Chapter 10Properties of Solutions

17.1 Solution Composition 17.2 The Thermodynamics of Solution Formation (skip)17.3 Factors Affecting Solubility 17.4 The Vapor Pressures of Solutions 17.5 Boiling-Point Elevation and Freezing-Point Depression 17.6 Osmotic Pressure 17.7 Colligative Properties of Electrolyte Solutions 17.8 Colloids

GAS

SOLID LIQUID

Freezing

MeltingSu

blim

atio

nD

epos

ition

Condensation

Evaporation

Colligative Properties of Solutions

• For Colligative properties, the difference between a pure solvent and dilute solution depends only on the number of solute particles present and not on their chemical identity.

• Examples– Vapor Pressure Depression– Boiling Point Elevation– Melting Point Depression– Osmotic Pressure

Lowering of Vapor Pressure– Vapor Pressure of a solvent above a dilute solution

is always less than the vapor pressure above the pure solvent.

Elevation of Boiling Point– The boiling point of a solution of a non-volatile

solute in a volatile solvent always exceeds the boiling point of a pure solvent

Boiling

• liquid in equilibrium with its vapor at the external pressure.

Boiling Point

• Vapor press = external pressure

Normal boiling point

• Vap press. = 1 atm

Phase diagrams for pure water (red lines)

and for an aqueous solution containing a nonvolatile

solution (blue lines).

Elevation of Boiling Point & Vapor Pressure Depression

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ΔT = Kb

msolute

ΔT is the boiling point elevation

Kb is molal boiling - point elevation constant

msolute

is the molality of the solute in solution

Solution Composition

Solution: A homogeneous mixture (mixed at level of atoms molecules or ions

Solvent:Solute:

The major componentThe minor component

The solute and solvent can be any combination of solid (s), liquid (l), and gaseous (g) phases.

Dissolution: Two (or more) substances mix at the level of individual atoms, molecules, or ions.

Solution Composition

Mass percentage (weight percentage):

mass percentage of the component =

X 100%mass of component

total mass of mixture

Mole fraction: The amount of a given component (in moles) divided by the total amount (in moles)

X1 = n1/(n1 + n2) for a two component system

X2 = n2/(n1 + n2) = 1 – X1 or X1+X2=1

Mass Fraction, Mole Fraction, Molality and Molarity

Molalitymsolute =

moles solute per kilogram solvent

= moles per kg or (mol kg-1)

Molarity (biochemists pay attention)

csolute =

moles solute per volume solution

= moles per liter of solution (mol L-1)

Factors Affecting Solubility

1. Molecular Interactions– Review chapter 4– Polar molecules, water soluble, hydrophilic (water

loving)• E.g., Vitamins B and C; water-soluble

– Non-polar molecules, soluble in non-polar molecules, hydrophobic (water fearing)• E.g., Vitamins A, D, K and E; fat-soluble

Factors Affecting Solubility of Gases

1. Structure Effects

2. Pressure Effects

Henry’s Law (for dilute solutions)

The mole fraction of volatile solute is proportional to the vapor pressure of the solute.

P = kH XkH = Henry’s Law constant, X = mole fraction.

Increasing the partial pressure of a gas over a liquid increases the amount of gas disolved in the liquid.

kH depends on temperature.

When the partial pressure of nitrogen over a sample of water at 19.4°C is 9.20 atm, the concentration of nitrogen in the water is 5.76 x 10-3 mol L-1. Compute Henry’s law constant for nitrogen in water at this temperature.

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Given

PN2= 9.20 atm

cN2= [N2] = 5.76x10−3mol/L

Henry's Law

PN2= kN2

XN2

XN2=

nN2

nN2+ nH2O

≈nN2

nH2O

Liter 1 assumeNext

FindGiven

X

Pk

rearrange

XkP

2N

2N

2N

2N

2N

2N

==

=

When the partial pressure of nitrogen over a sample of water at 19.4°C is 9.20 atm, then the concentration of nitrogen in the water is 5.76 x 10-3 mol L-1. Compute Henry’s law constant for nitrogen in water at this temperature.

