## General Chemistry (4th Edition)

$[H_3O^+] = 0.150M$ $[Cl{O_4}^-] = 0.150M$ $[OH^-] = 6.67 \times 10^{-14}M$ The solution is acidic.
$HClO_4$ 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.