Chapter 17Chapter 17

Properties of SolutionsProperties 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

17.6 Osmotic Pressure

17.7 Colligative Properties of Electrolytic Solutions

17.8 Colloids

3

The left beaker contains copper sulfate solution. In the right beaker, ammonia is

being added to create a precipitate of copper(II) hydroxide.

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!

Hydration shells around an aqueous ion

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 solutegrams of solution

moles of soluteliter of solution

moles of Solute Kg of solvent

nA

nA + nB

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!

Like dissolves Like: Solubility of methanol in water

Fig 13.4 (P 486) The structure and function of a soap

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 determinewhich forces will occur. A solute tends to be more soluble in a solventwhich has the same type of forces binding its molecules.

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 hydroxylgroup, and propanol’s longer hydrocarbon chain would form only weakforces with the ions, so it would be less effective at replacing the ionicattractions of the solvent.(b) Water Ethylene glycol molecules have two -OH groups, and themolecules interact with each other through H bonding. They would bemore soluble in water, whose H bonds can replace solute H bonds betterthan can the dispersion forces in hexane.(c) Ethanol Diethyl ether molecules interact with each other throughdipole and dispersion forces and could form H bonds to both water and ethanol. the ether would be more soluble in ethanol because the solventcan 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.

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 to5 mega watts of power.

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 roomfor the solute: (expanding the solvent)

Step #3 :

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

Figure 17.1: The formation of a liquid solution can be divided into three steps

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.

H of solution for Sodium Chloride

NaCl(s) Na +

(g) + Cl –(g) Ho

1 = 786 kJ/mol

H2O(l) + Na+(g) + Cl –

(g) Na+(aq) + Cl –

(aq)

Hohyd = Ho

2 + 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 isNaCl so soluble? The answer is in the Gibbs free energy equation from chapter 10, G = H – T S The entropy term – T S isNegative, 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.

Source: AP/Wide World Photos

Solution Cycle

Step 1: Solute separates into Particles - overcoming attractions Therefore -- EndothermicStep 2: Solvent separates into Particles - overcoming intermolecular attractions Therefore -- Endothermic solvent (aggregated) + heat solvent (separated) Hsolvent > 0Step 3: Solute and Solvent Particles mix - Particles attract each other Therefore -- Exothermic solute (separated) + solvent (separated) solution + heat

Hmix < 0The Thermochemical Cycle Hsolution = Hsolute + Hsolvent + Hmix

If HEndothermic Rxn < HExothermic Rxn solution becomes warmer

If HEndothermic Rxn > HExothermic Rxn solution becomes colder

Solution Cycles and the Enthalpy

Components ofthe Heat of

Solution

Multistage supercritical fluid extraction apparatus

Source: USDA

Chicken fat obtained by using supercritical carbon dioxide as the extracting agent.

Source: USDA

Figure 17.3: Vitamin A, C

A Fat – Soluble Vitamin A Water – Soluble VitaminA Hydrophobic Vitamin A Hydrophilic Vitamin

Carbonation in a bottle of soda

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

Henry’s Law of Gas solubilities in Liquids

P = kHX

P = Partial pressure of dissolved gasX = mole fraction of dissolved gaskH = Henry’s Law Constant

Henry’s Law of Gas SolubilityProblem: The lowest level of oxygen gas dissolved in water that willsupport life is ~ 1.3 x 10 - 4 mol/L. At the normal atmospheric pressure ofoxygen 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:

Soxygen = kH x PO2 = 1.3 x 10 -3 mol x ( 0.21 atm) liter atm SOxygen = mol O2 / liter

.

The Henry’s law constant for oxygen in water is 1.3 x 10 -3 mol liter atmand the partial pressure of oxygen gas in the atmosphere is 21%, or 0.21 atm.

.

This is adaquate to sustain life in water!

Figure 17.5: The solubilities of several solids as a function of temperature.

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 intemperature. (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 includesheat 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 onthe 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

Predicting the Effect of Temperature on Solubility - II

(a) continued

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

(aq) + Heat

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

H2O

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

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

(aq)

H2O

A higher temperature increases the solubility of CsI.

(c) When KF dissolves, heat is on the product side, and is given offso the reaction is exothermic.

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

(aq) + 17.7 kJH2O

A higher temperature decreases the solubility of KF

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)

Source: Visuals Unlimited

Lake Nyos in Cameroon

Source: Corbis

Figure 17.8: An aqueous solution and pure water in a closed environment

Figure 17.9: The presence of a nonvolatile solute inhibits the escape of solvent

molecules from the liquid

Figure 17.10: For a solution that obeys Raoult’s law, a plot of Psoln versus xsolvent yields

a straight line.

Fig. 13.15

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 sucrose342.30g 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

Vapor Pressure Lowering - II

Xsucrose = mole sucrosemoles of water + moles of sucrose

Xsurose = = 0.2629 0.51125 mole sucrose18.935 mol H2O + 0.51125 mol sucrose

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

Like Example 17.1 (P 841-2)

A solution was prepared by adding 40.0g of glycerol to 125.0g of waterat 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 be22.36 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 torr23.76 torr

mol H2Omol gly + mol H2O

125.0 g 18.0 g/mol

6.94 molmol gly + 6.96 mol

6.94 mol – (6.94 mol)(0.9411) 0.9411

40.0 g0.4357 mol

= g/mol (MMglycerol = 92.09 g/mol)

Figure 17.11: Vapor pressure for a solution of two volatile liquids.

