Calculate the $\mathrm{pOH}$ of a solution $25^{\circ} \mathrm{C}$ that contains $1 \times 10^{-10}…
Calculate the $\mathrm{pOH}$ of a solution $25^{\circ} \mathrm{C}$ that contains $1 \times 10^{-10} \mathrm{M}$ of hydronium ions i.e. $\mathrm{H}_3 \mathrm{O}^{+}$:
4.000
9.000
1.000
7.000
Solution
Given $\left[\mathrm{H}_3 \mathrm{O}^{+}\right]=1 \times 10^{-10} \mathrm{M}$ at $25^{\circ}\left[\mathrm{H}_3 \mathrm{O}^{+}\right]\left[\mathrm{OH}^{-}\right]=10^{-14}$
$\therefore \quad\left[\mathrm{OH}^{-}\right]=\frac{10^{-14}}{10^{-10}}=10^{-4}$
Now, $\quad\left[\mathrm{OH}^{-}\right]=10^{-p^{\text {OH }}}$
$=10^{-4}=10^{-P^{\mathrm{OH}}}$
$\therefore \quad p^{\mathrm{OH}}=4$.