Answer:
pOH = -log[OH-]
Explanation:
Starting from the equation of [OH-], we can obtain the pOH formula clearing pOH, where 10 passes to the other side of the equation as the base of log, and then the negative sign is passed:
[tex]\begin{gathered} \lbrack OH^-\rbrack=10^{-pOH} \\ log\lbrack OH^-\rbrack=-pOH \\ -log\lbrack OH^-\rbrack=pOH \end{gathered}[/tex]
