This is optional for Chem 110. - Pennsylvania State...

Chem 110 Concentration Dilution Module

1

Instructions This is optional for Chem 110. This module is designed to provide students with the framework needed to solve a variety of problems. Specifically, this should help students to successfully complete the Concentration Dilution Skill Check Test. You can work on it alone or with a partner. If you have more than two people working together, the value of the effort tends to diminish. We strongly encourage you to do this alone or with only one partner. NOTE: the estimated time need to complete this module is ~ 60-75 minutes. There is no “quick and easy” way to learn how to do this. If you put in the time intentionally, it will save you a lot of time in the long run! Please do the module completely. If you skip sections of the module, it is more likely that you will have problems understanding the intention in later sections. If you are not sure how to proceed, please re read the instructions carefully. Working with another student may also be helpful in this respect. No answer key is available and it will not be made available. Please do NOT e-mail your instructor or TA for the answers. If you need help or want to check your answers, do your best to complete the questions and then bring your work to the Chemistry Resource Room (211 Whitmore) during Chem 110 TA office hours. The TAs will be happy to go over the material with you and answer your questions. Finally, there are several fully worked out questions at the end. After working through the module, they may be helpful in answering questions if you still have any. However, keep in mind that the purpose of the worked out problems is to illustrate the method. READING or IMITATING the solutions of problems will not help you.


2

Concentration and Dilution Activity 1: Solvation at the Molecular Level Explore the solvation process using a strong electrolyte and a nonelectrolyte as our examples. Example 1: NaCl (Table Salt) NaCl is an ionic compound that dissolves and completely dissociates in water.

Q1: Based on the picture and your chemistry knowledge, is NaCl a strong electrolyte or

nonelectrolyte? ______________________________________

Q2: What ions and/or molecules are present in a solution of NaCl and water?

___________________________________________________________

Write a balanced chemical equation for this dissolution process.

water

_____________________ → _________________________


3

There are several different acceptable ways a balanced chemical equation can be written. For example, to express the process of magnesium sulfate (MgSO4) dissolving in water, one could write

MgSO4 (s) + H2O (l) Mg2+ (aq) + SO42(aq)

Here liquid water (as noted by the (l) following water) is included as one of the reactants, or the components found on the left-hand side of the arrow. Another acceptable way to write out this process is

MgSO4 (s) Mg2+ (aq) + SO42(aq)

Notice the water is not shown in this case, but it is known to be present because the magnesium sulfate is now present in the aqueous form (indicated by the (aq) notation). Remember the definition of an aqueous solution is a solute dissolved in water.

Typically we will use the second form when writing reactions that correspond to the dissolution process.


4

Concentration and Dilution Activity 1: Solvation at the Molecular Level Example 2: C6H12O6 (glucose) C6H12O6 is a molecular compound that completely dissolves but does not dissociate in water.

Q3: Looking at the picture and the process represented, is glucose a strong electrolyte or

nonelectrolyte? ___________________________________________________

Q4: What ions and/or molecules are present in a solution of glucose and water?

________________________________________________________________

.

Write a balanced chemical equation for this dissolution process.

water

_____________________________ → _____________________________________


5

Concentration and Dilution Activity 1: Solvation at the Molecular Level

Let’s now compare the drawings for NaCl and C6H12O6.

Q5: Give one key similarity between these two compounds when water is added.

Give one key difference between these two compounds when water is added. (How are strong

electrolytes different from nonelectrolytes?)


6

Concentration and Dilution Activity 2: Relative Ion Concentrations

Now that we are more familiar with what strong electrolytes and non-electrolytes look like on the molecular level, let’s use this knowledge to investigate relative concentrations of both ionic and molecular species in a solution. GIVEN STOCK SOLUTION:

1 mole of calcium chloride (CaCl2) is dissolved in 1,000 mL of water.

Like NaCl from Activity 1, calcium chloride is a strong electrolyte, meaning it breaks up into

ions upon dissolving.

STEP 1: Write a balanced chemical equation for calcium chloride dissolving in water.

Use Activity 2 as a guide.

For 1 calcium chloride, how many ions are present after dissociation?

