Tang 08 equilibrium and solubility

Solubility Solids form from solution when: a solution is saturated two solutions are mixed to form a precipitate Both are dependent on an equilibrium between the solution and the solid.

K sp K sp is the solubility product constant for solutions that can form a solid. This is true for saturated solutions . Why? Only at saturation point is equilibrium achieved.

Example #1 Determine the K sp expressions for the following equilibria: Ag 2 CrO 4(s) <===> 2 Ag + (aq) + CrO 4 2- (aq) BaCrO 4(s) <===> Ba 2+ (aq) + CrO 4 2- (aq) Ag 3 PO 4(s) <===> 3 Ag + (aq) + PO 4 3- (aq) K sp = [Ag + (aq) ] 2 [CrO 4 2- (aq) ] K sp = [Ba 2+ (aq) ][CrO 4 2- (aq) ] K sp = [Ag + (aq) ] 3 [PO 4 3- (aq) ]

Molar Solubility The molar solubility for a compound is the concentration that is necessary for a solution to become saturated. Ex. One litre of water can dissolve 7.1x10 -7 mol of AgBr K sp values may be used to calculate molar solubility and vice versa.

Example #2 Silver bromide, AgBr, is the light sensitive compound in nearly all photographic film. At 25°C, one litre of water can dissolve 7.1x10 -7 mol of AgBr. Calculate the K sp of AgBr at 25°C.

Example #2 Silver bromide, AgBr, is the light sensitive compound in nearly all photographic film. At 25°C, one litre of water can dissolve 7.1x10 -7 mol of AgBr. Calculate the K sp of AgBr at 25°C. AgBr (s) <=> Ag + (aq) + Br - (aq) I 0 0 C +7.1x10 -7 +7.1x10 -7 E 7.1x10 -7 7.1x10 -7 Changes when it dissolves, but it is a solid. K sp = [7.1x10 -7 ][7.1x10 -7 ] K sp = 5.0x10 -13 .: K sp = 5.0x10 -13

Example #3 At 25°C, the molar solubility of PbCl 2 in a 0.10 M NaCl solution is 1.7x10 -3 M. Calculate the K sp for PbCl 2 .

Example #3 At 25°C, the molar solubility of PbCl 2 in a 0.10 M NaCl solution is 1.7x10 -3 M. Calculate the K sp for PbCl 2 . PbCl 2(s) <=> Pb 2+ (aq) + 2Cl - (aq) I 0 0.1 C +1.7x10 -3 +2(1.7x10 -3 ) E 1.7x10 -3 0.1034 NOTE: 0.1M Na + does not affect equilibrium K sp = [1.7x10 -3 ][0.1034] 2 K sp = 1.8x10 -5 .: K sp = 1.8x10 -5

Example #4 The solubility of iron (II) hydroxide, Fe(OH) 2 , is found to be 1.4x10 -3 g/L. What is the K sp value?

Example #4 The solubility of iron (II) hydroxide, Fe(OH) 2 , is found to be 1.4x10 -3 g/L. What is the K sp value? n = m M = (1.4x10 -3 g) (89.861g/mol) = 1.557961741x10 -5 mol Fe(OH) 2(s) <=> Fe 2+ (aq) + 2OH - (aq) I 0 0 C +1.55796x10 -5 +2(1.55796x10 -5 ) E 1.55796x10 -5 3.11592x10 -5 K sp = [1.55796x10 -5 ][3.11592x10 -5 ] 2 K sp = 1.5x10 -14 .: K sp = 1.5x10 -14

Example #5 What is the molar solubility of AgCl in pure water at 25°C when K sp = 1.8x10 -10 .

Example #5 What is the molar solubility of AgCl in pure water at 25°C when K sp = 1.8x10 -10 . AgCl (s) <=> Ag + (aq) + Cl - (aq) I 0 0 C +x +x E x x K sp = [x][x] K sp = x 2 1.8x10 -10 = x 2 1.34x10 -5 M = x .: the molar solubility is 1.34x10 -5 mol/L

Example #6 The K sp for magnesium fluoride, MgF 2 , has a value of 6.4x10 -9 . What is its solubility in g/L?

