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Chapter 17. Properties of Solutions. Chapter 17: Properties of Solutions. 17.1 Solution Composition 17.2 The Thermodynamics of Solution Formation 17.3 Factors Affecting Solubility 17.4 The Vapor Pressure of Solutions 17.5 Boiling-Point Elevation and Freezing-Point Depression

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chapter 17

Chapter 17

Properties of Solutions

slide2

Chapter 17: Properties of Solutions

17.1 Solution Composition

17.2 The Thermodynamics of Solution Formation

17.3 Factors Affecting Solubility

17.4 The Vapor Pressure of Solutions

17.5 Boiling-Point Elevation and Freezing-Point Depression

17.6 Osmotic Pressure

17.7 Colligative Properties of Electrolytic Solutions

17.8 Colloids

slide3

The left beaker contains copper sulfate solution. In the right beaker, ammonia is being added to create a precipitate of copper(II) hydroxide.

slide5

Definitions for Solutions

Solute- The smaller (in mass) of the components in a solution, the

material dispersed into the solvent.

Solvent - The major component of the solution, the material that the

solute is dissolved into.

Solubility - The maximum amount that can be dissolved into a

particular solvent to form a stable solution at a specified

temperature.

Miscible - Substances that can dissolve in any proportion, so that it is

difficult to tell which is the solvent or solute!

slide7

Hydration

shells around

an aqueous

ion

slide10

Solution Composition

Mass percent:

weight percent Mass percent = x 100%

Mole fraction: symbolized by the Greek letter, chi = X

mole fraction of component A = XA =

Molarity: M ; (chapter 4)

M =

Molality: m

m =

grams of solute

grams of solution

nA

nA + nB

moles of solute

liter of solution

moles of Solute

Kg of solvent

like dissolves like
Like Dissolves Like
  • Polar molecules - dissolve best in Polar solvents.
  • Polar molecules can hydrogen bond with polar solvents, such as water, hence increasing their solubility.
  • Non-polar molecules - dissolve best in non - polar solvents.
  • Hydrocarbons, non - polar molecules, do not dissolve, or mix with water!
slide17

Predicting Relative Solubility's of Substances -I

Problem: Predict which solvent will dissolve more of the given solute.

(a) Sodium Chloride in methanol (CH3OH) or in propanol

(CH3CH2CH2OH).

(b) Ethylene glycol (HOCH2CH2OH) in water or in hexane

(CH3CH2CH2CH2CH2CH3).

(c) Diethyl ether (CH3CH2OCH2CH3) in ethanol (CH3CH2OH) or in

water.

Plan: Examine the formulas of each solute and solvent to determine

which forces will occur. A solute tends to be more soluble in a solvent

which has the same type of forces binding its molecules.

slide18

Predicting Relative Solubility's of Substances - II

Solution:

(a)Methanol - NaCl is an ionic compound that dissolves through ion-

dipole forces. Both methanol and propanol contain a polar hydroxyl

group, and propanol’s longer hydrocarbon chain would form only weak

forces with the ions, so it would be less effective at replacing the ionic

attractions of the solvent.

(b) Water Ethylene glycol molecules have two -OH groups, and the

molecules interact with each other through H bonding. They would be

more soluble in water, whose H bonds can replace solute H bonds better

than can the dispersion forces in hexane.

(c) Ethanol Diethyl ether molecules interact with each other through

dipole and dispersion forces and could form H bonds to both water and

ethanol. the ether would be more soluble in ethanol because the solvent

can form H bonds and replace the dispersion forces in the solute,

whereas the H bonds in water must be partly replaced with much weaker

dispersion forces.

slide19

An Energy Solution? – Solar Ponds

Solar ponds are shallow bodies of salt water ( a very high salt content)

designed to collect solar energy as it water the water in the ponds, and

then it can be used for heating, or converted into other forms of energy.

