Find out the solubility of `Ni(OH)_2` in `0.1 M` NaOH Given that the ionic product of `Ni(OH)_2` is `2xx10^(-15)`.

Options

A. `2xx10^(-8)` M
B. `1xx10^(-13)` M
C. `1xx10^(8)` M
D. `2xx10^(-13)` M

Answer

D

Explanation

Let the solubility of `Ni(OH)_2` is s
`underset(s)(Ni(OH)_2) harr underset(s)(Ni^(2+))+underset(2s)(2OH^(-))`
`underset(0.1)(NaOH)rarr underset(0.1)(Na) + underset(0.1)(OH^(-)`
As `K_(sp)` is small `2s lt lt 0.10`
therefore (0.10 + 2s) =0.10
so Total `[OH]^(-) = 0.10`
Ionic product = `[Ni^(2+)][OH^(-)]^2`
`2 xx 10^(-15) = s(0.10)^2`
`s=2xx10^(-13)` M