Example #6 The K sp for magnesium fluoride, MgF 2 , has a value of 6.4x10 -9 . What is its solubility in g/L? MgF 2(s) <=> Mg 2+ (aq) + 2F - (aq) I 0 0 C +x +2x E x 2x K sp = [x][2x] 2 = 6.4x10 -9 4x 3 = 6.4x10 -9 x = 6.4x10 -9 4 x = 1.16x10 -3 mol/L  m = nxM = 1.16x10 -3 mol x 62.301g/mol = 0.07226916g .: the solubility is 7.2x10 -2 g/L 3

Predicting Precipitation Similar to K eq problems, calculating Q can be used to determine whether a precipitate will form when provided with K sp . Three possibilities: Q < K sp (no ppt, unsaturated) Q = K sp (no ppt, saturated) Q > K sp (ppt forms, supersaturated)

Example #7 A student wished to prepare 1.0 L of a solution containing 0.015 mol of NaCl and 0.15 mol of Pb(NO 3 ) 2 . Knowing from the solubility rules that the chloride of Pb 2+ is insoluble, there was a concern that PbCl 2 might form. If the K sp for this reaction is 1.7x10 -5 , will a ppt form?

Example #7 A student wished to prepare 1.0 L of a solution containing 0.015 mol of NaCl and 0.15 mol of Pb(NO 3 ) 2 . Knowing from the solubility rules that the chloride of Pb 2+ is insoluble, there was a concern that PbCl 2 might form. If the K sp for this reaction is 1.7x10 -5 , will a ppt form? PbCl 2(s) <=> Pb 2+ (aq) + 2Cl - (aq) 0.15 0.015 K sp = [Pb 2+ (aq) ][Cl - (aq) ] 2 Q = [Pb 2+ (aq) ][Cl - (aq) ] 2 Q = [0.15][0.015] 2 Q = 3.375x10 -5 Q > K sp , so a precipitate WILL form .: a precipitate will form

Example #8 What possible precipitate might form by mixing 50.0 mL of 0.0010 M CaCl 2 with 50.0 mL of 0.010 M Na 2 SO 4 ? Will the precipitate form? (K sp = 7.1x10 -5 )

Example #8 What possible precipitate might form by mixing 50.0 mL of 0.0010 M CaCl 2 with 50.0 mL of 0.010 M Na 2 SO 4 ? Will the precipitate form? (K sp = 7.1x10 -5 ) CaSO 4 has low solubility.

Example #8 What possible precipitate might form by mixing 50.0 mL of 0.0010 M CaCl 2 with 50.0 mL of 0.010 M Na 2 SO 4 ? Will the precipitate form? (K sp = 7.1x10 -5 ) CaSO 4(s) <=> Ca 2+ (aq) + SO 4 2- (aq) C 1 V 1 =C 2 V 2 C 2 = C 1 V 1 V 2 C 2 = 0.0010M x 0.050L 0.100L C 2 = 5x10 -4 M C 2 = C 1 V 1 V 2 C 2 = 0.010M x 0.050L 0.100L C 2 = 5x10 -3 M Q = [Ca 2+ (aq) ][SO 4 2- (aq) ] Q = [0.0005][0.005] Q = 2.5x10 -6 Q < K sp , so a precipitate will NOT form (unsaturated) .: the possible precipitate, CaSO 4 , will not form

Example #9 Will a precipitate form if 20.0 mL of 0.010 M CaCl 2 are mixed with 30.0 mL of 0.0080 M Na 2 SO 4 ? (K sp = 2.45x10 -5 )

Example #9 Will a precipitate form if 20.0 mL of 0.010 M CaCl 2 are mixed with 30.0 mL of 0.0080 M Na 2 SO 4 ? (K sp = 2.45x10 -5 ) CaSO 4(s) <=> Ca 2+ (aq) + SO 4 2- (aq) C 2 = C 1 V 1 V 2 C 2 = 0.010M x 0.020L 0.050L C 2 = 4x10 -3 M C 2 = C 1 V 1 V 2 C 2 = 0.0080M x 0.030L 0.050L C 2 = 4.8x10 -3 M Q = [Ca 2+ (aq) ][SO 4 2- (aq) ] Q = [0.004][0.0048] Q = 1.92x10 -5 Q < K sp , so a precipitate will NOT form (unsaturated) .: a precipitate will not form

Tang 08 equilibrium and solubility

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Tang 08 equilibrium and solubility