1.

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`

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`


Discussion

No Comment Found

Related InterviewSolutions