Calculate pH at which `Mg(OH)_(2)` begins to precipitate from a solution containing `0.10M Mg^(2+)` ions. `(K_(SP)of Mg(OH)_(2)=1xx10^(-11))`

Explanation

`[Mg^(2+)]=0.1 M, K_(sp)=1.0xx10^(-11)`
`Mg(OH)_2 hArr Mg^(2+) + 2OH^(-)`
`K_(sp)=[Mg^(2+)][OH^-]^2`
`1.0xx10^(-11)=(0.1)[OH^-]^2`
or `[OH^-]^2=(1.0xx10^(-11))/0.1=10^(-10),[OH^-]=10^(-5)` M
Now, `[H_3O^+][OH^-]=K_w=10^(-14)`
`[H_3O^+]=10^(-14)/([OH^-])=10^(-14)/10^(-5)=10^(-9)` M
pH=-log `[H_3O^+]=-log (10^(-9))` =9