The solubility product of `Ni(OH)_(2)` is `2.0xx10^(-15)`. The molar solubility of `Ni(OH)_(2)` in `0.1 M NaOH` solution is

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 from `NaOH`.
`K_(sp) = 2.0 xx 10^(-15) = [Ni^(2+)][OH^(-)]^(2)`
`= (S) (0.10 + 2S)^(2)`
As `K_(sp)` is small, `2S lt lt 0.10`
thus, `(0.10 + 2S) = 0.10`
Hence,
`2.0 xx 10^(-18) = S (0.10)^(2)`
`S = 2.0 xx 10^(-13) M = [Ni^(2+)]`