Question

Determine [Zn2+], [CN−], and [HCN] in a saturated solution of Zn(CN)2 with a fixed pH of 3.530. The ?sp for Zn(CN)2 is 3.0×10−16. The ?a for HCN is 6.2×10−10.

[Zn2+]=

[HCN]=

[CN−]

Answer 1

Given pH = - log[H⁺(aq)] = 3.530

=> log[H⁺(aq)] = - 3.530

=> [H⁺(aq)] = 10^-3.530

=> [H⁺(aq)] = 2.95×10^-4 M

The balanced equations involved are:

Zn(CN)₂(s) < -----> Zn²⁺(aq) + 2CN^-(aq) ; Ksp = 3.0×10^-16

2CN^-(aq) + 2H⁺(aq) <-----> 2HCN(aq) ; K = (1/Ka)² = (1/6.2×10^-10)² = 2.60*10¹⁸

---------------------------------------------------------------------------------------------------------------

Adding the above two equation gives:

Zn(CN)₂(s) + 2CN^-(aq) + 2H⁺(aq)< -----> Zn²⁺(aq) + 2CN^-(aq) + 2HCN(aq) ; K" = Ksp * K

cancelling spectator ions:

--Zn(CN)₂(s) + 2H⁺(aq)< -----> Zn²⁺(aq) + 2HCN(aq) ; K" = Ksp * K = 3.0×10^-16 * 2.60*10¹⁸ = 780.44

I: ---------------- 2.95×10^-4 M --------------------------------

C: ------------------------------------- +X --------- +2X

E: ---------------2.95×10^-4 M ------ X M ------- 2X M

K" = 780.44 = [Zn²⁺(aq)] * [HCN(aq)]² / [H⁺(aq)]²

=> K" = 780.44 = X * X² / [2.95×10^-4]²

=> 780.44 * [2.95×10^-4]² = X³

=> X = cube root (6.79×10^-5) = 0.0408 M

Hence [Zn²⁺(aq)] = X = 0.0408 M (Answer)

[HCN(aq)] = 2X = 2*0.0408 M = 0.0816 M (Answer)

For HCN dissociation:

--HCN(aq) <----> CN^-(aq) + H⁺(aq) ; K = 6.2×10^-10

E: 0.0816 M ------ ?? --------2.95×10^-4

K = 6.2×10^-10 = [CN^-(aq)] *[H⁺(aq)] / [HCN(aq)]

=> 6.2×10^-10 = [CN^-(aq)] * 2.95×10^-4 / 0.0816

=> [CN^-(aq)] = 1.71*10^-7 M (Answer)