Sufficient salt must be added to establish a salt gradient in a pool

2-3 meters deep, with a dark bottom. A salt gradient will be established

in which the upper layer, called the conductive layer, has a salt content

of about 2% by mass. The bottom layer, called the heat storage layer has

a salt content of about 27%. The middle layer called the nonconvective

layer has an intermediate salt content, and acts as an insulator between

the two layers. The water in the deepest layer can reach temperatures of

between 90 and 100oC, temperatures as high as 107oC have been

reported. A 52 acre pond near the Dead sea in Israel can produce up to

5 mega watts of power.

slide20

Three Steps in making a Solution

Step #1 :

Breaking up the solute into individual components:

(expanding the Solute)

Step #2 :

Overcoming intermolecular forces in the solvent to make room

for the solute: (expanding the solvent)

Step #3 :

Allowing the solvent and solute to interact and form the solution.

slide22

Figure 17.2: (a) Enthalpy of solution Hsoln has a negative sign (the process is exothermic) if Step 3 releases more energy than is required by Steps 1 and 2. (b) Hsoln has a positive sign (the process is endothermic) if Steps 1 and 2 require more energy than is released in Step 3.

slide23

H of solution for Sodium Chloride

NaCl(s) Na+(g) + Cl –(g) Ho1 = 786 kJ/mol

H2O(l) + Na+(g) + Cl –(g) Na+(aq) + Cl –(aq)

Hohyd = Ho2 + Ho3 = _____________ kJ/mol

Hhyd = enthalpy (heat) of hydration

Hosoln = 786 kJ/mol – _______ kJ/mol = _______ kJ/mol

The dissolving process is positive, requiring energy. Then why is

NaCl so soluble? The answer is in the Gibbs free energy equation

from chapter 10, G = H – T S The entropy term – T S is

Negative, and the result is that G becomes negative, and as a

Result, NaCl dissolves very well in the polar solvent water.

french navy ship l ailette equipped with vacuum pumps approaches an oil slick
French Navy ship L'Ailette equipped with vacuum pumps approaches an oil slick.

Source: AP/Wide World Photos

slide27

Solution Cycle

Step 1: Solute separates into Particles - overcoming attractions

Therefore --Endothermic

Step 2: Solvent separates into Particles - overcoming intermolecular

attractions Therefore -- Endothermic

solvent (aggregated) + heat solvent (separated)

Hsolvent> 0

Step 3: Solute and Solvent Particles mix - Particles attract each other

Therefore -- Exothermic

solute (separated) + solvent (separated) solution + heat

Hmix< 0

The Thermochemical Cycle

Hsolution = Hsolute + Hsolvent + Hmix

If HEndothermic Rxn< HExothermic Rxn solution becomes warmer

If HEndothermic Rxn> HExothermic Rxn solution becomes colder

figure 17 3 vitamin a c
Figure 17.3: Vitamin A, C

A Fat – Soluble Vitamin A Water – Soluble Vitamin

A Hydrophobic Vitamin A Hydrophilic Vitamin

slide33

Figure 17.4: (a) a gaseous solute in equilibrium with a solution. (b) the piston is pushed in, which increases the pressure of the gas and the number of gas molecules per unit volume. (c) greater gas

slide34

Henry’s Law of Gas solubilities in Liquids

P = kHX

P = Partial pressure

of dissolved gas

X = mole fraction

of dissolved gas

kH = Henry’s Law

Constant

slide35

Henry’s Law of Gas Solubility

Problem: The lowest level of oxygen gas dissolved in water that will

support life is ~ 1.3 x 10 - 4 mol/L. At the normal atmospheric pressure of

oxygen is there adaquate oxygen to support life?

Plan: We will use Henry’s law and the Henry’s law constant for oxygen

in water with the partial pressure of O2 in the air to calculate the amount.

Solution:

The Henry’s law constant for oxygen in water is 1.3 x 10 -3 mol

liter atm

and the partial pressure of oxygen gas in the atmosphere is 21%,

or 0.21 atm.

.

Soxygen = kH x PO2 = 1.3 x 10 -3 mol x ( 0.21 atm)

liter atm

SOxygen = mol O2 / liter

.

This is adaquate to sustain life in water!

slide37

Predicting the Effect of Temperature

on Solubility - I

Problem: From the following information, predict whether the

solubility of each compound increases or decreases with an increase in

temperature.

(a) CsOH Hsoln = -72 kJ/mol

(b) When CsI dissolves in water the water becomes cold

(c) KF(s) K+(aq) + F -(aq) + 17.7 kJ

Plan: We use the information to write a chemical reaction that includes

heat being absorbed (left) or released (right). If heat is on the left, a

temperature shifts to the right, so more solute dissolves. If heat is on

the right, a temperature increase shifts the system to the left, so less

solute dissolves.

Solution:

(a) The negative H indicates that the reaction is exothermic, so when

one mole of Cesium Hydroxide dissolves 72 kJ of heat is released.