H O HH-C-C-C-H H H

--

----

H-O-H

Acetone + Water

H H H H H HH-C-C-C-C-C-C-H H H H H H H Hexane H H +H-C-C-O-H Ethanol H H

-

------ -

- -

- - -- -

Figure 17.12: Phase diagrams for pure water (red lines) and for an aqueous solution

containing a nonvolatile solution (blue lines).

Like Example 17.2 (P 845-6)

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

Tkb

0.61oC

0.51oC Kg msolute

mol solute kg solvent

62g0.18 mol

(MMsucrose = 342.18 g/mol)

Figure 17.13: Ice in equilibrium with liquid water.

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

Tkf

-26.1oC1.86 oC kg

mol

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 willbe the Freezing Point and Boiling Points of the resultant solution?Plan: We find the molality of the sucrose solution by calculating themoles of sucrose and dividing by the mass of water in kg. We then apply the equations for FP depression and BP elevation using the constantsfrom table 12.4.Solution: Sucrose C12H22O11 has a molar mass = 342.30 g/mol

475g sucrose342.30gsucrose/mol

= 1.388 mole sucrose

molality = = 2.313 m1.388 mole sucrose 0.600 kg H2O

Tb = Kb x m = (2.313 m)= 1.180C BP = 100.000C + 1.180C BP = 101.180C

0.5120C m

Tf = Kf x m = (2.313 m) = 4.300C FP = 0.000C - 4.300C = -4.300C1.860C m

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 ofa 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 bycalculating moles of each material, then we can apply the FP and BPchange equations and the contants for chloroform.Solution: napthalene = 128.16g/mol chloroform = 119.37g/mol

molesnap = =2.0053 mol nap 257g nap128.16g/mol

molarity = = = 4.01 mmoles napkg(CHCl3)

2.0053 mol 0.500 kg

Tb = Kb m = (4.01m) = 14.560C normal BP = 61.70C new BP = ______0C

3.630C m

Tf = Kf m = (4.01m) =18.850C normal FP = - 63.50C new FP = _______0C

4.700C m

Figure 17.14: A tub with a bulb on the end is covered by a semipermeable membrane.

Figure 17.15: The normal flow of solvent into the solution (osmosis) can be prevented by

applying an external pressure to the solution.

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.

Osmotic pressure calculation

Calculate the osmotic pressure generated by a sugar solution made up of5.00 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

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

1 kg2.205 lbs

342.3g/mol= 6.63 mol sucrose

1.00 gallon 8 pints

3.7854 L1.00 gallon

6.63 mol 2.36 L

L atmmol K

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. Theosmotic pressure of this solution was determined to be 1.28 torr at 25oC.Calculate the molar mass of the protein.Solution: = 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 xg6.87 x 10-5mol mol x = _______________g/mol

1 atm760 torr

1.68 x 10-3 atm 0.08206 L atm(298 K) mol K

=

Determining Molar Mass from Osmotic Pressure - I

Problem: A physician studying a type of hemoglobin formed during afatal disease dissolves 21.5 mg of the protein in water at 5.00C to make1.5 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 fromtorr to atm and T from 0C to K and use the osmotic pressure equation tosolve for molarity (M). Then we calculate the moles of hemoglobin fromthe known volume and use the known mass to find M.Solution:

P = 3.61 torr x = 0.00475 atm 1 atm760 torr

Temp = 5.00C + 273.15 = 278.15 K

Molar Mass from Osmotic Pressure - II

M = = = 2.08 x 10 - 4 MRT

0.00475 atm0.082 L atm (278.2 K) mol K

Finding moles of solute:

n = M x V = x 0.00150 L soln = 3.12 x 10 - 7 mol2.08 x 10 - 4 mol L soln

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

M = = ________________ g/mol0.0215 g3.12 x 10-7 mol

Figure 17.17: Representation of the functioning of an artificial kidney

Example 17.5 (P 850)

What concentration of sodium chloride in water is needed to producean aqueous solution isotonic with blood (= 7.70 atm at 25oC).Solution: RT 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-

RT

7.70 atm(0.08206 L atm) (298 K) mol K

Figure 17.18: Reverse osmosis

Reverse Osmosis for Removal of Ions

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

Source: Recovery Engineering, Inc.

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.

Colligative Properties of Volatile Nonelectrolyte Solutions

From Raoult’s law, we know that:

Psolvent = Xsolvent x P0solvent and Psolute = Xsolute x P0

solute

Let us look at a solution made up of equal molar quantities of acetoneand chloroform. Xacetone = XCHCl3 = 0.500, at 350C the vapor pressure ofpure acetone = 345 torr, and pure chloroform = 293 torr. What is vaporpressure of the solution, and the vapor pressure of each component. Whatare 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

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

PTotal

Xacetone = = = 0.541Pacetone

PTotal

172.5 torr172.5 + 146.5 torr

XCHCl3 = = = 0.459PCHCl3

PTotal

146.5 torr172.5 + 146.5 torr

Total Pressure = 319.0 torr

Colligative Properties

I ) Vapor Pressure Lowering - Raoult’s Law

II ) Boiling Point Elvation

III ) Freezing Point Depression

IV ) Osmotic Pressure

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.

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 ifrom 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.

MRT

8.45 atm(0.10 )(0.08206 )(298 K)mol

LL atmmol K

Figure 17.22: The Tyndall effect

Source: Stock Boston

Figure 17.23: Representation of two colloidal particles

Magnetic liquid seal

Figure 17.24: Cottrel precipitator installed in a smokestack.

Fig. 13.24

A Cottrell Precipitator for Removing Particulates from Industrial Flue Gases