Chemistry 102(01) Spring 2012

15-1CHEM 102, Spring 2012 LA TECH

CTH 328 9:30-10:45 am

Instructor: Dr. Upali Siriwardane

e-mail: [email protected]

Office: CTH 311 Phone 257-4941

Office Hours: M,W 8:00-9:00 & 11:00-12:00 am;

Tu,Th,F 8:00 - 10:00 am..

Exams: 9:30-10:45 am, CTH 328.

March 26 , 2012 (Test 1): Chapter 13

April 18 , 2012 (Test 2): Chapter 14 &15

May 14 , 2012 (Test 3): Chapter 16 &18

Optional Comprehensive Final Exam: May 17, 2012 :

Chapters 13, 14, 15, 16, 17, and 18

Chemistry 102(01) Spring 2012


Chapter 15. The Chemistry of Solutes and Solutions 15.1 Solubility and Intermolecular Forces 15.2 Enthalpy, Entropy, and Dissolving Solids 15.3 Solubility and Equilibrium 15.4 Temperature and Solubility 15.5 Pressure and Dissolving Gases in Liquids:

Henry's Law15.6 Solution Concentration: Keeping Track of Units15.7 Vapor Pressures, Boiling Points, and Freezing Points

of Solutions15.8 Osmotic Pressure of Solutions15.9 Colloids 15.10 Surfactants15.11 Water: Natural, Clean, and Otherwise


Solution Concentration Units a) Molarity (M) b) Molality (m) c) Mole fraction (Ca) d) Mass percent (% weight) e) Volume percent (% volume) f) "Proof" g) ppm and ppb


1) Define following solution concentration units:a) Molarity (M) b) Molality (m)

c) Mole fraction (Ca) d) Mass percent (% weight)

e) Volume percent (% volume) f) "Proof" g) ppm and ppb


MolarityThe number of moles of solute per liter of solution.

molarity M moles of solute

M = liter of solution

units molar = moles/liter = M


An aqueous solution 58.5 g of NaCl and 2206g water has a density of 1.108 g/cm3. Calculate the Molarity of the solution.

58.5 g 1 mole Solution volume 58.5 g + 2206 g in L

1.00 mole NaCl Molarity of NaCl solution = ------------------------- = 0.489

M 2.044 L solution

2264.5 g solution 1 cm3 solution 1 L solution= 2.044 L solution

1.108 g solution 1000 cm3 solution

Molarity Calculation


Molalitynumber of moles of solute particles (ions or

molecules) per kilogram of solvent#moles solute

m = #kilograms of solvent

Calculate the molality of C2H5OH in water solution which is prepared by mixing 75.0 mL of C2H5OH and 125 g of H2O at 20oC. The density of C2H5OH is 0.789 g/mL.


125 g of H2O = 0.125 kg H2O

1.284 mole C2H5OH Molality(m) = ------------------------ = 10.27 m 0.125 kg H2O

75.0 mL C2H5OH 0.789 g C2H5OH 1 mole C2H5OH= 1.284 C2H5OH

1 mL 46.08 g C2H5OH

Molarity Calculation


Mole Fraction#moles of component i

Xi = total number of moles

Calculate the mole fraction of benzene in a benzene(C6H6)-chloroform(CHCl3) solution which contains 60 g of benzene and 30 g of chloroform.M.W. = 78.12 (C6H6) M.W. = 119.37 (CHCl3)


moles of a na Mola Fraction(ca) = ------------------- = -------------- moles of na + moles nb na + nb a = C6H6 b = CHCl3 nC6H6 Mola Fraction(ca) = ------------------ nC6H6 + nCHCl3 m.w (C6H6) = 78.12 g/mole m.w (CHCl3) = 119.37 g/mole 60/78.12 = 0.768 mole C6H6 30/119.37 = 0.251 mole CHCl3

ca(C6H6) = 0.768/ 0.786+ 0.251 =0.754 Ca(CHCl3) = 0.0.251/ 0.786+ 0.251 = 0.246

1.000

Mole Fraction Calculation


Weight Percent #g of solute

wt % = 102

#g of solution

Volume Percent #L of solute

Vol % = 102

#L of solution

Proof

proof = Vol % x 2


Problem

What is the mole fraction of ethanol, C2H5OH, in ethanol solution that is 40.%(w/w) ethanol, C2H5OH, by mass?

a. 0.40 b. 0.46 c. 0.21 d. 0.54


Parts per Million #g of solute #mg of solute

ppm = 106 = #g of solution #kg of solution

#mL solute ppm =

#L of solution

Parts per Billion

#g of solute #micro-g of solute

ppb = 109

=

#g of solution #kg of solution


ppm and ppb conversions1 ppm = (1g/ 1x 106g) 1x 106

= (1/1000 g) x 1x 106

1x 106 / 1000g = mg/ 1x 103 g = mg/ L1 ppb = (1g/ 1x 109g) 1x 109

= (1/1000000 g) 1x 109/1000000g = mg/ 1x 103 g = mg/ L


A solution of hydrogen peroxide is 30.0% H2O2 by mass and has a density of

1.11 g/cm3. The MOLARITY of the solution is:

a) 7.94 M b) 8.82 M c) 9.79 M d) 0.980

e) none of these

M.W. = 34.02 (H2O2)

Problem


Comparison of Concentration Terms


Effect of Solutes on SolutionColligative Properties

Colligative Properties: Depend on the number of particles not on the identity of the particles

Solution Colligative Propertiesa) Vapor Pressure Loweringb) Freezing Point Depressionc) Boiling Point Elevationd) Osmotic Pressure

Two types of solutes affect colligative properties differentlya) Volatile solutes (covalent)b) nonvolatile solutes (ionic)


Vapor Pressure ofPure Water vs. Sea Water


Vapor Pressure LoweringRaoult’s Law

P1 = X1P1o

Psol = csolvent Psolvent

Psol = (1-csolute) Psolvent

The vapor pressure above a glucose-water solution at 25oC is 23.8 torr. What is the mole fraction of glucose (non-dissociating solute) in the solution. The vapor pressure of water at 25oC is 30.5 torr.


