Chapter 11. Solutions and Their Properties

Solutions

Solute: material present in least amount

Solvent: material present in most amount

Solution = solvent + solute(s)

What criteria must be satisfied in order to form a solution?

An artist’s conception of how a salt dissolves in water

How could we identify the cation and anion?

Dissolving of sugar

Solution energy = M+X- (s) + H2O M+(aq) + X-(aq)

Lattice energy = M+(g) + X-(g) M+X- (s)

Soluteaggregated

SolventaggregatedE

nth

alp

y, H

A Exothermic solution process

Hinitial

Soluteaggregated

Solventaggregated

SolutionHsoln < 0

En

thal

py

, H

A Exothermic solution process

Hfinal

Hinitial

Solventseparated

Soluteaggregated

Solventaggregated

SolutionHsoln < 0

En

thal

py

, H

Hso

lven

t

A Exothermic solution process

Hfinal

Hinitial

Solventseparated

Soluteaggregated

Solventaggregated

Soluteseparated

SolutionHsoln < 0

En

thal

py

, H

Hso

lute

Hso

lven

t

A Exothermic solution process

Hfinal

Hinitial

Solventseparated

Soluteaggregated

Solventaggregated

Soluteseparated

SolutionHsoln < 0

En

thal

py

, H

Hso

lute

Hsolute

+Hsolvent

Hso

lven

t

A Exothermic solution process

Hfinal

Hinitial

Solventseparated

Soluteaggregated

Solventaggregated

Soluteseparated

SolutionHsoln < 0

En

thal

py

, H

Hso

lute

Hsolute

+Hsolvent

Hmix

Hso

lven

t

A Exothermic solution processHfinal

Hinitial

Soluteaggregated

Solventaggregated

En

thal

py

, H

Hinitial

B Endothermic solution process

Soluteaggregated

Solventaggregated

En

thal

py

, H

Hfinal

Hinitial

Hsoln > 0

Solution

B Endothermic solution process

Solventseparated

Soluteaggregated

Solventaggregated

En

thal

py

, H

Hfinal

Hinitial

Hsoln > 0

Solution

B Endothermic solution process

Hso

lven

t

Solventseparated

Soluteaggregated

Solventaggregated

Soluteseparated

En

thal

py

, H

Hfinal

Hinitial

Hsoln > 0

Solution

B Endothermic solution process

Hso

lute

Hso

lven

t

Solventseparated

Soluteaggregated

Solventaggregated

Soluteseparated

En

thal

py

, H

Hfinal

Hinitial

Hsolute

+Hsolvent

Hsoln > 0

Solution

B Endothermic solution process

Hso

lute

Hso

lven

t

Solventseparated

Soluteaggregated

Solventaggregated

Soluteseparated

En

thal

py

, H

Hfinal

Hinitial

Hsolute

+Hsolvent

Hmix

Hsoln > 0

Solution

B Endothermic solution process

Hso

lute

Hso

lven

t

Interactions that must be overcome and those that form

Why do salts with a positive enthalpy of solution form solutions?

Variation of solubility of solids and liquids with temperature

Solubility of gases with temperature

Solvent : H2O

Solubility of gases as a function of pressure

Le Chatelier’s principle

Pure solvent A

Solution of A and B

Vapor pressure of a pure componentB is a non-volatile component

air + vapor

Mole fraction of a = na/(na +nb +nc + ...)

in the presence of a non-volatile solute

Colligative property: a physical property that depends on how the amount present

Distillation of volatile materials

Raoult’s Law: PA = xAPAo; PB = xBPBo

PT obs = xA PAo + xBPBo

for two volatile components where PAo and PBo are the vapor pressures of the pure components

Raoult’s Law: PA = XAPAo; PB = XBPB

o PT obs = XA PAo + XBPB

o

for two volatile components

25 °C

mol fraction

760

30

0

450

76

0

100

0

93 °C

200

560

Boiling occurs when the total pressure = 760 mm

For a 1:1 mol mixture at 93 ° C

PA = XAPAo; PB = XBPBo PT obs = XA PAo + XBPBo

Two volatile liquids

For an initial 1:1 mixture of benzene-toluene, the initial composition of the vapor based on its vapor pressure is approximately

Ptoluene = 200 mm

Pbenzene = 560 mm

Treating the vapor as an ideal gas, if we were to condense the vapor:

PbV/PtV = nbRT/ntRT

Pb/Pt = nb/nt

nb/nt = 56/20; the vapor is enriched in the more volatile component

xB = 56/(56+20) = 0.73

Starting with a 50:50 mixture: (the composition of the first drop)

A Volatile Liquid and Non-volatile Solid

A recipe for making maple syrup requires adding two cups sugar (sucrose, mw 342 g/mol) for every cup of water. Assuming a cup of water contains 300 mL of water and a cup of sugar contains 300 g of sucrose (C12H22O11), at what temperature would you expect the syrup to boil given the following vapor pressures of pure water?

