Question

The solubility of Ag₂Cro₄ at 85°c is 8.0 ×10.5 m. Find the solubility product of Ag₂Cro₄.

Answer 1

Ag₂Cro₄ → 2 Ag⁺ + cro₄^-2

using law at mass action

K = \(\frac {[Ag^+]^2[Cro_4^{-2}}{Ag_2Cro_4}\)

= K [Ag₂Cro₄] = [Ag⁺]² [Crp₄^-2]

= Ksp = [Ag⁺]² [Cro₄^-2] (Ksp = K [Ag₂Cro₄])

Let say solubility of Ag₂Cro₄ = S

M = 8.0 x 10^-5m

∴ Ksp = [2s]² [s]

Ksp = 4s³

Ksp = 4 x (8.0 x 10^-5)³

Ksp = 2048 x 10^-15

Ksp = 2.048 x 10^-12

Hence the solubility product at Ag₂Cro₄ will be 2.048 x 10^-12