Determining the molar solubility of a substance (Chemistry)

Question

What is the molar solubility of silver sulfite in:

a.) Pure water

b.) .020 M aqueous solution of sodium sulfite

Can you please explain and show how it was solved, thank you.

Answer 1

The first thing you need to answer this question is the Ksp value for silver sulfite, Ag₂SO₃.

Looking it up on the internet, I find a value of 1.5x10^-14.

In water, you have the following equilibrium:

Ag₂SO₃(s) ==> 2Ag⁺(aq) + SO₃^2-(aq)

The Ksp expression is:

Ksp = 1.5x10^-14 = [Ag⁺]^2[SO₃^2-] and if we let x = [SO₃^2-], the [Ag⁺] = 2x b/c there are 2Ag⁺ for each SO₃^2-

1.5x10-14 = (2x)²(x) = 4x³

x³ = 3.75x10^-15

x = 1.55x10^-5 M = molar solubility in pure water

If you have an aqueous solution of sodium sulfite (Na₂SO₃), you now have a common ion in SO₃^2-.

Since the [SO₃^2-] is now 0.020 M, we substitute that value in the above equation and solve for [Ag⁺]

1.5x10^-14 = [Ag⁺]²[0.02]

[Ag⁺]² = 7.5x10^-13

[Ag⁺] = 8.66x10^-7 M (NOTE: the presence of a common ion DECREASES the solubility compared to that in pure water. This is to be expected based on Le Chatelier's principle)