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Calculate the pOH of a solution at $25^\circ C$ that contains $1 \times 10^{-10}$ M of hydronium ions, i.e. $H_3O^+$.
A
4.000
B
9.000
C
1.000
D
7.000
Detailed Solution
Given $[H_3O^+] = 1 \times 10^{-10}$ M
At $25^\circ C$: $[H_3O^+][OH^-] = K_w = 10^{-14}$
$\therefore [OH^-] = \dfrac{10^{-14}}{10^{-10}} = 10^{-4}$ M
Now $[OH^-] = 10^{-pOH}$, so $10^{-4} = 10^{-pOH}$
$\therefore pOH = 4$
Check: $pH = 10$ and $pH + pOH = 14$, so $pOH = 14 - 10 = 4$.
At $25^\circ C$: $[H_3O^+][OH^-] = K_w = 10^{-14}$
$\therefore [OH^-] = \dfrac{10^{-14}}{10^{-10}} = 10^{-4}$ M
Now $[OH^-] = 10^{-pOH}$, so $10^{-4} = 10^{-pOH}$
$\therefore pOH = 4$
Check: $pH = 10$ and $pH + pOH = 14$, so $pOH = 14 - 10 = 4$.