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1. Chapter 12 - Solutions A) Online HW 12 Due by 11:30 pm Thursday, March 8.
B) End-of-Chapter Problems.
C) Final exam on Monday, March 12 from 1-3 pm. There will be questions from all the chapters we covered this quarter.
2. End-of-Chapter ProblemsPages 516 - 522 3 4 5 14 15 16 19 21 22 23
24 28 32 35 40 41 43 47 49
51 56 59 69 73 75 77 79 81 99
3. I. Solutions A. Types Definitions:
Solution Homogeneous mixture of two or more substances.
Solvent Single solution component present in greatest amount.
Solute(s) Solution component(s) present in smaller amounts.
Solubility Maximum amount of solute that can dissolve at a given temperature (Note: Is a dynamic equilibrium)
Solid Solute Dissolved Solute
Saturated Solution Solution with maximum amount of dissolved solute.
Hydration - Attraction of solute ions for water molecules.
Solvation - Attraction of solute ions for any solvent molecules.
Miscible - Fluids that completely dissolve in each other.
4. I. Solutions A. Types Types: Solute Solvent Examples
1 Gas Gas Air (O2 Ar CO2 in N2 )
2 Gas Liquid Soda Water (CO2 in H2O)
3 Liquid Liquid Vinegar (Acetic A. in H2O)
4 Solid Liquid Salt Water (NaCl in H2O)
5 Liquid Solid Hg in Ag old dental filling
6 Solid Solid Brass Zn, Sn, Pb, Fe in Cu
5. I. Solutions B. Rules General rule for solutions is like dissolves like in terms of polarity. Polar solvent can dissolve polar solutes and non-polar solvent can dissolve non-polar solutes.
Polar CH3OH will dissolve as a molecule in polar H2O.
Ionic KI will dissolve as ions in polar H2O.
Non-polar C6H14 will not dissolve in polar H2O.
Non-polar C6H14 will dissolve in non-polar CCl4.
6. I. Solutions B. Some Solubility Rules for ionic compounds from 121. (Know These Rules) All group IA (1), ammonium, acetate & nitrates are soluble. Examples:
NaI NH4Cl Ca(C2H3O2)2 Pb(NO3)2 K3AsO4 = soluble
All Chlorides - Iodides are soluble except silver, mercury & lead. Examples:
CrCl3 = soluble PbBr2 = insoluble AgCl = insoluble
All sulfides are insoluble except for first rule above. Examples:
(NH4)2S = soluble CdS = insoluble Ag2S = insoluble
All hydroxides insoluble except for first rule above & Ca, Sr, Ba. Examples:
Ni(OH)2 = insoluble Al(OH)3 = insoluble Ba(OH)2 = soluble
7. I. Solutions B. Some Solubility Rules for ionic compounds cont. - Why are some ionic & polar compounds insoluble and why are some dissolutions in water endothermic or exothermic?
- Dissolving involves two factors:
A - Break lattice structure = Energy required/added.
B - Solvation of released parts = Energy released.
Dissolving many compounds is exothermic: B > A
Some compounds are insoluble & some endothermic: A > B
Examples: 1) CuSO4 or H2SO4 (caution) in water = Exothermic (B >> A)
2) NH4NO3 in water = Endothermic (A > B)
3) AgCl, BaSO4, CuS are insoluble ionic compounds (A >> B)
Note: Heating a mixture of an insoluble ionic compound & water usually increases its solubility.
8. I. Solutions B. General Rules Polar CH3OH dissolves in H2O; new H bonds formed & E released.
Hydration energy may not be enough to overcome lattice energy Insoluble AgCl in water.
Ionic LiF dissolves in H2O since polar H2O solvates ions (Hydration) which releases enough E to pull ions out of LiF lattice.
9. I. Solutions B. General Rules - Ivory Soap in water: CH3 - (CH2)16 - C O- Na+ =
- 1) Water dissolves ionic/polar compounds & 2) Non-polar part of soap dissolves non-polar substances like grease. Soap forms micelles.
10. I. Solutions B. Rules A saturated gaseous solution is in equilibrium as illustrated:
CO2 (aq) CO2 (gas)
Le Chateliers Principle: A system at equilibrium will rearrange to relieve any stress. (Know this)
Apply pressure & above equilibrium will rearrange to remove extra pressure (dissolve gas).
11. I. Solutions B. Rules - Henrys Law In 1803 William Henry studied the effect of pressure on solubility.
Henrys Law: Solubility of a gas (S) a to the partial pressure of the gas above the solution: S1 = k x P1 or S2 = P2 (Know this)
Important to a scuba diver; Henry's Law tells us that at higher pressure our blood will absorb more gas.
Example: A diver is at 3.0 atm (P1) where solubility of N2 in blood is 53 ug/L (S1). What is the solubility (S2) of N2 in blood when the diver comes to the surface at 1.0 atm (P2) ?
S2 = P2 S2 = P2 x S1 = 1.0 atm x 53 ug/L = 18 ug/L (18 ppb)
S1 P1 P1 3.0 atm
12. II. Solution Concentration Terms A. Review - (know these) 1. W / W % = g solute x 100 % (also have W / V & V / V %)
- 18.0 g NaCl is added to 200. g of water. What is W / W % NaCl?