OO n

n

nn

nX

X

2H

2N

2H

2N

2N

2N

2N

2N

2N

32

2N

2N

kP

Law sHenry'

mol/l5.76x10][Nc

atm 9.20P

Given

≈+

=

=

==

=

−

When the partial pressure of nitrogen over a sample of water at 19.4°C is 9.20 atm, then the concentration of nitrogen in the water is 5.76 x 10-3 mol L-1. Compute Henry’s law constant for nitrogen in water at this temperature.

FindGiven

X

Pk

2N

2N

2N ==

atm 10 x 8.86

1.0378x10

atm 9.20

Find

Given

X

Pk

10 x 1.0378

18g/mol1000g/l

mol/l5.76x10

n

nX

4

4

2N

2N

2N

4

3

O2H

2N

2N

=

===

=

⎟⎟⎠

⎞⎜⎜⎝

⎛=≈

−

−

−

Factors Affecting Solubility

1. Structure Effects

2. Pressure Effects

3. Temperature Effects for Aqueous Solutions

The solubility of somesolids as a function of

temperature.

The aqueous solubilities of most solids increase with increasing temperature, some decrease with temp.

Endothermic – heat is absorbed by the system (think evaporation of water, or melting of ice)

Exothermic – heat is evolved by the system (think fire, or freezing of water).

Factors Affecting Solubility

1. Structure Effects

2. Pressure Effects

3. Temperature Effects for Aqueous Solutions

The solubility of some gases in water as a function of

temperature at a constant pressure of 1 atm.

The greatest gas solubility for a gas in solution is predicted under what conditions?

  1) low T, low P

  2) low T, high P

  3) high T, low P

  4) high T, high P

 5) solubility of gases does not depend

upon temperature

The greatest gas solubility for a gas in solution is predicted under what conditions?

  1) low T, low P

  2) low T, high P

  3) high T, low P

  4) high T, high P

 5) solubility of gases does not depend

upon temperature

According to Henry's Law, the solubility of a gas in a liquid

  1) depends on the polarity of the liquid

  2) depends on the liquid's density

  3) remains the same at all temperatures

 4) increases as the gas pressure above the

solution increases

 5) decreases as the gas pressure above the

solution increases

According to Henry's Law, the solubility of a gas in a liquid

  1) depends on the polarity of the liquid

  2) depends on the liquid's density

  3) remains the same at all temperatures

 4) increases as the gas pressure above

the solution increases

 5) decreases as the gas pressure above the

solution increases

The Person Behind the Science

Francois-Marie Raoult (1830-1901)

Highlights– 1886 Raoult's law , the partial pressure

of a solvent vapor in equilibrium with a solution is proportional to the ratio of the number of solvent molecules to non-volatile solute molecules.

– allows molecular weights to be determined, and provides the explanation for freezing point depression and boiling point elevation.

Moments in a Life– Raoult was a prominent member of the

group which created physical chemistry, including Arrhenius, Nernst, van t'Hoff,

Planck.

Psoln = XsolventP°solvent

For ideal solutions

Raoult’s Law, non-volatile solute

• Consider a non-volatile solute (component 2) dissolved in a volatile solvent (component 1).

• X1 = the mole fraction of solvent

P1=X1 P°1

P°1 = the vapor pressure of pure component 1

Raoult’s Law

Raoult’s Law, volatile solute

• Volatile solute (component 1)• Volatile solvent (component 2)

P1 = X1 P°1

Ptot = P1+ P2

P2 = X2 P°2

Vapor pressure for a solution of two volatile liquids.

Negative deviation

= solute-solvent attractions > solvent-solvent attractions

Positive deviation

= solute-solvent attractions < solvent-solvent attractions

For non-ideal Solutions

€

boiling point : ΔT = Kb

msolute

€

freezing point : ΔT = Kfm

solute

Osmotic Pressure

Fourth Colligative Property• Important for transport of

molecules across cell membranes, called semipermeable membranes

• Osmotic Pressure = ΠΠ = M RTΠV = n RT

Molarity (M) = moles/L or n/V

PV = nRT

Osmotic Pressure

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

Osmotic Pressure useful for

Determining the Molar Mass of protein and other macromolecules

small concentrations cause large osmotic pressures

Can prevent transfer of all solute particles

Dialysis at the wall of most plant and animal cells

Dialysis: Representation of the functioning of an artificial kidney

A cellophane (polymeric) tube acts as the semi-permeable membrane

Purifies blood by washing impurities (solutes) into the dialyzing solution.

