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If the solubility product of MOH is 1xx1...

If the solubility product of `MOH` is `1xx10^(-10) mol^(2) dm^(-6)` then `pH` of its aqueous solution will be

Text Solution

Verified by Experts

The correct Answer is:
9
`[OH]^(-)=(K_(sp))^(1//2)=1xx10^(-5)`
`:.pOH=-log1xx10^(-5)=5`
The `pH=14-5=9`
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