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If the solubility product of `MOH` is `1xx10^(-10) mol^(2) dm^(-6)` then `pH` of its aqueous solution will beA. `12`B. `9`C. `6`D. `3` |
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Answer» Correct Answer - B `s=sqrt(K_(sp))=sqrt(1xx10^(-10))=1xx10^(-5) M` `[OH^(-)]=[MOH]=1xx10^(-5)` `[H_(3)O^(+)]=(1xx10^(-14))/(1xx10^(-5))=1xx10^(-9)` `pH= -log[H_(3)O^(+)]= -log 10^(-9) =9` |
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