H2O

slide38

Predicting the Effect of Temperature

on Solubility - II

(a) continued

H2O

CsOH(s) Cs+(aq) + OH -(aq) + Heat

A higher temperature (more heat) decreases the solubility of CsOH.

(b) When CsI dissolves, the solution becomes cold, so heat is absorbed.

H2O

CsI(s) + Heat Cs+(aq) + I -(aq)

A higher temperature increases the solubility of CsI.

(c) When KF dissolves, heat is on the product side, and is given off

so the reaction is exothermic.

H2O

KF(s) K+(aq) + F -(aq) + 17.7 kJ

A higher temperature decreases the solubility of KF

slide39

Figure 17.6: The solubilities of several gases in water as a function of temperature at a constant pressure of 1 atm of gas above the solution.

figure 17 7 pipe with accumulated mineral deposits left lengthwise section right
Figure 17.7: Pipe with accumulated mineral deposits (left) lengthwise section (right)

Source: Visuals Unlimited

lake nyos in cameroon
Lake Nyos in Cameroon

Source: Corbis

slide43
Figure 17.9: The presence of a nonvolatile solute inhibits the escape of solvent molecules from the liquid
slide44
Figure 17.10: For a solution that obeys Raoult’s law, a plot of Psoln versus xsolvent yields a straight line.
slide46

Vapor Pressure Lowering -I

Problem: Calculate the vapor pressure lowering when 175g of sucrose

is dissolved into 350.00 ml of water at 750C. The vapor

pressure of pure water at 750C is 289.1 mm Hg, and it’s

density is 0.97489 g/ml.

Plan: Calculate the change in pressure from Raoult’s law using the

vapor pressure of pure water at 750C. We calculate the mole

fraction of sugar in solution using the molecular formula of

sucrose and density of water at 750C.

Solution:

molar mass of sucrose ( C12H22O11) = 342.30 g/mol

175g sucrose

342.30g sucrose/mol

= 0.51125 mol sucrose

350.00 ml H2O x 0.97489g H2O = 341.21g H2O

ml H2O

341.21 g H2O

18.02g H2O/mol

= ______ molH2O

slide47

Vapor Pressure Lowering - II

mole sucrose

moles of water + moles of sucrose

Xsucrose =

0.51125 mole sucrose

18.935 mol H2O + 0.51125 mol sucrose

Xsurose = = 0.2629

P = Xsucrose x P 0H2O = 0.2629 x 289.1 mm Hg = ________ mm Hg

slide48

Like Example 17.1 (P 841-2)

A solution was prepared by adding 40.0g of glycerol to 125.0g of water

at 25.0oC, a temperature at which pure water has a vapor pressure of

23.76 torr. The observed vapor pressure of the solution was found to be

22.36 torr. Calculate the molar mass of glycerol!

Solution:

Roults Law can be rearranged to give:

XH2O = = = 0.9411 =

mol H2O = = 6.94 mol H2O

0.9411 =

mol gly = = 0.4357 mol

Psoln

PoH2O

22.36 torr

23.76 torr

mol H2O

mol gly + mol H2O

125.0 g

18.0 g/mol

6.94 mol

mol gly + 6.96 mol

6.94 mol – (6.94 mol)(0.9411)

0.9411

40.0 g

0.4357 mol

= g/mol

(MMglycerol = 92.09 g/mol)

slide50

-

-

-

-

-

-

-

-

-

-

-

H O H

H-C-C-C-H

H H

-

-

-

-

-

-

-

-

-

-

-H

H H H H H H

H-C-C-C-C-C-C-H

H H H H H H Hexane

H H +

H-C-C-O-H Ethanol

H H

H-O

Acetone + Water

slide51
Figure 17.12: Phase diagrams for pure water (red lines) and for an aqueous solution containing a nonvolatile solution (blue lines).
slide54

Like Example 17.2 (P 845-6)

A solution is prepared by dissolving 62g of sucrose in 150.0g of water

the resulting solution was found to have a boiling point of 100.61oC.

Calculate the molecular mass of sucrose.

Solution:

T = kbmsolute kb = 0.51

T = 100.61oC – 100.00 oC = 0.61oC

msolute = = = 1.20 mol/Kg

Msolute = mol solute = (0.150 kg)(1.2 mol/kg)

mol solute = 0.18 mol MM = = _________g/mol

oC Kg

msolute

0.61oC

T

kb

oC Kg

msolute

0.51

mol solute

kg solvent

62g

0.18 mol

(MMsucrose = 342.18 g/mol)

slide57

Like Example 17.3 (P847)

What mass of ethanol (C2H6O) must be added to 20.0 liters of water to

keep it from freezing at a temperature of -15.0oF?