Temp (°C) Vapor pressure (mm) pure H2O100 760101 787102 815103 845104 875105 906

Raoult’s Law Pobs = xH2OPoH2O

Pobs = 760 mm for boiling to occur

300g/18 g/mol = 16.7 mol H2O

600g/342 g/mol = 1.75 mol sugar

xH2O =16.7/(16.7 + 1.75)= 0.91

PoH20 = Pobs/xH20; 760 /0.91 = 835 mm

Freezing point depression and boiling point elevation (using non-volatile solutes)

T = Km

K is a constant characteristic of each substance

m = molality (mols/1000g solvent)

T = Km m = molality, mol/1000g solvent

The boiling point elevation is for non-volatile solutes

Boiling Point Elevation for A Volatile Liquid and Non-volatile Solid

A recipe for making maple syrup requires adding two cups sugar (sucrose, mw 342 g/mol) for every cup of water. Assuming a cup of water contains 300 mL of water and a cup of sugar contains 300 g of sucrose (C12H22O11), at what temperature would you expect the syrup to boil given the following vapor pressures of pure water?

T = Kf m; T = 0.51*5.83m; T = 3 °C

Boiling point of water = 103 °C

sucrose, 600/342 g/mol = 1.75 mol

1.75 mol/300 g H2O = x mol/1000gH2O

x = 5.83 m

It is found that 1 g of naphthalene in 100 g of camphor depresses the freezing point of camphor by 2.945 °C. What is the molecular weight of naphthalene?

T = Kf m T= 37.7 °C kg/mol*m

2.945 °C = 37.7 °C /molal*m; m = 0.0781molal

If 1 g of naphthalene is dissolved in 10 g of the solvent camphor, 10 g of naphthalene is dissolved in 1000 g camphor

molality is equal to the number of moles/1000 g of solvent

therefore, 10g/MW = 0.0781 moles; MW = 10/0.0781 = 128 g /mol

Osmosis and osmotic pressure

A simple example of osmosis: evaporation of water from a salt solution

What acts as the semipermeable membrane?

Osmotic pressure

Osmotic Pressure

π = MRT

where π is the osmotic pressure between a pure solvent and a solution containing that solvent; M is the molarity of the solution (mols /liter), R is the gas constant, and T = temperature (K)

The total concentration of dissolved particle in red blood cells in 0.3 M. What would be the osmotic pressure between red cells and pure water if the blood cells were to be placed in pure water at 37 °C?

π = 0.3moles/liter*0.0821liter atm/(mol K)*(273.15+37 )K

π = 0.3 mol/l*0.0821l atm/(K mol)*298 K

π = 7.6 atm

In what direction would the pressure be directed?

What would happen to the cells?

What is the osmotic pressure developed from a solution of 0.15 M NaCl in contact with a semipermeable membrane of pure water at T = 37 °C?

NaCl + H2O = Na+ + Cl-

π = nRT

π = 0.3 mol/l*0.0821l atm/(K mol)*310.2 K

π = 7.6 atm

A B C

Solution

Semipermeablemembrane

SolutemoleculesSolventmolecules

Osmoticpressure

Applied pressure needed toprevent volume increase

Puresolvent

Netmovementof solvent

Reverse Osmosis

De-salination of sea water

What happens when polar molecules meet non-polar ones?

What happens when there are incompatible interactions are attached to the same molecule?

A typical soap molecule:CH3CH2CH2CH2CH2CH2CH2CH2CH2CH2CH2CH2CH2CO2

- K+

How does soap clear your clothes?

CO2-

-O2C CO2-

-O2C CO2-

-O2C CO2-

-O2C CO2-

CO2-

micelle

Soap in water forms micelles, small spherical objects suspended in water

- - - -

- -

- - - -

non-polar

interior

ionic exterior

Other examples of self assembly: