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Calculate [H^(+)] if pOH = 9.23....

Calculate `[H^(+)]` if pOH = 9.23.

Text Solution

Verified by Experts

`pH+pOH=14" "pH=14-pOH=14-9.23=4.77`
`pH=-log_(10)[H^(+)]=-4.77=log[H^(+)]`
Taking antilog on both the sides
`[H^(+)]` = anti log (-4.77) = anti log[+5 - 4.77 - 5] = anti log[0.23]`xx10^(-5)`
= `1.698xx10^(-5)mol//dm^(3)`.
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