroon) (1)

Ba 2 + (aq) + Cr0 4

2 " (aq) -> BaCr0 4 (s ) (yellow) (3)

The resulting precipitate w i l l be brownish in color, and the test wi l l be indeterminate. Here, silver and barium are said to "interfere" wi th each other. To confirm the presence of either ion using the chromate test, the two ions would have to be separated, and the chromate test would have to be performed on each. It is easy to see, then, that the identif ication of ions in solution requires (1) the separation of interfering ions, and (2) the conf i rmat ion of each ion using a separate test reaction for each (usually the formation of a precipitate).

C. Group Separation

To make it easier to detect the 28 cations, they are separated into more manageable groups that have some solubil i ty similarities; each ion in a group is then separated from the others. As an example, all the ions in Group I ( A g \ H g 2

2 + , and P b 2 + ) form insoluble chlorides. A solution of all 28 cations can be treated w i t h hydrochloric acid; this treatment immediately precipitates the insoluble chlorides of Group I cations, leaving the ions of all the other groups in solution. The white chloride precipitate can then be tested for Ag + , H g 2

2 + , and P b 2 + by separating each ion from the others and then confirming the presence of each ion individually.

Group II ions all form sulfides that are insoluble in acidic solutions. Thus, treating an acidic solution containing ions in Groups II—V wi th hydrogen sulfide

1 .. 1

287 Inorganic Qualitative Analysis

produces a black precipitate of Group II sulfides while the ions in Groups I I I , I V , and V remain in solution. Ions in Groups I I I , I V , and V can be separated from one another by similar precipitation techniques.

D. Technique

Your laboratory work in qualitative analysis wi l l be carried out on a somewhat smaller scale than in most of the other experiments in this manual. In addition to the obvious saving in quantities o f reagents, working on a smaller scale wi l l allow you to complete the qualitative experiments in less t ime, because you wi l l not have to wait for large quantities of precipitate to form, transfer large volumes of l iqu id , and so for th . This small-scale work is called semimicro qualitative analysis. The special laboratory techniques and equipment used in semimicro work wi l l be described in this section.

1. Equipment. Your set of desk equipment includes a micro burner, several semimicro test tubes in a rack, a metal test-tube support, and a set of dropping bottles for reagent solutions. Volume measurements are conveniently made by counting numbers of drops instead of by ml when working on a semimicro scale. Unless the dropping bottles have been filled in the s tockroom, your instructor wi l l tell you which reagent solutions to place in these dropping bottles. Be sure that the bottles have been washed and thoroughly rinsed twice w i th distilled water. Label each bott le .

In addit ion to the.droppers provided in these bottles, you should have a few droppers for withdrawing liquids from small test tubes. The most convenient k ind of dropper is a capillary pipet, a piece of glass tubing drawn to a very long and narrow constriction on one end and equipped wi th a rubber bulb for suction (see Figure i i -1) . These may be supplied by your instructor, or you can make your own. To make a capillary pipet, heat the center of a 5-in. length of glass tubing. Rotate the tubing and heat it unt i l it begins to sag in the middle; then remove the tubing from the flame and pull it out to a small diameter. When the glass is cool , cut it into two pi pets.

You should also make four or five stirring rods, each not longer than 4 in . (the technique for breaking and fire-polishing glass is described on page 9). Y o u w i l l be

288 Introduction II

using these stirring rods continual ly for several weeks, so be sure to fire-polish the ends to avoitl injury.

2. Avoiding Contamination. The anions and the cations that you w i l l work w i t h in qualitative analysis wi l l be supplied in concentrations of only < 10 mg/ml. I f you do not clean your glassware carefully and completely, you can introduce quantities of contaminants that are almost this large in themselves into your analysis. Added reagents could then react w i t h contaminants to produce spurious results.

Tap water is a particularly common source of contaminant ions. After washing any glassware to be used in qualitative analysis, rinse the equipment at least twice w i th distilled water. Capillary pipets should be rinsed twice w i th distilled water after each use; if you develop the h a b i f of automatically washing your glassware after each use, you w i l l be certain to pick up a clean pipet whenever you need one. Keeping a beaker of distilled water on your desk w i l l provide a convenient way to rinse your pipets.

It is almost unnecessary to say that any water you add to solutions must be distilled water.,. ^

A clean towel on your working area w i l l provide a place to lay stirring rods and pipets so that they w i l l not pick up any contaminants from the desk surface.

As before, remember never to introduce your pipets in to communi ty vessels. Most o f these vessels wi l l be equipped wi th their own droppers for removing solutions. The droppers you use to add reagents to test tubes should be held above the tube and not inserted in to i t .

3. Adding Reagents. When working on a semimicro scale, l iquid reagents are added from droppers or capillary pipets, and volumes are measured by counting the number of drops. A good rule of thumb is that 1 drop = 0.05 ml ; this means, of course, that 1 ml = 20 drops.

Occasionally, you wi l l need to measure small volumes accurately. For this purpose, you w i l l need to calibrate two capillary pipets. To do this, f i l l a 10-ml graduated cylinder to the 5.0-ml mark w i t h distilled water, and count the number of drops of distilled water required from the pipet to bring the level up 2 ml to the 7.0-ml mark on the cylinder. Obviously, half this number o f drops wi l l equal 1 ml . Label the pipet w i th this calibration informat ion.

4. Mix ing Solutions. Several techniques are available for mixing solutions in semimicro test tubes. The most obvious one is to use a stirring rod. This is really not an efficient stirring technique, however. It is more satisfactory ( i f the tube is not too ful l) to hold the test tube in the thumb and forefinger of one hand and flick the bo t tom of the tube w i th the forefinger of the other hand. Alternatively, the contents may be partially sucked up in a capillary pipet (but not up to the bulb) and then squirted back into the tube. This procedure should be repeated two or three times for each mixing.

5. Centrifuging. Many tests in qualitative analysis produce precipitates that must be separated from the remaining solut ion (called the supernatant liquid). This is commonly done by placing the test tube in the head of a centrifuge and spinning it at high speed. The centrifugal force packs the precipitate at the bo t t om


Centrifuge tube.

of the tube many times faster than it would take if you allowed the precipitate to settle naturally.

When you centrifuge, be sure to balance the revolving head by placing a similar test tube that is filled to the same level wi th distilled water opposite your test tube in the centrifuge. I f you do not do this, the centrifuge w i l l be damaged and fewer centrifuges w i l l be available for your laboratory work i f some must be removed to be repaired.

Once the centrifuge is turned on, it should spin at top speed for 30 seconds or so. Then turn o f f the switch and let the centrifuge come to a stop by itself. Do not attempt to slow it down or stop it w i th your hand.

Sometimes special test tubes called centrifuge tubes are supplied. These tubes have a narrow bot tom that helps to pack the precipitate (sec Figure i i -2) . However, an ordinary semimicro test tube is adequate for centrifuging.

6. Testing for Completeness of Precipitation. After you remove a tube from the centrifuge, you should always test to be sure the precipitat ion is complete. To do this, add an additional drop of the precipitating reagent and note whether more

290 Introduction II

precipitate forms. If it does, add a few more drops of the reagent unt i l precipitation seems to stop. Then centrifuge and test again.

7. Separating the Supernatant Liquid from a Precipitate. A capillary pipet can be used to draw o f f the solution above a precipitate (see Figure i i -3) . Before you insert the pipet in to the test tube, be sure to squeeze the air out of the bulb.

Some precipitates flocculate (remain partly dispersed in solution) even after centrifuging. If this happens, you can insert a t iny plug of cot ton in to the capillary pipet, and then fil ter the supernatant l iquid through the cot ton plug.

8. Washing Precipitates. After a supernatant l iquid has been wi thdrawn as completely as possible w i th a pipet, some o f the l iquid wi l l st i l l cling to the precipitate. This l iquid wi l l contain ions that can interfere w i th further tests you wi l l make on the precipitate. Thus, this solution must be removed by washing.

The wash l iquid is usually distilled water; sometimes another solution may be specified. The wash l iquid is added to the precipitate, and the packed precipitate is broken up by stirring it un t i l it is completely dispersed in the l iquid . The tube is then centrifuged, and the l iqu id is drawn off. This wash l iquid is usually discarded, but if it contains an appreciable quanti ty of ions, you may be directed to add it to the first solution you removed from the precipitate.

9. Heating Solutions. Test tubes containing solutions are usually heated in a water bath, as shown in Figure i i -4 . Some test-tube holders are constructed to sit inside the beaker instead of to f i t over the top (as shown in the figure here).

f

I

I


10. Evaporating Solutions. When the volume of a solution is to be reduced appreciably or when a solution is to be taken to dryness, it is placed in a porcelain casserole or crucible and heated under a hood, so that any toxic fumes can be expelled from the laboratory. This heating must be conducted in such a way that the solution does not spatter. This can be accomplished by using a y - in . flame from the micro burner; such a flame is referred to as a "microf lame." The gas level can be adjusted properly by using a screw clamp on the burner hose.

During heating, the crucible (held w i th tongs) or casserole should be continuously swirled in the microflame. As dryness approaches, the bo t tom o f the vessel w i l l dry first. Keep bringing fresh solution to the dry bo t tom by swirling and t i l t ing the vessel. Just before complete dryness is reached, remove the casserole or crucible from the flame and let the heat of the vessel complete the evaporation.

If the residue is baked in the flame after it is dry, volatile substances may be lost or the residue may be converted to an insoluble form.

E. Your Laboratory Work in Qualitative Analysis

After so many general instructions, it would probably be helpful to comment more specifically on exactly what you wi l l do in the laboratory.

1. First Qualitative Analysis Period. Before you can work in the laboratory, you wi l l need to:

(a) Clean your semimicro test tubes (and dropping bottles, if they are empty) , rinsing each twice wi th distilled water.

(b) Make five or six capillary pipets (unless these arc to be supplied), and obtain squeeze bulbs for them.

(c) Make four or five stirring rods, each 4 in . or so in length; fire-polish the ends. (d) Calibrate two capillary pipets. (e) Label and fi l l your dropping bottles wi th the solutions provided by your

instructor. If the stockroom provides these reagents, be sure to obtain a complete set.

2. Subsequent Laboratory Periods. You w i l l analyze each group o f ions separately. Before coming to class, carefully study the In t roduct ion to each qualitative experiment. You wi l l be expected to know the analytical procedure for each group and to be able to write an equation for each separation and confirmatory test. (Naturally, you do not need to memorize quantities of reagents to be used.) In the laboratory, you can obtain a so-called " k n o w n " solution containing ~ 1 0 mg/ml of each o f the group ions. You wi l l conduct the group analysis procedure using this known solution to observe the reactions and to practice your skills. After you complete your analysis o f the known solution, your instructor w i l l provide you w i t h an unknown solution that may contain any of the group ions. By fol lowing the analytical procedure, you w i l l decide and report which o f the possible ions arc present in your unknown solution and which are absent.

Finally, after all individual groups have been studied, you w i l l receive a general unknown that contains cations and anions from any o f the groups. You w i l l report all ions present in this unknown, which may be a solution or a solid.

292 I n t r o d u c t i o n I I

3. Keeping a Notebook. Unless you are specifically directed to do otherwise by your instructor, you are to keep a record of each step in the analysis of both your known and your unknown in a notebook, fol lowing the example below:

Solution or precipitate Operation Observation Conclusion

Group I known

Precipitate 1.1

Added HCl

Added hot water

White precipitate

Some precipitate dissolved

A g + , Pb 2 + , H g 2

2 +

possibly present

P b 2 + probably present

A complete record w i l l help your instructor to surmise where you may have made any errors and w i l l assure you that you have reached warranted conclusions.

If your notes are to be of any value, they must be recorded during the experiment—no/ at the end of the period or later in the day.

You may find it helpful to consult more complete treatments of qualitative analysis. [For example, see E.J. King, Ionic Reactions and Separations: Experiments in Qualitative Analysis (New Y o r k : Harcourt Brace Jovanovich, Inc., 1973); and C.H. Sorum, Introduction to Semimicro Qualitative Analysis, Four th Edit ion (Engle-wood Cliffs, N.J: Prentice-Hall, 1967).]

Experiment 29 Qualitative Analysis of the Group I Cations

Introduction

The separation of the three Group I cations A g + , Pb 2 + , and H g 2

2 + from other cations in later groups depends on the fact that chlorides of Group I cations are insoluble in acidic solution. This is a rather unusual occurrence, as you can see from the general solubil i ty rules given on page 374. A l l other cations (Groups II—V) form soluble chlorides and remain in solution.

S E P A R A T I O N OF G R O U P I C A T I O N S

The precipitate consisting of white, insoluble AgCl , PbCl 2 , and H g 2 C l 2 forms when 6M HC1 is added to 10 or so drops of the solut ion. Only a slight excess of hydrochloric acid is added to obtain the precipitate.

A g f + CP - AgCl(s) (29-1)

Al though Equation (29-1) and similar equations that can be wri t ten for P b 2 + and H g 2

2 + suggest that the equi l ibr ium can be shifted as far to the right as desired by adding sufficient C I " ions, the solubilities of the chlorides begin to increase after a large enough [ C P ] is reached, because of complex ion format ion:

AgCl(s)+ CI" - AgCl 2 "

PbCl 2(s) + C I " ^ P b C l 3 " (29-2)

H g 2 C l 2 ( s ) + 2C1" * H g C l 4

2 " + Hg(l)

Even a slight excess o f C I " w i l l cause the essentially complete removal o f these cations as their chlorides. In the solutions used here, [ A g + ] = 0.02 before the addit ion of HC1. At the conclusion of the Group I precipi tat ion, sufficient C I " is s t i l l present so that [ C P ] = 0.35. The maximum amount of A g + that can remain in solution then is given by

[Ag + 1 (0.35) = K,p = 1.8 X 1 0 " 1 0 (29-3)

which gives [ A g + ] = 5 X 1 0 " 1 0 . Thus, [ A g + ] has been reduced from 0.02A/ to 5 X \0~l0M; this is certainly complete enough removal of A g + for practical purposes.

Al though H g 2

2 + i s also completely precipitated under these conditions, PbCl 2 is sufficiently soluble that a high enough [ P b 2 + ] remains in solution to give a PbS precipitate w i th Group I I .

294 Experiment 29

S E P A R A T I O N AND CONFIRMATION OF THE CATIONS

The PbCU can be separated From the other chloride precipitates by taking advantage of a diFFerence in solubil i ty characteristics. A l l three chlorides increase in solubil i ty as the temperature oF the water is raised; however, since PbCl 2 is much more soluble than the other chlorides, even at room temperature, a given quanti ty oF hot water wi l l dissolve sufficient PbCl 2 to separate it From AgC'l and H g 2 C l 2 :

|i

P b C l 2 ( s ) - ^ P b 2 + + 2C1" v ' " * ' O ( 2 9 " 4 ) H 2 0

After the precipitates are heated and stirred and the PbCl 2 has dissolved, the AgCI and H g 2 C I 2 must be quickly centrifuged to the bo t tom oF the test tube. The supernatant l iquid containing P b 2 + is drawn o f f before the solution cools and the PbCl 2

reprecipitates. The presence of P b 2 + is confirmed by adding K 2 C r 0 4 . A yellow precipitate of

lead chromate (Ksp = 2 X 10" l f ' ) forms if P b 2 + i s present:

P b 2 + + C r 0 4

2 " ^ P b C r 0 4 ( s ) \ * V i • 1 (29-5)

The precipitate of AgCI and H g 2 C I 2 is treated w i th cone. (15/17) N H 3 , and the silver and mercurous ions are confirmed by the different behaviors of these chlorides wi th N H 3 .

Silver chloride (AgCI) dissolves in N H 3 by forming the complex ion A g ( N H 3 ) 2

+ :

AgCl(s) + 2NH 3 ( aq ) ^ A g ( N H 3 ) 2 * + C I " (29-6)

This same effect was noted in Experiment 28 when insoluble compounds were dissolved by forming complexes.

