Gases & colligative properties Ch.14

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)







Increase partial pressure ⇒ increase solubility

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 = 26.037 g/mol

1. 2. 3.

27.0 g x (mol/26.037 g) x (1/1.00 L) = 1.04 M Sg = kHPg 1.04 M = kH x 1.00 atm

4.

1. kH = 1.04 M/atm Sg = (1.04 M/atm) x 6.00 atm 1. = 6.24 M




Could also solve this by: – (Sg1/Pg1) = (Sg2/Pg2) – How did I come up with this?

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.

Solution −3

S2 1.38 ×10 M = 1.0atm 0.21atm −4 S2 = 2.9 ×10 M

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

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?

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

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

Solubility of solids based on temperature


In general, solubility increases w/ increasing temp – But exceptions – No general behavior pattern noted

Crystals


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

Colligative properties
Vapor

and osmotic pressures, bp, and mp are colligative properties – Depend on relative # of solute and solvent particles

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

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

Problem
The

vapor pressure of pure acetone (CH3COCH3) at 30°C is 0.3270 atm. Suppose 15.0 g of benzophenone, C13H10O (MW = 182.217 g/mol), is dissolved in 50.0 g of acetone (MW = 58.09 g/mol). – Calculate the vapor pressure of acetone above the resulting solution.

Solution mol solute :15.0g × = 0.0823mol 182.217g mol solvent : 50.0g × = 0.861mol 58.09g 0.861mol X solvent = = 0.913 0.861mol + 0.0823mol
Psolution = X solvent × P solvent = 0.913 × 0.3270atm = 0.2986atm

Problem
The

vapor pressure of pure liquid CS2 is 0.3914 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 0.3868 atm. – Determine the molecular formula of rhombic sulfur.

Solution Psolution = X solvent × P
solvent 0.3868atm = X solvent × 0.3914atm X solvent = 0.9882 mol solvent :1.00kg = 1.00 ×10 g × = 13.1mol 76.143g 13.1mol 0.9882 = 13.1mol + mol rhombic sulfur 3

mol rhombic sulfur = 0.156 40.0g 256g = 0.156mol mol 256g mol × = 7.98 ≈ 8 mol 32.066g sulfur S8

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

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?

Solute-solvent < solv-solv


Increases vapor pressure – increased volatility







Get higher vapor pressure than calculated Ex: – C2H5OH and H2O – Former disrupts Hbonding of latter  Does it increase or decrease the latter’s IMF?

Nonvolatile solute added to solvent
Salts

– Lower vapor pressure of solvent – Make solvent less volatile

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)

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

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

Example





Pure toluene (C7H8) has a normal boiling point of 110.60°C. A solution of 7.80 g of anthracene (C14H10) in 100.0 g of toluene has a boiling point of 112.06°C. – Calculate Kb for toluene.

1. 2. 3. 4. 5.

∆Tbp = Kbp • msolute ∆Tbp = 112.06°C 112.06 110.60°C = 1.46°C 7.80g x (mol/178.23g) = 4.38 x 10-2 mol (4.38 x 10-2 mol/0.1000 kg) = 0.438 m 1.46°C/0.438 m = 3.33°C/m

Freezing point depression ∆Tfp = Kfp • msolute
Kfp = molal fp depression constant (°C/m)
Antifreeze & CaCl2
Similarly,

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.

Solution ∆T = 962 C − 937 C = 25 C






108 C
∆T = 25 C = ×m m m = 0.23 moles solute 0.23 = 0.562g solvent moles solute = 0.13

12.0g 92g = 0.13mol mol




Solutions containing ions: their colligative properties



Colligative properties based on amount of solute/solvent Molality of ions depend on number of constituents in





Different for ionic vs. covalent cmpds Ex: NaCl ionizes into two ions

1.

cmpd

– So 0.5 m NaCl has 0.5 x 2 m = 1 mtot 2.

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

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

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 = 212.9962 g/mol

solution ∆T = 105.0
C - 100.0
C = 5.0
C
0.51 C
5.0 C = × m× 4 m m = 2.5 moles solute 2.5 = 1.00kg solvent moles solute = 2.5

212.9962g 2.5mol × = 530g Al(NO3 ) 3 needed mol

Osmosis
Net

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

More…









Pressure of column of soln = pressure of water moving through membrane Osmotic pressure = pressure made by column of soln = diff of heights Π = cRT c = mol/L = M R = 0.08206 L • atm/(mol • K) ⇒ ideal gas law T = in Kelvin Π = atm Useful for measuring MM of biochemical macromolecules – Proteins and carbs