Question

Determine [Zn2 ], [CN–], and [HCN] in a saturated solution of Zn(CN)2 in a buffer with a pH of 2.620. The Ksp for Zn(CN)2 is 3.0 × 10–16. The Ka for HCN is 6.2 × 10–10. Because a buffer is present, you cannot know the charge balance equation.

Answer 1

we have Zn(CN)₂  ---------> Zn⁺² + 2CN^-

- S 2S moles/L where S is the solubility of salt

and H⁺ + CN- -------> HCN

The solution is saturated with zinc cyanide thus

solubility of salt = s = cubeth root of (Ksp /4) =1.956x10^-5

the CN- obtained from Zn(CN)2 aq. reacts with H⁺ (pH = 2.62) to give HCN.

Thus [HCN] + [CN-] = 2 [Zn+2] = 2x1.956x10^-5

From the pH given we calculate [H+] as 2.4 x10^-3 mole/L

HCN - -------------> H+ + CN-

   Ka = [H+][CN-] /[HCN] Substituting the known values

6,2 x 10^-10 = [2.4x10^-3] [CN-]/[HCN]

Thus [CN-]/[HCN]  = 2.583x 10 ^-7

or [CN-] = 2.583x 10 ^-7 [HCN}

NOw substituting this [cn-] in the previous relation

2.583x 10 ^-7 [HCN} + [HCN] = 2x 1.956x 10^-5 we get

[HCN] = 3.887x 10^-5 mol/L and [CN-] = 1.04x 10^-12 mol/L

and [Zn+2] = 1.956x 10^-5mol/L