Question

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

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

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.

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