Question

Calculate the molar solubility of Ag2CrO4 (Ksp = 1.1 × 10-12) at 25 °C in various aqueous solutions.

Calculate the solubility in 0.249 M AgNO₃

Calculate the solubility in 0.134 M Na₂CrO₄

Answer 1

Ag2CrO4 (s) <------------> 2Ag+ + CrO4^-2

x 0 0 initial moles

- 2x x at equilibrium

Thus Ksp = [Ag+]²[CrO4-2] = 1.1x10^-12

Now

1) in 0.249M AgNO3

Ag2CrO4 (s) <------------> 2Ag+ + CrO4^-2

s 0.249 0 initial moles

- 0.249+x s at equilibrium

= 0.249 s as s is very very small compared to 0.249 due to common ion

Thus  Ksp = [Ag+]²[CrO4-2] = 1.1x10^-12

= (0.249)² [s] = 1.1x10^-12

Thus s = 1.774 x10^-11 M

Thus the solubility is very much decreased due to common ion effect.

2) in 0.134 M Na2CRO4

Ag2CrO4 (s) <------------> 2Ag+ + CrO4^-2

s 0 0.134 initial moles

- s 0.134 +s =0.134 at equilibrium

Thus  Ksp = [Ag+]²[CrO4-2] = 1.1x10^-12

= s² (0.134) = 1.1x10^-12

Hence s = 2.86x 10^-6 M

Thus the solubility is decreased than in pure water , but not as much with silver nitrate, due to common ion effect.