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`Mg(OH)_(2)` is precipitated when NaOH is added to a solution of `Mg^(2+)` . If the final concentration of `Mg^(2+)` is `10^(-10)` .M, the concetration of `OH^(-)` (M) is the solution is b [Solubility product for `Mg(OH)_(2)=5.6xx10^(-12)`]A. 0.056B. 0.12C. 0.24D. 0.025 |
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Answer» Correct Answer - C `K_(sp) Mg(OH)_(2)=[Mg^(+2)][OH^(-)]^(2)` `5.6xx10^(-12)=[10^(-10)][OH^(-)]^(2)` `[OH^(-)]=sqrt(5.6xx10^(-2))=0.24 M` |
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