Calculate the molar solubility of `Ni(OH)_2` in 0.10 M NaOH. The ionic product of `Ni(OH)_2` is `2.0xx10^(-15)` .

Explanation

0.1 M NaOH is strong base , so complete ionised
Thus NaOH `to Na_((aq))^(+) + OH_((aq))^(-)`
`therefore [NaOH]=[OH^-]=0.1`M , NaOH
Suppose solubility of `Ni(OH)_2 ="S mol L"^(-1)`
So, At equilibrium in this salt `[Ni^(2+)] = "S mol L"^(-1)`
`[OH^-] = "2 S mol L"^(-1)` produce
`{:(Ni(OH)_(2(s)) hArr, Ni_((aq))^(2+) + , 2OH_((aq))^(-)),("At equilibrium ", SM , 2SM):}`
concentration
So, total `[OH^-]`=(0.1+2S)M
(Addition of `HO^-` of NaOH and `Ni(OH)_2` )
but the value of S is much less , So ineligible in compare to 0.1 in solution.
`[OH^-]=(0.1 +2S) approx "0.1 mol L"^(-1)`
`K_(sp) = [Ni^(2+)] [OH^-]^2 = 2xx10^(-15)`
`therefore (S)(0.1)^2 =2xx10^(-15)`
`therefore S=2.0 xx 10^(-13) "mol L"^(-1) = [Ni^(2+)]`