Question

Consider the insoluble compound nickel(II) hydroxide , Ni(OH)2 . The nickel ion also forms a complex with cyanide ions . Write a balanced net ionic equation to show why the solubility of Ni(OH)2 (s) increases in the presence of cyanide ions and calculate the equilibrium constant for this reaction. For Ni(CN)42- , Kf = 1.0×1031 . Use the pull-down boxes to specify states such as (aq) or (s).

Answer 1

Answer: Equilibrium constant for this reaction is $2.8 \times 10^{15}$.

Explanation:

Chemical reaction equation for the formation of nickel cyanide complex is as follows.

$Ni(OH)_{2}(s) + 4CN^{-}(aq) \rightleftharpoons [Ni(CN)_{4}^{2-}](aq) + 2OH^{-}(aq)$

We know that,

      K = $K_{f} \times K_{sp}$

We are given that, $K_{f} = 1.0 \times 10^{31}$

and,    $K_{sp} = 2.8 \times 10^{-16}$

Hence, we will calculate the value of K as follows.

     K = $K_{f} \times K_{sp}$

     K = $(1.0 \times 10^{31}) \times (2.8 \times 10^{-16})$

        = $2.8 \times 10^{15}$

Thus, we can conclude that equilibrium constant for this reaction is $2.8 \times 10^{15}$.