Mercurous chloride undergoes a disproport ionat ion reaction wi th N H 3 . In a disproportionation reaction, there is simultaneous oxidat ion and reduction of different atoms of one k ind . In this case, mercurous ions H g 2

2 + are oxidized to the mercuric ion H g 2 + and reduced to metallic mercury H g ° ( l ) :

H g 2 C l 2 ( s ) + 2 N H 3 ^ Hg(NH 2 )Cl ( s ) + Hg(l) + NH 4 C1 (29-7)

The H g 2 + forms a white precipitate of H g ( N H 2 ) C l (mercuric amidochloride), while the metallic mercury in finely divided form is black. The mixture of these two precipitates appears gray. The appearance of this gray color is taken as confirmation of the presence of H g 2

2 + . The supernatant l iquid contains A g ( N H 3 ) 2 * ions, and their presence is confirmed

by acidifying this solution w i th H N 0 3 . Protons combine w i t h N H 3 t o form N H 4

+ , destroying the silver ammine complex in the process:

A g ( N H 3 ) 2

+ + 2 H 3 0 + - A g + + 2 N H / + 2 H 2 0 (29-8)

The Ag* ions then form a precipitate of white AgCI (which confirms the presence of Ag*) by reacting w i t h the C I " ions in solut ion:

A g + + C l " - AgCl(s) (29-9)

295 Qualitative Analysis of the Group I Cations

A possible complication can result if a large amount of mercury is present. Th metallic mercury formed via the reaction given in Equation (29-7) may reduce A :

to metallic silver:

2AgCl(s) + 2Hg(l) -* 2Ag(s) + Hg 2 Cl 2 ( s ) (29

I f this happens, the silver wi l l remain w i t h the gray mercury precipitate, instead o f dissolving via the reaction of Equation (29-6). Then no positive confirmatory test for A g + w i l l be seen. I f you do not detect the presence o f Ag* via the reactions o f Equations (29-8) and (29-9), you should try to find it in the gray precipitate.

The precipitate is treated wi th aqua regia, a mixture of cone. HCl and cone. H N 0 3 , which dissolves almost any salt. The H g ( N H 2 ) C l and Hg are oxidized and converted to chloro complexes:

2Hg(NH 2 )Cl(s ) + 6 N 0 3 " + 6 C P + 8 H 3 0 + - > -

2 I I g C l 4 - - + N 2 ( g ) + 6NO, (g ) + 141 UO (29-1 1 )

Hg(l) + 2 N 0 3 - + 4CI" + 4 I I 3 0 + -» H g C I 4

2 " + 2 N 0 2 + 6 I I 2 0 (29-1 2)

Ag is oxidized and converted to a chloro complex by the large excess of cone. HCl:

Ag + NO3-+ 2CT + 2 H 3 0 + ^ AgCI 2 " + N 0 2 ( g ) + 3 I I 2 0 (29-13)

The solution is evaporated to remove any excess aqua regia. When the remaining solution is di luted wi.th water, [ C I " I is decreased sufficiently that white AgCl precipitates; equi l ibr ium [Equat ion (29-14)] is shifted to the right when [CI") is lowered.

AgCl 2 "(aq) -> AgCl(s) + Cl"(aq) (29-14)

The H g C l 4

2 _ i o n remains in solution. If desired, its presence can be confirmed after centrifuging by adding Sn(I I ) chloride solution, which produces a gray precipitate by first reducing H g C l 4

2 " to white H g 2 C l 2 and then to black Hg:

2HgCl 4

2 " + S n 2 + - > Hg 2 Cl 2 ( s ) + SnCl 6

2 " (29-15)

Hg 2 Cl 2 ( s ) + 4C1" + S n 2 + -» 2Hg(l) + SnCl 6

2 " (29-16)


(Solutions without specified concentration are prepared as indicated in Appendix V.)

A g + ( 2 m g / m l ) K 2 C r 0 4 solution 6A/ H C 2 H 3 0 2 (acetic acid) litmus paper 6M HCl" 6M N H 3

12.1/ HCl 15yV/NH 3

H g 2

2 + (10 mg/ml) P b 2 + (15 mg/ml) 6M H N 0 3 SnCl 2 solution 16M H N 0 3

296 Experiment 29

P R O C E D U R E

Analysis of Known and Unknown Solutions

1. Use 5 drops of a known or practice solution that contains about 15 mg/ml Pb 2 + , 10 mg/ml H g 2

2 + , and 2 mg/ml Ag + . 2. The unknown solution may contain the ions of this and possibly other

groups. If it is basic, acidify w i t h 6M HC1. If the unknown is a solid, take a representative sample and dissolve about 20 mg in a few drops of 6M H N 0 3 .

The directions for the analysis are given in Outline 1. The superscript numbers in the instructions direct your at tention to the notes that fol low the outline for explanations, precautions, and further instructions. Consult your instructor for the type of notebook record that you should keep for the analysis.

298 Experiment 29

Notes to Outline 1

1. Small quantities of reagents are usually required. One drop (0.05 ml) of 6A1 HC1 contains 6 X 0.05 = 0.3 mmol . If a sample contains 20 mg of Ag (a very large amount) , this is equivalent to 0.19 mmol . Thus, 1 drop of 6A1 HC1 supplies more than enough Cl~ for complete precipitation.

2. Consult the technique section if you are in doubt as to how to perform any operations.

3. Thorough mix ing is d i f f icul t in a centrifuge tube because it has a narrow bo t tom. Work the glass rod up and down, and rub the inner wall to induce the crystallization of P b Q 2 , which often forms supersaturated solutions.

4. If no precipitate forms, A g + a n d H g 2

2 + ions are definitely absent, but P b 2 + ions may be at too low a concentration to precipitate w i th Group I (and w i l l then appear in Group I I ) or PbCl 2 may have supersaturated. Supcrsaturation can be overcome by vigorous stirring and rubbing the inner walls of the tube w i th a rod. A l l o w five minutes before you conclude that Group I is absent.

5. A large excess of reagent may dissolve the chlorides. 6. In these directions, "a few" is interpreted as 1 to 3 drops; "several," as 3 to 6

drops. Learn to judge for yourself from the size of the precipitate how much reagent to add. Use the least amount of reagent necessary to produce the required result.

7. The residue is washed wi th very dilute HQ to minimize the loss of PbCI 2 . Dilute 1 drop of 6M HQ wi th about 10 drops (0.5 ml ) of water in a test tube and mix well .

8. Use distilled water from a tube suspended in the water bath. Start heating this tube before beginning the analysis, so that it w i l l be ready when you need i t . Do not use water directly from the water bath.

9. Acetic acid prevents the precipitation of other chromates, such as C u C r 0 4 or ( B i O ) 2 C r 0 4 , that may appear at this point if the group precipitate was not washed carefully enough.

10. Centrifuge to determine the btdk of the precipitate. It is useful to observe the size of tests on known solutions, for these help you to distinguish between normal and trace amounts in unknown solutions. Traces are generally caused by contamination and should not be reported. When large amounts of bismuth are present, it may sometimes carry through to this point because of careless technique. If you suspect the presence of bismuth, test the solubil i ty of the yel low precipitate in 6AI NaOH; P b C r 0 4 w i l l dissolve, but ( B i O ) 2 C r 0 4 w i l l not.

1 1. If the lead is not almost completely removed from the precipitate, the chlorides can become coated w i th insoluble Pb (OH) 2 when ammonia is added; this w i l l prevent their reaction wi th the reagent.

12. It is better to use a test tube when acidifying Solution 1.3, because it is easier to mix in such tubes.

13. This is a more di f f icul t operation than the beginning student may realize. Silver is often missed because the solution and the nitr ic acid are not mixed well enough to obtain a definite acidic reaction. If you see a cloudy white layer floating on top of a clear one, you have not mixed the solution vigorously enough. Note particularly that a false acidic reaction can be obtained if the stirring rod touches an upper part of the tube that is wet wi th acid. Fol low the mixing procedure described in the technique section. Test for acidity by suspending a drop of the stirred solution on your stirring rod and touching it to wet l i tmus.


14. The purpose of evaporation is to remove excess HC1 and to destroy nitr ic acid that would otherwise interfere w i th the test for mercury. But if you permit too much solution to evaporate, some HgCl 2 can be lost. Use a microflame, and hold the crucible in a pair of tongs. Wave the crucible back and for th above, not in , the flame; withdraw the crucible momentarily if boiling becomes vigorous. Stop heating when a drop or two o f solution remains; further evaporation w i l l occur as the hot solution stands.

R E S U L T S

Submit a 3 X 5 in . card for your unknown solut ion, giving your name, desk number, ions present, and ions absent. Have your notebook available for your instructor to examine on request.

R E F E R E N C E

King, Edward J. Ionic Reactions and Separations: Experiments in Qualitative Analysis. New York : Harcourt Brace Jovanovich, Inc., 1973, pp. 129-42.

Experiment 30 Qualitative Analysis of the Group II Cations

Introduction

Group I I has many cations: H g 2 + , Pb 2+ B i 3 + , C u 2 + , C d 2 + , A s ( l I I ) , As(V) , Sb(IH) , Sb(V), Sn 2 + , and Sn( IV) . You wi l l undoubtedly find it more d i f f icul t to separate and confirm the Group II cations than you did the Group I cations. If you are to obtain good experimental results, you should be thoroughly familiar wi th the material in this section before coming to the laboratory. Your work must be done carefully, and you should be sure that you understand exactly what is happening as you perform each step.

You w i l l notice that some o f the cations are identified by their oxidation states (in Roman numerals) rather than w i t h values of the positive ionic charge. In these cases, the free cations do not exist in solution. Rather, some more complex species is present; for example, As(V) is found as the arsenate ion ASO4 3 in basic solutions and as H2ASO4 in acidic solutions.

Some of the cations may be present in more than one oxidat ion state, depending on what compounds are used to prepare solutions or solid unknowns [ for example, As ( I I I ) and As(V)] . :Tn the course o f the analysis, both oxidat ion states wi l l be converted to the same compound, which can then be reported merely as B i 3 + , As (V) , Sb( I I I ) , or Sn 2 + . Note that the two oxidation states of mercury can be separately identified: the mercurous ion H g 2

2 + appears in Group I , while the mercuric ion H g 2 +

appears in Group I I .

SEPARATION OF GROUP II CATIONS

The Group II cations have the common property that they all form insoluble sulfides in acidic solutions. Table 30-1 below shows the solubil i ty products for several insoluble sulfides, including those in Group I I .

302 Experiment 30

Other cations that form insoluble sulfides include several from Group I I I . The important question to be considered here is how the Group II cations can be cleanly separated by precipitation as sulfides from the Group I I I cations that also form slightly soluble sulfides. The answer is that solution acidity must be carefully controlled. Suppose that the solution of cations has [ M 2 + ] = 0.01 for the divalent cations. If C d 2 + is to be precipitated as CdS, then the [S 2~] must be large enough that

[ C d 2 + ] [S 2-] = ( O . O l ) l S 2 - ] > Ksp = 1 X 1 0 " 2 8 (30-1)

or, solving Equation (30-1), [ S 2 " [ must be > 1 X 10^ 2 6 to precipitate CdS. Any concentration greater than 1 X 10~ 2 6 is sufficient. If [S 2~] were 1 X 10~21yV/ and the solution also contained [ C o 2 + ] = 0 .01 , then

[ C o 2 + ] [ S 2 _ ] = (0.01)(1 X l C T 2 1 ) = 1 X 10~ 2 3 (30-2)

Since this is only one-tenth as large as the Ksp for CoS, no cobalt precipitates. Thus, by control l ing [S 2 ~[ , a large enough concentration can be attained that CdS (and all the less soluble Group I I sulfides) w i l l precipitate. This concentration can also be small enough that CoS (and all the more soluble Group I I I sulfides) wi l l remain in solution.

The sulfide concentration can be controlled by adjusting the pH. Sulfide ions for precipitations are supplied by continually saturating the solution w i th H 2 S, a weak acid. This gas dissolves in water to give [ H 2 S ] = 0.1 and participates in the same kind of equilibria as any diprot ic acid:

H 2S(aq) + H 2 0 - H 3 0 + + US" (30-3)

HS" + H 2 0 - H 3 0 + + S2" (30-4)

The equi l ibr ium constants are

[H3O+HHS-] = R ] Q x 1 Q - 7 ( 3 0 5 )

[ H 2 S ]

^ H s ' r 1 = K i = i j x ' ° ~ ' 3 ( 3 o _ 6 )

These two equilibria take place in the same solution. Hence, only one concentration of each species can be characteristic of the solution. If equations (30-5) and (30-6) are mult ipl ied together, [ H S - ] is cancelled in both the numerator and the denominator:

[ H 3 o - ] [ H s - i [H 3o*ns 2-i = = [ H 3 O * ] 2 [ S 2 - ] = 3 X 2 0

[ H 2 S ] [ H S - ] * [ H a S ]

Equation (30-7) can be considered merely a statement of the mathematical relationship among concentrations that must hold because of Equations (30-5) and (30-6).

303 Qualitative Analysis of the Group II Cations

If the solution remains saturated w i th H 2 S during the precipitation, [ H 2 S ] = 0.1 and

[ H 3 0 + ] 2 [ S 2 _ ] = (0.1) X 1.3 X 10~ 2 ° = 1.3 X 10~ 2 1 (30-8)

As in any equi l ibr ium, the species involved do not all need to come from the same source. Here, the [ H 3 0 + 1 of the solution can be adjusted by adding HC1. If [ H 3 0 + ] of the solution is made eqtial to 0.3/W and the solution is kept saturated w i th H 2 S , then

(0 .3 ) 2 [ S 2 _ ] = 1.3 X 10~ 2 1

[ S 2 _ ] = 1.4 X 1 0 " 2 0 (30-9)

This sulfide concentration is large enough that the ion product of the Group II sulfides exceeds their solubil i ty product constants, causing them to precipitate. At the same time, the concentration is sufficiently small that no ion product of the Group I I I sulfides exceeds its respective Ksp; therefore no Group II sulfides precipitate.

You should note from the information in Table 30-1 that SnS is relatively soluble and could precipitate wi th Group I I I . To avoid this, S n 2 + is oxidized to Sn(IV) before the precipitat ion, because SnS 2 , like the other Group II sulfides, is very insoluble.

Hydrogen sulfide is supplied to the reaction by generating it in the test tube. A solution of the organic compound thioacetamide ( T A ) is added, and the solution is heated. Thioacetamide breaks down on heating to generate H 2 S :

A C H 3 C ( S ) N H 2 + H 2 0 • H 2 S(g) + C H 3 C ( 0 ) N H 2 (30-10)

Thus, the solution can be kept saturated w i t h H 2 S during the course of the precipitat ion. During the precipitat ion, As(V) is reduced to As ( I I I ) by thioacetamide solution.

S U B D I V I S I O N OF G R O U P I I C A T I O N S

There are so many ions to test for in Group II that this group is separated into two subgroups after precipitation. The separation is possible because of the amphoteric nature of some of the metal sulfides. As you may recall, an amphoteric compound can function both as a base and as an acid. When treated w i t h K O H , A s 2 S 3 , S b 2 S 3 , and SnS 2 behave as acids and dissolve w i t h the formation of the complex ions

As 2 S 3 ( s ) + 6 0 H - ^ AsS 3

3~ + A s 0 3

3 " + 3 H 2 0 (30-1 1)

Sb 2 S 3 (s ) + 6 0 H " - * S b S 3

3 _ + S b 0 3

3 ~ + 3 H 2 0 (30-12)

SnS 2(s) + O H - ^ SnS 2 OH" (30-13)

304 Experiment 30

HgS is weakly acidic, and some HgS may also dissolve. The remaining precipitate consists of the very weakly acidic sulfides HgS, PbS, B i 2 S 3 , CdS, and CuS. These subgroups are referred to as the arsenic subgroup and the copper subgroup of Group I I , respectively.

Separation and Confirmation of Copper Subgroup Ions

Mercury. A l l the copper subgroup sulfides except HgS can be dissolved by heating them w i t h H N 0 3 . Ni t r ic acid oxidizes sulfide to elemental sulfur:

S2'+ 2 N 0 3 " + 4 H 3 0 + -»• S(s) + 2 N 0 2 ( g ) + 6 H 2 0 (30-14)

The oxidat ion of sulfide shifts the solubil i ty equilibria to the right as S2~ is cont inuously removed and the ion product drops below the value of Kip. For example

CdS(s) ^ Cd 2 + (aq) + S 2 _ (aq) (30-15) J N0 3 ~ S

HgS does not dissolve in H N 0 3 , because its solubil i ty product is so small (~ 1 0 ~ 5 2 ) that sufficient S2~ is not removed by oxidat ion to S.

After separating the solution containing Pb 2 + , B i 3 + , C d 2 + , and C u 2 + , the black precipitate of HgS can be dissolved in aqua regia. This reagent not only oxidizes S2~, but also removes H g 2 + by converting it to HgCl 4

2 ~ and shifting the solubil i ty equil ibr ium to the right:

HgS - Hg 2 + (aq) + S 2" (30-16) J C I" | N 0 3 "

H g C l 4

2 - S

HgS + 2 N 0 3 - + 4C1 + 4 H 3 0 + - > I l g C l 4

2 ~ + S(s) + 2NO,(g) + 6 H 2 0 (30-17)

The solution is evaporated to remove excess HC1 and H N 0 3 , and the residue is dissolved. To this solut ion, SnCl 2 is added to confirm the presence of H g 2 + :

2 H g C l 4

2 ~ + S n 2 + ^ Hg 2CT 2(s) + SnCl 6

2 ~ (30-18)

H g 2 C l 2 + 4CT + S n 2 + - » - 2Hg(l) + SnCl 6

2 " (30-19)

The gray precipitate of H g 2 C l 2 and Hg confirms the presence of Hg.

