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Equilibrium and Solubility
Solubility ,[object Object],[object Object],[object Object],[object Object]
K sp ,[object Object],[object Object],[object Object],Only at saturation point is equilibrium achieved.
Example #1 ,[object Object],[object Object],[object Object],[object Object],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 ,[object Object],[object Object],[object Object]
Example #2 ,[object Object]
Example #2 ,[object Object],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 ,[object Object]
Example #3 ,[object Object],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 ,[object Object]
Example #4 ,[object Object],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 ,[object Object]
Example #5 ,[object Object],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 ,[object Object]
Example #6 ,[object Object],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 ,[object Object],[object Object],[object Object],[object Object],[object Object]
Example #7 ,[object Object]
Example #7 ,[object Object],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 ,[object Object]
Example #8 ,[object Object],CaSO 4  has low solubility.
Example #8 ,[object Object],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 ,[object Object],[object Object]
Example #9 ,[object Object],[object Object],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

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