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If pH of a saturated solution of `Ba(OH)_(2)` is 12, the value of its `K_((sp))` isA. `5.00 xx 10^(-7) M^(3)`B. `4.00 xx 10^(-6) M^(3)`C. `4.00 xx 10^(-7) M^(3)`D. `5.00 xx 10^(-6) M^(3)` |
Answer» Correct Answer - A pH = 12 means `[H^(+)]=10^(-12)M`. Hence, `[OH^(-)]=10^(-2)M` `Ba(OH)_(2) hArr Ba^(2+)+2OH^(-)` `[OH^(-)]=10^(-2)M` `:. Ba^(2+)=(10^(-2))/(2) M=5xx10^(-3)M` `K_(sp)=[Ba^(2+)][OH^(-)]^(2)` `=(5xx10^(-3))(10^(-2))^(2)=5xx10^(-7) M^(3)` |
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