Answer
$[H_3O^+] = 3.02 \times 10^{- 11}$
$[OH^-] = 3.311 \times 10^{- 4}$
- It is a basic solution.
Work Step by Step
1. Calculate the hydronium concentration:
$[H_3O^+] = 10^{-pH}$
$[H_3O^+] = 10^{- 10.52}$
$[H_3O^+] = 3.02 \times 10^{- 11}$
2. Use "Kw" to calculate the hydroxide concentration:
$[H_3O^+] * [OH^-] = Kw = 10^{-14}$
$ 3.02 \times 10^{- 11} * [OH^-] = 10^{-14}$
$[OH^-] = \frac{10^{-14}}{ 3.02 \times 10^{- 11}}$
$[OH^-] = 3.311 \times 10^{- 4}$
- Since $[OH^-] > [H_3O^+]$, this solution is basic.