Question 9.5: Calculate the [H3O^+] and [OH^-] of a sodium hydroxide solut......
Calculate the \left[\mathrm{H}_{3} \mathrm{O}^{+}\right] and \left[\mathrm{OH}^{-}\right] of a sodium hydroxide solution with a \rm pH=10.00.
Sodium hydroxide solution
pH = 10.00
Final Answer
First, calculate \left[\mathrm{H}_{3} \mathrm{O}^{+}\right]:
\begin{aligned}\mathrm{pH} & =-\log \left[\mathrm{H}_{3} \mathrm{O}^{+}\right] \\10.00 & =-\log \left[\mathrm{H}_{3} \mathrm{O}^{+}\right] \\-10.00 & =\log \left[\mathrm{H}_{3} \mathrm{O}^{+}\right] \\\text {antilog }-10 & =\left[\mathrm{H}_{3} \mathrm{O}^{+}\right] \\1.0 \times 10^{-10} \mathrm{M} & =\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\end{aligned}
To calculate the \left[\mathrm{OH}^{-}\right], we need to solve for \left[\mathrm{OH}^{-}\right] by using the following expression:
\begin{aligned}{\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\left[\mathrm{OH}^{-}\right] } & =1.0 \times 10^{-14} \\{\left[\mathrm{OH}^{-}\right] } & =\frac{1.0 \times 10^{-14}}{\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]}\end{aligned}
Substituting the \left[\mathrm{H}_{3} \mathrm{O}^{+}\right] from the first part, we have
\begin{aligned}{\left[\mathrm{OH}^{-}\right] } & =\frac{1.0 \times 10^{-14}}{\left[1.0 \times 10^{-10}\right]} \\& =1.0 \times 10^{-4} \mathrm{M}\end{aligned}