At `25^(@)C,` the solubility product of `Mg(OH)_(2)` is `1.0xx10^(-11)`. At which of 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. 9
B. 10
C. 11
D. 8

Answer

B

Explanation

`Mg(OH)_(2) to Mg^(2+) + 2OH^(-)`
`K_(sp)=[Mg^(2+)][OH^(-2)]`
`[OH^(-)]=sqrt(K_(sp)/([Mg^(2+)])) = sqrt((1xx10^(-11))/0.001) = 10^(-4)`
`pOH = -log[OH^(-)] = -log[10^(-4)]`
`pOH=4 therefore pH=14-pOH =14-4=10`