Question

Determine [Zn2 ], [CN–], and [HCN] in a saturated solution of Zn(CN)2 in a buffer with a pH of 2.840. 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

pH = 2.840 = -log[H+]

[H+] = antilog (- pH) = 10^-2.840 = 0.00144 = 1.44 x 10^-3 M

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

[H+] = [CN-] = 1.44 x 10^-3 M

6.2 x 10^-10 = (1.44 x 10^-3)² / [HCN]

[HCN] = (1.44 x 10^-3)² / 6.2 x 10-10

          = 0.33 x 10⁴ M

Ksp = [Zn²⁺] [CN^-]²

[Zn²⁺] = Ksp / [CN^-]²

            = 3 x 10^-16 / (1.44 x 10^-3 )²

           = 1.446 x 10^-10 M