Gases & colligative properties

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. 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 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 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? Example

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 N2then 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 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.

17. Solution

18. 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.

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. 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. Tbp = Kbp  msolute Tbp = 112.06°C - 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 Example

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 = 212.9962 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 = 0.08206 L  atm/(mol  K)  ideal gas law • T = in Kelvin •  = atm • Useful for measuring MM of biochemical macromolecules • Proteins and carbs