# Gases & colligative properties

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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 = 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?

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

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 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.

Solution

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.

Solution

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 H-bonding of latter Does it increase or decrease the latter’s IMF?

Salts Lower vapor pressure of solvent Make solvent less volatile

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

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 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

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

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

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

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 = g/mol

solution

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  = 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