Question

Determine [Zn2 ], [CN–], and [HCN] in a saturated solution of Zn(CN)2 in a buffer with a pH of 1.060. 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

assuming the pH = 1 ( approx) so

[H+] = 10^-1 M

ka1 = ([H⁺][CN^-])/[HCN] = 6.2 x 10^-10

         [10^-1 ] x [CN^-])/[HCN] = 6.2 x 10^-10

       [CN^-])/[HCN] = 6.2 x 10^-9 => [HCN] = 10 ⁹x [CN^-] /(6.2)

solubility = s = [CN^-] + [HCN]   

          s = [CN^-] + 10 ⁹x [CN^-] /(6.2) ~ 10 ⁹x [CN^-] /(6.2)

so   s =10 ⁹x [CN^-] /(6.2) = [CN^-] = s x 6.2 x 10^-9

[Zn+ ] = s

again substituting in below equation    

ksp = [Zn⁺²][CN^-]² = 3 x 10^-16

             s   x ( s x 6.2 x 10^-9   )² = 3 x 10^-16

               s³ = (3 x 10^-16 )/ (6.2 x6.2 x 10^-18 ) = [3 /(6.2x6.2)] x 10²

   s³   = 7.8 04

s = 1.98 M

so the concentration are as follows

[ Zn⁺² ]        = s = 1.98 M

[CN- ] = s x 6.2 x 10^{-9 =} 1.98 x 6.2 x 10^-9   = 1.227 x 10^-8

[HCN] = 10 ⁹x [CN^-] /(6.2) =( 10⁹ x 1.227 x 10^-8 )/ 6.2 = 1.98 M