What are these ions? (Give chemical symbol, charge, and quantity)

STEP 2: What is the calcium chloride stock solution concentration?

STEP 3: (a) What is the concentration in molarity of Ca2+(aq) in this solution? (Show work)

(b) What is the concentration of Cl (aq)? (Show work)

(c) Do these concentration values differ? Why or why not?


7

Concentration and Dilution Activity 3: Dilution Process We have now seen how molecular and ionic concentrations can be determined from given information of a stock solution. However, it is common to take a small portion of a stock solution and dilute it to make a new solution of a different concentration. We shall explore this process now. GOAL: to find the concentration of a dilute solution GIVEN:

A stock solution Kool Aid® (KA) can be prepared by dissolving 1 packet of Kool Aid® mix with 1

liter of water. (Let’s assume 1 packet contains 10 moles of Kool Aid®)

Stock Solution

MS

VS

STEP 1: Write a balanced chemical equation for Kool Aid® dissolving in water.

(Assume Kool Aid® is a nonelectrolyte and use KA to represent Kool Aid®.)

STEP 2: Calculate the concentration (MS) of the stock solution of Kool Aid®.

1 L


8

Concentration and Dilution Activity 3: Dilution Process 125 mL of this Kool Aid® stock solution is removed and placed into another container. It is then

diluted with water to a final volume of 1 liter.

Solution 1 Solution 2

M1=MS M2

V1 V2

(a) How many moles of Kool Aid®

are contained in 125 mL of stock solution (solution 1)?

(Show work including units)

(b) Using the symbols shown for molarity and volume, write a general equation for the calculation

of moles in a solution.

(c) How does the number of moles of Kool Aid® compare for solution 1 and solution 2?

You should have observed that the number of moles removed from a stock solution equals the number of moles contained in the diluted solution. This concept can be represented by a relationship you should have seen before in your textbook:

M1·V1 = M2·V2

add water


9

Concentration and Dilution Activity 3: Dilution Process

Reminder: our goal is to find the concentration of the dilute solution (M2) STEP 1 was to write the chemical equation for the process that was occurring KA (s) + H2O (l) → KA (aq)

STEP 2 was to get the concentration of the stock solution MS = 10M STEP 3: What information do we know about each Kool Aid® solution?

Organize your data from the previous steps below. Circle the variable that is unknown. Variable Value (Units) Word Description M1=MS 10 M Concentration of the stock solution

V1

n1 =M1V1

M2

V2

n2=M2V2

STEP 4: Using your organized data from STEP 3, solve for the concentration of the dilute

solution. (Show all of your work before continuing.)


10

Concentration and Dilution Activity 4: Putting it All Together You have now seen questions relating to electrolyte strength, the solvation process, molecular and ionic concentrations, and the dilution process. Now we shall apply several of these concepts at the same time to answer a given question. A 1-L stock solution of AlCl3 has a concentration of 20.0 M. When 100 mL of this stock solution is diluted to 500 mL, what is the concentration of chloride ions (Cl(aq)) in the dilute solution? STEP 1: Write the balanced chemical equation for AlCl3 dissolving in water.

STEP 2: Draw how the diluted solution is being made (refer to the Kool Aid® example in Activity 2 for help if necessary)

M1 = M2 = V1 = V2 = Concentration and volume Concentration and volume of stock soln = [AlCl3] of dilute soln = [AlCl3] STEP 3: What information about each solution do you know? Organize the information

underneath the pictures. Circle the unknown variable.

Did you include units?

STEP 4: Using information from STEP 1 through 3, (a) Determine the [AlCl3] of the dilute solution. (Show ALL work)

Re-read the question: have you answered the question that was ASKED???


11

Concentration and Dilution Activity 4: Putting it All Together To get the [Cl−] the dissociation of AlCl3 must be considered (Step 1). STEP 4 <continued>: (b) Determine the [Al3+] of the dilute solution. (Show ALL work) (c) Determine the [Cl−] of the dilute solution. (Show ALL work)

What if the question asked for the moles of Cl−(aq) in the dilute solution? To answer this, an additional step (STEP 5) is necessary to calculate the number of moles. RECALL moles = M×V

STEP 5: (a) How many moles of aluminum ion (Al3+) are present in the dilute solution?