Vapor Pressure Lowering


Effect on Boling and Freezing point


Boiling Point Elevation


Boiling Point ElevationDTb = Tfinal - Tinitial

(DTb = bpsolution - bppure solvent)DTb = kb x mwhere kb => boiling point elevation constant

m => molality of all solutes in solutionFreezing Point Depression

(DTf = fppure solvent - fpsolution)

DTf = kf x m

where kf => freezing point depression constant

m => molality of all solutes in solution

For electrolytes multiply

i => number of particles per formula unit


Boiling Point Elevation & Freezing point Depression Constants


What is the freezing point of a 0.500 m aqueous solution of glucose? (Kf for H2O is 1.86 oC/m) (DTf = fppure solvent - fpsolution)DTf = kf x m

Freezing Point Depression Problem


Calculation of Molecular Weight

A 2.25g sample of a compound is dissolved in 125 g of benzene. The freezing point of the solution is 1.02oC. What is the molecular weight of the compound? Kf for benzene = 5.12 oC/m, freezing point = 5.5oC. DTf = kf x mm = moles/ kg of solventMW = g/moles


Solvent Freezing


Colligative Properties ofElectrolytesNumber of solute particles in the solution depends

on dissociation into ions expressed as Van’t Hoff facotor(i)

Van’t Hoff facotor (i) moles of particles in solution moles of

solutes dissolved


Colligative Properties of ElectrolytesIonic vs. covalent substances

vpwater > vp1M sucrose > vp1M NaCl > vp 1M CaCl2

1 mole sucrose = 1 mole molecules (i = 1)1 mole NaCl = 2 mole of ions (i = 2)1 mole CaCl2 = 3 moles ions (i = 3)

i => number of particles per formula unitPsol = (1- i csolute) Psolvent

DTf = i kf x mDTb = i kb x mP = i MRT


Osmosis


Measuring Osmotic Pressure


Osmosis and the Cell


Osmotic Pressure

P = MRTiwhere P => osmotic pressure

M => concentration R => gas constant T => absolute Kelvin temperature

i => number of particles per formula unit


Calculate the osmotic pressure in atm at 20oC of an aqueous solution containing 5.0 g of sucrose (C12H22O11), in 100.0 mL solution.M.W.(C12H22O11)= 342.34P = MRT R = 0.0821 L-atm/mol K = 62.4 L-torr/mol K

Calculation


Calculate the osmotic pressure in torr of a 0.500 M solution of NaCl in water at 25oC. Assume a 100%dissociation of NaCl.

Calculation


Define the Van't Hoff factor (i). Which of the following solutions will show the highest osmotic pressure: a) 0.2 M Na3PO4 b) 0.2 M C6H12O6 (glucose)c) 0.3 M Al2(SO4)3 d) 0.3 M CaCl2 e) 0.3 M NaCl

Which one has higher Osmotic Pressure


Normal vs. Reverse Osmosis


Deviations from Raoult’s LawIntermolecular forces between components in a

dissolved solution cause deviations from the adjustment to vapor pressure.

Vapo

r Pre

ssur

e

cA

Pvap A

Pvap B


Ideal, Negative, Positive Behavior of Vapor Pressure

of Two Volatile Liquids


Predict the type of behavior (ideal, negative, positive) based on vapor pressure of the following pairs ofvolatile liquids and explain it in terms of intermolecular attractions: a) Acetone/water(CH3)2CO/H2Ob) Ethanol(C2H5OH)/hexane(C6H14) c) Benzene (C6H6)/toluene CH3C6H5.

Ideal, Negative, Positive Behavior of Vapor Pressure


Acetone/water(CH3)2CO/H2O


Ethanol(C2H5OH)/hexane(C6H14)


Benzene (C6H6)/toluene CH3C6H5


a) True solutionsb) Colloids (Tyndall effect)c) Suspensions.

Types of Solutions


Solution vs. Dispersion vs. SuspensionSmaller particles => Larger particles

Colloidal True solution dispersion Suspension

Particles Ions & molecules Colloids Large-sized particles

Particle size 0.2-2.0 nm 2-2000 nm >2000 nmProperties * Don’t settle out * Don’t settle out * Settle out on

on standing on standing on standing* Not filterable * Not filterable * Filterable

Example Sea water Fog River silt


Tyndall Effect


Surfactants


Soaps and Detergents

CH 3CH 2CH2CH 2CH 2CH 2CH2CH 2CH 2CH 2CH2CH 2CH 2CH 2CH2CH 2CH 2

Hydrophobic end Hydrophilic end

sodium stearate

C O-Na+

O

CH 3CH 2CH2CH 2CH 2CH 2CH2CH 2CH 2CH 2CH2CH 2OS O-3Na +

sodium lauryl sulfate


Cleaning Action


Earth’s Water Supply


Treatment of Drinking Water


Hard Waternatural water containing relatively high

concentrations of Ca+2, Mg+2, Fe+3, or Mn+2 cations and CO3

-2 and HCO3-1 anions


Common HazardousHousehold Chemicals

Chemistry 102(01) Spring 2012

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