Answer : The solubility of [tex]Ni(OH)_2[/tex] in water is, [tex]5.3\times 10^{-6}M[/tex]
Explanation :
The solubility equilibrium reaction will be:
[tex]Ni(OH)_2\rightleftharpoons Ni^{2+}+2OH^{-}[/tex]
Let the molar solubility be 's'.
The expression for solubility constant for this reaction will be,
[tex]K_{sp}=[Ni^{2+}][OH^{-}]^2[/tex]
[tex]K_{sp}=(s)\times (2s)^2[/tex]
[tex]K_{sp}=4s^3[/tex]
Given:
[tex]K_{sp}[/tex] = solubility constant = [tex]6.0\times 10^{-16}[/tex]
Now put all the given values in the above expression, we get:
[tex]K_{sp}=4s^3[/tex]
[tex]6.0\times 10^{-16}=4s^3[/tex]
[tex]s=5.3\times 10^{-6}M[/tex]
Therefore, the solubility of [tex]Ni(OH)_2[/tex] in water is, [tex]5.3\times 10^{-6}M[/tex]