Lead. The lead ion is separated from the solution of Pb 2 + , B i 3 + , C d 2 + , and C u 2 +

by precipitation as white PbS0 4 . This sulfate is somewhat soluble in H N 0 3 , so the nitric acid must be destroyed. Because H N 0 3 is volatile, it can be removed by evaporation after the addit ion of I I 2 S 0 4 :

N 0 3 " + H 2 S 0 4 H N 0 3 ( g ) + HS0 4 - (30-20)

305 Qua l i t a t i ve Analys is of the G roup I I Cat ions

When all the H N 0 3 has evaporated, dense white S 0 3 w i l l appear and the heating can be discontinued.

H 2 S 0 4 - ^ - * S 0 3 ( g ) + H 2 0 ( g ) (30-21)

The lead sulfate residue is then separated from the solution containing B i 3 + , C d 2 + , and C u 2 + . Because the residue may be contaminated wi th small quantities of bismuth and copper sulfates, it is dissolved in hot ammonium acetate:

PbS0 4 + 3 C 2 H 3 0 2 " •* P b ( C 2 H 3 0 2 - ) 3 " + S 0 4

2 - (30-22)

Impurit ies are left behind as solids. Potassium chromate is added to the solution to produce a precipitate of yel low lead chromate, which confirms the presence of Pb:

P b ( C 2 H 3 Q 2 - ) 3

- + C r 0 4

2 - •* P b C r 0 4 + 3 C 2 H 3 0 2 " (30-23)

Bismuth. The separation of B i 3 + from C d 2 + and C u 2 + can be achieved by the precipitation of B i ( O H ) 3 . Both C d 2 + and C u 2 + also form insoluble hydroxides; but i f the base used is N H 3 , these ions wi l l remain in solution as ammine complexes:

C d 2 + + 4 N H 3 ^ C d ( N H 3 ) 4

2 + A ' = 3 X 1 0 7 (30-24)

C u 2 + + 4 N H 3 - C u ( N H 3 ) 4

2 + AT= 1.1 X 1 0 1 3 (30-25)

The formation of these complex ions reduces [ C u 2 + ] and [ C d 2 + ] to the point where

[ C u 2 + | [ O H " l 2 < Ktp (30-26)

[ C d 2 + ] [ O f T ] 2 < Ksp (30-27)

The precipitate of white B i ( O H ) 3 can be reduced to black metallic Bi by a basic stannite solut ion:

• 2Bi (OH) 3 ( s ) + 3Sn(OH) 3 " + 3 0 H " -> 2Bi(s) + 3Sn(OH) 6

2 " (30-28)

The black Bi confirms the presence of bismuth.

Cadmium and copper. The copper tetraammine complex C u ( N H 3 ) 4

2 + is deep blue. Thus, the presence of C u 2 + can usually be detected from the color of the solution after the precipitation of B i ( O H ) 3 .

If this test seems inconclusive, a port ion of the solution can be used to form a maroon copper ferrocyanide precipitate. The ammine complex is first destroyed

C u ( N H 3 ) 4

2 + + 4 H C 2 H 3 0 2 •* C u 2 + + 4 N H 4

+ + 4 C 2 H 3 0 2 " (30-29)

2 C u 2 + + Fe(CN) 6

4 " •+ C u 2 Fe (CN) 6 (30-30)

Cadmium is confirmed by reprecipitating yel low CdS:

C d ( N H 3 ) 4

2 + + H 2 S -* CdS(s) + 2 N H 3 + 2 N H 4

+ (30-31)

306 Experiment 30

If no C u 2 + is present, this test can be made by using the solution from the B i ( O H ) 3

precipitat ion. If C u 2 + is present, the black precipitate of CuS that forms when the solution is treated wi th H 2 S w i l l mask the yel low CdS. In this case, the C u 2 + is first removed from solution by reduction w i t h d i th ioni te ion S 2 0 4

2 ~. This ion wi l l not reduce C d 2 + :

C u ( N I I 3 ) 4

2 + + S 2 0 4

2 - + 4 N H 3 + 2 H 2 0 -> Cu(s) + 2 S 0 3

2 _ + 4 N H 3 + 4 N H 4

+ (30-32)

Separation and Confirmation of Arsenic Subgroup Ions

The cations of the arsenic subgroup are first reprecipitated as A s 2 S 3 , S b 2 S 3 , and SnS 2 by acidifying the solution containing complex ions. This simply reverses the reaction that occurs when these cations are dissolved in K O H :

AsS 3

3 " + A s 0 3

3 " + 6 H 3 0 * -* As 2 S 3 ( s ) + 911 2 0 (30-33)

S b S 3

3 _ + S b 0 3

3 " + 6 H 3 ( y ^ Sb 2 S 3 (s ) + 9 H 2 0 (30-34)

S n S 2 O i r + H 3 0 + ^ SnS 2(s) + 2 H 2 0 (30-35)

The mixture of sulfide precipitates is then reacted w i th dilute HC1 to dissolve S b 2 S 3

and SnS 2 as the chloro complexes:

Sb 2 S 3 (s) + 6 H 3 0 + + 8Cl~-> 2SbCl4~ + 3H 2 S(g) + 6 H 2 0 (30-36)

SnS 2(s) + 4 I I 3 0 + + 6CP -* SnCl 6

2 ~ + 2H 2 S(g) + 4 I I 2 0 (30-37)

Note that here these sulfides display the basic side of their amphoteric behavior in the reactions given by Equations (30-33)-(30-35) and the acidic side in the reactions given by Equations (30-36) and (30-37). A s 2 S 3 does not dissolve in dilute HC1.

Arsenic. The precipitate of A s 2 S 3 (after removal of the supernatant solution containing SbCl 4 and SnCl 6

2 ~) is oxidized w i t h hydrogen peroxide in ammonia solut ion:

A s 2 S 3 + 5 0 2 f T + O H " -> 2 A s 0 4

3 " + 3S + 3 H 2 0 (30-38)

The arsenate ion A s 0 4 " is precipitated by the addit ion of a magnesia mixture , a solution of M g ( N 0 3 ) 2 and N H 4 N 0 3 in aqueous ammonia:

A s 0 4

3 _ + M g 2 + + N H 4

+ + 6 H 2 0 -> M g N H 4 A s 0 4 • 6 H 2 0 ( s ) (30-39)

This solid is redissolved in acetic acid, which converts much of the basic A s 0 4

3 ~ ion to H 2 A s 0 4

_ and H A s 0 4

2 " :

M g N H 4 A s 0 4 - 6 H 2 0 ( s ) + 2 H C 2 H 3 0 2 -> H 2 A s 0 4 ' + M g 2 + + N H 4

+ + 2 C 2 H 3 0 2 - + 6 H 2 0 (30-40)

Arsenic is finally confirmed by adding silver nitrate solution to precipitate reddish-


brown silver arsenate: ff & c <A- T A o . //§• f * \ % \

• H 2 A s 0 4 " + 3 A g + + 2 H 2 0 - A g 3 A s 0 4 ( s ) + 2 H 3 0 + ( f c l ) , : \ ] j | j

An t imony . The solution containing SbCl 4~and SnCl 6

2 ~ is divided into two v> parts: one part for the ant imony test and one part for the tin test. The a n t i m o n y \ > r } ; ^ test solution is allowed to react wi th oxalic acid; both Sb(I I I ) and Sn(IV) form oxalate complexes:

SbCl 4 "+ 3 H 2 C 2 0 4 + 6 H 2 0 - > S b ( C 2 0 4 ) 3

3 ~ + 4 C r + 6 H 3 0 + (30-42)

SnCl 6

2 " + 3 H 2 C 2 0 4 + 6 H 2 O ^ S n ( C 2 0 4 ) 3

2 " + 6 C l " + 6 H 3 0 + (30-43) U g r o ^ A * ,

The formation constant for the reaction given in Equation (30-43) is so large that when thioacetamide is added and the solution is heated, no SnS 2 w i l l precipitate because of the low [ S n ( I V ) ] . Reddish-orange Sb 2 S 3 is the precipitate, and it confirms the presence of ant imony:

2 S b ( C 2 0 4 ) 3

3 - + 3 H 2 S •* Sb 2 S 3 (s) + 6 H C 2 0 4 (30-44)

T i n . The Sn( IV) in the second por t ion of the SbCl 4""/SnCl 6

2 _ solution is first reduced to S n 2 + by the reaction w i th Al wire

3SnCl 6

2 "+ 2 A l ( s ) - > 3 S n 2 + + 2 A J 3 + + 18C1" (30-45)

Sb w i l l be reduced to the metal

SbCl 4 + Al(s) -»• Sb(s) + A l 3 + + 4CP (30-46)

A side reaction w i l l also reduce some S n 2 + to Sn:

3 S n 2 + + 2Al(s) h . 3Sn(s) + 2 A 1 3 + (30-47)

However, Sn (but not Sb) can be reoxidized by hot HCl once the Al reduction is complete:

Sn(s) + 2 H 3 0 + ^ S n 2 + + H 2 (g ) + 2 H 2 0 (30-48)

When all the t in is in the form of Sn 2 + , it is confirmed by the familiar reaction w i th H g 2 +

S n 2 + + 2 H g 2 + + 8Cr-»> SnCl 6

4 " + H g 2 C l 2 ( s ) (30-49)

This reaction was used earlier to confirm the presence of H g 2 + . Because S n 2 + is not in excess this t ime, H g 2 + w i l l be reduced only as far as the white precipitate H g 2 C l 2 .


A g N 0 3 A s ( l I I ) (2 mg/ml) A l wire B i 3 + ( 2 mg/ml)

308 E x p e r i m e n t 30

P R O C E D U R E

Analysis of Known and Unknown Samples

1. Use 5 drops of a known or a practice solution that contains 2 mg/ml each of H g 2 + , B i 3 + , C u 2 + , A s ( I l I ) , and Sb( I I I ) , 4 mg/ml o f Pb 2 + . and 8 mg/ml each o f C d 2 + and Sn( IV) . If the solution contains a sediment, shake it well and then quickly withdraw the sample solution.

2. The unknown may be (1) a solution containing the ions of this and possibly other groups, or (2) a solid mixture . If it is Solution 1.1 from Outline 1 (see Experiment 29), use all of i t . If it is a special unknown solution containing only Group II ions, treat it as you would the practice solution. If it is a solid, dissolve about 20 mg of a representative sample in 10 drops of 6/W HC1 and use half of this solution for the analysis.

Preliminary Preparation

Calibration of a test tube. Measure 2.5 ml of water into a clean, dry test tube. Mark the position of the bo t tom of the meniscus w i t h a strip of label.

Oxidat ion of t i n ( I I ) . Transfer the solution to be analyzed (which should be acidic) to a small casserole or crucible. Add 2 drops of 3% H 2 0 2 , and boi l the solution un t i l only a few drops remain. Do this by swirling the solution constantly in a microflame and wi thdrawing the dish from this flame just before the final volume is reached (the solution wi l l continue to evaporate as i t cools). The chlorides o f As(V) , Sb(V), Sn( IV) , and H g 2 + are sufficiently volatile that some of these compounds may be lost if evaporation is excessive.

Dilute the remaining solution wi th 3-5 drops of water, and transfer this solution to the calibrated test tube. Rinse the casserole walls and bo t tom twice w i th 3-5 drops of water, and transfer both rinses to the test tube. Do not be concerned if the

C d 2 + (8 mg/ml) magnesia mixture C u 2 + ( 2 m g / m l ) 6A/ NaOH 6M H C 2 H 3 0 2 N a 2 S 2 0 4 (solid) 3 % H 2 0 2 6 M N H 3

1 8 A / H 2 S 0 4 1 5 A / N H 3

2A/HC1 N H 4 C 2 H 3 0 2

6M HC1 N H 4 C 1 1 2M HC1 N H 4 N 0 3

H g 2 + (2 mg/ml) oxalic acid HgCl 2 pH indicator paper 6A/HNO3 P b 2 + ( 4 m g / m l ) 1 6 A / H N 0 3 Sb( I I I ) ( 2 mg/ml) K 2 C r 0 4 Sn( IV) (8 mg/ml) O.SM K O H SnCl 2

litmus paper thioacetamide ( T A )


solution is tu rb id . This is usually caused by SbOCl or BiOCl, which are readily converted to the sulfides in a later step.

Adjustment of Acidity

Add 6M N H 3 to the solution unt i l it is basic to litmus or wide-range indicator paper. If C u 2 + ion is present, the formation of the deep-blue ammonia complex also acts as an indicator, but the blue color may be obscured by colored precipitates or complex ions of Group I I I cations. To the basic solution, add 2M HC1 dropwise unt i l the pH is between 2 and 3, according to wide-range indicator paper. If the solution is too acidic, adjust it w i th a fraction of a drop of 6M N H 3 . Then use your calibrated pipet to add exactly 0.35 ml of 2M HC1. Then add 4 drops of 13% thioacetamide solution (hereafter abbreviated T A ) and enough water to make the total volume 2.5 ml . Mix well . Check the acidity of the solution wi th short-range indicator paper. It should have a pH of 0.5, corresponding to a hydrogen ion concentration of 0.3A/. To treat the solution further, see Outline 2, which follows.

310 Experiment 30


4. Use water from a test tube in the hot water bath or add cold water and heat in the water bath for a few minutes. Washing is necessary to remove any excess H 2 S , which increases the solubil i ty of HgS in K O H , as well as ions of other groups, if present.

5. A m m o n i u m chloride prevents the sulfides from turning to a colloidal suspension.

6. This evaporation is necessary to remove H 2 S and T A , which are slowly oxi dized by atmospheric oxygen to sulfur and even to sulfate. Sulfate would precipitate Ba 2 + and S r 2 + ions, causing them to be lost.

7. If a colloidal suspension persists despite the presence of N H 4 N 0 3 , add a few crystals of the solid salt. Sometimes it is necessary to repeat this treatment; if so, add only the m i n i m u m amount of N H 4 N 0 3 that is required to coagulate the col loid.

8. If you cannot analyze the copper subgroup of Group II cations immediately, cover the sulfides wi th water and a drop of TA and stopper the tube w i th a cork or the bulb of a medicine dropper. If possible, it is better to conduct the analysis as far as the separation of Precipitate 3.1 from Solut ion 3.1 in Outline 3 before storing the sulfides.

9. The complete separation of solution and precipitate is diff icult to achieve in a test tube. Therefore, a second centrifugation and separation are made in a centrifuge tube. If a colloid persists, add a few crystals of N H 4 N 0 3 or a drop of 0.2M N I I 4 N 0 3 solution and warm to coagulate the precipitate.

1 0 . I f this solution cannot be analyzed immediately, add a drop of TA and reserve it in a t ight ly stoppered tube. A precipitate sometimes appears if the solution stands for a day or more. If the precipitate is colored, it is a sulfide. Its formation is not harmful; you can keep it wi th the solution.

312 Experiment 30


Notes to Outline 3

1. There is an induction period during which no apparent change occurs. When bubbling starts, the bubbles often carry sulfur and sulfides to the surface. Absence of bubbling does not mean there is no reaction.

2. As these formulas and colors imply, a black precipitate at this stage is not necessarily an indication of the presence of mercury, nor is the formation of a white or a yel low precipitate proof of the absence of mercury. The precipitate must always be dissolved, and the mercury must be sought in solution.

3. In drawing o f f the solution, leave behind any particles of sulfur. 4. Mercury( I I ) chloride is somewhat volatile and should not be heated too

strongly. ••' 5. Evaporation destroys nitrate and removes excess HC1. The concentration of

acid is further decreased by d i lu t ion , because the reaction between HgCl 2 and the S n 2 + ion is slow if the concentration of HC1 is high.

6. The formation of S 0 3 indicates that nitrate has been removed as molecular H N 0 3 . Once seen, S 0 3 fumes are easy to recognize thereafter. They are heavier than the fumes of nitric acid that are given o f f first. Failure to evaporate un t i l copious fumes are evolved may cause a sufficient amount of lead to remain in solution to spoil subsequent tests. Yet evaporation should not be carried to dryness, because the anhydrous sulfates return to solution slowly. If dryness does occur, add 2 drops of 6A/ H 2 S 0 4 and 3 drops of water, and then warm the casserole on the water bath for several minutes. Stir to disperse the residue in the solution.

7. The formation of a precipitate at this point is not a certain indication of the presence of Pb 2 + . Occasionally, other sulfates such as ( B i O ) 2 S 0 4 precipitate here. The bismuth compound is less soluble than PbS0 4 in N H 4 C 2 H 3 0 2 ; ( B i O ) 2 C r 0 4 is more soluble than P b C r 0 4 in H C 2 H 3 0 2 , and is insoluble in NaOH.

