1. Gases & colligative properties
Ch.14

2. Gases dissolving in liquids
Pressure and temperature influence gas solubility
Solubility directly proportional to gas pressure
Henry’s Law:
Sg = kHPg
Sg = gas solubility (M = mol/L)
kH = Henry’s law constant (unique to each gas; M/mm Hg)
Pg = partial pressure of gaseous solute (mm Hg)

3. Increase partial pressure  increase solubility

4. Example
27.0 g of acetylene gas dissolves in 1.00 L of acetone at 1.00 atm partial pressure of acetylene.
If the partial pressure of acetylene is increased to 6.00 atm, what is the solubility of acetylene in acetone in mol/L? MW of acetylene = g/mol
27.0 g x (mol/ g) x (1/1.00 L) = 1.04 M
Sg = kHPg
1.04 M = kH x 1.00 atm
kH = 1.04 M/atm
Sg = (1.04 M/atm) x 6.00 atm
= 6.24 M
Could also solve this by:
(Sg1/Pg1) = (Sg2/Pg2)
How did I come up with this?

5. Problem
The partial pressure of oxygen gas, O2, in air at sea level is 0.21 atm.
Using Henry’s Law, calculate the molar concentration of oxygen gas in the surface water (at 20°C) of a lake saturated with air given that the solubility of O2 at 20°C and 1.0 atm pressure is 1.38•10-3 M.

6. Solution

7. They call it “pop” in the Midwest
Drinks carbonated under high pressure
Above 90 atm
Under CO2 atmosphere
Once bottle opened, partial pressure of gas above soda plummets
CO2 solubility decreases drastically
Gas bubbles out of soln
Once the fizz is gone, it can never be regained
Truly, one of the existential tragedies of this universe

8. The bends Deeper diving has higher pressures
Must use breathing tank
If it contains N2 then higher pressure forces N2 to dissolve in higher amounts in blood
If ascension too fast, lower pressure causes N2 to start bubbling out of blood too quickly
Rupturing of arteries
Excruciatingly painful death
Must be rushed to hyperbaric chamber
Tanks now don’t use N2, but He
Why?

9. Effects of temp on solubility
Obviously, as temp increases, solubility decreases
Since increasing heat causes gases to dissolve out (endothermic)
 dissolving gases is an exothermic process

10. Another look at gas solubility: Le Châtelier’s Principle
Explains temperature relevance of solubility
For systems in equilibrium, change in one side causes system to counteract on other side:
Gas + liquid solvent  sat. soln + heat
So add heat, rxn goes to left by kicking out gas
Add gas, rxn goes to right by saturating soln & giving off heat

11. Solubility of solids based on temperature
In general, solubility increases w/ increasing temp
But exceptions
No general behavior pattern noted

12. Crystals
One can separate impure dissolved salts by reducing temperature
Impurity or desired product crystallizes out at specific temp as solubility collapses

13. Colligative properties
Vapor and osmotic pressures, bp, and mp are colligative properties
Depend on relative # of solute and solvent particles

14. Vapor Pressure Remember:
Equilibrium vapor pressure
Pressure of vapor when liq and vapor in equilibrium at specific temp
Vapor pressure of soln lower than pure solvent vapor pressure
Vapor pressure of solvent  relative # of solvent molecules in soln
i.e., solvent vapor pressure  solvent mole fraction

15. Raoult’s Law Psolution = Xsolvent  P°solvent
So if 75% of molecules in soln are solvent molecules (0.75 = Xsolvent)
Vapor pressure of solvent (Psolvent) = 75% of P°solvent

16. Problem
The vapor pressure of pure acetone (CH3COCH3) at 30°C is atm. Suppose 15.0 g of benzophenone, C13H10O (MW = g/mol), is dissolved in 50.0 g of acetone (MW = g/mol).
Calculate the vapor pressure of acetone above the resulting solution.

17. Solution

18. Problem
The vapor pressure of pure liquid CS2 is atm at 20°C. When 40.0 g of rhombic sulfur (a naturally occurring form of sulfur) is dissolved in 1.00 kg of CS2, the vapor pressure falls to atm.
Determine the molecular formula of rhombic sulfur.