A dilute aqueous solution of a non-dissociating compound contains 1.19 g of the compound per liter of solution and has an osmotic pressure of 0.0288 atm at a temperature of 37°C. Compute the molar mass of the compound.

€

Strategy

1) use Π = MRT to find M (mol/L)

2) Recall that # of moles = mass

mwt

3) Rearrange mwt = g

mole=

gL

moleL

=1.19

M

A dilute aqueous solution of a non-dissociating compound contains 1.19 g of the compound per liter of solution and has an osmotic pressure of 0.0288 atm at a temperature of 37°C. Compute the molar mass of the compound

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Solution

1) use Π = MRT or M =Π

RT

c =Π

RT=

0.0288 atm

(0.0820 L atm mol−1K−1)(37 + 273.15K)

M = 1.132x10−3 mol/L

2) Rearrange M = g

mole=

gL

moleL

M =1.19g

L1.132x10−3 mol/L

= 1.05x103g/mol

The Person Behind the Science

J.H. van’t Hoff (1852-1901)

Highlights– Discovery of the laws of chemical

dynamics and osmotic pressure in solutions

– Mathematical laws that closely resemble the laws describing the behavior of gases.

– his work led to Arrhenius's theory of electrolytic dissociation or ionization

– Studies in molecular structure laid the foundation of stereochemistry.

Moments in a Life– 1901 awarded first Noble Prize in

Chemistry

van’t Hoff Factor (i)

dissolved solute of moles

solutionin particles of moles =i

ΔT = − i m K

Colligative Properties of Electrolyte Solutions

Elevation of Boiling Point

ΔTb = m Kb

Where m = molality

(Molality is moles of solute per kilogram of solvent)

The Effect of Dissociation

ΔTb = i m Kb

i = the number of particles released into the solution per formula unit of solute

e.g., NaCl dissociates into i = 2

e.g., Na2SO4 dissociates into i = 3

(2 Na+ + 1 SO4-2)

e.g., acetic acid (a weak acid and weak electrolyte) does not dissociate i = 1

also

Depression of Freezing Point

ΔTf = − m Kf

ΔTf = − i m Kf

Which aqueous solution would be expected to have the highest boiling point?

  1) 0.100 m NaCl

  2) 0.100 m CaCl2

  3) 0.080 m Fe(NO3)3

  4) 0.080 m Fe(NO3)2

  5) 0.080 m Co(SO4)

Which aqueous solution would be expected to have the highest boiling point?

  1) 0.100 m NaCl ΔTb = (2)(0.100) Kb = 0.200 Kb

  2) 0.100 m CaCl2 ΔTb = (3)(0.100) Kb = 0.300 Kb

  3) 0.080 m Fe(NO3)3 ΔTb = (4)(0.080) Kb = 0.320 Kb

  4) 0.080 m Fe(NO3)2 ΔTb = (3)(0.080) Kb = 0.240 Kb

  5) 0.080 m Co(SO4) ΔTb = (2)(0.080) Kb = 0.160 Kb

Elevation of Boiling Point

The Effect of Dissociation

ΔTb = i m Kb

Colloids: Colloidal Dispersions• Colloids are large particles dispersed in

solution– 1nm to 1000 nm in size– E.g., Globular proteins 500 nm

• Examples– Opal (water in solid SiO2)– Aerosols (liquids in Gas)– Smoke (solids in Air)– Milk (fat droplets & solids in water)– Mayonnaise (water droplets in oil)– Paint (solid pigments in liquid)– Biological fluids (proteins & fats in water)

• Characteristics– Large particle size colloids: translucent, cloudy,

milky)– Small particle size colloids: can be clear

Colloidal Dispersions

– Tyndall Effect • Light Scattering