The 𝐾sp of Ag2CrO4 is 1.12×10−12 . What is the solubility (in mol/L) of silver chromate in 1.20 M potassium chromate aqueous solution?

Answer 1

Hi Madison,

This is a problem involving the "common ion" effect. So, here we go ...

Ag₂CrO₄(s) ==> 2Ag⁺(aq) + CrO₄^2-(aq) ... Ksp = 1.12x10^-12

Ksp = [Ag⁺]²[CrO₄^2-]

The common ion would be CrO₄^2- since you have 1.20 M potassium chromate (K₂CrO₄). Because the [CrO₄^2-] is so large from the potassium chromate (1.2 M) we can ignore the [CrO₄^2-] contributed by the Ag₂CrO₄ as it will be very small. Now we can write the Ksp expression as follows:

Ksp = [Ag⁺]²[1.2 M]

1.12x10^-12 = (x)²(1.2)

x2 = 9.33x10^-13

x = 9.66x10^-7 M = [Ag⁺]

Since there are 2 mols of Ag⁺ ions for each 1 mol of Ag₂CrO₄...

the molar solubility of Ag₂CrO₄ = 9.66x10^-7 M ÷ 2 = 4.83x10^-7 M