% = 18 g NaCl x 100 % = 8.26 %
218 g soln
2. ppm, ppb, ppt ppm = ug/mL ppb = ng/mL ppt = pg/mL
(Note: g = mL for dilute, aqueous solutions; so, ppm also = ug/g)
- 4.2 x 10-8 g of Pb+2 is dissolved in 1000 mL H2O. What is ppt Pb+2 ?
4.2x10-8 g 1012 pg = 42 pg = 42 ppt
1000 mL g mL
13. II. Solution Concentration Terms A. Review 3. Mole Fraction, X = Xa / Xt Note: X is always = 1.00
- 0.12 mol N2 is mixed with 0.33 mol O2 and 0.01 mol Ar. What is X of N2
XN2 = 0.12 mole N2 = 0.26 (units cancel)
0.12 + 0.33 + 0.01 mole Total
Note: Sum of all Xas in a solution = 1.0000
4. Molarity, M = moles solute / L solution M = m / L
- 0.80 g of NaOH is dissolved in 200. mL. Calculate the M
M = 0.80 g NaOH x 1 mol / 40. g NaOH = 0.10 mol = 0.10 M
0.200 L L
14. II. Solution Concentration Terms B. Molality, m 5. molality, m = moles Solute
- This concentration term is strange, but useful for colligative property calculations; its the only one dealing with amount / Solvent.
Example: 5.7 g of glucose (MW=180) is in 30.9 g of solution. Calculate m.
- 5.7 g glucose x (1 mol / 180 g) = 0.032 mol glucose (solute)
- 30.9 g soln - 5.7 g solute = 25.2 g solvent = 0.0252 kg solvent
m = moles glucose = 0.032 mol glucose = 1.3 m
kg solvent 0.0252 kg solvent
15. III. Colligative Properties A. Vapor Pressure Lowering (VPL) - Colligative Property = Property that depends only on # of solute particles (not on the type of the solute particles).
- Examples: VPL, BP elevation, FP depression, Osmosis
- VPL = a nonvolatile solute lowers the vapor pressure of a solvent. Why?
- VP relationship = Raoults Law: Pa = Psa x Xa (Know this)
Pa is VP of solvent in the solution
Psa is VP of pure solvent
Xa is the mole fraction of the Solvent.
- Note: are other forms of Raoults Law.
16. III. Colligative Properties A. Vapor Pressure Lowering (VPL) Example: What is VP & change in VP of 1.00 mol H2O that contains 0.100 mole NaI at 20. oC? Given: VP of H2O (Psa) at 20. oC = 17.5 Torr
Pa = Psa x Xa = 17.5 torr x Xa Need mole fraction of water:
**0.100 mol NaI yields 0.200 mols of particles (Na+ & I-) when it dissolves.
**Total mols of particles = 1.00 mol H2O + 0.200 mol ions = 1.20 mols
Xa = 1.00 mole H2O / 1.20 mole = 0.833 [mole fraction of water]
Pa = Psa x Xa
Pa = 17.5 Torr x 0.833 = 14.5 Torr
?VP = 17.5 - 14.5 = 3.0 Torr
17. III. Colligative Properties B. BP Elevation & FP Depression an added nonvolatile solute lowers MP & raises BP of a solvent; the change in BP or MP depends on the type of solvent and the number of particles of solute dissolved (colligative property).
For MP depression: ?Tf = Kf x m
For BP elevation: ?Tb = Kb x m
- ?Tf & ?Tb are changes in BP or MP & m = molality of SOLUTE particles.
- Kf & Kb are constants in units of oC/m; are unique for each solvent:
Kf = 1.86 oC/m for water Kf = 40.0 oC/m for camphor
Kb = 0.512 oC/m for water Kb = 3.12 oC/m for acetic acid
18. III. Colligative Properties B. BP Elevation & FP Depression Example: Calculate the freezing point of a 50.0 % (W/W) ethylene glycol (C2H6O2 or EG ) solution in your car radiator? 50.0% - assume 100. g of ethylene glycol is added to 100. g of H2O. Kf = 1.86 oC/m for H2O
?Tf = Kf x m = (1.86 oC/m) x ? m
Need m of ethylene glycol (MW = 62.0 g/mole) in 50.0 % aqueous soln:
( moles EG/kg H2O = moles EG / 0.100 kg H2O )
- 100. g C2H6O2 x 1 mol EG / 62.0 g EG = 1.61 moles C2H6O2
- m = 1.61 moles C2H6O2 / 0.100 kg H2O = 16.1 mols EG/kg water = 16.1 m
?Tf = Kf x m = (1.86 oC/m) x 16.1 m = 30.0 oC .
MP = (0.0-30.0)oC = -30.0oC. Car engine protected to -30 oC or -22 oF
19. III. Colligative Properties B. BP Elevation & FP Depression FP or BP changes can be used to determine MW of low MW molecules.
