Answer: Equilibrium constant for this reaction is [tex]2.8 \times 10^{15}[/tex].
Explanation:
Chemical reaction equation for the formation of nickel cyanide complex is as follows.
[tex]Ni(OH)_{2}(s) + 4CN^{-}(aq) \rightleftharpoons [Ni(CN)_{4}^{2-}](aq) + 2OH^{-}(aq)[/tex]
We know that,
K = [tex]K_{f} \times K_{sp}[/tex]
We are given that, [tex]K_{f} = 1.0 \times 10^{31}[/tex]
and, [tex]K_{sp} = 2.8 \times 10^{-16}[/tex]
Hence, we will calculate the value of K as follows.
K = [tex]K_{f} \times K_{sp}[/tex]
K = [tex](1.0 \times 10^{31}) \times (2.8 \times 10^{-16})[/tex]
= [tex]2.8 \times 10^{15}[/tex]
Thus, we can conclude that equilibrium constant for this reaction is [tex]2.8 \times 10^{15}[/tex].
