What is pOH of a solution prepared by adding a 15.8*mg mass of potassium hydroxide to 11*mL of solution?

1 Answer
May 9, 2017

pOH=1.60

Explanation:

Now..........pOH=-log_10[HO^-]...........

And we have 15.8*mg KOH dissolved in 11*mL of solution.

[HO^-]=((15.8xx10^-3*g)/(56.11*g*mol^-1))/(11*mLxx10^-3*mL*L^-1)

=2.56xx10^-2*mol*L^-1.

But by definition, pOH=-log_10(2.56xx10^-2)=1.60

What is pH of this solution? You should be able to answer immediately. Why? Because in aqueous solution, pH+pOH=14.

Also see this old problem here for more treatment.