Properties of Solutions Chapter 11. Solutions... the components of a mixture are uniformly intermingled (the mixture is homogeneous).

Properties of Solutions

Chapter 11

Solutions

. . . the components of a mixture . . . the components of a mixture are uniformly intermingled are uniformly intermingled (the mixture is (the mixture is homogeneoushomogeneous).).

Solution Composition

1.1. Molarity (Molarity (MM) =) =

2.2. Mass (weight) percent =Mass (weight) percent =

3.3. Mole fraction (Mole fraction (AA) =) =

4.4. Molality (Molality (mm) =) =

moles of soluteliters of solution

m ass o f so lu te

m ass o f so lu tio n1 0 0 % )(

molestotal moles in solution

A

moles of solutekilograms of solvent

Molarity Calculations

Mg

mol

L

mL

mL

gethanol215.0

07.46

1

1

1000

101

00.1

Mass % Calculations

OHHC

g

g

onmasssoluti

lmassethanoOHHC

52

52

%990.0

%1000.101

00.1

%100%

Mole Fraction

00389.0

58.5

1017.2

56.51017.2

1017.2

2

2

2

2

x

molmolx

molx

nnnonethanolMoleFracti

OHethanol

ethanol

Molality Calculations

mg

mol

kg

g

g

gethanol217.0

07.46

1

1

1000

0.100

00.1

Molarity & Molality

For dilute solutions, molarity (M) and For dilute solutions, molarity (M) and molality(m) are very similar.molality(m) are very similar.

In previous example, M = 0.215 M and In previous example, M = 0.215 M and m = 0.217 m.m = 0.217 m.

Normality

Acid-Base Equivalents = (moles) (total (+) Acid-Base Equivalents = (moles) (total (+) charge)charge)

Redox Equivalents = (moles)(# eRedox Equivalents = (moles)(# e-- transferred) transferred)

ionLitersolut

ssoluteequivalentNNormality )(

Normality Calculations

.250 M H3PO4 =______N

N = M(total(+) charge)

N = (0.250)(3)

N = 0.750 N H3PO4

Concentration & Density Calculations

See Example 11.2 on pages 517-518.

Know how to do this problem!!

Steps in Solution Formation

Step 1 -Step 1 - Expanding the solute (endothermic)Expanding the solute (endothermic)

Step 2 -Step 2 - Expanding the solvent (endothermic)Expanding the solvent (endothermic)

Step 3 -Step 3 - Allowing the solute and solvent to Allowing the solute and solvent to interact to form a solution (exothermic)interact to form a solution (exothermic)

HHsolnsoln = = HHstep 1step 1 + + HHstep 2step 2 + + HHstep 3step 3

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Expanded solute

Solute Solvent

Expanded solvent

Step 1 Step 2

Step 3

Directformationof solution

H3

H1

Hsoln

H2

Three steps of a liquid solution: 1) expanding the solute,2) expanding the solvent, & 3) combining the expandedsolute and solvent to form the solution.

a) Hsoln is negative and solution process is exothermic.b) Hsoln is positve and solution process is endothermic.

Processes that require large amounts of energy tend not to occur. Solution process are favored by an increase in entropy.

Structure & Solubility

Like dissolves like.Like dissolves like.

Hydrophobic --water-fearing. Fat soluble Hydrophobic --water-fearing. Fat soluble vitamins such as A, D, E, & K.vitamins such as A, D, E, & K.

Hydrophilic --water-loving. Water soluble Hydrophilic --water-loving. Water soluble vitamins such as B & C.vitamins such as B & C.

Hypervitaminosis--excessive buildup of vitamins Hypervitaminosis--excessive buildup of vitamins A, D, E, & K in the body.A, D, E, & K in the body.

Henry’s Law

PP = = kkCC

PP = = partial pressure of gaseous solute above the partial pressure of gaseous solute above the solutionsolution

CC = = concentration of dissolved gasconcentration of dissolved gaskk = = a constanta constant

The amount of a gas dissolved in a solution is The amount of a gas dissolved in a solution is directly proportional to the pressure of the gas directly proportional to the pressure of the gas above the solution.above the solution.

