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)`.

Answer

`2.0xx10^(-13)M`

Explanation

Suppose the solubility of `Ni(OH)_(2)` in 0.10 M NaOH = s mol `L^(-1)`
`Ni(OH)_(2)` in the solution dissociates as :`Ni(OH)_(2) hArr Ni^(2+)+2OH^(-)`.
Thus, s mol `L^(-1)` of `Ni(OH)_(2)` in the solution gives s mol `L^(-1) ` of `Ni^(2+)` ion and 2 s mol `L^(-1)` of `OH^(-)` ions. But `OH^(-)` ions are also produced from NaOH . As NaOH dissociates completely , `OH^(-)` from NaOH = 0.1 M . Hence, `[Ni^(2+)]=s "mol" L^(-1) and [OH^(-)]=2 s + 0.1 "mol " L^(-1) ~~ 0.1 "mol" L^(-1)` (`:'` 2 s is very small compared to 0.1 )
`K_(sp) ` of `Ni(OH)_(2) = [Ni^(2+)][OH^(-)]^(2)`
`:. s (0.1)^(2) = 2.0 xx 10^(-15 ) or s = 2.0 xx 10^(-13) "mol" L^(-1)`