Question

Determine [Zn2+] [CN-] and [HCN] in a saturated solution of Zn(CN)2 with a fixed ph of 2.900. The Ksp for Zn(CN)2 is 3.0*10^-16. The Ka for HCN is 6.2*10^-9

Answer 1

Zn(Cn)₂ ------------> Zn⁺² + 2 CN^-
K_sp = [Zn⁺² ] [ CN^- ]²
let the conc of Zn⁺² be s and conc of CN^- be 2s

HCN ----------------> H⁺ +CN^-
K_a = [H⁺] [CN^-] / [HCN]
6.2*10^-9 = [H⁺] [CN^-] / [HCN]
If P^H = 2.9 then P^H = -log₁₀[H⁺]
[H⁺] = 10^-2.9 = 1.258925*10^-3 M
1.258925*10^-3 * [CN^-] / [HCN] = 6.2*10-9
[HCN] =203048.3871 [CN^-]
[Zn⁺²] = s ,
[CN^-] + [HCN^-] = 2s
[CN^-] + 203048.3871 [CN^-] = 2s
[CN^-] = 9.84982*10^-6 s

K_sp = [Zn⁺² ] [ CN^- ]² , solve for s
3*10^-16 = [s] * [ 9.84982*10^-6 s ]²
3.092179*10^-6 = s³
s = 0.0145687 M
Therefore [Zn⁺²] = 0.01456M ; [CN^-] =2s = 0.029137M ;