(b) How many moles of chloride ion (Cl−) are present in the dilute solution?

(c) How does this amount of aluminum and chloride ions in the dilute solution compare to the amount of aluminum and chloride ions removed from the stock solution? (Recall Activity 3)


12

Concentration and Dilution Activity 5: More Practice There are several ways questions relating to concentrations and dilutions can be presented. The same problem solving approach as outlined in Activities 3 and 4, however, still applies.

STEP 1 Write out the chemical process (what is happening on the atomic scale?)

STEP 2 –Draw pictures (what is happening on the macroscopic level?)

STEP 3 – Organize information given in the problem. (Put this information directly under the pictures but be sure it is clear what M refers to, i.e., is M the concentration of salt? Or the concentration of one of the ions?)

STEP 4 – Solve for unknowns by using M1V1 = M2V2 or other relationships

STEP 5 –What are you asked for? The process outlined in step 2 will help you clarify what you want to know. Use the stoichiometry of the chemical process (step 1) or MV = moles to obtain desired answer.

Let’s look at another example question: 10 mL of a stock solution of potassium sulfate (K2SO4) is removed and diluted with water to a final volume of 2 liters. The dilute solution has 0.5 moles of K+ ions. What is the concentration of the stock solution used to make the dilute solution?

STEP 1: Write out a balanced chemical equation for K2SO4 dissolving in water

STEP 2 Draw how the dilute solution is being made and label the pictures with the information given.

V1 = V2 = M1 = M2 = Conc. [K2SO4] dilute [K2SO4] BUT, also know

nK+= Moles of K+ in dilute solution

STEP 3: Organize all of the information given below the pictures. (be sure to include units).


13

Concentration and Dilution Activity 5: More Practice Reminder: this is the original problem we are trying to solve. Be sure to circle or otherwise indicate what you want to know.

10 mL of a stock solution of potassium sulfate (K2SO4) is removed and diluted with water to a final volume of 2 liters. The dilute solution has 0.5 moles of K+ ions. What is the concentration of the stock solution used to make the dilute solution?

Unlike the previous examples, you should notice that a concentration value for the stock solution and the dilute solution are not given directly in the problem. However, you ARE given the moles of potassium ions in the dilute solution. How can this be used to find the concentration? Recall the definition of concentration from Activity 1:

M = moles per liter. STEP 4: Using the information in STEP 1 through 3,

(a) Determine the [K+] for the dilute solution. (b) Is this [K+] the same as the [K2SO4] of the dilute solution? (Hint: Look at STEP 1!) Show why or why not below and calculate [K2SO4] for the dilute solution. (c) Use M1·V1 = M2·V2 to determine the concentration of K2SO4 in the original solution. (Show all work including units. Be sure to answer the question being asked)


14

Concentration and Dilution Activity 6: Practice The following questions apply several of the concepts you have learned from this module. You should be able to answer these questions using both the skills you acquired in the previous Activities as well as your previous chemistry knowledge. When solving these problems, be sure to write out all of the STEPS as shown in the prior Activities. Also, when applicable, circle any “extra” information not needed to solve the problem. Pract Q1: Calculate the concentration of the sodium ions (Na+) in a solution made by diluting 90.0

mL of a 0.750M solution of sodium sulfate (Na2SO4) to a total volume of 450.0 mL. Pract Q2: Determine the number of moles of bromide ions (Br −) in 0.500 L of a 0.300 M solution of

potassium bromide (KBr) that has been diluted with water to a final volume of 1.500 L.


15

Concentration and Dilution Worked Through Examples The module you just completed attempted to cover the basic framework for concentration and dilution problems. However, since there are many different ways you could see these questions being asked, we have provided additional pages to demonstrating how some more advanced problems are solved using the same step-by-step process shown in the module. For ease of understanding, each step to solving the problem has been made a different color. Also, since these problems are more advanced, additional steps beyond Step 5 may be necessary.