8. If a large quanti ty of bismuth is found, check the solubil i ty of this precipitate in NaOH.

9. If the blue color is faint and further confirmation is required, acidify a drop of the solution wi th H C 2 H 3 0 2 and add a drop of K 4 F e ( C N ) 6 solution. Copper(II) forms a red ferrocyanide, whereas cadmium forms a white ferrocyanide.

1 0 . Beginning students may find it hard to see a small, gelatinous precipitate of B i ( O H ) 3 , particularly when it is suspended in a blue solution.

314 Experiment 30

11. To a drop of SnCl 2 reagent, add 6M NaOH dropwise w i th careful mixing un t i l the precipitate Sn (OH) 2 redissolves. This w i l l require at least 2 drops o f NaOH solution. If you doubt the effectiveness of the SnCl 2 solution, which does gradually oxidize in air, prepare some stannite from the solution and test its effect on B i ( O H ) 3

made from the standard solution of B i 3 + i o n . 12. Sodium dithionite is a powerful reducing agent and is unstable. Keep it dry

and away from heat. Use a dry spatula to measure the solid. One-half of a spatula-full , or about 50 mg, should reduce 12.5 mg of C u 2 + ion. The blue color o f the copper ammonia complex should permanently disappear when di thioni te is added.

13. Precipitated copper is red. Black or brown residues frequently obtained here are colored by other metals. This is not harmful; in fact, it is advantageous to remove traces of other metals before the cadmium test.

14. If the solution remains in contact w i th copper too long, some of the copper can redissolve.

15. After prolonged treatment w i th di thioni te , a yellow color may appear; this is cadmium sulfide. If this occurs, add TA at once and warm. The sulfide apparently results from the decomposition of the di thioni te after the copper is removed.

16. Occasionally a buff, green, or black precipitate is obtained at this point . This is a restdt of careless separations in previous steps of the analysis. If the precipitate is large, remove the solution and discard i t . Dissolve the precipitate as far as possible in one drop of 6M HC1. Centrifuge and transfer the clear solution to a test tube. Dilute wi th 1 ml of water, add T A , and warm to reprecipitate the sulfide.

17. Cadmium may also be identified during the precipitation of the group (see note 3 to Outline 2).

316 Experiment 30

Notes to Outline 4

1. Thioacetamide is required to supply the H 2 S that is lost by vaporization. 2. If the solution becomes too acidic, some SnS 2 may be lost. Thus, 24/ rather

than 647 HC1 is used. Mix thoroughly after each addition of HC1. 3. The sulfides you are working w i th in this section are all highly colored. A

white tu rb id i ty that does not settle readily upon centrifugation is sulfur in its col-lodial form. When you obtain only this, the arsenic subgroup is absent.

4. The sulfides are precipitated in a test tube to permit careful neutralization and thorough mix ing . Because sulfides do not pack well , the separation of precipitate and solution is improved if a centrifuge tube is used.

5. The extraction wi th 1247 HC1 in the next step w i l l fail i f the acid becomes diluted wi th excess water in the precipitate.

6. Note any change in the character of the precipitate during this treatment. A bright yel low residue is usually A s 2 S 3 ; a black precipitate may contain HgS. A colorless solution and a white turb id i ty of S indicate the absence of arsenic. Do not heat the mixture for a prolonged period of time, for some A s 2 S 3 may dissolve slowly.

7. Draw o f f the supernatant solution through a wisp of cot ton wrapped around the tip of a pipet. If the pipet or the casserole contains water, this may decrease the concentration of HC1 to such an extent that S b 2 S 3 reprecipitates. This is not harmful.

8. A black residue is probably HgS. It can be identified by the procedure given in Outline 3 for Precipitate 3 .1 . Ammonia readily dissolves A s 2 S 3 , but does not dissolve HgS; the H 2 0 2 oxidizes A s ( I I I ) to As(V) .

9. Solutions of M g N H 4 A S 0 4 - 6 H 2 0 often become supersaturated. Give the precipitate time to form before you report a negative result.

10. A washing here removes traces of chloride that would precipitate wi th A g N 0 3 in the next step.

11 . If a white precipitate of AgCl forms, add more A g N 0 3 . 12. If the solution is permitted to evaporate beyond this point, some SnCl 4 may

be lost. 13. A l u m i n u m can reduce Sn( lV) and Sn(II) to gray Sn that wi l l dissolve in HC1

after the A l has been consumed; Sb wi l l not dissolve. 14. If the regular antimony test is indecisive, you may make a confirmatory test

on this residue. Wash it wi th water and dissolve it in a drop of 647 H N 0 3 . Dilute wi th water, add T A , and warm the solution to precipitate Sb 2 S 3 .

15. Since S n 2 + is oxidized by oxygen in the air, the addition of HgCl 2 should not be delayed.

317 Qualitative Analysis of the G oup II Cations

16. The precipitate is frequently orange at first and then darkens because of the postprecipitation of impurities. If the precipitate is yellow, it may be A s 2 S 3 because the treatment of Precipitate 4.1 w i th HC1 was too prolonged. Test the solubil i ty of the precipitate in 12/W HC1 and in N H 3 + H 2 0 2 . On the other hand, if the precipitate in the ant imony test is very dark, it may be contaminated w i th HgS. In this case, remove the solution, add 6/V/ HC1 to the precipitate, and warm the solution for one or two minutes. Centrifuge and discard the residue. Then dilute the solution, add T A , and warm to reprecipitate Sb 2 S 3 .

R E S U L T S

Submit a 3 X 5 in . card for your unknown solution, giving your name, desk number, ions present, and ions absent. Have your notebook available for your instructor to examine on request.

R E F E R E N C E

King, Edward J. Ionic Reactions and Separations: Experiments in Qualitative Analysis. New Y o r k : Harcourt Brace Jovanovich, Inc., 1973, pp. 143-62.

Experiment 31 Qualitative Analysis of the Group III Cations

Introduction

The cations o f Group I I I ( A l 3 \ C r 3 + , Fe 3 + , Fe 2 + , M n 2 + , C o 2 + , N i 2 + , and Z n 2 + ) form insoluble hydroxides or sulfides in basic solutions. A l l of these cations except A l 3 + and Z n 2 + are transition metals; thus (except for A l 3 + and Z n 2 + ) , they form many highly colored complex ions that can be easily identified and that indicate the presence of Group I I I ions in some preliminary tests even when ions from other groups are present in solution.

S E P A R A T I O N O F G R O U P III C A T I O N S

Group I I I cations can be separated from the cations of Groups IV and V by precipitation as sulfides or hydroxides. The precipitation is carried out w i t h thio-acetamide in an ammonia—ammonium chloride buffer. This buffer maintains the pH of the solution at ~ 9 . 3 ; this means, of course, that [ H 3 0 + ] = 5.3 X 1 0 " 1 0 and [ O H - ] = 1.8 X 10~ 5 . From Equation (30-8) in the preceding experiment, we can compute the sulfide concentration in this buffered solution when it is saturated w i th H 2 S:

[ H 3 0 + ] 2 [ S 2 " ] = (5.3 X 1 0 - ' ° ) 2 [ S 2 - ] = 1.3 X 10" 2 1 (31-1) [ S 2 _ ] - 5 X 10~ 3

If the metal ions are 0.0!/>/, then

[ M 2 + ] I S 2 ' ] = 5 X 10" 5 (31-2)

and the ion product is larger than Ksp for CoS, FeS, MnS, NiS, and ZnS. Thus, all these ions w i l l precipitate as sulfides. A l 3 + , C r 3 + , F e 3 + , F e 2 + , and M n 2 + also form insoluble hydroxides, as Table 31-1 indicates. Because the sulfides of F e 2 + and M n 2 +

are less soluble than the hydroxides, these ions wi l l precipitate as FeS and MnS even though the [ O F T ] is fairly substantial.

Fe 3 + , which has a very insoluble hydroxide, is reduced to F e 2 + by H 2 S , and precipitates as FeS:

2 F e 3 + + I I 2 S + 2 H 2 0 -> 2 F e 2 + + S + 2 H 3 0 + (31-3)

The purpose of the buffer is to maintain a sufficiently high concentration of O H " to precipitate A l ( O H ) 3 , C r ( O H ) 3 , and M n ( O H ) 2 , but a low enough concentration to prevent the Group V cation M g ( O H ) 2 from precipitating. The ion products for

319

320 Experiment 31

divalent and trivalent hydroxides are

[ M 2 + ] [ O r T ] 2 = (0.01)(1.8 X 1 0 ~ 5 ) 2 = 2 . 2 X 10" 1 2 (31-4)

[ M 3 + ] [ O H " ] 3 = (0.01)(1.8 X 1 0 " 5 ) 3 = 3.0 X 1 0 " 1 7 (31-5)

respectively. These values are obviously larger than A^ s p for Group I I I hydroxides, yet smaller than Ksp for M g ( O H ) 2 , so that its precipitation is prevented.

An additional reason for maintaining [ O H - ] at a reasonably low level is that C r ( O H ) 3 and A l ( O H ) 3 are amphoteric and would dissolve as C r ( 0 H ) 4

_ and Al (OH) 4 ~ i f [ O H - ] were too large.

Separation and Confirmation of Group III Ions

Manganese. After the supernatant l iquid containing Group IV and Group V ions is removed, the Group I I I precipitate is dissolved in H N 0 3 . The amphoteric hydroxides display basic behavior. For example

A l ( O H ) 3 ( s ) + 3 H 3 0 + + A l 3 + + 6 H 2 0 (31-6)

The sulfides dissolve because H N 0 3 oxidizes the sulfide ion to sulfur and shifts the solubil i ty equi l ibr ium to the right (as in Experiment 30 w i th the copper subgroup ions in Group I I ) :

FeS(s) + 3NO-T+ 6 H 3 0 + -> F e 3 + + S + 3 N 0 2 ( g ) + 9 H 2 0 (31-7)

Note that F e 2 + is reoxidized to F e 3 + by N 0 3

_ . The solution is then treated w i t h C103". M n 2 + is oxidized to insoluble brown M n 0 2 , which can be separated.

M n 2 + + 2C10J -> M n 0 2 ( s ) + 2C10 2 (g) (31-8)

321 Qualitative Analysis of the Group I I I Cations

In addit ion, the chlorate ion oxidizes C r 3 + to yellow-orange dichromate ( C r 2 0 7

2 - ) , and C I " is oxidized to C l 2 by N 0 3

_ :

2 C r 3 + + 6 C 1 0 3 - + 3 H 2 0 - > C r 2 0 7

2 - + 6C10 2 + 2 H 3 0 + (31-9)

3 C r + N 0 3 - + 4 H 3 0 * - * C l 2 + NOC1 + 61UO (31-10)

The precipitated M n 0 2 is dissolved in acidic hydrogen peroxide and again reduced t o M n 2 + :

M n 0 2 ( s ) + I I 2 0 2 + 2 I l 3 0 + •* M n 2 4 + 0 2 ( g ) + 411 2 0 (31-1 1 )

The M n 2 + is oxidized to deep-purple permanganate w i th sodium bismuthate to con firm manganese:

4 M n 2 + + 5 B i 2 0 5 + 1 8 H 3 0 + ^ 4 M n 0 4 - + 1 0 B i 3 + + 2 7 H 2 0 (31-12)

Division of Group III Ions

Further advantage can be taken of the amphoteric behavior of some of the hydroxides to separate the remaining Group I I I ions into two subgroups: (1) Fe 3 + , C o 3 + , N i 2 + , and (2) A l 3 + , Z n 2 + , C r ( V I ) .

When the solution is treated w i th NaOH, hydroxides precipitate for all ions except C r 3 + , wh ich has been oxidized to C r 2 0 7

2 - . As excess base is added, the amphoteric Z n ( O H ) 2 and A l ( O H ) 3 redissolve:

Zn(OH) 2 ( s ) + 2 0 F T -* Z n ( O H ) 4

2 _ (31-13)

A I ( O M ) 3 + O F T ^ A l ( O H ) 4 - (31-14)

Dichromate is converted to chromate:

C r 2 0 7

2 " + 2 0 F T -> 2 C r 0 4

2 " + H 2 0 (31-15)

N i ( O H ) 2 and Fe (OH) 3 are not affected by the addition of excess base. Because C o ( O H ) 2 is somewhat soluble, it is oxidized w i t h H 2 0 2 to less soluble C'o(OH) 3 .

Confirmation of i ron, cobalt, and nickel. The precipitate of F e (OH) 3 , C o (OH) 3 , and N i ( O H ) 2 is then dissolved in HC1. The hydroxides act as bases, and the H 3 0 + ions of the acid shift the solubil i ty equilibria to the right by removing O H " :

N i ( O I I ) 2 ( s ) ^ N i 2 + + 2 0 t r ( 3 M 6 )

| H 3 O +

I I 2 0

322 Experiment 31

The C o 3 + ion is unstable and spontaneously oxidizes water to 0 2 , so that C o 3 + is reduced to C o 2 + :

4 C o 3 + + 6 H 2 0 -»• 4 C o 2 + + 0 2 ( g ) + 4 H 3 0 * (31-17)

One por t ion of the solution can be used to confirm the presence of C o 2 + and F e 3 + ; another por t ion can be used to conf i rm N i 2 + .

To half of the solution, a crystal of N H 4 N C S is added. A blood-red F e ( N C S ) 2 +

complex confirms the presence of F e 3 + :

F e 3 + + N C S " - F e ( N C S ) 2 + (31-18)

If F e 3 + is present, the color of F e ( N C S ) 2 + must be destroyed before testing for C o 2 + . This is done by adding NaF, which converts Fe (NCS) 2 + to colorless FeF 6

3 ~:

F e ( N C S ) 2 + + 6 F - ^ F e F 6

3 " + NCS (31-19)

A solution of N H 4 N C S in a lcohol - ether is added and C o 2 + , if present, forms a blue complex in this solvent, confirming the presence of Co:

C o 2 + + 4NCS" > Co(NCS) 4

2 ~ (3 1-20) alcohol ether

To test for N i 2 + , all F e 3 + and C o 2 + must be removed from the other por t ion o f solut ion. Adding N H 3 precipitates F e ( O H ) 3 and C o ( O H ) 2 , which are discarded. N i 2 +

remains in solution as the blue-purple complex ion N i ( N H 3 ) 6

2 + . Dimethylglyoxime is added to the solution, and a red precipitate whose structure appears in Figure 3 1-1 confirms the presence of nickel.

Chromium. The basic solution containing C r 0 4

2 A 1 ( 0 H ) 4 ~ , and Z n ( O H ) 4

2 " can now be tested for the presence of these ions. The presence of C r 0 4

2 " is indicated by the yel low color of the solut ion C r 0 4

2 ~ a n d can be confirmed by making one port ion of the solution acidic and adding P b 2 + to precipitate yellow lead chromate:

P b ( C 2 H 3 0 2 ) 2 + C r 0 4

2 ' - * P b ( C r 0 4 ) ( s ) + 2 C 2 H 3 0 2 " (31-21)

323 Qualitative Analysis of the Group III Cations

S

FIGURE 31-2

Zinc complex with dithizone.

Aluminum. An ammonia-ammonium chloride buffer is added to a second port ion of the Cr0 4

2 ~ , Al (OH) 4 " , Zn (OH) 4

2 ~ solution, as in the ini t ia l precipitation of the group. As shown earlier, this buffer provides a sufficient [ O H " | to precipitate A l ( O H ) 3 . The presence of N H 3 retains Z n 2 + in solution as the complex ion Z n ( N H 3 ) 4

2 + . The A l ( O H ) 3 precipitate is separated by centrifugation and then dissolved in an acidic buffer of acetic acid and ammonium acetate. Aluminon reagent is added, and A l 3 + ( i f present) produces a red precipitate A l ( C 2 2 H 1 3 0 9 ) 3 . This reaction occurs only around the pH provided by the buffer.

Zinc. The solution that remains after the precipitation of A l ( O H ) 3 must be freed of C r 0 4

2 _ , so that it cannot interfere w i t h the Z n 2 + test. This is done by adding BaCl] to precipitate B a C r 0 4 , which is discarded.

Zinc sulfide is then precipitated by the addit ion of T A . The white color characteristic of ZnS confirms the presence of Z n 2 + .

A final confirmatory test is the formation of a purple-red complex between Z n 2 +

and dithizone. The ZnS precipitate is dissolved in H N 0 3 , which oxidizes S 2" to S. The resulting solution is made basic and applied to a piece of dithizone paper. The purple-red complex whose structure appears in Figure 31-2 confirms the presence of Z n 2 + .

