1. Effect of Temperature and Pressure on Solubility
College Chemistry
John A. Schreifels
Chemistry 212

2. Solubility: Temperature Dependence
All solubilities are temperature dependent; must report temperatures with solubilities.
Most solids are more soluble at higher temperatures. Exceptions exist.
All gases are less soluble at higher temperatures.
Temperature related to sign of Hsoln;
negative means less soluble at high temperatures
positive means more soluble
E.g. Predict the temperature dependence of the solubility of Li2SO4, Na2SO4 and K2SO4 if their Hsoln are 29.8 kJ/mol, 2.4 kJ/mol and kJ/mol, respectively.
John A. Schreifels
Chemistry 212

3. Fractional Crystallization – Using Solubility at Temperatures to Our Advantage
Fractional crystallization – separation of a mixture of substances into pure compounds on the basis of their differing solubilities
Suppose we have impure KNO3 (10% is NaCl)
We gradually cool the sample and KNO3 will crystallize out of the solution before NaCl b/c KNO3 is less soluble at this temperature

4. Gas Solubility
Solubility of gases almost always decreases with increasing temperature
Ex: when you boil water, bubbles of air try to escape and “boil out” long before the solution begins to bubble
Fishing and Gas solubility – most fishers like fishing early in the morning or pick a deep spot in the river
Why? Fish need oxygen and the temperature is less there  more oxygen

5. Gas Solubility

6. Solubility: Pressure Dependence
Pressure has little effect on the solubility of a liquid or solid, but has dramatic effect on gas solubility in a liquid.
Henry’s law c = kHP. Allows us to predict the solubility of a gas at any pressure.
States that the solubility of a gas in a liquid is proportional to the pressure of the gas over the solution
c = molar concentration (molarity) P = pressure
kH = constant that depends only on temperature
Because of higher concentration, there are more gas particles moving around and hitting each other  increases pressure
John A. Schreifels
Chemistry 212

7. Solubility of Gases
Ex: soft drinks are pressurized with a mixture of air and CO2
Because of high partial pressure of CO2, the amount dissolved in the soft drink is many times the amount that would dissolve in the atmosphere
When the cap is removed,
the pressurized gases escape
and the pressure falls to the
atmospheric pressure

8. Exceptions of Henry’s Law
If gases dissolved in water end up reacting with the water, solubility will be much higher than you would expect
An example is oxygen in blood
Normally, O2 isn’t very soluble in water
However, its solubility in blood is great because of the high content of hemoglobin (Hb) that can bind oxygen)
Hb + 4 O2 ⇌Hb(O2)4

9. Physical Behavior of Solutions: Colligative Properties
Compared with the pure solvent the solution’s:
Vapor pressure is lower
Boiling point is elevated
Freezing point is lower
Osmosis occurs from solvent to solution when separated by a membrane.
All the above are considered colligative properties – properties that depend only on the # of solute particles in solution and not on the nature of the solute particles (no difference between NaCl and KCl)
John A. Schreifels
Chemistry 212

10. Vapor-Pressure Lowering of Solutions: Raoult’s Law
Raoult’s Law: Psoln = PsolvxXsolv
Non–volatile solute: vapor pressure decreases upon addition of solute.
Non-volatile solute: does NOT have a measurable vapor pressure (Ex: salt water)
Linear for dilute solutions
Vapor pressure lowering : P = Po  P = Po(1Xsolv).
Decrease in vapor pressure, DP, is proportional to the solute concentration (measured in mole fraction)
John A. Schreifels
Chemistry 212

11. Example 12.7
Calculate the vapor pressure of a solution made by dissolving 218 g of glucose (molar mass = g/mol) in 460 mL of water at 30°C. What is the vapor-pressure lowering? The vapor pressure of pure water at 30°C is mmHg. Assume the density of 1.00 g/mL.
We know: P1 = X1P1° (trying to find P1, need to find X1)
1. # of moles of glucose and water
n1(water) = 460 mL x (1.00 g/1 mL) x (1 mol/18.02 g) = 25.5 mol
n2(glucose) = 218 g x (1 mol/180.2 g) = 1.21 mol
2. mole fraction of water
X1 = n1 / total = 25.5 mol/ ( ) = 0.955
3. Find vapor pressure of water
P1 = (0.955) x (31.82 mmHg)
P1 = mmHg
4. Vapor pressure lowering
31.82 mmHg – mmHg = 1.4 mmHg

12. Vapor-Pressure Lowering of Solutions: Volatile Solute
In ideal solutions each substance in the solution obeys Raoult’s law. Thus,
PA = XAP°o and Ptotal = PA + PB.
Combining gives: .
Notice the equation predicts the linear variation in total vapor pressure with variations in the mole fraction of one of the components in the liquid phase.

13. Fractional Distillation
Fractional distillation – separates compounds based on their different boiling points
Similar to fractional crystallization
I used this a LOT in the lab
In grad school

14. Colligative Properties
Properties of a solvent that depend only on the amount of solute particles present (i.e.: freezing point)
Why is vapor pressure decreased and boiling point raised?
The solute particles take away space from the solvent and block the solvent from being evaporated
This means less solvent particles near the surface and therefore less “pressure”
Boiling point raises because more energy is needed

15. Colligative Properties
This same argument can be used to explain why freezing point is lowered
The solute particles get in the way of the pure solvent
It takes a lower temperature to get(stronger “push”) the solvent to freeze

16. Colligative Properties
We will be dealing with freezing point depression today
Can be found by the equation
DTf=Kfm
DTf= freezing point of pure solvent – f.p. of solution
Kf= freezing point constant for the solvent (Table 12.1)
m= molality
m= moles of solute/ kg of solvent

