A solution has #[H^+] = 2.0 * 10^-5# #M#. How do you find the pOH of this solution?

1 Answer

pOH = 9.30


You could find the answer one of two ways. The first is using the ion-product constant for water:

#K_w = ["H"_3"O"^(+)]["OH"^(-)] = 1.0xx10^(-14)#

Since Kw and the concentration of hydrogen ions in solution are known we can rearrange the equation to solve for the concentration of hydroxide ions. We can do this by dividing Kw by [H+] to solve for [OH-] like so:

Kw / [H+] = [OH-]

#(1.0xx10^(-14))/(2.0xx10^(-5) "M") = 5xx10^(-10) "M"#

Now we have the concentration of hydroxide ions in solution. To find the pOH, you have to take the -log of that concentration like this:
-log(#5xx10^(-10)# M) = 9.30. Thus, the pOH = 9.30.

The second method is much easier. Just take the -log of the concentration of hydrogen ions: -log(#2.0xx10^(-5# M) = 4.70

That value represents the pH of the solution. To obtain the pOH from the pH just do 14 - pH = pOH

Then you should arrive at the same answer: 14 - 4.70 = 9.30