Handout Chapter 17

8/13/2019 Handout Chapter 17

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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!


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


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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.

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 bemore 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.


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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.

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.


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Figure 17.1: The formation of a liquid

solution can be divided into three steps

Figure 17.2: (a) Enthalpy of solution Hsoln has anegative 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.


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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 = - 783 kJ/mol

Hhyd = enthalpy (heat) of hydration

Hosoln = 786 kJ/mol – 783 kJ/mol = 3 kJ/mol

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

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


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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

Solution Cycles

and the Enthalpy

Components of

the Heat of

Solution


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Figure 17.3: Vitamin A, C

A Fat – Soluble Vitamin A Water – Soluble Vitamin

A Hydrophobic Vitamin A Hydrophilic Vitamin

Figure 17.4: (a) a gaseous solute inequilibrium 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


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Henry’s Law of Gas solubilities in Liquids

P = k HX

P = Partial pressure

of dissolved gas

X = mole fraction

of dissolved gas

k H = Henry’s Law

Constant

Henry’s Law of Gas SolubilityProblem: 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:

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

liter atm

SOxygen = 2.7 x 10- 4 mol O2 / liter

.

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.

.

This is adaquate to sustain life in water!


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Figure 17.5: The solubilities of several

solids as a function of temperature.

Predicting the Effect of Temperatureon 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 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 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


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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 off

so the reaction is exothermic.

KF(s) K +

(aq) + F-(aq) + 17.7 kJ

H2O

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 of1 atm of gas

above the

solution.


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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 obeysRaoult’s law, a plot of Psoln versus xsolvent

yields a straight line.


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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 750

C.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= 18.935 molH2O


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Vapor Pressure Lowering - II

Xsucrose =mole sucrose

moles of water + moles of sucrose

Xsurose = = 0.26290.51125 mole sucrose

18.935 mol H2O + 0.51125 mol sucrose

P = Xsucrose x P0H2O = 0.2629 x 289.1 mm Hg = 7.600 mm Hg

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:

X H2O = = = 0.9411 =

mol H2O = = 6.94 mol H2O

0.9411 =

mol gly = = 0.4357 mol

Psoln

P

o

H2O

22.36 torr

23.76 torr

mol H2O

mol gly + mol H2O125.0 g

18.0 g/mol6.94 mol

mol gly + 6.96 mol6.94 mol – (6.94 mol)(0.9411)

0.9411

40.0 g

0.4357 mol= 91.81 g/mol (MMglycerol = 92.09 g/mol)


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Figure 17.11: Vapor pressure for a solution

of two volatile liquids.

H O H

H-C-C-C-H

H H -

-

-

- - -

H-O - H

Acetone + Water

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

-

- - - - -

- -

- -

- - - - -


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Figure 17.12: Phase diagrams for pure water

(red lines) and for an aqueous solution

containing a nonvolatile solution (blue lines).


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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 = k bmsolute k b = 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 = = 344g/mol

oC Kg

msolute

T

k b

0.61oC

0.51oC Kg

msolutemol solute

kg solvent62g

0.18 mol(MMsucrose = 342.18 g/mol)


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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 = k f 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) = 12.9 kg ethanol

T

k f

-26.1oC

1.86oC kg

mol


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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

molality = = 2.313 m1.388 mole sucrose

0.600 kg H2O

T b = K b x m = (2.313 m)= 1.180C BP = 100.000C + 1.180CBP = 101.180C

0.5120

Cm

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

1.860C

m

Determining the Boiling Point Elevation and FreezingPoint 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/molmolesnap = =2.0053 mol nap

257g nap

128.16g/molmolarity = = = 4.01 m

moles nap

kg(CHCl3)

2.0053 mol

0.500 kg

T b = K b m = (4.01m) = 14.560C normal BP = 61.70C

new BP = 76.30C

3.630C

m

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

new FP = - 82.40C

4.700C

m


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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.


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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

Π = MRT = ( )(0.08206 )(298 K) = 68.7 atm

1 kg2.205 lbs

342.3g/mol= 6.63 mol sucrose

1.00 gallon

8 pints3.7854 L

1.00 gallon

6.63 mol

2.36 L

L atm

mol 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. The

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

Calculate the molar mass of the protein.

Solution:

Π = 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 = 2.47 x 104g/mol

1 atm

760 torr

1.68 x 10-3 atm

0.08206 L atm(298 K)

mol K

=


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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 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 (π),R, and T. We convert π 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:

P = 3.61 torr x = 0.00475 atm1 atm760 torr

Temp = 5.00C + 273.15 = 278.15 K

Molar Mass from Osmotic Pressure - II

M = = = 2.08 x 10 - 4 Mπ

RT

0.00475 atm

0.082 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 = = 6.89 x 104 g/mol0.0215 g

3.12 x 10-7 mol


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Example 17.5 (P 850)

What concentration of sodium chloride in water is needed to produce

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

Solution:

Π = ΜRT 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-

ΠRT

7.70 atm

(0.08206 L atm) (298 K)

mol K

Figure 17.18: Reverse osmosis


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Reverse Osmosis for Removal of Ions

Colligative Properties of Volatile Nonelectrolyte Solutions

From Raoult’s law, we know that:

Psolvent = Xsolvent x P0

solvent and Psolute = Xsolute x P0

solute

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 P

0CHCl3 = 0.500 x 293 torr = 146.5 torr

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

Xacetone = = = 0.541PacetonePTotal

172.5 torr

172.5 + 146.5 torr

XCHCl3 = = = 0.459PCHCl3PTotal

146.5 torr

172.5 + 146.5 torr

Total Pressure = 319.0 torr


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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 XsoluteP0

solvent

For boiling point elevation: T b = i K b m

For freezing point depression: Tf = i K f m

For osmotic pressure: π = i MRT


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Figure 17.21: In a aqueous solution a few ions

aggregate, forming ion pairs that behave as a

unit.


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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.

Π= iMRT or i = =

i = 3.45 This is less than the value expected of 4 so there must be

some ion paring occurring in the solution.

Π

MRT

8.45 atm

(0.10 )(0.08206 )(298 K)mol

LL atm

mol K

Figure 17.22: The Tyndall effect

Source: Stock Boston


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Figure 17.23: Representation of two

colloidal particles

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Handout Chapter 17