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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))`A. 4B. 6C. 9D. 7 |
Answer» Correct Answer - C `K_(sp)" for "Mg(OH)_(2)=1.0xx10^(-11)` `[Mg^(2+)][OH^(-)]^(2)=(0.1)xx10[OH^(-)]^(2)=1.0xx10^(-11)` `[OH^(-)]^(2)=(1xx10^(-11))/(0.1)=10^(-10)` `[OH^(-)]=10^(-5),[H^(+)]=(10^(-14))/(10^(-5))=10^(-9)` `pH=9` |
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