Question

What concentration of SO32– is in equilibrium with Ag2SO3(s) and 5.20 × 10-3 M Ag ? The Ksp of Ag2SO3 can be found here.

Answer 1

Here, the reaction can be represented as,

Ag₂SO₃(s) 2 Ag⁺ + SO₃^2-

Ksp of Ag₂SO₃ = 1.50 x 10^-14

We have,

Ksp = [Ag⁺]² [SO₃^2-]

Thus, if x moles of Ag₂SO₃(s) dissolve, we will have 2x moles of Ag⁺ and x moles of SO₃^2-

Therefore, here [Ag⁺ ] = (2x + 5.20 x 10^-3 M)

Ksp = 1.50 x 10^-14 = (2x + 5.2 x 10^-3)² x (x)

Here, since Ksp is very small, x <<< 5.20 x 10^-3 and x can be neglected in comparison with 5.20 x 10^-3

Therefore,

1.50 x 10^-14 = x (5.20 x 10^-3)²

x = 1.50 x 10^-14 / (5.20 x 10^-3)² = 5.54 x 10^-10 M

Concentration of SO₃^2- is in equilibrium with Ag₂SO₃(s) and 5.20 x10^-3 M Ag⁺ = 5.54 x 10^-10 M