Calculate the molar solubility of lead (ll) chloride in 0.100 M aqueous solution of sodium chloride. The Ksp of lead (II) chloride is 1.6 x 10 ^-5
NaCl here is Strong electrolyte
It will dissociate completely to give [Cl-] = 0.100 M
At equilibrium:
PbCl2 <----> Pb2+ + 2 Cl-
s 0.100 + 2s
Ksp = [Pb2+][Cl-]^2
1.6*10^-5=(s)*(0.1+ 2s)^2
Since Ksp is small, s can be ignored as compared to 0.100
Above expression thus becomes:
1.6*10^-5=(s)*(0.100)^2
1.6*10^-5= (s) * 10^-2
s = 1.6*10^-3 M
Answer: 1.6*10^-3 M
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