Question

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

Answer #1

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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