Solution:

oC = (oF – 32)5/9 = (-15 – 32)5/9 = -26.1oC

T = kf msolute msolute = = = 14.0 mol/kg

14.0 mol/kg(20 kg H2O) = 280 mol ethanol

Ethanol = 2x12.01 + 6x 1.008 + 1x16.0 = 46.07

280 mol ethanol (46.07 g ethanol/mol) = ___________ kg ethanol

T

kf

-26.1oC

1.86

oC kg

mol

slide58

Determining the Boiling Point Elevation and Freezing

Point Depression of an Aqueous Solution

Problem: We add 475g of sucrose (sugar) to 600g of water. What will

be the Freezing Point and Boiling Points of the resultant solution?

Plan: We find the molality of the sucrose solution by calculating the

moles of sucrose and dividing by the mass of water in kg. We then apply

the equations for FP depression and BP elevation using the constants

from table 12.4.

Solution: Sucrose C12H22O11 has a molar mass = 342.30 g/mol

475g sucrose

342.30gsucrose/mol

= 1.388 mole sucrose

1.388 mole sucrose

0.600 kg H2O

molality = = 2.313 m

0.5120C

m

Tb = Kb x m = (2.313 m)= 1.180C BP = 100.000C + 1.180C

BP = 101.180C

1.860C

m

Tf = Kf x m = (2.313 m) = 4.300C FP = 0.000C - 4.300C = -4.300C

slide59

Determining the Boiling Point Elevation and Freezing

Point Depression of a Non-Aqueous Solution

Problem: Calculate the effect on the Boiling Point and Freezing Point of

a chloroform solution if to 500.00g of chloroform (CHCl3) 257g of

napthalene (C10H8, mothballs) is dissolved.

Plan: We must first calculate the molality of the cholorform solution by

calculating moles of each material, then we can apply the FP and BP

change equations and the contants for chloroform.

Solution: napthalene = 128.16g/mol chloroform = 119.37g/mol

257g nap

128.16g/mol

molesnap = =2.0053 mol nap

moles nap

kg(CHCl3)

2.0053 mol

0.500 kg

molarity = = = 4.01 m

3.630C

m

Tb = Kb m = (4.01m) = 14.560C normal BP = 61.70C

new BP = ______0C

4.700C

m

Tf = Kfm = (4.01m) =18.850C normal FP = - 63.50C

new FP = _______0C

slide61
Figure 17.15: The normal flow of solvent into the solution (osmosis) can be prevented by applying an external pressure to the solution.
slide62

Figure 17.16: A pure solvent and its solution (containing a nonvolatile solute) are separated by a semipermeable membrane through which solvent molecules (blue) can pass but solute molecules (green) cannot.

slide63

Osmotic pressure calculation

Calculate the osmotic pressure generated by a sugar solution made up of

5.00 lbs of sucrose per 5.00 pints of water.

Solution:

5.00 lbs ( ) = 2.27 kg

Molar mass of sucrose = 342.3 g/mol 2,270g

5.00 pints H2O ( )( ) = 2.36 liters

P = MRT = ( )(0.08206 )(298 K) = __________ atm

1 kg

2.205 lbs

= 6.63 mol sucrose

342.3g/mol

1.00 gallon

8 pints

3.7854 L

1.00 gallon

L atm

mol K

6.63 mol

2.36 L

slide64

Like example 17.4 (P 848-9)

To determine the molar mass of a certain protein, 1.7 x 10-3g of the

protein was dissolved in enough water to make 1.00 ml of solution. The

osmotic pressure of this solution was determined to be 1.28 torr at 25oC.

Calculate the molar mass of the protein.

Solution:

P = 1.28 torr( ) = 1.68 x 10-3 atm

T = 25 + 273 = 298 K

M = = 6.87 x 10-5 mol/L

1.70 g xg

6.87 x 10-5mol mol x = _______________g/mol

1 atm

760 torr

1.68 x 10-3 atm

0.08206 L atm(298 K)

mol K

=

slide65

Determining Molar Mass from

Osmotic Pressure - I

Problem: A physician studying a type of hemoglobin formed during a

fatal disease dissolves 21.5 mg of the protein in water at 5.00C to make

1.5 ml of solution in order to measure its osmotic pressure. At

equilibrium, the solution has an osmotic pressure of 3.61 torr. What is

the molar mass(M) of the hemoglobin?

