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 

  ${\displaystyle \underset{S}{A}g2CrO4\rightleftharpoons \underset{2s}{2}A{g}^{+}+\underset{2s}{C}r{O}_4^{-}}$

${\displaystyle K_{sp\left(A{g}_{2}CrO4\right)}={[Ag+]}^2\left[Cr{O}_4^{2-}\right]}$

                        =[2s]² [s]

                        = 4s³

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

                 K_sp = 2.048 × 10^-12