Consider the following equilibria for lead. Calculate the total concentration of

Question

Consider the following equilibria for lead. Calculate the total concentration of lead for a system containing a fixed concentration (0.01 M) of KI. Note (s) denotes solid and (aq) denotes aqueous. Equation ?2 is not the reverse of the Ksp. PbI2 (s) ? Pb2+(aq) + 2I- (aq) - Ksp = 7.9 x 10-9 Pb2+ (aq) + I- (aq) ? PbI+ (aq) K1 = 1.0 x 102 Pb2+ + 2I- ? PbI2 (aq) ?2 = 1.4 x 103

Explanation / Answer

PbI2 (s) ------> Pb2+(aq) + 2I- (aq) Ksp = 7.9 x 10^-9 Ksp = [Pb+] [I-]^2 let x = [Pb2+] and let 2x = [I-] 7.9 x 10^- 9 = (x) (2x)^2 7.9 x 10^- 9= (x) (4x^2) 7.9 x 10^-9 = 4x^3 1.3 x 10^- 3 = x = [Pb^2+] Pb2+ (aq) + I- (aq) --------> PbI(aq) K1 = 1.0 x 10^2 PbI(aq) -------->Pb2+ (aq) + I- (aq) k1=1*10^-2 [I^?] = 1*10^-2 before - 0 .01 after - x x equilbrium x x+.01 K1=x*(x+.01) 1*10^-2=x^2+.01x x=9.512*10^-2 Pb2+(aq) + 2I-(aq) ---------> PbI2 (aq) K2 = 1.4 x 10^3 PbI2 (aq)--------->Pb2+(aq) + 2I-(aq) K2 = 7.14*10^-4 [I^?] = 1*10^-2 before - 0 2* .01 after - x 2* x equilbrium x 2*(x+.01) K2=x*2(x+.01) 7.14*10^-4=x*2(x+.01) we get x=1.45*10^-2 the total concentration of lead is:1.3 x 10^- 3+95.12*10^-3+14.5*10^-3=110.92*10^-3 is answer

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