STEP 1&2 – Writing out the chemical process and drawing pictures

STEP 3 – Organizing information given in the problem

STEP 4 – Solving for unknowns by using M1V1 = M2V2 or other relationships

STEP 5 – Manipulating information solved for to obtain desired answer

ADDITIONAL STEPS NOT ADDRESSED IN THE MODULE

FEEDBACK FOR CLARIFICATION


16

Concentration and Dilution Worked Through Examples EX1: Determine the number of moles of chloride ions (Cl–) in 1.800 L of a 0.650 M solution

of magnesium chloride (MgCl2) that has been diluted with water to a final volume of 2.500 L.

STEP 1 MgCl2 (s) + H2O (l) Mg2+ (aq) + 2Cl– (aq) STEP 2 Add water V1 = 1.800L V2 = 2.500L M1 = 0.650M M2 = ????

MgCl2 solution MgCl2 solution n (Cl−)= ???

STEP 3

M1 = 0.650 M MgCl2 V1 = 1.800 L V2 = 2.500 L M1 = ??? Moles of Cl−(aq) = ???

STEP 4

M1V1 = n1

0.650 moles MgCl2 × 1.800 L = 1.17 moles MgCl2 1 L

STEP 5

1.17 moles MgCl2 × 2 moles Cl– = 2.34 moles Cl– 1 mole MgCl2

The final volume of 2.500 L the solution was diluted to was NOT necessary to answer the question. This dilution did not change the number of moles present in the 1.800-L of 0.650 M solution. If the question had asked for the new CONCENTRATION (or the amount of moles per unit volume), then this volume would have been important.


17

Concentration and Dilution Worked Through Examples EX2:(a) What is the concentration of potassium ions (K+) present in

980 mL of a 0.45 M solution of potassium sulfate (K2SO4) diluted to 2.0 L? STEP 1 K2SO4 (s) + H2O 2K+ (aq) + SO4

2– (aq)

STEP 2 K2SO4 solution Add water V1 = 980mL V2 = 2.00L M1 = 0.450M M2 = ???? K2SO4 solution K2SO4 solution

[K+]= ??? STEP 3

M1 = 0.45 M K2SO4 V1 = 980 mL (0.980L) V2 = 2,000 mL (2.0L)

STEP 4

M1V1=M2V2

(0.45M) (0.980L) = M2 (2.0L)

M2 = (0.45M) (0.980L) 2.0L M2 = 0.2205 M K2SO4

STEP 5 0.2205 moles K2SO4 × 2 moles K+ ions = 0.4410 moles K+ = 0.4410 M K+

1 L 1 mole K2SO4 1L

The final volume of 2.0 L that the solution was diluted to WAS necessary to answer this question because a concentration was asked for. Recall that concentration is dependent on the volume.


18

Concentration and Dilution Worked Through Examples (b) How many sulfate ions (SO4

2–) ions are present in 500 mL of the dilute solution? STEP 1 K2SO4 (s) + H2O 2K+ (aq) + SO4

2– (aq)

STEP 2 shown on previous page STEPS 3 and 4 (above)

M2 = 0.2205M K2SO4 STEP 5 To get number of sulfate ions, first we need moles of sulfate ions in the dilute solution. Use reaction shown in step 1 0.2205 moles K2SO4 × 1 mole SO4

2– = 0.2205 moles SO42– = 0.2205 M SO4

2–

1 L 1 mole K2SO4 1L This is the concentration of the SO4

2– ion for the ENTIRE dilute solution. However, the question is asking for the number of SO4

2– ions in a specified volume, NOT the entire dilute solution. Moles = M×V

= 0.2205 moles SO42– × 0.500L = 0.1103 moles SO4

2– 1 L

We still have NOT answered the question! We need to use our known relationship between a mole and the number of objects in a mole (Avogadro’s Number = 6.02 × 1023). 0.1103 moles SO4

2– × 6.02 × 1023 SO42– ions = 6.64 × 1022 SO4

2– ions 1 mole SO4

2–


19

Concentration and Dilution Worked Through Examples EX3: (a) 13 mL of a dilute solution of calcium nitrate (Ca(NO3)2) has 1.61× 1023 NO3

– ions. If 0.450 L of a stock solution of calcium nitrate was used to prepare a 1.0 L dilute solution, what is the concentration of the stock solution?

