Question

What is the lead concentration of a saturated solution of lead (II) sulfate (ksp=6.3x10^-7) containing 0.020 molar Na₂SO₄? Include in your answer the reaction and mathematical equation that represents solubility product.

Answer 1

At equilibrium:

PbSO4 <----> Pb2+ + SO42-

   s s

Ksp = [Pb2+][SO42-]

6.3*10^-7=(s)*(s)

6.3*10^-7= 1(s)^2

s = 7.937*10^-4 M

Na2SO4 here is Strong electrolyte

It will dissociate completely to give [SO42-] = 0.02 M

At equilibrium:

PbSO4 <----> Pb2+ + SO42-

   s 2*10^-2 + s

Ksp = [Pb2+][SO42-]

6.3*10^-7=(s)*(2*10^-2+ s)

Since Ksp is small, s can be ignored as compared to 2*10^-2

Above expression thus becomes:

6.3*10^-7=(s)*(2*10^-2)

6.3*10^-7= (s) * 2*10^-2

s = 3.15*10^-5 M

Answer: 3.15*10^-5 M