CH 11 c

Report
Chapter 11c: Solutions and Their
Properties
Some Factors Affecting Solubility
Solubility
The amount of solute per unit of solvent needed to form
a saturated solution
Miscible
Mutually soluble in all proportions
Effect of Temperature on Solubility
1. Most solid substances become more soluble as
temperature rises
2. Most gases become less soluble as temperature rises
Some Factors Affecting Solubility
Effect of Pressure on Solubility
1. No effect on liquids or solids
2. The solubility of a gas in a liquid at a given
temperature is directly proportional to the partial
pressure of the gas over the solution, @ 25°C
Henry’s Law
solubility = k x P
k = constant characteristic of specific gas, mol/Latm
P = partial pressure of the gas over the sol’n
Some Factors Affecting Solubility
a) Equal numbers of gas molecules escaping liquid and
returning to liquid
b) Increase pressure, increase # of gas molecules
returning to liquid, solubility increases
c) A new equilibrium is reached, where the #’s of
escaping = # of returning
Example 9
Which of the following will become less
soluble in water as the temperature is
increased?
1) NaOH(s)
2) CO2(g)
Example 10
The solubility of CO2 in water is 3.2 x 10-2 M @
25°C and 1 atm pressure. What is the Henry’sLaw constant for CO2 in mol/L atm?
solubility = k x P
Physical Behavior of Solutions:
Colligative Properties
• H2O
• 1.00 m NaCl
o
b.p. 100.0o C
b.p. 101.0o C
f.p. 0.0o C
o
of.p. -3.7 C
Colligative properties:
Properties that depend on the amount of a dissolved solute but not
on its chemical identity
There are four main colligative properties:
1. Vapor pressure lowering
2. Freezing point depression
3. Boiling point elevation
4. Osmosis:
The vapor pressure of a solution is different from the
vapor of the pure solvent.
Two different cases:
1. solute is non-volatile
solute has no vapor pressure of its own
example: dissolving a solid
vapor pressure of the solution is always lower than that of the
pure solvent
2. solute is volatile
solute has its own vapor pressure
example: mixing 2 liquids
vapor pressure of the mixture is intermediate between the
vapor pressures of the two pure liquids
Vapor-pressure Lowering of Solutions: Raoult’s Law
• Solutions with a Nonvolatile Solute
If the solute is nonvolatile and has no appreciable vapor pressure
of its own (solid dissolved) the vapor pressure of the solution is
always lower that that of the pure solvent.
Solutions with a Nonvolatile Solute!!!
Raoult’s Law:
Psoln = Psolv · Xsolv
Psoln = vapor pressure of the solution
Psolv = vapor pressure of the pure solvent
Xsolv = mole fraction of the solvent in the solution
Vapor pressure lowering is a colligative property (only
dependent on amount of solute and not on its chemical identity!)
For ionic substances calculate the total moles of solute particles,
1 mol NaCl will result in 1 mol Na+ and 1 mol
Cl- = 2 moles of particles
1 mol Na2SO4 will give 3 moles of particles
Raoult’s Law applies to only Ideal solutions
1. Law works best when solute concentrations are low an d
when solute and solvent particles have similar intermolecular
forces.
2. Further complication is that at higher concentrations
ionic compounds are not 100% dissociated.
Example
1 mol NaCl is only 90% dissociated
10% is undissociated
resulting in less particles in solution than expected
Example 9
What is the vapor pressure (in mm Hg) of a solution
prepared by dissolving 5.00 g of benzoic acid
(C7H6O2) in 100.00 g of ethyl alcohol (C2H6O) at
35°C? The vapor pressure of the pure ethyl
alcohol at 35°C is 100.5 mm Hg
Psoln = Psolv · Xsolv
Psolv = 100.5 mm Hg
MM C7H6O2 = 122.12 g/mol
MM C2H6O = 46.07 g/mol
1mol
a) 5 g C7H6O2 x ---------- = 0.0409 mol
122.12g
b)
100 g C2H6O x 1 mol/ 46.07 g = 2.17 mol
c)
Xsolv = 2.17 mol / (0.0409 + 2.17 mol) = 0.982
Psoln = Psolv · Xsolv
= 100.5 mm Hg · 0.982 = 98.7 mm Hg
Solutions with a Nonvolatile Solute
Close-up view of part of the vapor pressure curve for
a pure solvent and a solution of a nonvolatile
solute. Which curve represents the pure solvent,
and which the solution?
Why?
Reason for vapor pressure lowering
DG = DHvap - T DS
DHvap = positive, disfavored
DS
= positive, favored
DHvap is (nearly) the same for a pure solvent and a solvent in
a solution
DS is different
solvent in a solution has more disorder than pure solvent
entropy of a solution is higher than the pure solvent
entropy of the vapor in both cases the same
Entropy increase for vaporization from a solution is smaller than
vaporization from a pure solvent
Less entropy increase means less favored
Solutions with a Volatile Solute!!
For a mixture of 2 volatile liquids A and B the overall vapor
pressure is the sum of the vapor pressure of the 2
components (Dalton’s law)
Ptotal = PA + PB
The vapor pressure for each component is calculated by
Raoult’s law: vapor pressure is equal to the mole fraction of
A times the vapor pressure of pure A
Ptotal = PA + PB = (P0A · XA) + (P0B · XB)
P°A = vapor pressure of pure A
P°B = vapor pressure of pure B
XA = mole fraction of A
XB = mole fraction of B
Solutions with a Volatile Solute
Close-up view of part of the vapor pressure curves for
two pure liquids and a mixture of the two. Which
curves represent the pure liquids, and which the
mixture?
