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

`Ks` for `Mg(OH)_(2)=[Mg^(2+)][OH^(-)]^(2)`
`:. 1.0 xx 10^(-11) =(0.001)[OH^(-)]^(2)`
or `|OH^(-)|^(2)=10^(-8)` or `|OH^(-)|=10^(-4)`
`:. pOH =4` or `pH =14-4=10`