Question

Solubility product of Ag₂CrO₄ is 1 × 10⁻¹². What is the solubility of Ag₂CrO₄ in 0.01 M AgNO₃ solution?

Answer 1

Solubility product of Ag₂CrO₄,

K_sp = 1 × 10⁻¹²

\[\ce{\underset{(s)}{Ag2CrO4_{(s)}} ⇌ \underset{(2s)}{2Ag^+_{( aq)}} + \underset{(s)}{CrO^{2-}_{4(aq)}}}\]

\[\ce{\underset{(0.01 M)}{AgNO3_{(s)}} ⇌ \underset{(0.01 M)}{Ag^+_{( aq)}} + \underset{(0.01 M)}{NO^-_{3(aq)}}}\]

[Ag⁺] = 2s + 0.01

0.01 >> 2s

∴ [Ag⁺] = 0.01 M

\[\ce{[CrO^{2-}_4]}\] = s

K_sp = \[\ce{[Ag^+]^2 [CrO^{2-}_4]}\]

1 × 10⁻¹² = (0.01)² (s)

(s) = `(1 xx 10^-12)/((10^-2)^2)`

 = `(10 xx 10^-12)/(10^-4)`

= 1 × 10⁻⁸ M