Question

Determine [Zn2 ], [CN–], and [HCN] in a saturated solution of Zn(CN)2 with a fixed pH of 2.240. The Ksp for Zn(CN)2 is 3.0 × 10–16. The Ka for HCN is 6.2 × 10–10..

Answer 1

pH = 2.240

pH = -log[H+]

[H+] = 5.75 x 10^-3 M

let x be the solubility of Zn(CN)2 in solution

Ksp = [Zn2+][CN-]^2

3.0 x 10^-16= (x)(2x)^2

some CN- formed also combines with H+ in solution to give HCN, so,

2x = [CN-] + [HCN] ---- (1)

Ka = [H+][CN-]/[HCN]

6.2 x 10^-10 = (5.75 x 10^-3)[CN-]/[HCN]

[HCN] = 5.75 x 10^-3[CN-]/6.2 x 10^-10 = 9.27 x 10^6[CN-] ---- (2)

Substitute (2) in (1),

2x = [CN-] + 9.27 x 10^6[CN-] = 9.27 x 10^6[CN-]

[CN-] = x/2.16 x 10^-7 ----- (3)

Substitute (3) in Ksp equation,

Ksp = 3.0 x 10^-16 = (x)(x/2.16 x 10^-7)^2

3.0 x 10^-16 = x^3/4.66 x 10^-14

x^3 = 1.40 x 10^-29

x = 2.41 x 10^-10 M

So the concentration of,

[Zn2+] = 2.41 x 10^-10 M

[CN-] = 2(2.41 x 10^-10) = 4.82 x 10^-10 M

[HCN] = 9.27 x 10^6(2.41 x 10^-10) = 2.23 x 10^-3 M