324 Experiment 31

P R O C E D U R E


1. Use 5 drops of a known or practice solution that contains 2 mg/ml each of A l 3 + , C r 3 + , N i 2 + , C o 2 + , and M n 2 + , and 4 mg/ml o f Z n 2 + . Note that the practice solution contains no i ron . Carry out the tests for iron, and note the faint colorations that may appear caused by impurities. Then add 1 drop of F e 3 +

solution to the test solut ion, and observe the appearance of a bona fide test. 2. If the unknown is Solut ion 2.1 from Outline 2 (Experiment 30), use all of

the solution in the systematic analysis. If the unknown solution contains only Group I I I ions, use 5 drops in the systematic analysis (more must be available for preliminary tests). If the unknown is a solid, take a representative sample and dissolve about 50 mg in water or in 64/ HC1; use about a third of this solution in the systematic analysis.

Always note the color of the k n o w n or unknown sample. Compare the colors of the unknown samples w i t h the colors of known ions. Some of the most common ions and colors are: C r 3 + , violet; N i ( N H 3 ) 6

2 + , blue-purple; C o C l 2 and C o C l 4

2 _ , blue; C r C l 2

+ a n d C r ( O H ) 4 , deep green; Fe 2 + , pale green; FeCl 2 + , yellow; F e ( I I I ) - O H complexes, yellow to red; C o 2 + , red; M n 2 + , very pale pink (almost colorless); A l 3 +

and Z n 2 + , colorless. In the solid state, combinations of colors must be considered; green nickel and red cobalt salts produce a gray mixture.

Preliminary Tests

It is possible to make specific tests for all of the ions in this group (except aluminum and chromium) on the original sample, even when it contains ions from other groups. These tests are given here. It is wise to confirm each test by systematic analysis, according to Outline 5.

Tests for i ron. In a general analysis, all F e 3 + ioiis are reduced to F*e2+ by H 2 S or T A . Tests for the oxidat ion state of iron must be made on the original sample.

Test for Fe2+: Dissolve one small crystal of K 3 F e ( C N ) 6 in about 20 drops of water. Add a drop of this solution to a drop of the test solution. A blue color or a blue precipitate of K F e F e ( C N ) 6 indicates the presence of Fe 2 + . If the solution turns green, you may have used too much ferricyanide so that its yellow color masks the blue. In this case, add 1 drop of ferricyanide reagent to 5 drops of distilled water, mix well , and use this reagent to try the test again. Other ions produce colored precipitates w i t h this reagent, but only F e 2 + forms a blue precipitate.

N H 4 N C S (solid) sodium bismuthatc (solid) N i 2 + ( 2 mg/ml) thioacetamide P b ( C 2 H 3 0 2 ) 2 Z n 2 + ( 4 m g / m l ) SnCl 2

325 Qualitative Analysis of the Group III Cations

Test for Fe3*: Dissolve a few crystals of N H 4 N C S in a l i t t le water, and add a few drops of this reagent to the test solution. The dense red color of FeNCS 2 + {not a precipitate) indicates the presence of F e 3 + ion. Pink or light red colorations are caused by traces of i ron, a common impur i ty in reagents. A l l aged solutions of F e 2 +

contain F e 3 + , because the i ron( I I ) is slowly oxidized by the oxygen in the air. Do not report traces. If you are not sure that the color of your test solution is deep enough to indicate the presence of Fe 3 + , test a few drops of a solution that contains 0.1 mg/ml of F e 3 + (dilute one drop of the standard ion solution w i t h 5 ml of water and mix well.) An indication of the presence of iron is also obtained in the test for cobalt.

Test for nickel. Dilute one drop of the test solution w i t h a few drops of water; then add 154/ N H 3 unt i l the solution is strongly alkaline. Centrifuge and draw o f f the clear solution to another tube. To the solution add several drops of d imethyl -glyoxime solution. A deep-red precipitate of nickel dimethylglyoximate indicates the presence o f N i 2 + . I f a brown coloration caused by C o 2 + forms instead, add more reagent. This test wi l l fail when only a small amount o f nickel is present, because N i ( O H ) 2 coprecipitates w i t h the larger precipitate of the gelatinous hydroxides of aluminum and iron ( I I I ) , which is centrifuged and discarded.

Test for manganese. Test a drop of the solution for chloride by acidifying the solution w i th nitric acid ( i f necessary) and then adding A g N 0 3 solution.

1. If chloride is absent, add a drop of 164/ H N 0 3 and one-quarter spatula-full of sodium bismuthate. Centrifuge to see the color. The purple color of M n 0 4

_ indicates the presence of manganese. 2. If chloride is present, add A g N 0 3 solution un t i l no more precipitate forms.

Then test the solution w i t h H N 0 3 and sodium bismuthate, as before.

Test for cobalt. To a drop of the acidic solution, add a few drops of water and several crystals of N H 4 N C S . If a red color (FeNCS 2 + ) appears, add solid NaF, a l i t t le at a time, un t i l the color disappears. Then add 5-10 drops of a solut ion of N H 4 N C S in alcohol-ether. A blue upper layer containing Co( N C S ) 4

2 _ indicates the presence of Co. A green upper layer indicates the presence of low concentrations of either C o 2 + or NCS - . If the color is indistinct or if it fades when the solution is shaken, add a spatula-full of solid N H 4 N C S . If the color is red, add more NaF. A dark, muddy-green color may be caused by the presence of C u 2 + ; add a drop of SnCl 2 to reduce Cu( I I ) to Cu(I ) .

Test for zinc. To 1 drop of the test solution, add 64/ NaOH unt i l the solution is basic and then add 1 or 2 drops in excess. Touch a pipet to this mixture without squeezing the bulb. Some o f the solution w i l l rise in the t ip o f the pipet by capillary action. Then bring the t ip down vertically on the center of a square of dithizone paper. A purple-red spot indicates the present of Z n 2 + . An orange ring is caused by NaOH alone. Mercury(I or I I ) and lead(II) produce purple-blue spots, and Sn(II) produces a pink spot. Now touch the pipet to some water and bring the t ip down on the center o f the spot. The zinc spot wi l l spread as the water flows out o f the pipet; a spot produced by another ion would not spread.

326 Experiment 31

328 Experiment 31

)


6. If a purple color appears but then fades rapidly, chloride is present and you should add more bismuthate.

7. The oxidat ion of M n ( I I ) to M n 0 2 by chlorate requires boil ing, the presence of 16/17 H N 0 3 , and the absence of chloride. Avoid adding too much KC10 3 at one t ime, or the evolution of C10 2 may become too violent.

8. I f the side test shows the absence of manganese, no precipitate wi l l be obtained. Even so, the potassium chlorate treatment is used to oxidize C r 3 + to C r 2 0 7

2 7 9. The first bismuthate that is added wi l l oxidize 11 2 0 2 to 0 2 . An excess is

present when the layer of brown or black, solid on the bo t tom is several millimeters thick and no bubbles are evolved.

10. The hydrogen peroxide oxidizes C o ( O I l ) 2 to C o ( O H ) , , and a better separat ion o f C o 3 + from A l 3 + and Z n 2 + results. Warming accelerates the reaction and decomposes the excess I I 2 0 2 .

1 1. A double centrifugation is advisable to ensure the complete separation of Fe 3 + , C o 3 + , and N i 2 + from C r ( V I ) , A l 3 + , and Z n 2 + .

12. The thiocyanate reagent contains ether and amyl alcohol, which are volatile solvents. I f the solution is warm when the reagent is added, most of the solvents wi l l evaporate.

13. The preliminary test is a more reliable indicator of F e 3 + than a red color at this stage. A certain amount of iron is frequently picked up during the analysis from impurities in the reagents or through careless technique. Therefore, even when no iron is detected in the preliminary test, a pink color is sometimes observed here.

14. If the blue color fades after mixing , add a spatula-full of solid N H 4 N C S . A large water layer extracts so much thiocyanate from the upper layer that the sensit iv i ty of the test is reduced.

15. A large, false nickel test may be found here if the solution was stirred wi th a nickel spatula at any earlier stage. Glass rods should always be used for stirring.

16. If cobalt is present, it combines w i th dimethylglyoxime to produce a brown solution before any dimethylglyoxime reacts wi th nickel. If a brown color forms, be sure to add a sufficient excess of reagent.

17. If the aluminum hydroxide is brownish in color, it may contain iron. Dissolve the A l ( O H ) 3 in a drop or two of 6A7 NaGTI and centrifuge to remove the brown Fe (OH) 3 . Neutralize the solution w i t h 6A/ HC1, and add NH 4 C1 and N H 3 to reprecipitate A l ( O H ) 3 .

18. The best lake is formed if the pH of the.acidic solution containing the aluminon is raised to between 5 and 7.2 by the addition of ammonium acetate. A precipitate may form in this weakly acidic solution i f large amounts o f A l 3 * are present. Otherwise, the precipitate wi l l not appear un t i l the solution is made weakly basic. An excess of N H 3 is harmful. Silica from glass storage vessels does not produce a red lake, but Fe 3 + , C r 3 + , Pb 2 + , and other metals do give a red lake. Only silica could be present here if earlier separations were made carefully.

R E S U L T S

For your unknown solution, turn in a 3 X 5 in . card giving your name, desk number, ions present, ions absent. Have your notebook available for your instructor to examine on request.

330 E x p e r i m e n t 31

R E F E R E N C E

King, Edward J. Ionic Reactions and Separations: Experiments in Qualitative Analysis. New York : Harcourt Brace Jovanovich, Inc., pp. 163-82.

Experiment 32 Qualitative Analysis of the Group IV Cations

Introduction

The cations of Groups IV and V are those of the alkali and alkaline-earth families of the periodic table. Many of the cations of Groups IV and V can be identified by the flame-test method you used in Experiment 8. However, flame tests are less satisfactory for analyzing samples in which several different ions may be present. This is because some flames tend to mask those of other ions. For this reason, a method of chemical analysis that can be used to identify these cations is presented in Experiments 32 and 33. You may still wish to use the procedure in Experiment 8 to make some flame tests in the preliminary analysis of your unknown solution.

SEPARATION OF THE GROUP

The Group IV cations, Ca 2 + , Sr 2 + , and B a 2 + have very similar chemical behaviors in that each forms only a +2 ion that has the noble gas configuration. Thus, no redox reactions occur to permit separation and ident if icat ion. The analytical scheme depends primari ly on solubil i ty differences between salts of oxo acids.

The ions are precipitated as insoluble carbonates from a buffer mixture of ammonia, ammonium chloride, and ammonium carbonate. Carbonate precipitation is particularly advantageous, because carbonates can easily be redissolved in an acidic solution. The acid dissociation constants given in Table 32-1 show that C 0 3

2 ~ and HC0 3 ~are both strong enough bases to react essentially completely wi th protons and to shift the solubil i ty equilibria to the right by the removal of C 0 3

2 _ ions:

331

332 Experiment 32

Before precipitating the group, the ammonium salts that you have added to the sample as you separated the Group I I I cations are removed by evaporation and baking. Then the concentrations of N H 3 , N H 4 C 1 , and ( N H 4 ) 2 C 0 3 are adjusted precisely. After this adjustment, [ C 0 3

2 ~ ] w i l l be about 3 X 10 - 2/W. Again i f you suppose that all the metal ions are present in 0 . 0 \ M concentrations, you can see from Table 32-2 that the ion products wi l l be large enough to precipitate all the Group IV carbonates, but not the fairly insoluble M g C 0 3 from Group V.

Separation and Confirmation of the Ions

Barium. After removing the supernatant solution that contains Group V ions, the carbonate precipitate is dissolved in an ammonium acetate-acetic acid buffer that has a pH of ~ 4 . 7 . Some chromate ion is added to precipitate yellow B a C r 0 4 .

In an acidic solution, the chromate ion participates in two important equilibria:

C r 0 4

2 - + I 1 3 0 + ^ H C r 0 4 + I 1 2 0 (32-2)

2 C r 0 4 + 2 H 3 0 + - C r 2 0 7

2 " + 3 H 2 0 (32-3)

At the pH of the buffer used, only about 1% of the chromate is present as C r0 4

2 ~ ; the other 99% is converted to H C r 0 4 " or C r 2 0 7

2 " by the reactions of Equations (32-2) and (32-3). Thus, [ C r 0 4

2 _ ] is kept small enough that B a C r 0 4 (A' S | ) = 1 X 1 0 ' 1 0 ) precipitates and C a C r 0 4 ( A ' s p = 7 X 10" 4 ) and S r C r 0 4 ( A ' s p = 3 X 10~ s ) do not.

The B a C r 0 4 precipitate can be dissolved in the strong acid H Q , which w i l l shift the equilibria reactions in Equations (32-2) and (32-3) far to the right:

BaCr0 4 ( s ) # B a 2 + + C r 0 4

2 _ (32-4)

I H 3 0 +

H Q C V + C r 2 0 7

2 _

B a 2 + is confirmed by precipitat ion from this solution as insoluble BaS0 4 .

Calcium and s t ron t ium. The chromate that remains after the precipitation of B a C r 0 4 must be removed. This can be done by precipitating C a C 0 3 and S r C 0 3 and

333 Qualitative Analysis of the Group IV Cations

then washing the precipitate. Since there is no longer a possibility of precipitating the M g 2 + that was carried in to the Group V solution, no buffer is needed to control [ C 0 3

2 ~ ] . N a 2 C 0 3 can be used as the carbonate source. The carbonate ion is el iminated when the precipitate is allowed to react wi th concentrated H N 0 3 :

CaC0 3 (s ) + 2 H 3 0 + -+ C a 2 + + H 2 C 0 3 + 2 H 2 0 (32-5) ^ C a 2 + + C 0 2 ( g ) + 3 i l 2 0

Stront ium carbonate ( S r C 0 3 ) dissolves similarly; but S r ( N 0 3 ) 2 is insoluble in cone. H N 0 3 and precipitates immediately. Since C a ( N 0 3 ) 2 is soluble, this procedure separates C a 2 + from Sr 2 + .

The S r ( N 0 3 ) 2 precipitate can then be dissolved in water:

S r ( N 0 3 ) 2 ( s ) Sr 2 + (aq) + 2 N 0 3

_ ( a q ) (32-6)

St ront ium is confirmed by the addit ion of ( N H 4 ) 2 S 0 4 to precipitate white S rS0 4 . Any traces o f C a ( N 0 3 ) 2 that have precipitated wi th S r ( N 0 3 ) 2 w i l l form a precipitate of CaS0 4 , as the A^p values given in Table 32-2 show. Thus, before precipitat ion , some triethanolamine is added to form a complex w i th Ca 2 + :

C a 2 + + 2 N ( C 2 H 4 O H ) 3 -»• C a [ N ( C 2 H 4 O H ) 3 1 2

2 + (32-7)

This reduces [ C a 2 + ] by converting it to the complex, so that [ C a 2 + ] [ S 0 4

2 _ ] < 2 X 1CT5, and no precipitate o f CaS0 4 w i l l form.

Finally, Ca 2* is confirmed by the precipitation of C a C 2 0 4 T I 2 0 from a hot , weakly alkaline solution. The solution must be weakly alkaline because C 2 0 4

2 ~ is a good base (see Table 31-1) and would be converted to H C 2 0 4 " in an acidic solution, thus reducing [ C 2 0 4

2 ~ ] to the point where [ C a 2 + ] [ C 2 0 4

2 ~ ] < A ' s p and preventing precipitation. The solution is heated so that a precipitate o f larger particle size w i l l form.

334 Experiment 32

P R O C E D U R E


1. A known or practice solution for Group IV should contain 3 mg/ml of Ba 2 + , Sr 2 + , Ca 2 + , and M g 2 + ions, and about 30 mg/ml of N H 4 * ion. Use 10 drops for the analysis. Remove the ammonium salts, and analyze the group according to Outline 6. To check your technique, add 1 drop of ammonium sulfate and 1 drop of ammonium oxalate to Solution 6 .1 . No more than a trace of precipitate should appeur, even after warming. To this solution, add N a 2 H P 0 4 solution to precipitate magnesium ammonium phosphate and to verify that magnesium did not precipitate w i t h Group IV .

2. The unknown may contain only the ions of Group IV or a combination of Group IV and Group V ions. If it is a solut ion, use 10 drops for the analysis, according to Outline 6. If it is a solid, take a representative sample, dissolve 20-30 mg in water or in 6M HC1, and use this solution for the analysis. Remove the ammonium salts only if a positive test for the ammonium ion is obtained (see the Preliminary Test, Outline 7, Experiment 33) . If the unknown is Solution 5.1 (Experiment 3 1 , Outline 5, Cation Group I I I ) , be sure it has been evaporated w i t h hydrochloric acid as directed in Outline 5 (Experiment 3 1) to remove thioacetamide and hydrogen sulfide before any nitric acid is added. (Otherwise, ni t r ic acid can oxidize sulfide to sulfate and cause the premature precipitation of barium and s t ront ium.)