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0

20

60

100

140

180

220

260

300Sugar(C12H22 O11)

KNO 3

NaNO 3

NaBr

KBr

KClNa 2SO4

Ce2 (SO 4)3

0 20 40 60 80 100

Temperature (C)

Sol

ubi

lity

(g s

olu

te/1

00 g

H2O

)

Solubility of several solids as a function of temperature.

The solubility of various gases at differenttemperatures.

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WaterVapor

Water Aqueoussolution

Aqueoussolution

(a) (b)

When an aqueous solution and pure water are in a closedenvironment, the water is transferred to the solution because of the difference in vapor pressure.

Raoult’s Law

PPsolnsoln = = solventsolventPPsolventsolvent

PPsolnsoln = vapor pressure of the solution= vapor pressure of the solution

solventsolvent = mole fraction of the solvent = mole fraction of the solvent

PPsolventsolvent = vapor pressure of the = vapor pressure of the purepure solvent solvent

The presence of a nonvolatile solute lowers The presence of a nonvolatile solute lowers the vapor pressure of a solvent.the vapor pressure of a solvent.

Raoult’s Law Calculations

Sample Exercise 11.6 on page 532.Sample Exercise 11.6 on page 532.

NaNa22SOSO4 4 forms 3 ions so the number of moles of forms 3 ions so the number of moles of solute is multiplied by three.solute is multiplied by three.

PPsolnsoln = = waterwaterPPwater water

Psoln = (0.929)(23.76 torr)

PPsolnsoln = = 22.1 torr

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Vapor pressure of pure B

Vapor pressureof pure A

Vap

or

pre

ssure

B

Vapor pressureof solution Vapor pressure

of solution

(a) (b) (c)

AB

AB

A

Vapor pressure for a solution of two volatile liquids.a) Ideal(benzene & toluene) -- obeys Raoult’s Law, b) Positive deviation (ethanol & hexane) from Raoult’s Law, & c) Negative deviation (acetone & water). Negative deviation is due to hydrogen bonding.

Liquid-Liquid Solutions

PPtotaltotal = P = PAA + P + PBB

= = AAPPooAA + + BBPPoo

BB

Raoult’s Law CalculationsSample Exercise 11.7 on page 535.Sample Exercise 11.7 on page 535.

AA= n= nAA/(n/(nAA+n+nCC))

AA= 0.100 mol/(0.100 mol + 0.100 mol)= 0.100 mol/(0.100 mol + 0.100 mol)

A A = 0.500 = 0.500 CC = 0.500 = 0.500

PPtotaltotal = = AAPPooAA + + CCPPoo

CC

PPtotaltotal = (0.500)(345 torr) + (0.500)(293 torr) = (0.500)(345 torr) + (0.500)(293 torr)

PPtotaltotal = = 319 torr

Colligative Properties

Depend only on the number, not on the Depend only on the number, not on the identity, of the solute particles in an ideal identity, of the solute particles in an ideal solution.solution.

- Boiling point elevationBoiling point elevation

- Freezing point depressionFreezing point depression

- Osmotic pressureOsmotic pressure

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atm

Pre

ssur

e (a

tm)

Tf Tb

Freezingpoint ofsolution

Freezing pointof water

Boiling pointof water

Boiling pointof solution

Temperature (C)

Vapor pressureof pure water

Vapor pressureof solution

Phase diagrams for pure water and for an aqueous solution containing a nonvolatile solute -- liquid rangeis extended for the solution.

Boiling Point ElevationA nonvolatile solute elevates the boiling point of the A nonvolatile solute elevates the boiling point of the solvent. The solute lowers the vapor pressure of the solvent. The solute lowers the vapor pressure of the solution.solution.

