Question

 The solubility of Ag₂CrO₄ at 25°C is 8.0 × 10^-5 moles litre^-1. Calculate its solubility product.

Answer 1

Let the solubility of Ag₂CrO₄ in s mol L^-1 

  `underset(S)Ag2CrO4 ⇌ underset(2s)2Ag^+ + underset(2s) CrO_4^-`

`K_(sp (Ag_(2)CrO4))=[Ag+]^2[CrO_4^(2-)] `

                        =[2s]² [s]

                        = 4s³

                       = 4 × (8.0 × 10^-5)³

                 K_sp = 2.048 × 10^-12