336 Experiment 32


9. The precipitates of this group are more densely packed than the sulfides and hydroxides of preceding groups; although they appear to be small, it is possible to obtain a satisfactory test w i th them.

10. Disregard the formation of any precipitate after the addit ion of NaOH. Sodium hydroxide solution is frequently contaminated w i t h carbonate, so any additional precipitate may be the carbonates of s trontium and calcium or it may be magnesium hydroxide if some magnesium carbonate was carried into the Group IV solution.

11. Dissolve a heaping spatula-full (~ 0.1 5 g) of solid N a 2 C 0 3 in 10 drops of water. When such solutions are stored in glass bottles, they etch the glass and dissolve some silica. This would precipitate wi th the carbonates of strontium and calcium and make it d i f f icul t to judge the quantity of these substances. It is best to use a fresh solution of the reagent.

12. The solid nitrate settles rapidly. Unless the mixture is poured quickly , only the supernatant solution is decanted; the precipitate is left behind in the casserole. Examine the casserole if no precipitate is obtained, to see if any crystals are adhering to the walls.

13. For a satisfactory separation of S r ( N 0 3 ) 2 , the concentration of H N 0 3 must not be below 70% (16/W); ideally, it should be close to 80%. Thus, the in t roduct ion of water must be avoided by (1) using a dry tube, (2) pouring instead of pipetting, and (3) using a dry rod. Dry the tube either w i th a clean towel or in an oven. C A U T I O N : 16A/ H N 0 3 is very corrosive. I t w i l l produce yellow stains on skin that take a week to wear (or peel) off. Flood all spills of ni tr ic acid w i t h water, whether they arc on the skin or on the desk. The use of plastic gloves is recommended.

14. S t ron t ium nitrate tends to form supersaturated solutions. Stirring, rubbing, and letting the solution stand are the usual ways of coping w i t h supersaturation. Cooling also he lps-not because the solubili ty is appreciably decreased, but because the supersaturated solution becomes more unstable.

15. Strontium sulfate also tends to form supersaturated solutions. A l l o w at least 5 minutes for a precipitate to form.

16. If desired, a flame test can be made for confirmation. The sulfate must be converted to carbonate and dissolved in hydrochloric acid. Remove the solution above the precipitate, and treat the precipitate w i t h ammonium carbonate solution. Stir up the caked precipitate and warm it briefly. Centrifuge and wash the precipitate several times to remove sulfate. Then dissolve the precipitate in a few drops of 12:1/ HC1 and make the flame test.

17. Since 10 drops o f 16A/ H N 0 3 were used to separate S r ( N 0 3 ) 2 , i t w i l l take at least this much 1 5M N H 3 to achieve neutralization. The nitric acid was diluted wi th water to make the reaction less violent. Failure to make the solution basic is a common error at this stage. Be particularly careful when mixing, and test the solution w i th litmus.

R E S U L T S

Submit a 3 X 5 in . card for your unknown, giving your name, desk number, ions present, and ions absent. Have your notebook available for your instructor to examine on request.

338 Experiment 32

R E F E R E N C E

King, Edward J. Ionic Reactions and Separations: Experiments in Qualitative Analysis. New York : Harcourt Brace Jovanovich, Inc., 1973, pp. 183-97.

Experiment 33 Qualitative Analysis of the Group V Cations

Introduction

The four Group V cations (Na + , K + , M g 2 + , and N H 4

+ ) present a special problem in qualitative analysis. With the exception of M g 2 + . practically all the compounds of these ions are soluble. Therefore, no single reagent precipitates the Group. The salts that do precipitate are often complex ones that are subject to the interference of other ions.

Flame tests produce somewhat more definitive results w i th Group V than w i th Group IV ions, especially when used in conjunction w i th chemical evidence from the analysis scheme. This is because only two ions remain that color a flame: Na + and K + . The Na + flame test is so sensitive that even the minute amounts of Na + that leach into solution from glass bottles produce a yellow flame. When t ry ing to determine whether your unknown contains Na + , you should compare the color intensity of the flame w i t h that of a known solution containing about 4 mg/ml of Na + . The shortlived violet color of the K + flame is masked by the yel low of Na + and should be viewed through two thicknesses of blue cobalt glass.

Separation and Confirmation of the Ions

Sodium and potassium. Use two-thirds of the solut ion left from the Group IV precipitation to test for Na + and K + . The N H 4

+ ion that may have been present in the original solution and that has been added in great quantities during analysis is first destroyed by evaporating the solution to dryness. N F I 4

+ would interfere w i t h the K* test by forming a precipitate wi th the reagent used, if ammonium salts were not first thermally decomposed.

The residue is then redissolved and divided into two parts. One part is treated w i t h magnesium uranyl acetate to give a greenish-yellow crystalline precipitate that confirms the presence of Na + :

N a + + M g 2 + + 3 U 0 2

2 + + 9 C 2 H 3 0 2 - + 9 H 2 0 -> N a M g ( U 0 2 ) 3 ( C 2 H 3 0 3 ) - 9 H 2 0 ( s ) (33-1)

It is important that the volume of solution be very small; otherwise, [ N a + ] w i l l be too small for this precipitate to form. Therefore, only a min imum volume of water should be used to dissolve the residue after evaporation.

The other por t ion of the solution is treated w i t h N a 3 [ C o ( N 0 2 ) 6 ] to give a yellow precipitate of dipotassium sodium cobal t ini t r i te that confirms the presence of K + :

2 K + + N a + + C o ( N 0 2 ) 6

3 " + H 2 0 - + K 2 N a [ C o ( N 0 2 ) 6 ] - H 2 0 ( s ) (33-2)

339

340 Experiment 33

Magnesium. The second port ion of the supernatant l iquid from the Group IV precipitat ion is used to test for M g 2 + . In the presence of N H 4

+ i o n and N H 3 , some N a 2 M P 0 4 solution is added to form a white precipitate of magnesium ammonium phosphate:

M g 2 + + N H 4

+ + P 0 4

3 " + 6 H 2 0 -> M g ( N H 4 ) ( P 0 4 ) - 6 H 2 0 ( s ) (33-3)

The N H 3 and N H 4

+ ion act as a buffer system to provide a sufficiently high I O H " ] to convert H P 0 4

2 " to P0 4

3 ~, but not enough [ O H - ] to precipitate M g ( O H ) 2 . The presence of M g 2 + is confirmed by the formation of a blue lake. The precipi

tate is first dissolved in acid:

M g ( N H 4 ) ( P 0 4 ) - 6 H 2 0 ( s ) + 2 I I 3 0 + -»• M g 2 + + N H 4

+ + H 2 P 0 4

_ + 8 H 2 0 (33-4)

A colored lake is then formed by precipitating M g ( O H ) 2 in the presence of a dye that the precipitate adsorbs, coloring it blue. The dye is called "S and O" reagent, ( H O ) 2 C 6 H 3 N = N C 6 H 4 ( N 0 2 ) . Its chemical name isp-nitrobenzenazoresorcinol.

A m m o n i u m ion. The test for ammonium ion (NH 4 *) is carried out on a por t ion of the original solution. This is done because many N H 4

+ salts are added in the course of group precipitations, and a solution would test positively for N H 4

+ after Groups I , I I , I I I , and IV had been removed.

The reaction of ammonia w i th water is the basis for the N H 4

+ test:

N H 3 + H 2 O ^ N H 4

+ + O i r (33-5)

A strong base is added to the solution, shifting the equil ibrium of l iquat ion (33-5) to the left. The solution is heated, and N H 3 gas is given off, because of the solubi l i ty of gases decreases at high temperatures. A watch glass w i t h a piece o f wet, red l i tmus paper is placed above the test solution. As the escaping ammonia hits the watch glass and dissolves, the reaction of Equation (33-5) occurs and the red litmus turns blue.


6M HC1 6M NaOH 16yV/HN0 3 Na 2 S K + ( 4 mg/ml) 6 M N H 3

magnesium uranyl acetate N H 4

+ ( 2 mg/ml) M g 2 + ( 2 m g / m l ) NH 4 C1 Na + ( 4 mg/ml) ( N H 4 ) 2 C 2 0 4

N a 3 C o ( N 0 2 ) 6 ( N H 4 ) 2 S 0 4

N a 2 H P 0 4 l i tmus paper N a N 0 2 solid "S and O" reagent

341 Qualitative Analysis of the Group V Cations

P R O C E D U R E


1. Use 10 drops of a known or practice solution that contains about 2 mg/ml of M g 2 + , 2 mg/ml of N H 4

+ , 4 mg/ml of K + , and 4 mg/ml of Na + . 2. The unknown may be:

(a) A solution containing only the ions of Group V: Use 10 drops of the solut ion to test for M g 2 + , K\ and Na +, and a separate port ion of the solution to test for NH 4 * .

(b) A solid mixture containing only the ions of Group V: Take a representative sample, and use 20 mg to test for M g 2 + , Na+, and K + . Dissolve in 8-10 drops o f water. I f the solution is acidic, make it slightly alkaline wi th 6M N I I 3

(check w i t h l i tmus) . If the solution is alkaline, acidify wi th 64/ HC1 and then make it barely alkaline wi th 64/ N H 3 .

(c) Solution 6.1 (Experiment 32, Outline 6, Cation Group IV) obtained in a general analysis after removal of Groups HIV: Use the entire solution to test for M g 2 + , K\ and Na*. Add 1 drop each of ( N H 4 ) 2 C 2 0 4 and ( N H 4 ) 2 S 0 4

solutions and 64/ N H 3 ( i f necessary) to make the solution alkaline. Centrifuge to settle the precipitate, which wil l contain traces o f the ions o f Group IV that escaped precipitation. Withdraw the solution, and use it for the analysis; discard the residue. The test for the ammonium ion must be made on a separate port ion of the original unknown.

342 Experiment 33

343 Qualitative Analysis of the Group V Cations

344 Experiment 33

10. These tests are based on the formation of fairly soluble precipitates and require that concentrated solutions and excesses of reagents be used. If you add too much water here, y o u may fail to obtain precipitates.

11 . The sodium cobalt ini tr i te reagent is unstable. It should have a deep reddish-amber color. Pale yel low or pink solutions should be discarded. Test the reagent w i t h a standard solution of K + i f its potency is doubted.

12. If the ammonium salts have not been completely removed, a voluminous precipitate at this point is often ( N H 4 ) 2 N a C o ( N 0 2 ) 6 ' H 2 0 , which is indistinguishable f rom the potassium compound. Warming w i l l remove a small amount o f the ammonium compound, but not a large quant i ty .

13. Dissolve the precipitate in 1 drop of 6M HC1. View the flame through two thicknesses of cobalt-blue glass to fil ter out the sodium color. The potassium flame appears red.

14. Both the potassium and the sodium precipitates supersaturate easily (see note 6 above).

R E S U L T S

Submit a 3 X 5 in . card for your unknown, giving your name, desk number, ions present, and ions absent. Have your notebook available for your instructor to examine on request.

R E F E R E N C E

King, Edward J. Ionic Reactions and Separations: Experiments in Qualitative Analysis. New Y o r k : Harcourt Brace Jovanovich, Inc., 1973, pp. 198-209.

Experiment 34 Qualitative Analysis of Anions

Introduction

In this experiment, you w i l l learn to test for seven different anions (Cl~, Br", I " , N0 3 " , N0 2 ~, S0 4

2 ~, and C 0 3

2 " ) by chemical analysis. In general, anion tests are subject to interference, and each test should be performed on a fresh por t ion of the original sample unless you are instructed otherwise. Directions as to how to el iminate any possible interfering ions are given w i t h each test.

You may recall from Experiment 9 that some of these anions can be detected by infrared spectroscopy. This technique can also be applied to a solid sample. However, in a mixture of anions overlapping bands can cause confusion. Moreover, CI". Br", and 1" cannot be detected in this way.

Identification of Anions

Chloride, bromide, and iodide ions. A preliminary test w i l l show whether any of these ions are present. An ammoniacal solution of A g N 0 3 is added, which forms insoluble AgCl , AgBr, and A g l w i th any halide ions present. Of course, if no precipitate forms, then no halides are present.

The A g N 0 3 . s o l u t i o n contains ammonia, so that most of the silver ion is present as A g ( N H 3 ) 2

+ . [Ag + 1 is so small that only the least soluble salts of silver precipitate and A g 2 S 0 4 remains in solution, since

[ A g + l 2 [ S 0 4

2 " | < A ' s p = 2 X 10" 5 (34-1)

The precipitate of AgCl, AgBr, and Agl is dissolved by heating w i th Zn metal, which reduces A g + to Ag and forms soluble zinc halides. For example

2AgCl(s) + Zn •+ 2Ag(s) + Z n 2 + ( a q ) + 2Cl"(aq) (34-2)

The Br" and I" ions are then oxidized one by one to halogens. The reduction potent i a l

I 2 + 2<f -> 21" E° = +0.54 (34-3)

Br 2 + 2e •+ 2Br" E° = +1.06 (34-4)

C l 2 + 2 « f -* 2C1" E ° = + 1 . 3 6 (34-5)

show that I" is most easily oxidized. Thus an oxidizing agent (N0 2 ~) is chosen that can oxidize I" but not Br". The I 2 i s extracted into CC1 4 and identified by its violet color. After adding additional N 0 2 " to be sure that all the I" is removed, Br" is

345

346 Experiment 34

oxidized to Br 2 by Mn0 4 ~, which is not a strong enough oxidizing agent to react w i th CI". B r 2 is also extracted into CC1 4 and identified by its orange-brown color. A i l the remaining Br" is oxidized w i t h additional Mn0 4 " , and the C I " is identified by its format ion of a white AgCl precipitate w i t h A g N 0 3 .

Ni t r i te ion . Ni t r i te (N0 2 " ) is identified by its reaction in acid solution w i th sulfamic acid:

N 0 2 " + N H 2 S 0 3 " » N 2 ( g ) + S 0 4

2 " + H 2 0 (34-6)

The N 2 gas can be seen bubbling off , and the S 0 4

2 " produced can be detected by the addit ion of BaCl 2 , w i th which S 0 4

2 " forms a white precipitate of BaS0 4 . Obviously, any sulfate already present in the original solution would interfere

wi th this test for N 0 2 " . Therefore, any sulfate present is removed by precipitating it wi th BaCl 2 and discarding the precipitate before testing for nitr i te .

Nitrate ion. Nitrate ( N 0 3 " ) is identified by detecting the ammonia produced when it is reduced w i t h Al in basic solut ion:

3 N 0 3 " + 8A1 + 5 0 K T + 1 8 H 2 0 -> 3 N H 3 ( g ) + 8A1(0H) 4 " (34-7)

The N H 3 is detected by allowing it to contact a piece of wet red li tmus, which the the N H 3 turns blue because of the equi l ibr ium:

N H 3 + H 2 0 =^ N H 4

+ + O H " (34-8)

Al is supplied in the form of Devarda's alloy (45% A I , 50% Cu, 5% Zn) . The ammonium ion would interfere w i t h this test by producing N H 3 via the

reverse of the reaction given in Equation (34-8) in basic solution. N 0 2 " is also reduced by Al to N H 3 . If these two ions are present, they are removed before the N 0 3 " test is begun.

Sulfate ion. Sulfate ( S 0 4

2 " ) is easy to detect because it forms a white precipitate o f B a S 0 4 w i th BaCl 2 . No other ions you deal w i th here wi l l interfere wi th this reaction.

Carbonate ion. When C 0 3

2 " is acidified, C 0 2 ( g ) is evolved:

C 0 3

2 " + 2 H 3 0 + ^ C 0 2 ( g ) + 3 H 2 0 (34-9)

If the carbon dioxide is allowed to contact a basic solution that contains B a 2 + ions, a white precipitate of B a C 0 3 forms, which confirms the presence of C0 3

2 ~ .

B a 2 + + 2 0 H " + C 0 2 ( g ) -> BaC0 3 (s ) + H 2 0 (34-10)

If your instructor has supplied a known or an unknown solution, this test can be made direct ly on the solution. Otherwise, it should be made on a por t ion of the solid sample.

347 Qualitative Analysis of Anions


0.2M A g N 0 3 K M n 0 4

BaCl 2 known solid (1 part KBr ; 1 part K I ; 2 parts K N 0 3

Ba(OH) 2 2 parts N a 2 C 0 3 , 1 part KCl) boil ing chips ' N a 2 ( C 0 3 ) anhydrous CC1 4 • N a N 0 2 solid Devarda's alloy 6M NaOH 6 A f H C 2 H 3 0 2 6 M N H 3

6M HCl i^ii-«»****^ r e d litmus paper H N H 2 S 0 3 (sulfamic acid) (solid) Solution A: 3.3/W A g N 0 3

6M H N 0 3 Solution B: ammoniacal ( N H 4 ) 2 C 0 3

3% H 2 0 2 Zn granules 6M H 2 S 0 4

P R O C E D U R E

Preparing a Solution for Analysis

Your instructor may supply known and unknown solutions for anion analysis. If a solid sample is supplied, you must prepare a solution. First, t ry dissolving about 20 mg of the solid in 1 ml of distilled water. If necessary, heat the mixture in a water bath. If all of the sample dissolves, you can prepare a larger quant i ty of solution for anion testing by dissolving 100 mg of the solid in 5 ml of distilled water.

