The solubility product of `Mg(OH)_2` is `1.2 xx 10^(-11)`. Calculate its solubility in 0.1 M NaOH solution.

Options

A. `1,4 xx 10^(-4)`
B. `8.16 xx 10^(-4)`
C. `0.816`
D. `1.4`

Answer

B

Explanation

`Mg(OH)_(2)hArr underset ((S))(Mg^(2+)) + underset ((2S)^(2)) (2OH^(-))`
`K_(sp) = 4S^(2) rArr S = root3 ((K_(sp))/4) = root 3 ((1.2 xx10^(-11))/4)`
`S = 1.44 xx 10^(-4) " mol L"^(-1) = 58 xx 1.44 xx10^(-4) " gL"^(-1)`
( mol .mass of `Mg(OH)_(2) = 58`)
`= 83.52 xx 10^(-4) " g L"^(-1) = 8.35 xx 10^(-4) " g/100cc"` .