TT = = KKbbmmsolutesoluteii

KKbb = molal boiling point elevation constant = molal boiling point elevation constant

mm = molality of the solute = molality of the solute

i i = van’t Hoff factor ( # ions formed)= van’t Hoff factor ( # ions formed)

Boiling Point Calculations

Sample Exercise 11.8 on page 537.Sample Exercise 11.8 on page 537.

TT = = KKbbmmsolutesoluteii

mmsolute solute = = T/(T/(KKbbi)i)

mmsolute solute = (0.34 C= (0.34 Coo)/[(0.51 C)/[(0.51 Cookg/mol)(1)]kg/mol)(1)]

mmsolute solute = 0.67 mol/kg= 0.67 mol/kg

Boiling Point Calculations(Continued)

mmsolutesolute = n = nsolutesolute/ kg/ kgsolventsolvent

nnsolute solute = = mmsolutesolute kg kgsolventsolvent

nnsolute solute = (0.67 mol/kg)(0.1500 kg)= (0.67 mol/kg)(0.1500 kg)

nnsolute solute = = 0.10 mol0.10 mol

Boiling Point Calculations(Continued)

n = m/M

M = m/n

M = 18.00 g/0.10 mol

M = 180 g/mol

Freezing Point Depression

A nonvolatile solute depresses the freezing point A nonvolatile solute depresses the freezing point of the solvent. The solute interferes with crystal of the solvent. The solute interferes with crystal formation.formation.

TT = = KKffmmsolutesoluteii

KKff = molal freezing point depression constant= molal freezing point depression constant

mm = molality of the solute = molality of the solute

ii = van’t Hoff factor ( # ions formed) = van’t Hoff factor ( # ions formed)

Freezing Point Calculations

Sample Exercise 11.10 on page 539.Sample Exercise 11.10 on page 539.

TT = = KKffmmsolutesoluteii

mmsolute solute = = T/(T/(KKffi)i)

mmsolute solute = (0.240 C= (0.240 Coo)/[(5.12 C)/[(5.12 Cookg/mol)(1)]kg/mol)(1)]

mmsolute solute = 4.69 x 10= 4.69 x 10-2-2 mol/kg mol/kg

Freezing Point Calculations(Continued)

msolute = nsolute/ kgsolvent

nsolute = msolute kgsolvent

nsolute = (4.69 x 10-2 mol/kg)(0.0150 kg)

nsolute = 7.04 x 10 -4 mol

Freezing Point Calculations(Continued)

n = m/M

M = m/n

M = .546 g/7.04 x 10-4 mol

M = 776 g/mol

Osmotic Pressure

OsmosisOsmosis: The flow of solvent into the : The flow of solvent into the solution through the semipermeable solution through the semipermeable membrane.membrane.

Osmotic PressureOsmotic Pressure: The excess hydrostatic : The excess hydrostatic pressure on the solution compared to the pressure on the solution compared to the pure solvent.pure solvent.

Due to osmotic pressure,the solution is diluted by water transferred through the semipermeable membrane. The diluted solution travels up the thistle tube until the osmotic pressure is balanced by the gravitational pull.

Osmosis

The solute particles interfere with the passage of the solvent, so the rate of transfer is slower from the solution to the solvent than in the reverse direction.

a) The pure solvent travels at a greater rate into the solution than solvent molecules can travel in the reverse direction. b) At equilibrium, the rate of travel of solvent molecules in both directions is equal.

Osmotic Pressure

= MRT= MRT

= osmotic pressure (atm)= osmotic pressure (atm)

M = Molarity of solutionM = Molarity of solution

R = 0.08206 Latm/molKR = 0.08206 Latm/molK

T = Kelvin temperatureT = Kelvin temperature

Osmotic Pressure Calculations

Sample Exercise 11.11 on pages 541-542.Sample Exercise 11.11 on pages 541-542.