If all of the solid does not dissolve in the preliminary test for solubi l i ty , place 100 mg of the sample in a small flask and add 5 ml of water, 500 mg of anhydrous N a 2 C 0 3 , and a boil ing chip. Boil the contents for at least 10 minutes, occasionally adding water to keep the tota l volume constant. Divide the contents of the flask between two semimicro test tubes and centrifuge. Combine the solutions; wash each precipitate, and add the wash water to the solution.

Combine the precipitates and add 6M H C 2 H 3 0 2 un t i l gas ceases to evolve. I f any residue remains, it must be BaS0 4 , AgCl, AgBr, or A g l . The test outl ine gives procedures for testing the residue.

Only the first solution (the one wi thou t acetic acid) and the residue are needed for the anion tests. The analysis should fol low Outline 8.

348 Experiment 34

349 Qualitative Analysis of Anions

volume of 6/W NaOH, being careful not to wet the upper walls of the tube w i th base. Have a small co t ton plug and a piece of wet, red li tmus paper ready. Add several granules of Devarda's alloy, and insert the cot ton plug about a third of the way into the tube. Bend the litmus paper in to a V-shape and place it in the mouth of the tube (see Figure 34-1). A l low the tube to stand for several minutes. The l i tmus turns blue from evolving N H 3 , indicating the presence oT N 0 3 " .

S0 4

2 ~ test. To 5 drops of the test solution, add BaCl 2 un t i l a precipitate no longer forms. Centrifuge and wash the precipitate. Add a few drops of 6M HC1. The white precipitate of BaS0 4 is insoluble in HC1 and indicates the presence of S 0 4

2 _ .

C 0 3

2 ~ test. To a 30-mg port ion of the original solid (or the solution supplied by your instructor) , add 2 drops of 6/W H 2 S 0 4 . Have ready a medicine dropper containing Ba(OH) 2

solution, and place the dropper, supported by a piece of rubber tubing, in the mouth of the tube (see Figure 34-2). Do not squeeze the bulb of the dropper. The turb id i ty in the dropper is due to B a C 0 3 and indicates the presence of C 0 3

2 - .

Note on Outline 8

1. Zn reduces A g + to the metal and yields a solution that contains Z n 2 + , CI", Brand P, and a gray Ag precipitate. Any insoluble residue that is present after making up the anion test solution can be treated by this method. Silver halides wi l l dissolve, but B a S 0 4 w i l l not. Even i f all the precipitate does not dissolve (indicating the presence of BaS0 4 ) , the solution should be tested for halide ions after combining it w i t h Solution 8 .1 .

350 Experiment 34

R E S U L T S

Submit a 3 X 5 in. card for your unknown, giving your name, desk number, ions present, and ions absent. Have your notebook available for your instructor to examine on request.

R E F E R E N C E

King, Edward J. Ionic Reactions and Separations: Experiments in Qualitative Analysis. New Y o r k : Harcourt Brace Jovanovich, Inc., 1973, pp. 59-106 .

Experiment 35 Analysis of a General Unknown

Introduction

You w i l l be issued a general unknown which may contain any o f the cations in Groups I—V and any of the seven anions in Experiment 34. You should carry out the group separations and ion identifications according to Outlines 1-8 given in Experiments 29 -34 .

If your unknown is supplied in solution, you may use this solution for both cation and anion analyses.

I f your unknown is a solid, you wi l l need to prepare separate solutions for the cation and the anion analyses. Directions for preparing a solution for anion analysis have already been given in Experiment 34.

To prepare the solution for cation analysis test the solubi l i ty of separate small portions of the sample successively in cold and hot portions of H 2 0 , 64/ HC1, 1247 HC1, 647 H N 0 3 , 1647 H N 0 3 , and aqua regia ( in this order). Thoroughly mix 5-10 drops of each solvent w i th about 10 mg of sample, and allow time for a reaction to occur. Centrifuge to determine if a small residue remains undissolved. When concentrated acids are used, dilute the mixture slowly w i t h an equal volume of water after the reaction ceases. Certain salts are only sparingly soluble in concentrated acid. When you find a solvent in the list that w i l l dissolve the solid, do not test any more solvents.

Weigh 100 mg of the sample, and treat it w i t h the solvent you select. If no solvent dissolves the sample completely, use the solvent that produced the greatest effect or, if there is l i t t le difference among the solvents, use aqua regia. ( I f an insoluble residue forms, use the procedure described in the fol lowing discussion.)

If the solution was originally obtained by using one of the acid solvents, evaporate the solution to a volume of a few drops to remove any excess acid. Do not evaporate to dryness, because this may cause decomposition or the formation of basic salts that are d i f f icul t to redissolve. Take about one-fourth of the solution (the equivalent of 25 mg of the original sample) for the systematic analysis of the cations.

If there is a residue insoluble in acid, special treatment is required. The common substances the residue may contain are PbS0 4 , B a S 0 4 , S r S 0 4 , CaS0 4 ( i f a large-amount is present); the ignited sulfates o f F e ( I I I ) , A l ( I I I ) , or C r ( I I I ) ; AgCl, AgBr, A g l ; AUO3, C r 2 0 3 , S n 0 2 , S b 2 0 5 , S i 0 2 , or silicates; C; or S.

First the residue is treated w i th sodium carbonate solution to convert the sulfates to carbonates; the residue is then treated w i t h zinc and acid to reduce the silver halides to metallic silver.

1. Wash the residue twice w i th a moderately dilute solution of the acid used to dissolve the sample. Then suspend the residue in 1 ml of water and add two heaping spatula-fulls of sodium carbonate. Heat the mixture in boil ing water for 10 minutes, stirring frequently. Centrifuge, discard the solut ion, and wash the precipitate three

351

352 Experiment 35

times w i t h hot water to remove the sulfate. Add a few drops of 6M H N 0 3 to the residue and, after the reaction ceases, di lute the mixture wi th several drops o f water. Centrifuge and wi thdraw the solut ion; add it to the acidic solution of the sample to be used in the general analysis. If a residue sti l l remains, repeat the sodium carbonate treatment.

2. The residue from the conversion of the sulfates may be silver salts, oxides, silicates, C, or S. Add 1 ml of water, 1 drop of 1 8M H 2 S 0 4 , and some Zn granules to reduce the silver halides to free silver. Warm for several minutes, and break apart and disperse the residue w i t h a rod. Centrifuge and discard the solution. Wash the precipitate twice. Add several drops of 6M H N 0 3 to dissolve the silver and excess zinc. Identify silver in the solution by precipitat ing the chloride, dissolving it in ammonia, and reprecipitating the chloride w i t h ni t r ic acid (see Outline 1, Experiment 29) .

In your notebook, record your procedures and observations on dissolving the sample.

P R E L I M I N A R Y T E S T S

It is useful to test a solution of the sample w i t h various reagents that react w i th sizable numbers of ions. The formation of a precipitate or the absence of an apparent reaction indicates the presence or the absence of certain groups of ions. The simplici ty of the tests makes them wor th t ry ing , but the results must be confirmed by other tests. For each test, dilute 1 drop of the solution of the sample w i t h a few drops of water to obtain a convenient volume; then add the particular reagent dropwise, mix ing thoroughly after each addi t ion, un t i l no further change is observed.

A. Act ion of Excess Sodium Hydroxide

Add 6M NaOH unt i l an excess is present; the final mixture should be strongly basic (check wi th l i tmus) . Centrifuge to consolidate any precipitate; the white , gelatinous hydroxides of magnesium and aluminum are diff icul t to see when dispersed. If no precipitate remains, the fo l lowing cations are absent:

A « + C u 2 + H g 2

2 +

B i ( l I I ) F e 2 + M g 2 +

C d 2 + F e 3 + M n 2 +

C o 2 + H g 2 + N i 2 +

If there is a precipitate, its color may be significant; examples: blue C o ( O H ) 2 or C u ( O H ) 2 ; reddish-brown F e ( O H ) 3 ; yel low FlgO; brown A g 2 0 . Consult a chemical handbook for the colors of other oxides and hydroxides.

Other observations may or may not be made, depending on the circumstances. Oxides or hydroxides o f the fol lowing cations w i l l precipitate at first, and then redissolve in excess base:

353 Analysis of a General Unknown

A l 3 + P b 2 + Sn(IV) A s ( I I I ) Sb( I I I ) Z n 2 +

C r 3 + S n 2 +

Whether or not you observe this depends on the rate at which the base is added. It is easy to add too much base and overshoot the precipitation step. The chromium hydroxo complex is green:

C r 2 0 3 - 3 H 2 0 ( s ) + 2 0 H " •* 2Cr(OH) 4 - (35-1)

Copper(I l ) and cobal t ( I I ) hydroxides may dissolve if a large excess of base is used. Pink manganese(II) hydroxide w i l l not dissolve, but wi l l darken in t ime, since it oxidizes to brown manganese(lll) oxyhydroxidc :

0 2 + 4Mn(OH) 2 ( s ) -» 4MnO(OH)(s) + 2 H 2 0 (35-2)

B. Act ion of Excess Aqueous Ammonia

Use 15A/ N H 3 ; again, check w i t h l i tmus to be sure that the final solution is strongly alkaline. If no precipitate forms, all cations are absent except

Group A: A g + ; C u 2 + (deep-blue solut ion); C d 2 + ; C o 2 + (ye l low-brown solution); N i 2 + (blue solut ion); Z n 2 + .

Group B: Na + ; K*j NH 4 *.

The cations in Group A form weakly dissociated ammonia complexes:

A g + + 2 N H 3 ^ A g ( N H 3 ) 2

+ (35-3)

C o 2 + + 6 N H 3 +- C o ( N H 3 ) 6

2 + (35-4)

Z n 2 + + 4 N H 3 * - Z n ( N H 3 ) 4

2 + (35-5)

The cations in Group B do not form insoluble hydroxides. The hydroxides of Ba 2 + , Sr 2 + , and C a 2 + are also moderately soluble and should not precipitate, but ammonia solution often absorbs sufficient carbon dioxide from the air to produce small precipitates of the carbonates of these ions. The white, gelatinous hydroxides of magnesium and aluminum are d i f f icul t to see when they are dispersed in solution. Always centrifuge before you decide whether a precipitate is present or absent.

C. Act ion of Sulfuric Acid

Use one drop of 6A/ H 2 S 0 4 , and add several drops of 95% ethyl alcohol to decrease the solubil i ty of the sulfates, particularly calcium sulfate. Stir vigorously

354 Experiment 35

for a minute to overcome supersaturation. If a precipitate forms, Pb 2 + , Ba 2 + , Sr 2 + , or C a 2 + may be present.

Be sure to record the results of any preliminary tests in your notebook.

R E S U L T S

The unknown should be analyzed according to Outlines 1-8 given in Experiments 29 -34 .

Submit a 3 X 5 in . card for your unknown solution, giving your name, desk number, cations and anions present, and cations and anions absent. Have your notebook available for your instructor to examine on request.

R E F E R E N C E

King, Edward J. Ionic Reactions and Separations: Experiments in Qualitative Analysis. New Y o r k : Harcourt Brace Jovanovich, Inc., 1973, pp. 210-26 .

Appendix I Sample Laboratory Report: Density Determination

356 Appendix I

best, this estimation is accurate to ±0 .0001 g (and may be only to ± 0.001 g depending on the type of balance).

The last figure in the volume measurement must also be estimated; this estimation is accurate to ±0.1 ml .

The volumes are measured to only three significant figures; the weights are measured to five significant figures. The calculated densities can therefore be accurate to only three significant figures. This means that the weighings are really more accurate than they need to be. They could have been made on a triple-beam balance to only three significant figures; for these weights, this would have been to an accuracy of ±0.1 g. However, one purpose of the experiment is to gain experience using the analytical balance, which is why the measurements were made on the analytical balance rather than the triple-beam balance.

Considering the inherent uncertainties in the estimation procedures, the weight Of the sulfur sample is actually 24.597 ± 0.002 g and the volume is 1 2.0 ± 0.1 ml . Hence, the max imum value that the density could have would result if the true weight were 24.599 g and the actual volume were 11.9 ml . Then

_ 24.599 g _ n n . , p = - = 2.07 g/ml

11.9 ml

The m i n i m u m value for the density would result if the true weight were 24.595 g and the volume were 12.1 ml . Then

24.595 g - M , . p =———£ = 2.03 g/ml

12.1 ml

Thus, there is an uncertainty of ±0 .02 g/ml in the density of the sulfur. This uncertainty should be expressed as 2.05 ± 0.02 g/ml.

By the same method, the uncertainty of the l iquid density is p = 1.00 ± 0.01 g/ml.

According to the CRC Handbook of Chemistry and Physics, the density of sulfur is 2.07 g/ml. Therefore, the percentage of error in the experimental determination of the density of sulfur is

Correct value - Experimental value ^ Correct value

or

2 07 - 2 05 0 02 ^ — — — X 1 0 0 = ^ ^ X 100= 1 %

2.07 2.07

QUESTIONS

No questions to answer.

Appendix II Graphing Experimental Data

Many scientific experiments consist of measuring what happens to one property of a system as another property is varied. For example, it is possible to observe how the volume V of a sample of gas changes as the pressure P on the gas is increased. The fol lowing data could be obtained:

Pressure Volume (atm) (liters)

1.0 16.0 2.0 8.0 5.0 3.2

10.0 1.6

The data make it clear that the volume decreases as the pressure increases. However, a graph provides a more striking presentation of the data (see Figure I I - 1 , which illustrates just how fast the volume drops wi th increasing pressure).

In an experiment from which such data would be obtained, the experimenter can control the amount of pressure on the gas. Thus, P is said to be an independent variable; thevalues of the independent variable are usually plotted on the horizontal or x axis of a graph. The size of V depends on how large or small the experimenter chooses to make P. Thus, V is a dependent variable; the values of the dependent variable are usually plotted on the vertical ox y axis of a graph. Dots are placed at the points that correspond to measured values of P and V, and these points are connected by a smooth curve. We draw a continuous smooth curve to connect the points (rather than, for example, several straight-line segments) because we recognize that one property should vary smoothly and regularly as some other property on which it is dependent varies. Circles may be drawn around the experimental points

358 Appendix II

FIGURE 11-2

so that the values can be easily located and are not " los t" on the line or mistaken for normal imperfections in the paper. When two or more different sets of data are plotted on one graph, one set of points can be enclosed in small squares or triangles instead of circles, so that it is immediately apparent which data belong together and were used for a particular line.

Because of the smooth variation of a dependent variable wi th an independent variable, a graph also permits us to estimate what the value of one variable should be from a particular value of the other variable wi thout making laboratory measurements. For example, we might choose a particular pressure (say, 4.0 atm) and determine what volume corresponds to the point on the curve at this value of P. For P= 4.0 atm, the corresponding volume is 4.0 liters.

The preceding set o f data is more accurate than the data you wil l usually obtain in the laboratory. Actual laboratory measurements (whether made by you or by a professional chemist) are subject to experimental error. This simply means that in spite of efforts to be as accurate as possible, some random errors inevitably affect a measurement; this usually results from problems encountered in estimating the Final significant figure. For example, some laboratory balances measure a mass to the nearest mil l igram, but the last significant figure must be estimated. In a weighing, the optical scale might appear as shown in Figure I I -2 . Here, the pointer is between 1.02 and 1.03 g, and the experimenter must estimate how close the weight is to 1.02. A good guess would be 1.026. But estimates of 1.025 or 1.027 would also be acceptable; in fact, one of these weights might be recorded if the object were re-weighed.

Thus, the last figure in a weight or in any other measurement is subject to uncertainty, which is called experimental error. Of course, experimental error does not include blunders (such as spilling some of the substance to be weighed) or systematic errors (such as making an incorrect scale reading because the balance is broken). Random experimental errors are unavoidable and remain after accidents, mistakes, and systematic errors have been eliminated. It is useful to note that random errors are just as l ikely to be positive as negative.

Random errors produce a set of data points that are more scattered on the graph than ideal data points would be. Real data points for an experiment in which P and Fare measured for a gas sample might be similar to those plotted in Figure 11-3. The curve drawn is not the wiggly one that would connect all the points. Instead, the best smooth and continuous curve is drawn through the set of points, leaving an almost equal number of points above and below the curve. This counterbalances positive and negative experimental errors and produces a curve that represents an average of the experimental results. , .

