15. For Hg2I2 Ksp = 4.6 x 10^-29 a) Ve volume of KI added = 0.1 M x 25 ml/0.2 M
ID: 700294 • Letter: 1
Question
15. For Hg2I2 Ksp = 4.6 x 10^-29
a) Ve
volume of KI added = 0.1 M x 25 ml/0.2 M = 12.5 ml
[Hg2I2] formed = 0.1 m x 25 ml/(25 + 12.5) ml = 0.067 M
ICE chart
Hg2I2 <==> Hg2^2+ + 2I-
I 0.067 - -
C -x +x -2x
E 0.067-x x 2x
with x being a small change in precipitate concentration
Ksp = [Hg2^2+][I-]^2
4.6 x 10^-29 = (x)(2x)^2
x = 2.26 x 10^-10 M
p[Hg2] = -log(2.26 x 10^-10) = 9.65
b) 1.2 Ve
volume KI added = 12.5 x 1.2 = 15 ml
excess [I-] = 2.5 ml x 0.2 M/15 ml = 0.033 M
feed this value in Ksp equation,
[Hg2^2+] = Ksp/[I-]^2 = 4.6 x 10^-29/(0.033)^2 = 4.22 x 10^-26 M
p[Hg2] = -log[Hg2^2+] = 25.37
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For PbI2 Ksp = 7.9 x 10^-9
a) Ve
volume of KI added = 0.1 M x 25 ml/0.2 M = 12.5 ml
[PbI2] formed = 0.1 m x 25 ml/(25 + 12.5) ml = 0.067 M
ICE chart
PbI2 <==> Pb^2+ + 2I-
I 0.067 - -
C -x +x -2x
E 0.067-x x 2x
with x being a small change in precipitate concentration
Ksp = [Pb^2+][I-]^2
7.9 x 10^-9 = (x)(2x)^2
x = 1.25 x 10^-3 M
p[Pb] = -log(1.25 x 10^-3) = 2.90
b) 1.2 Ve
volume KI added = 12.5 x 1.2 = 15 ml
excess [I-] = 2.5 ml x 0.2 M/15 ml = 0.033 M
feed this value in Ksp equation,
[Pb^2+] = Ksp/[I-]^2 = 7.9 x 10^-9/(0.033)^2 = 7.25 x 10^-6 M
p[Pb] = -log[Pb^2+] = 5.14
Explanation / Answer
15. A 25.0 ml solution containing 0.10 M with respect to Pb,' and Hg ions are titrated with a 0.20 M KI solution. Calculate the pM values (where M-Pb or Hg) at a) V, and b) 1.2 Ve. Kap(Pbl2)-7.9x109
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