1. Making a solution of lead (II) bromide PbBr 2 Pb 2+ Br - PbBr 2(s) → Pb 2+ (aq) + 2Br - (aq) ← K = [Pb 2+ ] [Br - ] 2 00 x2x x (2x) 2 K = 4x 3 = 4.6 x 10 -6 sp = Solubility product x = 0.010 M = [Pb 2+ ] 0.020 M = [Br - ] 0.010 M = Molar Solubility of PbBr 2 (because of the 1:1 ratio of Pb 2+ to PbBr 2 )

2. Find the solubility product of calcium phosphate if the molar solubility of this ionic solid is 1.64 x 10 -7. Ca 3 (PO 4 ) 2(s) 3Ca 2+ (aq) + 2PO 4 3- (aq) K sp = [Ca 2+ ] 3 [PO 4 3- ] 2 4.92 x 10 -7 M or [3 x 1.64 x 10 -7 ] 3.28 x 10 -7 M or [2 x 1.64 x 10 -7 ] K sp = [4.92 x 10 -7 ] 3 [3.28 x 10 -7 ] 2 K sp = 1.3 x 10 -32

3. Heating a saturated solution of lead (II) bromide PbBr 2 Pb 2+ Br - PbBr 2(s) → Pb 2+ (aq) + 2Br - (aq) ← K = [Pb 2+ ] [Br - ] 2 sp = Solubility product Hot Water energy +  H = + Must increase to accommodate the higher concentrations > 0.010 M > 0.020 M

4. Cooling a solution of lead (II) bromide PbBr 2 Pb 2+ Br - PbBr 2(s) → Pb 2+ (aq) + 2Br - (aq) ← K = [Pb 2+ ] [Br - ] 2 sp = Solubility product Cold Water energy +  H = + Reverse reaction is favored as the solubility product decreases These go down to accommodate the lower K sp value

5. Adding water to a saturated solution of lead (II) bromide PbBr 2 Pb 2+ Br - PbBr 2(s) → Pb 2+ (aq) + 2Br - (aq) ← K = [Pb 2+ ] [Br - ] 2 sp = Solubility product energy +  H = + Forward reaction is favored as the concentration of the products were forced down and Q < K sp The system makes more of these ions in response.

6. Barium sulfate is a salt which is used to enhance x-rays taken of the gastrointestinal tract. There is a problem in that Ba 2+ ions are poisonous. Look up the K sp value for barium sulfate and make a case for whether or not large amounts of Ba 2+ will be put into the body when barium sulfate is used to enhance x-ray pictures of the intestines.

7. Write the reaction equation for the dissociation of barium sulfate in water. BaSO 4(s) Ba 2+ (aq) + SO 4 2- (aq) K sp = 1.5 x 10 -9 How can we ensure that the Ba 2+ will be kept to a minimum in the body? ADD SULFATE!!!!! Shifts to produce more reactant (as dictated by LCP!)

8. Zn(OH) 2(s) → Zn 2+ (aq) + 2OH - (aq) ← K sp = [Zn 2+ ] [OH - ] 2 = 4.5 x 10 -17 Zn 2+ (aq) + 4 OH - (aq) → Zn(OH) 4 2- (aq) K = 3 x 10 15 ← Zn 2+ (aq) + 4 NH 3(aq) → Zn(NH 3 ) 4 2+ (aq) K = 1 x 10 9 ← H + (aq) + OH - (aq) → H 2 O (l) K = 1 x 10 14 ← Adding an acid (removes OH - ) and causes the forward reaction to be favored and dissolve more Zn(OH) 2 Adding additional hydroxide (removes Zn 2+ ) and causes the forward reaction to be favored and dissolve more Zn(OH) 2 Adding ammonia (removes Zn 2+ ) and causes the forward reaction to be favored and dissolve more Zn(OH) 2 This reaction occurs when a few drops of sodium hydroxide sol’n is added to a sol’n of zinc nitrate.

9. What if 5.00 mL of 0.000043 M barium nitrate would be mixed with 2.50 mL of 0.000065 M potassium sulfate….Would BaSO 4 ppt out? Initially [Ba 2+ ] = 0.000029 M [SO 4 2- ] = 0.000022 M Q = [Ba 2+ ] [SO 4 2- ] = 6.38 x 10 -10 NO PPT !!!!