At `25^(@)C`, the solubility product of `Mg(OH)_(2)` is `1.0xx10^(-11)`. At which `pH`, will `Mg^(2+)` ions start precipitating in the form of `Mg(OH)_(2)` from a solution of `0.001 M Mg^(2+)` ions ?

Options

A. 11
B. 8
C. 9
D. 10

Answer

D

Explanation

`K_(sp) ` for `Mg(OH)_(2)=[Mg^(2+)][OH^(-)]^(2)`
`:. 1.0xx10^(-11)=(0.001)[OH^(-)]^(2)`
or `[OH^(-)]^(2)=10^(8) or [OH^(-)]=10^(-4)`
i.e, `pOH=4 or pH = 10`