Chapter 10 Properties 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 Melting Freezing LIQUID

• • 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

Elevation of Boiling Point & Vapor Pressure Depression

 T 

K

b

m

solute  T is the boiling point elevation

K

b is molal boiling - point elevation constant

m

solute is the molality of the solute in solution 

Phase diagrams for pure water (red lines) and for an aqueous solution containing a nonvolatile solution (blue lines).

Solution Composition 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: A homogeneous mixture (mixed at level of atoms molecules or ions Solvent: The major component Solute: The minor component

Solution Composition

Mass Fraction, Mole Fraction, Molality and Molarity

Mass percentage (weight percentage): mass percentage of the component = mass of component X 100% total mass of mixture Mole fraction:

The amount of a given component (in moles) divided by the total amount (in moles) X 1 = n 1

/

(n 1 + n 2 ) for a two component system X 2 = n 2

/

(n 1 + n 2 ) = 1

–

X 1 or X 1 +X 2 =1

Molality

m

solute = moles solute per kilogram solvent = moles per kg or (mol kg -1 )

Molarity

(biochemists pay attention)

c

solute = 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 =

k

H

X

k

H = 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.

k

H 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.

Given P N 2  9.20 atm c N 2  [N 2 ]  5.76x10

 3 mol/L Henry' s Law P N 2  k N 2

X

N 2

X

N 2 

n

N 2

n

N 2 

n

H 2

O



n n

N H 2 2

O

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.

Given P N 2



9.20

atm c N 2



[N 2 ]



5.76x10



3 mol/l Henry' s Law P N 2



k N 2

X

N 2

X

N 2



n

N 2

n

N 2



n

H 2

O



n

N 2

n

H 2

O

P

N 2



k

N 2

rearrange k

N 2



P X

N N 2 2

X



N 2

Given Find Next assume 1 Liter

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.

k N 2



P N 2 X N 2



Given Find

X N 2 k N 2  n N 2 n H 2 O  5.76x10

 3 mol/l 1000g/l 18g/mol  1.0378

x 10  4  P N 2  Given X N 2 Find  8.86

x 10 4 atm  9.20

atm 1.0378x10

 4

Factors Affecting Solubility 1. Structure Effects 2. Pressure Effects 3. Temperature Effects for Aqueous Solutions The solubility of some

solids

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) 2) 3) 4) 5) low T, low P low T, high P high T, low P high T, high P solubility of gases does not depend upon temperature

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

1)

2)

3) 4) 5) low T, low P

low T, high P

high T, low P high T, high P solubility of gases does not depend upon temperature

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

1) 2) 3) 4) 5) depends on the polarity of the liquid depends on the liquid's density remains the same at all temperatures increases as the gas pressure above the solution increases decreases as the gas pressure above the solution increases

According to Henry's Law, the solubility of a gas in a liquid 1) 2) 3) depends on the polarity of the liquid depends on the liquid's density remains the same at all temperatures

4)

5)

increases as the gas pressure above the solution increases

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 .

For ideal solutions P soln = Moments in a Life X solvent

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

P

°

solvent

Planck.

Raoult’s Law, non-volatile solute • Consider a

non-volatile

dissolved in a

volatile

solute (component 2) solvent (component 1).

• X 1 = the mole fraction of solvent Raoult’s Law P 1 =X 1 P ° 1 P ° 1 = the vapor pressure of pure component 1

Raoult’s Law, volatile solute • Volatile solute (component 1) • Volatile solvent (component 2) P 1 = X 1 P ° 1 P 2 = X 2 P ° 2 P tot = P 1 + P 2

Vapor pressure for a solution of two volatile liquids.

Positive deviation

= solute-solvent attractions < solvent-solvent attractions

For non-ideal Solutions Negative deviation

= solute-solvent attractions > solvent-solvent attractions

 

boiling point :



T



K

b

m

solute freezing point :



T



K

f

m

solute

PV = nRT

Osmotic Pressure

Fourth Colligative Property • Important for transport of molecules across cell membranes, called semipermeable membranes •

Osmotic Pressure = Π Π

=

M

RT

ΠV

=

n

RT

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

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   MRT to find M (mol/L) 2) Recall that # of moles 3) Rearrange mwt  g  mass mwt mole  g L mole L  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

S

olution 1) use   MRT or M   RT c M   RT  0.0288 atm (0.0820 L atm mol  1 K  1 )(37  1.132x10

 3 mol/L  273.15K) 2) Rearrange M  g mole  g L mole L

M

 1.19

g L  3 1.132x10

mol/L  1.05x10

3 g/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. van’t Hoff Factor (

i

)

Moments in a Life

– 1901 awarded first Noble Prize in Chemistry

i

 moles of particles in solution moles of solute dissolved

ΔT = − i

m

K

Colligative Properties of Electrolyte Solutions

Elevation of Boiling Point ΔT b =

m

K b Where

m

= molality (Molality is moles of solute per kilogram of solvent) The Effect of Dissociation ΔT b = i

m

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

e.g., NaCl dissociates into i = 2 e.g., Na 2 SO 4 dissociates into i = 3 (2 Na + + 1 SO 4 -2 ) e.g., acetic acid (a weak acid and weak electrolyte) does not dissociate i = 1 also Depression of Freezing Point

ΔT f = −

m

K f ΔT f = − i

m

K f

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

1) 0.100

m

NaCl 2) 0.100

m

CaCl 2 3) 0.080

m

Fe(NO 3 ) 3 4) 0.080

m

Fe(NO 3 ) 2 5) 0.080

m

Co(SO 4 )

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

1) 2)

3)

0.100

m

NaCl ΔTb = (2)(0.100) K b = 0.200 K b 0.100

m

CaCl 2 ΔTb = (3)(0.100) K b = 0.300 K b

0.080 m Fe(NO 3 ) 3 ΔTb = (4)(0.080) K b = 0.320 K b

4) 5) 0.080

m

Fe(NO 3 ) 2 ΔTb = (3)(0.080) K b = 0.240 K b 0.080

m

Co(SO 4 ) ΔTb = (2)(0.080) K b = 0.160 K b Elevation of Boiling Point The Effect of Dissociation ΔT b = i

m

K b

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 SiO 2 ) – 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