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?

1 Answer
May 13, 2017

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?