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 `Smol//L` of `Ni(OH)_(2)` provides `Smol//L` of `Ni^(2+)` and `2Smol//L` of `OH^(-)`, but the toal concentration of `OH^(-)=(0.10+2S)mol//L` because the solution already contains `0.10mol//L` of `OH^(-)` from `NaOH`.
`K_(sp)=2.0xx10^(-15)=[Ni^(2+)][OH]^(2)`
`=(S)(0.10xx2S)^(2)`
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+)]`