Calculate the molar solubility of `Ni(OH)_(2)` in `0.10M NaOH`. The ionic product of `Ni(OH)_(2)` is `2.0xx10^(-15)`

Explanation

Let the solubility of `Ni(OH)_(2)` be equal to S . Dissolution of S mol/L. of `Ni(OH)_(2)` provides S mol / L of `Ni^(2+)` and 2S mol/L of `OH^(-)`, but the total concentration of `OH^(-)=(0.10+2S)` mol/L because the solution already contains 0.10 mol/L of `OH^(-)` from NaOH.
`K_(sp)=2.0xx10^(-15)=[Ni^(2+)][OH^(-2)]`
(S) `(0.10+2S) AS K_(sp)` is small, `2S lt lt 0.10,`
thus `(0.10+2S)=0.10`
Hence, `2.0xx10^(-15)S(0.10)^(2)`
`S=2.0xx10^(-13)M=[Ni^(2+)]`