Question

The solubility of CdCO_3(s) in 0.500 M KI_(aq) is 3.0x10^-4 M. Given the K_f for [CdI₄]^2- is 2.8x10⁵ , what is the K_sp for CdCO₃?

Answer 1

Sol :-

Formation of [CdI₄]^2- is :

Cd²⁺ (aq) + 4I^- (aq) -----------------> [CdI₄]^2- (aq) , Formation constant (K_f) = 2.8 x 10⁵ ............(1)

Now, Solubility of CdCO₃ in 0.500 M KI solution given = 3.0 x 10^-4 M. It means

Concentration of CO₃^2- at equilibrium stage = [CO₃^2-] = 3.0 x 10^-4 M

So, ICE table is :

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

Initial (I).............................................0.500 M.....................0.0 M...............................0.0 M

Change (C)..........................................-4α...........................+α....................................+α

Equilibrium (E)....................................(0.500-4α) M...............α M................................α M

Here, α = Amount dissociated per mole = [CO₃^2-] = 3.0 x 10^-4

Expression of equilibrium constant (K_eq) is :

K_eq = [CdI₄]^2- .[CO₃^2-] / [I^-]⁴

=α²/(0.500-4α)⁴

=(3.0 x 10^-4)² / (0.500-4x3.0x10^-4)⁴

= 1.45 x 10^-6

Now,

Aim equation : CdCO₃ (s) <------------> Cd²⁺ (aq) + CO₃^2- (aq) which can be obtained by subtracting equation (2) from equation (1) :

So,

K_sp of aim equation = K_eq/K_f

K_sp = (1.45 x 10^-6) / (2.8 x 10⁵ )

K_sp = 5.18 x 10^-12

Hence, Solubility product of CdCO₃ = 5.18 x 10^-12