Example: 0.99 g sorbitol in 0.010 kg water has a MP of -1.0 oC. Calculate the MW of sorbitol.
- Problem solving method: MW = g / mole. Have g (0.99 g) & need moles
?Tf = Kf x m Given: Kf = 1.86 oC/m for water
m = ?Tf / Kf = 1.0 oC / 1.86 oC/m = 0.54 m (mole sorbitol per kg water)
- (0.54 mol sorbitol / kg water) x 0.010 kg water = 0.0054 mol sorbitol
- MW = g/mole = 0.99 g sorbitol / 0.0054 mole sorbitol = 180 g / mole (2 SF)
20. III. Colligative Properties B. BP Elevation & FP Depression Question: An aqueous 1.0 molal solution of ethanol (CH3-CH2-OH) lowers the mp of water by 1.86 oC, but an aqueous 1.0 molal solution of NaCl lowers the mp of water by 3.72 oC (twice that of ethanol). Why?
Answer: 1.0 mol of ethanol yields 1.0 mols of particles in water; 1.0 mol of NaCl in water yields 2.0 mols of particles (1.0 mol of Na+ & 1.0 mol of Cl-).
Better equation: ?Tf = i x Kf x m
i = vant Hoff factor (pg 508): i = # moles of ions per mole of compound i = 1.0 for nonelectrolyte (CH3OH)
i > 1.0 for electrolyte (2.0 for NaCl)
Question: Why is i = 1.02 for 1.00 M acetic acid?
21. III. Colligative Properties C. Osmosis - Jacobus Vant Hoff received 1st nobel prize
in chem. in 1901 for work on solns. Definition of
Vant Hoff Factor: i = # moles of ions per mole
of compound. (Know this)
- Summary of Colligative Property Equations using the Vant Hoff factor:
- Note: assume i = 1.00 unless problem dictates another value.
?Tf = i x Kf x m
?Tb = i x Kb x m
p = i x M x R x T (for osmotic pressure) (Know These)
Note: i is automatically included in Raoults Law.
22. III. Colligative Properties C. Osmosis - Osmosis is the net flow of solvent molecules from a pure solvent through a semipermeable membrane into a solution.
- Osmotic Pressure (p) is a colligative property & equals the pressure that, when applied to the solution, just stops osmosis.
23. III. Colligative Properties Examples & Notes Note: During osmosis, water molecules go from the purest water to the less pure water from less concentrated to more concentrated.
Skin gets pruney when you swim in ocean or take a long, soapy bath.
Red blood cells puff up & burst when placed in deionized water.
Osmotic pressure can be very large and help water to reach to the top of tall trees.
We can use reverse osmosis to purify sea water.
Can use p to obtain MW of large molecules.
24. III. Colligative Properties Osmosis & Colloids - p = i x M x R x T (Know this)
M = Molarity (mols / Lsoln) R = Gas Constant = 0.0821 La/Kmol T in K
- Example: What is the p of a 0.2 M (i=1.0) soln. at 300 K?
p = 1.0 x 0.2 mol/L x [0.0821 Latm/Kmol] x 300 K = 5 atm
(5 atm can support a 1.0 in2 column of water over 150 feet high)
- p can be used to determine M of dilute solutions due to large values of p. Useful for determining MW of dilute solutions of high MW molecules - proteins, carbohydrates, nucleic acids
25. III. Colligative Properties Osmosis MW example Problem: Hemocyanin (Hy) is a protein found in crabs. If 1.5 g of Hy dissolved in 0.250 L of water has a p of 0.00370 atm at 277 K, then what is the MW of Hy? Given i = 1.1 for Hy.
p = i x M x R x T MW Hy = g/mol = 1.5g/mol M = p / ( i x R x T)
1) M of Hy = (0.00370 a) = 1.48 x 10-4 mol/L
1.1 x (0.0821 La/Km) x 277 K
2) 1.48 x 10-4 mol/L x 0.250 L = 3.70 x 10-5 mol Hy
3) MW = g / mole = 1.5 g / 3.70 x 10-5 mol Hy = 41000 g/mol
Note: If used mp or bp, the ?T of this same solution would have been ~10-4 oC; much too small to measure accurately.
26. IV - Colloids - Colloid = dispersion (not solution) of particles of one substance throughout another substance. (Know this)
- Colloid differs from solutions in that dispersion particles are usually between 1 and 200 nm in diameter they are not in solution, but they are small enough to stay in suspension.
- Colloids will give Tyndall effect = the scattering of a light beam by the colloidal particles. (Know this)
-Three names for colloids:
1) aerosols = liquid/solid dispersion in a gas;
2) emulsion = liquid dispersed in a liquid;
3) sol = solid dispersed in a liquid
27. IV Colloids; common names 1) Aerosols = liquids or solids dispersed in a gas
Examples: Fog (liquid water in air)
Dust in air (solid dirt in air)
2) Emulsions = liquids dispersed in a liquid
Examples: Mayonnaise (oil in water)
Milk (fat in water)
3) Sols = solids dispersed in a liquid
Examples: Jell-O (collagens in water)
Milk (casein in water)
28. IV . Colloids - Tyndall Effect