Question

Use the rules for logarithms and exponents to solve for

Answer 1

pOH = -log[OH-]

Step-by-step 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:

`\begin{gathered} \lbrack OH^-\rbrack=10^(-pOH) \\ log\lbrack OH^-\rbrack=-pOH \\ -log\lbrack OH^-\rbrack=pOH \end{gathered}`