If $0.561 \, g$ of $KOH$ is dissolved in water to give $200 \, mL$ of solution at $298 \, K$. Calculate the concentrations of potassium,hydrogen and hydroxyl ions. What is its $pH$?

(N/A) $1$. Calculate the molarity of $KOH$ solution:
$M = \frac{\text{mass in } g}{\text{molar mass} \times \text{volume in } L} = \frac{0.561 \, g}{56.11 \, g/mol \times 0.200 \, L} = 0.05 \, M$
$2$. Since $KOH$ is a strong base,it dissociates completely:
$KOH_{(aq)} \to K^{+}_{(aq)} + OH^{-}_{(aq)}$
Therefore,$[K^{+}] = 0.05 \, M$ and $[OH^{-}] = 0.05 \, M$.
$3$. Calculate $[H^{+}]$ using the ionic product of water ($K_w = 10^{-14}$ at $298 \, K$):
$[H^{+}] = \frac{K_w}{[OH^{-}]} = \frac{10^{-14}}{0.05} = 2 \times 10^{-13} \, M$
$4$. Calculate $pH$:
$pH = -\log[H^{+}] = -\log(2 \times 10^{-13}) = 13 - \log(2) = 13 - 0.3010 = 12.699 \approx 12.70$