a)
At equilibrium:
PbCl2 <----> Pb2+ + 2 Cl-
s 2s
Ksp = [Pb2+][Cl-]^2
1.6*10^-5=(s)*(2s)^2
1.6*10^-5= 4(s)^3
s = 1.587*10^-2 M
Answer: 1.6*10^-2 M
b)
Pb(NO3)2 here is Strong electrolyte
It will dissociate completely to give [Pb2+] = 0.17 M
At equilibrium:
PbCl2 <----> Pb2+ + 2 Cl-
0.17 +s 2s
Ksp = [Pb2+][Cl-]^2
1.6*10^-5=(0.17 + s)*(2s)^2
Since Ksp is small, s can be ignored as compared to 0.17
Above expression thus becomes:
1.6*10^-5=(0.17)*(2s)^2
1.6*10^-5= 0.17 * 4(s)^2
s = 4.851*10^-3 M
Answer: 4.9*10^-3 M
c)
NaCl here is Strong electrolyte
It will dissociate completely to give [Cl-] = 0.017 M
At equilibrium:
PbCl2 <----> Pb2+ + 2 Cl-
s 1.7*10^-2 + 2s
Ksp = [Pb2+][Cl-]^2
1.6*10^-5=(s)*(1.7*10^-2+ 2s)^2
Since Ksp is small, s can be ignored as compared to 1.7*10^-2
Above expression thus becomes:
1.6*10^-5=(s)*(1.7*10^-2)^2
1.6*10^-5= (s) * 2.89*10^-4
s = 5.536*10^-2 M
Answer: 5.5*10^-2 M
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