---

Example 13.6: Calculation of \(\ce{pOH}\)

What are the \(\ce{pOH}\) and the pH of a 0.0125-M solution of potassium hydroxide, \(\ce{KOH}\)?

Solution

\([\ce{KOH}]\) \(= 0.0125\)


\(\mathrm{pH}\) = ?


\(\mathrm{pOH}\) = ?


Potassium hydroxide is a highly soluble ionic compound and completely dissociates when dissolved in dilute solution, yielding \(c_{\mathrm{\ce{OH-}}} = 0.0125\) M.

\([\ce{OH-}]\) \(= [\ce{KOH}]\)

\(\ \ \ =0.0125\)

\(\ \ \ =0.0125\)


Because we know \(c_{\mathrm{\ce{OH-}}}\), it makes sense to calculate the \(\ce{pOH}\) first:

\(\mathrm{pOH}\) \(= -\mathrm{log}([\ce{OH-}])\)

\(\ \ \ =-\mathrm{log}(0.0125)\)

\(\ \ \ =1.903\)


The pH can be found from the pOH:


\(\mathrm{pH}\) \(= 14.00 - \mathrm{pOH}\)

\(\ \ \ =14.00 - 1.903\)

\(\ \ \ =12.10\)