Answer
$[K^+] = 0.25M$
$[OH^-] = 0.25M$
$[H_3O^+] = 4 \times 10^{-14}M$
This solution is basic.
Work Step by Step
$KOH$ is a strong base; therefore:
The reaction: $1KOH(aq) -- \gt 1K^+(aq) + 1OH^-(aq)$ will completely occur, so:
$Initial[KOH] = [K^+] = [OH^-] = 0.25M$
- Now, use the $K_w$ value to calculate the $[H_3O^+]:$
$[OH^-] * [H_3O^+] = Kw = 10^{-14}$
$ 0.25 * [H_3O^+] = 10^{-14}$
$[H_3O^+] = \frac{10^{-14}}{ 0.25}$
$[H_3O^+] = 4 \times 10^{- 14}M$
- $KOH$ is a base, so its solution should be basic.
Other method:
- $[OH^-] > [H_3O^+]$, so the solution is basic.
