In aqueous solution, Cd 2+ will precipitate out of solution in the form of CdS i

Question

In aqueous solution, Cd2+will precipitate out of solution in the form of CdS in the presence of hydrogen sulfide (H2S). The Ksp of CdS = 1.0 x 10-28 and Ksp of H2S = 1.0 x 10-26 for the reaction: H2S (aq) = 2 H+(aq) + S2-(aq)

a.   Write the balanced net ionic equation for the reaction of cadmium ion with hydrogen sulfide to form cadmium sulfide solid.

b.   Write the equilibrium expression for the equation in part a.

c.   Calculate Koverall for the reaction in part a.

d.   Calculate the [Cd2+] in a solution saturated with H2S (0.10 M) at a pH = 2.00 ([H+] = 0.010 M).

Explanation / Answer

a) The balanced ionic equation for the formation of cadmium sulfide from cadmium ion and hydrogen sulfide is given as

Cd2+ (aq) + H2S (aq) <=======> CdS (s) + 2 H+ (aq) ……..(1)

b) The equilibrium constant for the reaction is given as

Koverall = [H+]2/[Cd2+][H2S] (ans)

Note that the concentration of a solid is not written in the expression for the equilibrium constant.

c) Write down the two ionization reactions as below.

CdS (s) <=======> Cd2+ (aq) + S2- (aq) ……..(2)

Ksp = [Cd2+][S2-]

H2S (aq) <=======> 2 H+ (aq) + S2- (aq) …….(3)

Ka = [H+]2[S2-]/[H2S]

Subtract (2) from (3) and obtain

Cd2+ (aq) + H2S (aq) <======> CdS (s) + 2 H+ (aq)

Koverall = [H+]2/[Cd2+][H2S] = ([H+]2/[H2S])*(1/[Cd2+])

= ([H+]2[S2-]/[H2S])*1/([Cd2+][S2-]

= Ka*1/Ksp

= (1.0*10-26)*1/(1.0*10-28) = 100 (ans).

d) We have [H2S] = 0.10 M and pH = 2.00.

We define pH as pH = -log [H+]

====> 2.00 = -log [H+]

====> [H+] = antilog (-2.00) = 0.01 M

Since Ka (H2S) is extremely small, we may assume the equilibrium concentration of H2S to be equal to 0.10 M; hence, we have

Koverall = [H+]2/[Cd2+][H2S]

====> 100 = (0.01)2/[Cd2+].(0.10)

====> [Cd2+] = (0.01)2/(100*0.10) = 1.0*10-5

The concentration of Cd2+ at equilibrium is 1.0*10-5 M (ans).

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