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At `25^(@)C K_(b)` for `BOH=1.0xx10^(-12).0.01M` solution of `BOH` has `[OH^(-)]`:A. `1.0xx10^(-6)M`B. `1.0xx10^(-7)M`C. `1.0xx10^(-5)M`D. `2.0xx10^(-6)M` |
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Answer» Correct Answer - B `[OH^(-)]=Calpha=Csqrt((K_(b))/(C ))` `= sqrt((K_(b).C))= sqrt(1.0xx10^(-12)xx0.01)` `=1.0xx10^(-7)M` |
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