359 Graphing Experimental Data

360 Appendix II

Assuming that a linear relationship exists between two properties, a straight line results when the numerical values of one property (say, volume) are plotted along the v axis, called the ordinate, and numerical values of the other property (say, temperature) are plotted along the x axis, called the abscissa. The m is constant for a particular line and is called the slope of the line. The b is also constant for a particular line and is called the y intercept.

The slope of the line shows how rapidly y changes when x is varied. Consider the straight lines shown in Figure I I -5 . The arrows on the axes indicate the directions of increasing values of x and y. It is obvious y increases more rapidly wi th x for line A than for line B; line A has a greater slope than B. For line C, y decreases as x increases; this line therefore has a negative slope. At the point where x = 0 (that is, on the y axis) the general equation for a straight line reduces to

y = (mX0) + b = b (11-2)

The intercept I) is exactly the value of y at which the line crosses the y axis. Note that in all the graphs in this Appendix a particular distance along the .v axis

does not have the same value as the same distance along the y axis. When you first learned to construct graphs in a mathematics class you probably made one square on the graph paper represent the same number of units along each axis. However, this is not at all necessary. A graph should f i l l as much o f the graph paper as possible, so that the experimental data can be plot ted w i th the greatest possible accuracy and so that any characteristic curvature or l inearity w i l l be obvious. To accomplish this, each axis can be scaled in the most convenient manner, perhaps hundreds of units per square along one axis and only a fraction of a unit per square along the other axis. Nor is it necessary for either scale to start at zero. For example, experimental data relating temperature changes to time changes might cover only the temperature range of 7 2 - 1 0 7 ° C . In this case, it might be convenient to scale the temperature axis f rom 7 0 ° C to 1 10°C; if the scale began at 0 °C , most of the graph would be a blank space containing no experimental points. If, on the other hand, the y intercept is to be determined, the x axis must include the point x = 0 and the y axis must extend far enough that they intercept can be found.

In experimental work, values are usually obtained in the laboratory for at least four or five points, these points are plot ted, and the best smooth curve is drawn. If

%

361 Graphing Experimental Data

this curve is a straight line, the constants m and b can be evaluated from the graph. The advantage of doing this is that we then have an algebraic expression which enables us to calculate the value of y corresponding to any x w i thou t having to make actual measurements of these values. Nor is it necessary to retain the graph and read it each t ime such informat ion is desired if the equation for the particular line has been recorded.

The slope of a line tells how rapidly y changes w i th x. To evaluate the slope, we choose two points on the line (say, points I and II on line A shown in Figure I I -6) . At point I, y = 2.1 and x = 1.0; at point I I , y - 4.3 and x = 3.0. Between points 1 and I I , the value of y changes by an amount Ay = 4 . 3 — 2.1 = 2.2, and x changes by an amount Ax = 3.0 - 1.0 —-2.0. The slope of this line is then given by

m = *y = ll=U ( I I -3 ) Ax 2.0

The value calculated for the slope of line A should be the same, regardless of which two values of x and y are chosen on the line. For example, t ry evaluating m using point I I I (y = 5.4; x = 4.0) and point I. Occasionally, using different sets of points to determine a slope may result in slightly different values; this is because of slight errors in reading the values for the points that are used and inaccuracies arising from loss of significant figures in the calculation. To minimize these problems, the two points should be as far apart as possible on the line; for example, the use of points y2 = 27.8 and>', = 1.2, so that Ay = 26.6 (rather than the use of points y2 = 27.8 and y x = 21.6, so that Ay = 6.2) allows for retention of more significant figures. It is also advantageous to choose points on the line that fall exactly at the intersection of lines on the graph paper, so that there is as l i t t le doubt as possible in assigning values.

T h e y intercept b can be easily read from the graph. In Figure I I -6 , the line crosses the y axis (that is, at the point x - 0) at the value y = 1.0; this, by defini t ion, is the y intercept.

Thus, the equation for line A in Figure II-6 is

y= \ .\x+ 1.0 ( I I -4)

362 Appendix II

With this equation, we can calculate the values of y corresponding to any value of x, or vice versa.

We wi l l now use this procedure wi th the straight line obtained earlier by p lo t t ing Vvs. T for a gas sample (Figure 11-4). We choose two points on the line w i t h which to evaluate the slope. Let point I be V- 16.0 liters, T= 2 0 0 ° K . Let point II be V = 36.0 liters, T = 4 5 0 ° K . Note that the points used to calculate m are chosen from the points on the line and are not experimentally determined points that may lie slightly above or below the best straight line. Using these points

Ay AV (36.0 - 16.0) liters , T T c . m=—-= = — (Ho)

Ax AT (450 - 200) degrees

= 2 ^ = 0.0800 liter/degree 250

The slope o f the line w i l l have units i f either or both o f the quantities plot ted along the x and y axes have units. The difference term that has the fewest significant figures l imits the accuracy of the slope. When the line is extrapolated (that is, when the line is extended beyond the region for which actual experimental values are plotted), we see that the point at which the line crosses the y axis is the origin. For this particular line, the intercept is 0 (however, this is often not the case). This means that V = 0 when T - 0. The equation for this line is

V= (0.0800 liter/degree) T+0 ( I I -6)

= (0.0800 liter/degree) T

Appendix III Infrared Spectrophotometers

A spectrophotometer for use in the infrared region consists of a source of infrared radiation (a glower), a monochromator for selecting a single wavelength at a time, and a detector for measuring the intensity of radiation before and after it passes through the sample. In addition, a recording device is provided for displaying the output of the detector.

This Appendix contains photographs and schematic diagrams of some infrared spectrophotometers that may be available for first-year laboratory use. The purpose of presenting this informat ion is to allow interested students to learn how these instruments work. Your instructor w i l l provide you w i t h operating instructions that pertain to the spectrophotometer available in your laboratory.

A schematic diagram of the optical system of the Beckman IR 33 appears in Figure I I I - l ; Figure 111-2 is a photograph o f the instrument itself.

Another spectrophotometer often found in first-year laboratories is the Perkin-Elmer 735, shown in Figure 111-3.

Appendix II I

FIGURE 111-2

Beckman IR 33 Spectrophotometer. (Courtesy of Beckman Instruments, Inc.)

FIGURE 111-3

Tlie Perkin-Elmer Model 735 Infrared Spectrophotometer. (Courtesy of the Vci^in-EÄter Corporation.) ,•];(: , ó»..UvU>'tA

Appendix IV Visible and Ultraviolet Spectrophotometers

The parts of a spectrophotometer for use in the visible and ultraviolet regions of the electromagnetic spectrum are similar in function to those of an infrared spectrophotometer. The principal differences lie in the nature of the light source (a tungsten lamp instead of a glower) and the systems for selecting monochromatic radiation (the grating) and for detecting intensity (phototubes), all of which must generate and interact wi th light of shorter wavelength (higher energy) than infrared.

Your instructor wi l l provide you wi th operating instructions that pertain to the instrument available in your laboratory.

Figure IV-1 is a schematic diagram of the Bausch and L o m b Spectronic 2 0 ® . A photograph of this instrument, which is capable of measuring the spectrum of solutions in the 3400-9500 A region, appears in Figure IV-2 . Various controls are labeled in the photograph.

A slightly more sophisticated instrument, the Bausch and L o m b Spectronic 70®, is shown in Figure IV-3 . The Spectronic 70® is capable of measuring 3250-9250 A and has a range that is further toward the ultraviolet regions of the spectrum than the Spectronic 20®.

The Sargent-Welch Chem Anal® appears in Figure IV-4 . The cover of this instrument may be opened to display its optical system.

365

Appendix IV

Visible and Ultraviolet Spectrophotometers

FIGURE IV-4

The Sargent-Welch Chcm-Anal System. (Courtesy of Sargent-Welch Scientific Company.)

Appendix V Reagents and Constants for Qualitative Analysis

R E A G E N T S O L U T I O N S

Reagent Use Bottle Cone. Method of Preparation

*Acid , acetic B 30 6M Dilute the cone. (17A/) acid. *Acid , hydrochloric C 30 2M Dilute the cone. (12A/) acid. *Acid , hydrochloric C 30 6M Dilute the cone. (12A/) acid. *Acid , hydrochloric ( ' 30gs \2M Use the cone, reagent. *Acid , ni tr ic B 30 6M Dilute the cone. (16A7) acid. *Ac id , ni tr ic C 30gs 164/ Use the cone, reagent. *Ac id , sulfuric A 8 6M Dilute the cone. (18A/) acid. * A c i d , sulfuric ( ' 30gs ISM Use the cone, reagent. * A l u m i n o n C 8 0 . 1 % 1.0 g/liter ammonium salt of

aurintricarboxylic acid. * A m m o n i a B 30 6M Dilute the cone. (15A/) base. * Ammonia C 30 15M Use the cone, reagent. * A m m o n i u m acetate C 8 3M 231.2 g/liter * A m m o n i u m carbonate c 8 See page 369. * A m m o n i u m chloride c 8 satd. 321 g/liter (approx. 6A/) * A m m o n i u m nitrate c 8 o . : . i / 16.0 g/liter * A m m o n i u m oxalate c 8 0.2M 28.4 g ( N H 4 ) 2 C 2 0 4 - H 2 0 per li ter * A m m o n i u m sulfate c 8 \M 132 g/liter * A m m o n i u m thiocyanate ( ' 8 satd. See page 369. : i : Barium chloride li 8 0.2,1/ 48.8 g B a C l 2 - 2 H 2 0 per liter;

store in bottle of borosilicate

'"Barium hydroxide A 8 0.2A/ glass.

Satd. solution [approx. 63 g B a ( O H ) 2 - 8 H 2 0 per l i t e r ] .

*Dimethylg lyoxime C 8 1% 10 g/liter of 95% ethyl alcohol. *] lydrogen peroxide 1? 8 am 3% Use USP solution. *Lead acetate c 8 0.24/ 76.0 g Pb(OAc) 2 - 3 H 2 0 per liter •Magnesia mixture ( ' 8 See page 369. •"Magnesium uranyl

acetate c 8 See page 369. *Mercury(II)chloride c 8 0.1 M 27.2 g/liter *Potassium Chromate c 8 0.5 M 97.1 g/liter * Po t assi u in f e rr o c y a n i d e ( 8 0.2M 84.5 g K 4 F e ( C N ) 6 - 3 H 2 0 per lifer •Potassium hydroxide c 30p 0.5 M 28.0 g / l i t e r VJ o?

\

368

369 Reagents and Constants for Qualitative Analysis

I

I

o

Reagent Use Bottle Cone. Method of Preparation

*Potassium permanganate A 8am 0.02/W 3.2 g/liter *Silver nitrate B 8 am 0.2 A/ 34.0 g/liter *S. and 0. reagent C 8 0.05% 0.500 g p-nitrobenzeneazoresorcinol

M-i, -Sodium cobalt ini tr i te »*3 :

in 1 li ter of 0 .025A/NaOH. M-i, -

Sodium cobalt ini tr i te »*3 : < C SS 0.3 3 M 135 g/liter; keep cool. 1 i ^ U ^ w -*Disodium hydrogen

phosphate ( 8 0.53/ Add 71.0 g anhyd. N a 2 H P 0 4

gradually to H 2 0 ; dilute to 1 liter. *Sodium hydroxide B 30p 6A/ 240 g/liter

Solution A A SSam 3.3 A/ 170 g A g N 0 3 in 300 ml 11,0. Solution B A SS 25-ml satd. ( N H 4 ) 2 C 0 3 + 10 ml

15A/ N l l 3 + 100 ml H 2 0 . "Thioacetamide C 30 13% See below. *T in ( I I ) chloride C 8 See below. *Triethanolamine c 8 20% 20% by volume wi th water.

Key: Asterisks indicate solutions suggested for the reagent k i t . A: solutions used only for anion analysis; B: solutions used in both anion and cation analysis; C: solutions used only in cation analysis. 8: 8-nil size; 30: 30-ml size;SS: side-shelf reagent (250-ml); am: amber; gs: glass-stoppered; p: polyethylene. Although 8 ml is an ample supply of most reagents, refill stock solutions should be available. Like side-shelf reagents, stock solutions may be kept in 250-ml bottles fitted wi th droppers. Unless otherwise indicated, the weight in the last column of the table is dissolved in distilled water and diluted to a final volume of 1 liter.

S O L U T I O N S : D I R E C T I O N S AND C O M M E N T S

Ammonium carbonate: Dissolve 192 g powdered reagent in 500 ml of water and 80 ml 15A/ N H 3 . Dilute to 1 liter.

Ammonium thiocyanate: Saturate a mixture of equal volumes of ethyl ether ami isoamyl alcohol w i t h solid N H 4 S C N .

Magnesia mixture: Dissolve 130 g M g ( N 0 3 ) 2 - 6 H 2 0 and 240 g N H 4 N 0 3 in 500 ml water. Add 150 ml 15A/ N H 3 , and dilute to 1 liter. Store in a polyethylene bottle.

Magnesium uranyl acetate: Dissolve 30 g U 0 2 ( C 2 H 3 0 2 ) 2 - 2 H 2 0 in a mixture of 120 ml of water and 100 ml glacial acetic acid. Also prepare a solution of magnesium acetate by adding 148.5 g M g ( C 2 H 3 0 2 ) 2 - 4 I I 2 0 slowly to a hot mixture of 40 ml of water and 320 ml glacial acetic acid. Mix the magnesium and uranyl solutions while they arc still warm, and allow the mixture to stand overnight. Filter through cotton or sintered glass.

Thioacetamide: Dissolve 130 g in a liter of water; the solution is almost saturated. After being stored for more than a year in a clear glass bott le , the solution is still effective in precipitating sulfides. Decomposition occurs on standing (indicated by the precipitation of sulfur); this is not harmful. More serious is the gradual formation of sulfate. If the reagent gives a precipitate w i t h barium chloride, a fresh olut ion should be prepared. — Tin(II) chloride: Let 45.0 g SnCl 2 - H 2 0 stand w i t h 1 70 ml 1 2M HC1 unt i l the SfiHipS disir grate. Dilute slowly to 1 liter. Keep t in shot in the solution.

370 Appendix V

371 Reagents ¿nd Constants for Qualitative Analysis

372 Appendix V

•ont il. C i

373 Reagents and Constants for Qualitative Analysis

374 Appendix V

Reaction K

F e 3 + + S C N - = F e ( N C S ) 2 + 1.1 X 10 2

H g 2 + + 4 C r = H g C l 4

2 - 5.0 X 10 1 5

H g 2 + + 4 I " = H g l 4

2 - 1.9 X 1 0 3 0

N i 2 + + 6 N H 3 ( a q ) = N i ( N H 3 ) 6

2 + 2.0 X 10 s

P b 2 + + Cl~ = P b C l + 1 1 Z n 2 + + 4NH 3(aq) = Z n ( N H 3 ) 4

2 + 2.8 X 10 9

G E N E R A L S O L U B I L I T Y R U L E S

The fol lowing generalizations indicate the relative solubilities of common inorganic salts in water. Cation refers to ammonium ion or to any of the monoatomic, common metal ions.

1. The nitrates, chlorates, and acetates of all cations are soluble. Silver acetate is only moderately soluble, however.

2. The chlorides, bromides, and iodides of all cations are soluble, except those of silver and mercury(I) . The halides of lead(II) are only slightly soluble in cold water, but are moderately soluble in hot water. Mercury(II) iodide is insoluble. The water-insoluble halides are also insoluble in dilute acid.

3. The sulfates of all cations are soluble, except those of barium and lead. The sulfates of silver, calcium, and mercury( l ) are slightly soluble. The water-insoluble sulfates are also insoluble in dilute acid.

4. The carbonates, phosphates, borates, chromates, and sulfites of all cations are insoluble, except those of sodium, potassium, and ammonium. Magnesium chromate is soluble, and magnesium sulfite is slightly soluble. Water-insoluble salts of these anions are generally soluble in dilute acid because of anion hydrolysis.

5. The hydroxides and sulfides of all cations are insoluble, except those of sodium, potassium, and ammonium. Those of magnesium, calcium, and barium arc moderately soluble.

6. • A l l sodium, potassium, and ammonium salts are soluble. 7. A l l silver salts are insoluble, except the nitrate and chlorate salts. Silver

acetate is moderately soluble. 8. A n t i m o n y and bismuth salts hyd ro l ' z e . i t e r . dv i iluble basic salts;

however, these salts are soluble in In e aci 9. I r o n ( I I I ) , lead(II) , t i n ( I I ) , t in ( IV. ~ r cu ry ( I I ) , coppe.(L ), ?r:}. a luminum

tend to hydrolyze in neutral or i i ; , . i n l y b sic solw.ic ~.s 1 3 givs insoluble hydroxides or basic salts; these SL -S are soluble in - i i re iX'yl

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