19. Solution

20. Limitations of Raoult’s Law
Doesn’t take into consideration attractive forces in solns
For ideal soln (to right), forces between solute/solvent molecules = forces w/in pure solvent
Thus, Ptot = PA + PB
Like graph to right
Fine for similarly constructed molecules (hydrocarbons)
London dispersion forces are weakest

21. Solute-solvent > solv-solv
Decreases vapor pressure
decreased volatility
Get lower vapor pressure than calculated
Ex:
CHCl3 & C2H5OC2H5
H on former H-bonds to latter
Does it increase or decrease the latter’s IMF?

22. Solute-solvent < solv-solv
Increases vapor pressure
increased volatility
Get higher vapor pressure than calculated
Ex:
C2H5OH and H2O
Former disrupts H-bonding of latter
Does it increase or decrease the latter’s IMF?

23. Nonvolatile solute added to solvent
Salts
Lower vapor pressure of solvent
Make solvent less volatile

24. Nonvolatile solute added to solvent
Raises bp
Lowers mp
Why?
Adding more nonvolatile solute or increasing solute molality
decreases vapor pressure even more
Phase diagram to right
Pure water (black)
Adulterated water (pink)

25. Bp and molality relationship
Tbp = Kbp  msolute
Kbp = molal boiling pt elevation constant for solvent (°C/m)
Bp elevation, Tbp, directly proportional to solute molality

27. Antifreeze Propylene glycol 1,2-propanediol
Formerly used ethylene glycol
Phased out
Poisonous
Lowers melting pt
Increases boiling pt
Reduces risk of radiator “boiling over”
Appreciated during the summer months in the desert

28. Example Pure toluene (C7H8) has a normal boiling point of 110.60°C.
A solution of 7.80 g of anthracene (C14H10) in g of toluene has a boiling point of °C.
Calculate Kb for toluene.
Tbp = Kbp  msolute
Tbp = °C °C = 1.46°C
7.80g x (mol/178.23g) = 4.38 x 10-2 mol
(4.38 x 10-2 mol/ kg) = m
1.46°C/0.438 m = 3.33°C/m

29. Freezing point depression
Similarly, Tfp = Kfp  msolute
Kfp = molal fp depression constant (°C/m)
Antifreeze & CaCl2

30. Problem
Barium chloride has a freezing point of 962°C and a Kf of 108 °C/m.
A solution of 12.0 g of an unknown substance dissolved in 562 g of barium chloride gives a freezing point of 937°C.
Determine the molecular weight of the unknown substance.

31. Solution

32. Solutions containing ions: their colligative properties
Colligative properties based on amount of solute/solvent
Molality of ions depend on number of constituents in cmpd
Different for ionic vs. covalent cmpds
Ex:
NaCl ionizes into two ions
So 0.5 m NaCl has 0.5 x 2 m = 1 mtot
Benzene doesn’t ionize
So 0.5 m benzene = 0.5 mtot
Using equation w/out above factor will lead to values that are off

33. How to correct for it: the van’t Hoff factor
i = the number of solute particles after dissolving
Colligative properties are larger for electrolytes than for nonelectrolytes of the same molality
Why? (Hint: solve the below)
Give the i-values for: methanol, CaSO4, BaCl2
Tfp (measured) = Kfp  m  i

34. Problem
How many grams of Al(NO3)3 must be added to 1.00 kg of water to raise the boiling point to 105.0°C
Kb = 0.51 °C/m
MW = g/mol

35. solution

36. Osmosis
Net movement of water (solvent) from area of lower solute concentration to area of higher solute concentration across a semi-permeable membrane
Bio101

38. More…
Pressure of column of soln = pressure of water moving through membrane
Osmotic pressure = pressure made by column of soln = diff of heights
 = cRT
c = mol/L = M
R = L  atm/(mol  K)  ideal gas law
T = in Kelvin
 = atm
Useful for measuring MM of biochemical macromolecules
Proteins and carbs