Question

K_sp of Ag₂CrO₄ is 1.1 × 10⁻¹². what is the solubility of Ag₂CrO₄ in 0.1 M K₂CrO₄?

Answer 1

\[\ce{\underset{x}{Ag2CrO4} ⇌ \underset{2x}{2Ag^+} + \underset{x}{CrO^{2-}_4}}\]

x is the solubility of Ag₂CrO₄ in 0.1 M K₂CrO₄

\[\ce{\underset{0.1 M}{K2CrO4} ⇌ \underset{0.2 M}{2K^+} + \underset{0.1 M}{CrO^{2-}_4}}\]

[Ag⁺] = 2x

\[\ce{[CrO^{2-}_4]}\] = (x + 0.1) ≈ 0.1 ..........(∵ x << 0.1)

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

1.1 × 10⁻¹² = (2x)² (0.1)

1.1 × 10⁻¹² = 0.4x²

x² = `(1.1 xx 10^-12)/0.4`

x = `sqrt((1.1 xx 10^-12)/0.4)`

x = `sqrt(2.75 xx 10^-12)`

x = 1.65 × 10⁻⁶ M