Answer
$[H_3O^+] = 0.150M$
$[Cl{O_4}^-] = 0.150M$
$[OH^-] = 6.67 \times 10^{-14}M$
The solution is acidic.
Work Step by Step
$HClO_4$ is a strong acid; therefore:
The reaction: $1HClO_4(aq) + 1H_2O(l) -- \gt 1H_3O^+(aq) + 1Cl{O_4}^-(aq)$ will occur completely, so:
$Initial [HClO_4] = [H_3O^+] = [ClO_4] = 0.150M$
- Now, use the $K_w$ value to calculate the hydroxide concentration:
$[H_3O^+] * [OH^-] = K_w = 10^{-14}$
$ 0.150 * [OH^-] = 10^{-14}$
$[OH^-] = \frac{10^{-14}}{ 0.150}$
$[OH^-] = 6.67 \times 10^{- 14}$
- $HClO_4$ is an acid, so its solution should be acidic.
Other method:
- $[H_3O^+] > [OH^-]$, which means it is an acidic solution.
