Question

What is the `"pH"` and `"pOH"` of a `2.34* 10^-5M` `["OH"^-]` solution?

Answer 1

`"pH"=9.37`
`"pOH"=4.63`

Explanation:

`"pOH"=-log(["OH"^-])`

`"pOH"=-log(2.34*10^-5)=4.63`

`"pH"=14-"pOH"=14-4.67=9.37`