Plan: We know the osmotic pressure (),R, and T. We convert  from

torr to atm and T from 0C to K and use the osmotic pressure equation to

solve for molarity (M). Then we calculate the moles of hemoglobin from

the known volume and use the known mass to find M.

Solution:

1 atm

760 torr

P = 3.61 torr x = 0.00475 atm

Temp = 5.00C + 273.15 = 278.15 K

slide66

Molar Mass from Osmotic Pressure - II



RT

0.00475 atm

0.082 L atm (278.2 K)

mol K

M = = = 2.08 x 10 - 4 M

Finding moles of solute:

2.08 x 10 - 4 mol

L soln

n = M x V = x 0.00150 L soln = 3.12 x 10 - 7 mol

Calculating molar mass of Hemoglobin (after changing mg to g):

0.0215 g

3.12 x 10-7 mol

M = = ________________ g/mol

slide69

Example 17.5 (P 850)

What concentration of sodium chloride in water is needed to produce

an aqueous solution isotonic with blood (p = 7.70 atm at 25oC).

Solution:

P = MRT M =

M = = 0.315 mol/L

Since sodium chloride gives two ions per molecule, the concentration

would be ½ that value, or 0.158 M

NaCl Na+ + Cl-

P

RT

7.70 atm

(0.08206 L atm) (298 K)

mol K

slide72
Family uses a commercially available desalinator, similar to those developed by the Navy for life rafts.

Source: Recovery Engineering, Inc.

slide73

Figure 17.20: Residents of Catalina Island off the coast of southern California are benefiting from a desalination plant that can supply 132,000 gallons of drinkable water per day, one-third of the island's daily needs.

slide74

Colligative Properties of Volatile Nonelectrolyte Solutions

From Raoult’s law, we know that:

Psolvent = Xsolvent x P0solvent and Psolute = Xsolute x P0solute

Let us look at a solution made up of equal molar quantities of acetone

and chloroform. Xacetone = XCHCl3 = 0.500, at 350C the vapor pressure of

pure acetone = 345 torr, and pure chloroform = 293 torr. What is vapor

pressure of the solution, and the vapor pressure of each component. What

are the mole fractions of each component?

Pacetone = Xacetone x P0acetone = 0.500 x 345 torr = 172.5 torr

PCHCl3 = XCHCl3 x P0CHCl3 = 0.500 x 293 torr = 146.5 torr

PA

PTotal

From Dalton’s law of partial pressures we know that XA =

Pacetone

PTotal

172.5 torr

172.5 + 146.5 torr

Xacetone = = = 0.541

Total Pressure = 319.0 torr

PCHCl3

PTotal

146.5 torr

172.5 + 146.5 torr

XCHCl3 = = = 0.459

slide75

Colligative Properties

I ) Vapor Pressure Lowering - Raoult’s Law

II ) Boiling Point Elvation

III ) Freezing Point Depression

IV ) Osmotic Pressure

slide76

Colligative Properties of Ionic Solutions

For ionic solutions we must take into account the number of ions present!

i = van’t Hoff factor = “ionic strength”, or the number of ions present

For vapor pressure lowering: P = i XsoluteP 0solvent

For boiling point elevation: Tb = i Kb m

For freezing point depression: Tf = i Kf m

For osmotic pressure:  = i MRT

figure 17 21 in a aqueous solution a few ions aggregate forming ion pairs that behave as a unit
Figure 17.21: In a aqueous solution a few ions aggregate, forming ion pairs that behave as a unit.
slide79

Like Example 17.6 (P 853)

  • The observed osmotic pressure for a 0.10M solution of Na3PO4 at 25oC
  • is 8.45 atm. Compare the expected and experimental values of i!
  • Solution:
  • Tri sodium phosphate will produce 4 ions in solution.
  • Na3PO4 3 Na+ + PO4-3
  • Thus i is expected to be 4, now to calculate the experimental value of i
  • from the osmotic pressure equation.
  • = iMRT or i = =

i = ______ This is less than the value expected of 4 so there must be

some ion paring occurring in the solution.

P

MRT

8.45 atm

(0.10 )(0.08206 )(298 K)

mol

L

L atm

mol K