Question

The solubility of CdCO3(s) in 0.500 M KI(aq) is 3.0x10-4 M. Given the Kf for [CdI4]2-is 2.8x105 , what is the Ksp for CdCO3?

Answer 1

CdCO₃ (s) <===> Cd²⁺ (aq) + CO₃^2- (aq) K_sp = [Cd²⁺][ CO₃^2- ]

Cd²⁺ (aq) + 4 I^- (aq) <====> [CdI₄]^2- (aq) K_f = 2.8 * 10⁵

The solubility of CdCO₃(s) in 0.500 M KI(aq) is 3.0x10^-4 M

CdCO₃ (s) + 4 I^- (aq) <====> [CdI₄]^2- (aq)+ CO₃^2- (aq)   

Initial (M) : - 0.500 0 0

Change (M) : - -4x + x +x

Equilibrium (M): - (0.500-4x) x x

x is the solubility of CdCO₃ in 0.500 M I^{- =} 3.0*10^-4 M

K_eq = [CO₃^2-][CdI₄]^2- /[ I^-]⁴

[CO₃^2-] = 3.0*10^-4 M

[CdI₄]^2- = 3.0*10^-4 M

[I^-] = 0.500-4x = 0.500- 4(3.0*10^-4) = 0.4988 M

K_eq = (3.0*10^-4 M)(3.0*10^-4 M) / 0.4988 M

= 1.8*10^-7

We know that

K_eq = K_sp . K_f

K_sp = K_eq / K_f

= 1.8*10^-7/2.8x10⁵

= 6.4 * 10^-13