Question

Show by calculations whether MnS (K_sp = 5.6 ✕ 10^-16) will precipitate when a 0.01 M solution of Mn⁺² containing 0.02 M H⁺¹ is saturated with H₂S (0.1 M). For H₂S, K₁xK₂ = 6.3 ✕ 10^-22

a) [Mn⁺²][S^-2] K_sp; no, a precipitate will not form

b) [Mn⁺²][S^-2] > K_sp; yes, a precipitate will form     

c) [Mn⁺²][S^-2] < K_sp; yes, a precipitate will form

d) [Mn⁺²][S^-2] > K_sp; no, a precipitate will not form

Answer 1

Write down the step-wise acid dissociation of H₂S as below:

H₂S (aq) <====> H⁺ (aq) +HS^- (aq); K₁ = [H⁺][HS^-]/[H₂S] …..(1)

HS^- (aq) <====> H⁺ (aq) + S^2- (aq); K₂ = [H⁺][S^2-]/[HS^-] …..(2)

For the equilibrium,

H₂S (aq) <======> 2 H⁺ (aq) + S^2- (aq)

K = [H⁺]²[S^2-]/[H₂S] = K₁K₂ = 6.3*10^-22

Given [H⁺] = 0.02 M and [H₂S] = 0.1 M, we have,

6.3*10^-22 = (0.02)²[S^2-]/(0.1)

===> 6.3*10^-23 = (4.0*10^-4).[S^2-]

===> [S^2-] = 1.575*10^-19

Concentration of S^2- is 1.575*10^-19 M; find out the ion product of the ions for the dissociation:

MnS (s) <=====> Mn²⁺ (aq) + S^2- (aq)

Ion product, Q = [Mn²⁺][S^2-] = (0.01).(1.575*10^-19) = 1.575*10^-21

The ion product I much smaller than the K_sp. Solubility product, K_sp is a measure of the maximum concentration of ions that a solution can hold without precipitating the ions. Since Q is much smaller than K_sp, the solution can hold more ions without precipitating MnS. Hence, a precipitate will not form.

Ans: (a) [Mn²⁺][S^2-] < K_sp, no precipitate will form

(You missed the sign in a and I believe you should have a < sign there).