Question

What is the molar solubility of lead (II) sulfate (Ksp=2.53 x 10^-8), when it is added to a solution that contains 0.08 M sodium sulfate.

Answer 1

PbSO_4(s) ------> Pb⁺²_(aq) + SO₄^2-_(aq)

Na₂SO₄ = 0.08 M . Na₂SO_4(s) -----> 2Na⁺_(aq) + SO₄^-2_(aq)

PbSO₄ ----> Pb⁺² + SO₄^2-

Initial . 0. 0.08 M

Change . +x. +x

Equilibrium. X . 0.08+x

Ksp = 2.53 x 10^-8 = [Pb⁺²] [SO₄^-2] = x (0.08+X)

2.53 x 10^-8 = 0.08X + X²

X² + 0.08X - 2.53 x10^-8 = 0

X1 = -b + b²-4ac)/2a = -0.08 + 0.08²-4×1×2.53 × 10^-8/2 ×1 = -0.08 + 0.0060001012/2 = +3.16249 x10^-7 ~ 3.165 x10^-7 M

X2 = -b - b²-4ac/2a = -0.08 - 0.08² -4 ×1×2.53 x 10^-8/2×1 =

= -0.08 - 0.0060001012/2 = -0.08

Molar solubility of lead sulphate = 3.165 x 10^-7 M

2) Or simply we can approximate the calculation

2.53 x 10^-8 = X (0.08+X)

0.08+X ~ = 0.08

2.53 x 10^-8 = 0.08 X

X = 2.53 x 10^-8 M/ 0.08 = 31.625 x 10^-8 M = 3.1625 x 10^-7 M

Molar solubility of lead sulphate = 3.1625 x 10^-7 M

There is no much difference in the above two calculations