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The `pH` of an aqueous solution of `Ba(OH)_(2)` is `10`. If the `K_(sp)` of `Ba(OH)_(2)` is `1xx10^(-9)`, then the concentration of `Ba^(2+)` ions in the solution in `mol L^(-1)` isA. `1xx10^(-2)`B. `1xx10^(-4)`C. `1xx10^(-1)`D. `1xx10^(-5)` |
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Answer» Correct Answer - C `pOH =14-10=4` `:. [OH^(-)] =1xx10^(-4) M` `K_(sp)=[Ba^(2+)][OH^(-)]^(2)` or `[Ba^(2+)]=K_(sp)//[OH^(-)]^(2)=(1xx10^(-9))/((1xx10^(-4))^(2))` |
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