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Calculate [OH^(-)] if pOH = 8.3....

Calculate `[OH^(-)]` if pOH = 8.3.

Text Solution

Verified by Experts

`pOH=-log_(10)[OH^(-)],8.3=-log_(10)[OH^(-)]`
Taking antilog on both the sides
`[OH^(-)]` = anti log(-8.3), `[OH^(-1)]` = antilog(-9-8.3+9) = antilog`(0.7)xx10^(-9)`
= `5.012xx10^(-9)mol//dm^(3)`.
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