Ptotal should be intermediate to A & B
Raoult’s law applies only to ideal solutions
Most real solutions show deviations
Example 10
What is the vapor pressure ( in mm Hg) of a sol’n
prepared by dissolving 25.0 g of ethyl alcohol
(C2H5OH) in 100.0 g of water at 25°C? The
vapor pressure of pure water is 23.8 mm Hg and
the vapor pressure of ethyl alcohol is 61.2 mm Hg
at 25°C
Example 10
P°H2O = 23.8 mm Hg
P°C2H5OH = 61.2 mm Hg
XH2O = mole fraction of A
25 g C2H5OH x 1 mol / 46.07 g = 0.543 mol
C2H5OH
100.0 g H2O x 1 mol/ 18 g = 5.56 mol H2O
XH2O = 5.56 /(5.56 + 0.543) = 0.91
XC2H5OH = mole fraction of B
XC2H5OH = 0.543 / (0.543 + 5.56) = 0.09
Ptot = (23.8 x 0.91) + (61.2 x 0.09) = 27.2 mm Hg
Boiling Point Elevation and Freezing Point Depression
of Solutions
A solution has a lower vapor pressure than the pure liquid.
To reach the atmospheric pressure (boiling point) the
temperature must be higher.
D Tb = Kb · m
D Tf = Kf · m
Boiling point elevation
Freezing point depression
Kb = molal boiling-point elevation constant
Kf = molal freezing-point depression constant
m = molality
Example 11
What is the normal boiling point in °C of a solution
prepared by dissolving 1.50 g of aspirin (C9H8O4)
in 75.00 g of chloroform (CHCl3)? The normal
boiling point of chloroform is 61.7 °C and Kb of
chloroform is 3.63 °C kg/mol
Example 11
DTb = Kb · m
m = mole solute / kg solvent
MM C9H8O4 = 180.16 g/mol
1.50 g C9H8O4 x 1 mol / 180.16 g = 0.00833 mol C9H8O4
75.00 g CHCl3 = 0.07500 kg CHCl3
m =0 .00833 mol C9H8O4 / 0.07500 kg CHCl3 = 0.111 m
DTb = 3.63 °C kg/mol · 0.111 mol/kg = 0.403 °C
Boiling point = 0.403 °C + 61.7 °C = 62.1 °C
Osmosis and Osmotic Pressure
Membranes are semipermeable materials
They allow water and other small molecules to
pass through, but they block the passage of
larger molecules or ions.
All living cells contain membranes
and osmosis is important in biological systems
Osmosis provides the primary means by which
water is transported into and out of cells
Osmosis is responsible for the ability of plant
roots to suck up water from the soil
Osmosis
Thermodynamic explanation
Every system wants to balance out the concentration
One side pure solvent
Other side solution
ordered system
The system tries to get into a more disordered
more randomness state
The entropy will increase
Osmosis is similar to diffusion
Osmosis Pressure
1. The amount of pressure necessary to achieve
equilibrium
2.  = MRT
 = osmotic pressure
M = molarity
R = gas constant, .08206 L atm/K mol
T = temperature in Kelvin
Isotonic sodium chloride solution
The total concentration of dissolved
particles inside
red blood cells is 0.30 M.
What is the osmotic pressure at body
temp (310 k) ?
Isotonic sodium chloride solution
The total concentration of dissolved particles inside
red blood cells is 0.30 M.
What is the osmotic pressure at body temp (310 k) ?
= MRT
= 0.30 mol/L x 0.08206 L atm/K mol x
310 K
 = 7.63 atm
All medical infusions must have the same osmotic
pressure
Otherwise the blood cells would burst!
Therefore isotonic NaCl solutions are injected
Q: What is the mass% of an isotonic NaCl solution?
Since the molarity in blood cells is 0.3 M, we need 0.15 M
NaCl (0.15 M Na+ and 0.15 M Cl-)
Na = 23.0 amu
Cl = 35.5 amu
NaCl = 58.5 amu
0.15 M NaCl = 58.5 x 0.15 = 9 g/L ; 0.9 mass%
Example 12
What osmotic pressure in atm would you expect for
a solution of 0.125 M C6H12O6 that is separated
from pure water by a semipermeable membrane at
310 K?
 = MRT
= (0.125 mol/L)(.08206 L atm/K mol)(310 K) = 3.18
atm
Example 13
A solution of unknown substance in water at 300 K
gives rise to an osmotic pressure of 3.85 atm.
What is the molarity of the solution?
 = MRT
M = /RT
M = 3.85 atm / [(.08206 L atm/K mol)(300 K)]
M = .156 mol/L
Some uses of colligative properties
1 Freezing-point depression
- sprinkling of salt to melt snow
- antifreeze in automobile cooling system
- de-icing of airplane wings
2 Osmosis
- desalination of seawater with reverse osmosis
3 Molar mass determination
can use any four colligative properties
most accurate is osmotic pressure, since the magnitude
of osmosis effect is so great
Example 14
• What is the molar mass of sucrose if a
solution prepared by dissolving 0.822 g of
sucrose in water and diluting to a volume of
300.0 mL has an osmotic pressure of 149
mm Hg at 298 K?
 = MRT
149 mm Hg x 1 atm / 760 mm Hg = 0 .196 atm
M =  /RT
= 0.196 atm / [(0.08206 L atm/K mol)(298 K)]
= 0.00802 mol/L
0.00802 mol/L x 1 L/1000 mL x 300 mL = 0.00241 mol
MM = mass of sucrose / moles of sucrose
= 0.822 g / 0.00241 mol = 341.08 g/mol
Summary

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