If the solubility product of `Ni(OH)_2` is ` 4.0 xx 10^(-15)` the solubility (in mol `L^(-1) `) is

Options

A. `5.0 xx 10^(-5)`
B. `4.0 xx 10^(-5)`
C. `2.0 xx 10^(-5)`
D. `1.0 xx 10^(-5)`

Answer

D

Explanation

Given , solubility product `(K_(sp))` of `Ni(OH)_2` is ` = 4.0 xx 10^(-15)`
Solubility (s) = ?
`NI(OH)_2 iff underset([s] "mol/L")(Ni^(2+) ) + underset("25 mol/L")(2OH^(-) )`
`K_(sp) = [s] [2s]^2`
`K_(sp) = 4s^3`
` s = 3 sqrt( (K_(sp))/(4)) = 3 sqrt( (4 xx 10^(-15) )/(4)) = 1 xx 10^(-5) ` mol / L