Question

Write a balanced net ionic equation to show why the solubility of Ni(OH)2(s) increases in the presence of a strong acid and calculate the equilibrium constant Knet for the reaction of this sparingly soluble salt with acid.

Answer 1

Answer : The value equilibrium constant $(K_{eq})$ for net ionic reaction is, $5.84\times 10^{12}$

Explanation :

The dissociation reaction of $Ni(OH)_2$ is :

$Ni(OH)_2(s)\rightleftharpoons Ni^{2+}(aq)+2OH^-(aq)$    $K_{sp}=5.84\times 10^{-16}$

The dissociation reaction of $H_2O$ is:

$2H_2O(l)\rightleftharpoons 2H^{+}(aq)+2OH^-(aq)$    $K_{w}=1.00\times 10^{-14}$

The balanced Net ionic equation is:

$Ni(OH)_2(s)+2H_3O^+(aq)\rightarrow 4H_2O(l)+Ni^{2+}(aq)$

Now we have to calculate the value equilibrium constant $(K_{eq})$ for net ionic reaction.

$K_{eq}=\frac{K_{sp}}{(K_w)^2}$

Now put all the given values in this expression, we get:

$K_{eq}=\frac{5.84\times 10^{-16}}{(1.00\times 10^{-14})^2}$

$K_{eq}=5.84\times 10^{12}$

Thus, the value equilibrium constant $(K_{eq})$ for net ionic reaction is, $5.84\times 10^{12}$