Question

The Ksp for Ag2SO4 = 1.20 X 10-5. What is the molar solubility of Ag2SO4 in a 2.20M solution of aluminum sulfate?

Answer 1

Solution:

The 2.2M solution of aluminum sulfate dissociates to produce 4.4M Al³⁺ and 6.6M SO₄^2-

Al₂(SO₄)₃ 2 Al³⁺ + 3SO₄^2-

When Ag₂SO₄ is added to 2.2M aluminum sulphate solution, it dissociates as follows

Ag₂SO₄ 2Ag⁺ + SO₄^2-

Let y be the molar solubility of Ag₂SO₄ in almunium sulphate solution.

At equilibrium

[Ag⁺] = 2y

[SO₄^2-] = (6.6 + y)

Therefore solubility product

K_sp = [Ag⁺] [SO₄^2-]

1.2 x 10^-5 = 2y(6.6 + y)

2y² + 13.2y - (1.2 x 10^-5) = 0

Solving this quadratic equation, we get

y = 9.1 x 10^-7 M