Determine the molar solubility for Ag₂CrO₄ (Ksp = 1.2 × 10⁻¹²).

Question

Determine the molar solubility for Ag₂CrO₄ (Ksp = 1.2 × 10⁻¹²).

Show ICE table and Ksp equation set up

Answer 1

Ag₂CrO₄(s) <==> 2Ag⁺(aq) + CrO₄^2-(aq) ... Ksp = 1.2x10^-12

..............................0.................0.........Initial

.............................+2x..............+x........Change

...............................2x................x.........Equilibrium

Ksp = [Ag^+]]²[CrO₄^2-]

Let x = [CrO₄^2-] and then [Ag⁺] = 2x

1.2x10^-12 = (2x)²(x)

4x³ = 1.2x10^-12

x³ = 3x10^-13

x = 6.69x10^-5 M = [CrO₄^2-] = molar solubility of Ag₂CrO₄