= MRT= MRT

M = M = /RT/RT

M = (1.12 torr)(1 atm/760 torr)/[(0.08206 M = (1.12 torr)(1 atm/760 torr)/[(0.08206 Latm/molK)(298K)]Latm/molK)(298K)]

M = 6.01 x 10M = 6.01 x 10-5 -5 mol/Lmol/L

Osmotic Pressure CalculationsContinued

Molar Mass = (1.00 x 10 -3g/1.00 mL)(1000 mL/1 L)(1 L/6.01 x 10-5 mol) =

1.66 x 104 g/mol protein

Crenation & LysisCrenationCrenation-solution in which cell is bathed is -solution in which cell is bathed is

hypertonic (more concentrated)-cell shrinks. hypertonic (more concentrated)-cell shrinks. Pickle, hands after swimming in ocean. Meat is Pickle, hands after swimming in ocean. Meat is salted to kill bacteria and fruits are placed in salted to kill bacteria and fruits are placed in sugar solution.sugar solution.

LysisLysis-solution in which cell is bathed is hypotonic -solution in which cell is bathed is hypotonic (less concentrated)-cell expands. Intravenous (less concentrated)-cell expands. Intravenous solution that is hypotonic to the body instead of solution that is hypotonic to the body instead of isotonic.isotonic.

If the external pressure is larger If the external pressure is larger than the osmotic pressure, than the osmotic pressure, reverse osmosis reverse osmosis occurs.occurs.

One application is One application is desalination of desalination of seawaterseawater..

Colligative Properties of Electrolyte Solutions

TT = = mKimKi

= = MRTiMRTi

i = moles of particles in solution

moles of solute dissolved

van’t Hoff factor, “van’t Hoff factor, “ii”, relates to the number of ”, relates to the number of ions per formula unit.ions per formula unit.

NaCl = 2, KNaCl = 2, K22SOSO44 = 3 = 3

Electrolyte SolutionsThe value of The value of ii is never quite what is expected is never quite what is expected

due to ion-pairing. Some ions stay linked due to ion-pairing. Some ions stay linked together--this phenomenon is most together--this phenomenon is most noticeable in concentrated solutions.noticeable in concentrated solutions.

Osmotic Pressure Calculation for Electrolyte

Sample Exercise 11.13 on page 548.Sample Exercise 11.13 on page 548.

Fe(NHFe(NH44))22(SO(SO44))2 2 produces 5 ions.produces 5 ions.

== MRTiMRTi

ii//MRTMRT

i i == 10.8 atm/[(0.10 mol/L)(0.08206 10.8 atm/[(0.10 mol/L)(0.08206 Latm/molK)(298 K)]Latm/molK)(298 K)]

i i == 4.44.4

Colloids

Colloidal Dispersion (colloid)Colloidal Dispersion (colloid): A suspension : A suspension of tiny particles in some medium.of tiny particles in some medium.

aerosols, foams, emulsions, solsaerosols, foams, emulsions, sols

CoagulationCoagulation: The addition of an electrolyte, : The addition of an electrolyte, causing destruction of a colloid. Examples causing destruction of a colloid. Examples are electrostatic precipitators and river deltas.are electrostatic precipitators and river deltas.

The eight types of colloids and examples of each.

Tyndall EffectThe scattering of light by particles of a colloid is called The scattering of light by particles of a colloid is called

the Tyndall Effect. Which of the glasses below the Tyndall Effect. Which of the glasses below contains a colloid?contains a colloid?

Calorimeter Problem

Add this problem to the Chapter 11 set of Add this problem to the Chapter 11 set of problems. problems. KNOWKNOW how to work this how to work this problem--show the appropriate formula!!problem--show the appropriate formula!!

When 8.50 g of sodium nitrate is dissolved in When 8.50 g of sodium nitrate is dissolved in 600. g of water, the temperature of the 600. g of water, the temperature of the solution rises 0.817 Csolution rises 0.817 Coo. What is the . What is the molarmolar heat of solution for sodium nitrate?heat of solution for sodium nitrate?