2.8 text{ g} of KOH is dissolved in a 500 tex | vedclass

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(D) Step $1$: Calculate the number of moles of $KOH$: $n = \frac{	ext{mass}}{	ext{molar mass}} = \frac{2.8 	ext{ g}}{56 	ext{ g/mol}} = 0.05 	ext

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$13$

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(D) Step $1$: Calculate the number of moles of $KOH$: $n = \frac{	ext{mass}}{	ext{molar mass}} = \frac{2.8 	ext{ g}}{56 	ext{ g/mol}} = 0.05 	ext{ mol}$.Step $2$: Calculate the molarity $(M)$ of the solution: $M = \frac{n}{V(	ext{in L})} = \frac{0.05 	ext{ mol}}{0.5 	ext{ L}} = 0.1 	ext{ M}$.Step $3$: Since $KOH$ is a strong base, $[OH^-] = 0.1 	ext{ M} = 10^{-1} 	ext{ M}$.Step $4$: Calculate $pOH$: $pOH = -\log[OH^-] = -\log(10^{-1}) = 1$.Step $5$: Calculate $pH$: $pH = 14 - pOH = 14 - 1 = 13$.

$2.8 \text{ g}$ of $KOH$ is dissolved in a $500 \text{ mL}$ solution at $298 \text{ K}$. What is the $pH$ of the solution? (Molar mass of $KOH = 56 \text{ g/mol}$)

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