STEP 1 Ca(NO3)2 (s) + H2O (l) Ca2+

(aq) + 2 NO3– (aq)

STEP 2 Ca(NO3)2 solution Add water remove some of dilute soln V1 = 0.450L V2 =1.0L V3 = 13 mL M1 = ??? M2 = ???? M3 = M2 Ca(NO3)2 solution Ca(NO3)2 solution Ca(NO3)2 solution

# of NO3−= 1.61× 1023

STEP 3 M1 = ? M Ca(NO3)2 V1 = 0.450 L M2 = ? M Ca(NO3)2 V2 = 1.0 L # NO3

– ions = 1.61× 1023 V3 = V of dilute examined = 13 mL (0.013L) Unlike previous examples, this question only gives you 2 of the variables in the expression M1V1 = M2V2 . We must use other pieces of information in order to solve for one of the concentrations (either the dilute solution or the stock solution) before we can use this expression. Since we know the number of nitrate ions present in a given volume of dilute solution, we can first determine the number of moles of nitrate ions present in this volume. 1.61× 1023 NO3

– ions × 1 mole = 0.2674 moles NO3– ions

6.02× 1023 NO3–

Using the balanced chemical equation, we can relate the number of moles of nitrate ions to the number of moles of calcium nitrate present in the specified volume of dilute solution. 0.2674 moles NO3

– × 1 mole Ca(NO3)2 = 0.1337 moles Ca(NO3)2 2 moles NO3

–


20

Concentration and Dilution Worked Through Examples Reminder: Problem being solved EX3: (a) 13 mL of a dilute solution of calcium nitrate (Ca(NO3)2) has 1.61× 1023 NO3

– ions. If 0.450 L of a stock solution of calcium nitrate was used to prepare a 1.0 L dilute solution, what is the concentration of the stock solution?

Goal is to get M1 = concentration of the stock solution STEP 1 Ca(NO3)2 (s) + H2O (l) Ca2+

(aq) + 2 NO3– (aq)

STEP 2 Ca(NO3)2 solution dilute soln soln 3 Add water remove some of dilute soln V1 = 0.450L V2 =1.0L V3 = 13 mL M1 = ??? M2 = ???? M3 = M2

# of NO3−= 1.61× 1023

We used # of NO3

− to find moles of Ca(NO3)2 in solution 3 =0.1337 moles Ca(NO3)2 Now we can use the definition of concentration to calculate M3 (concentration of solution 3) M3 = n3 = 0.1337 moles Ca(NO3)2= 10.2862 M Ca(NO3)2

V3 0.0013L Note: M2 = M3 so

M1V1 = M2V2 can be applied. M1V1 = M2V2

M1 (0.450L) = (10.2862M) (1.000L)

M1 = (10.2862M) (1.000L) 0.450 L

M1 = 22.9 M Ca(NO3)2


21

Concentration and Dilution Worked Through Examples (b) What is the total ion concentration of 50 mL of the dilute solution of calcium nitrate? Given/Determined Previously: M1 = 22.9 M Ca(NO3)2 M2 = 10.3 M Ca(NO3)2 V1 = 0.450 L V2 = 1.0L # NO3

– ions = 1.61× 1023 V3 = V of dilute examined = 13 mL (0.013L) Using the information we have, the number of moles of calcium nitrate in 50 mL of the dilute solution can be determined. Concentration of 50mL of dilute solution is the same as the concentration of the full liter of solution (Concentration is an intensive property.) M of 50 mL = M2 = 10.3M Based on the balanced chemical equation, for every 1 mole of Ca(NO3)2, 3 ions (1 Ca2+

and 2 NO3

–) result. [ions] = 10.3M Ca(NO3)2 × 3 ions = 30.9 M ion concentration 1 mole Ca(NO3)2

This is optional for Chem 110. - Pennsylvania State...

Documents

Transcript of This is optional for Chem 110. - Pennsylvania State...