17. BP Elevation and FP Depression of Solutions
The magnitude of the change in FP and BP is directly proportional to the concentration of the solute (molality) – expressed in terms of the total number of particles in the solution.
BP Elevation The magnitude of the BP increase is given by the equation:
where Kb has units of °Ckg/mol or °C/m
FP Depression: linear variation with composition and given by:
where the units for this constant are the same as for Kb
John A. Schreifels
Chemistry 212

18. Boiling Point Elevation and Freezing Point Depression
Freezing point is lowered when a solute is added
Ex: salt is added to icy sidewalks to get the freezing point to lower and not have ice
Boiling point of a solution is raised when a solute is added
For example, you add salt to water to help it boil
Contrary to popular belief, the salt does NOT help the water boil at a lower temperature (or faster)

19. BP Elevation and FP Depression

20. Example 12.8
Ethylene glycol (EG) is a common automobile antifreeze. It is water soluble and fairly nonvolatile (b.p 197C). Calculate the freezing point of a solution containing 651 g of this substance in 2505 g of water. Would you keep this substance in your car radiator during the summer. MM = g
651 g EG x (1 mol EG / g EG) = 10.5 mol EG
Molal = moles solute / mass solvent (kg)
m = 10.5 mol EG / kg H2O = 4.19 m
Remember, DTf = Kfm
DTf = (1.86°C/m)(4.19 m)
DTf = 7.79°C
Since water freezes at 0°C, the solution will freeze at °C

21. Example 12.8 Cont. To find the boiling point elevation: DTb = Kbm
= (0.52 °C/m)(4.19 m)
= 2.2°C
Since water normally boils at 100°C, it would be 102.2°C that this boils
It would be preferable to leave the antifreeze in your car to keep the water from boiling
Kb and Kf will either be given OR you will need to find it (and have all the information necessary)

22. Osmosis and Osmotic Pressure
Osmosis: the passage of solvent through a membrane from the less concentrated side to the more concentrated side.
Osmotic pressure: the amount of pressure necessary to stop Osmosis.
Small molecules such as water can move through certain types of materials (membranes). .
where M = is molarity of solute particles, R = gas constant, T = temperature, and p = osmotic pressure
John A. Schreifels
Chemistry 212

23. Example 12.9
The average osmotic pressure of seawater, is about 30.0 atm at 25°C. Calculate the molar concentration of an aqueous solution of sucrose (C12H22O11) that is isotonic (same osmotic pressure) with seawater.
p = MRT
Rearrange to M = p/RT
M = 30.0 atm / ( L*atm/K*mol)(298 K)
M = 1.23 M

24. Using Colligative Properties to Find Molar Mass
Let’s say you add an unknown substance to the solvent, how can you find what it is? Find molar mass!
You have a solution of g glucose in 100g water. Freezes at -1.53°C. What is the molar mass? 1.86°C is Kf
Start by finding molality
m=DTf/Kf
m=1.53°C/1.86°C
m=0.823

25. Helpful example
Remember molality is equal to moles/kg solvent and molar mass is in grams
Convert from moles/kg to grams
So we have molality or .823 moles per 1kg (1000g) of water
We only have 100g (0.1 kg)in this case with 15g of glucose
(0.100 kg) x (0.823 mol / 1 kg) = mol
Now we can solve for molar mass
M= 15g / m
M=182 g/mol

26. Colligative Properties of Ionic Solutions
Colligative properties of solutions depends upon the total concentration of particles.
Each equation describing colligative properties must be modified to account for this with ionic solutions since each ionic compound gives more than one mole of ions for every mole of compound.
BP elevation:
Freezing point depression:
Osmotic pressure:
Where i = van’t Hoff factor.
van’t Hoff factor can be something other than integer under certain circumstances, but for completely ionic solutions is equal to the number of ions/ionic compound to be found in solution:
E.g. NaCl: i = 2; Na2SO4: i = 3;
John A. Schreifels
Chemistry 212

27. Dissociation Equations and the Determination of i
NaCl(s) 
Na+(aq) + Cl-(aq)
i = 2
AgNO3(s) 
Ag+(aq) + NO3-(aq)
i = 3
MgCl2(s) 
Mg2+(aq) + 2 Cl-(aq)
i = 3
Na2SO4(s) 
2 Na+(aq) + SO42-(aq)
AlCl3(s) 
Al3+(aq) + 3 Cl-(aq)
i = 4

28. Ideal vs. Real van’t Hoff Factor
The ideal van’t Hoff Factor is only achieved in VERY DILUTE solution.

29. Example 12.12
Calculate the osmotic pressure of a M potassium iodide (KI) solution at 25°C.
KI dissolves into 2 ions, K+ and I-
p = iMRT
p = (2.0)(0.010 M)( L*atm/mol*K)(298 K)
p = atm

30. Colloids Colloids are not true solutions
Colloids – dispersion of particles of one substance throughout a dispersion medium made of another substance
Particle size is on the order of 10 to 200 nm
Might be super-sized molecules (e.g., proteins) or aggregates of ions
Colloidal particles cannot be filtered and do not settle out of solution
Colloids exhibit the Tyndall effect (whereas solutions don’t)
The particles in a colloid are large enough to scatter light passing through)
Examples:
blood, milk, jelly, propofol

32. Soaps and Detergents polar head greasy tail
A soap is prepared by hydrolyzing fat with alkali
A detergent is a synthetic chemical with a structure similar to soap
Both have a polar (hydrophilic) head and a non-polar (hydrophobic, greasy) tail
polar head
greasy tail

33. Monolayers The greasy tails stick out of the surface of water
This breaks down the surface tension of water