Question

What is the resultant `pOH` when a `5*mL` volume of `NaOH` of `0.165*mol*L^-1` concentration is diluted to a `450*mL` volume?

Answer 1

`pOH=2.74`

Explanation:

`pOH=-log_(10)[HO^-]`; i.e. it is the basic counterpart to the `pH` scale.

And so we find `[HO^-]` of the given solution...........

`=(5xx10^-3*Lxx0.165*mol*L^-1)/(0.450*L)=1.83xx10^-3*mol*L^-1.`

And so `pOH=-log_10(1.83xx10^-3)=2.74`.

And further we know that `pH+pOH=14` for aqueous solution......You should be able to tell `pH` of this solution `"pdq"`. What is `pH` here?