1.

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 0.0001 M `Mg^(2+)` ions ?A. 11B. 8C. 9D. 10

Answer» Correct Answer - D
`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`


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