4 g of NaOH is added in water to form 500 m | vedclass

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(D) Step $1$: Calculate the number of moles of $NaOH$. $n = \frac{	ext{mass}}{	ext{molar mass}} = \frac{4 \ g}{40 \ g \ mol^{-1}} = 0.1 \ mol$. Step

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$13.3010$

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(D) Step $1$: Calculate the number of moles of $NaOH$. $n = \frac{	ext{mass}}{	ext{molar mass}} = \frac{4 \ g}{40 \ g \ mol^{-1}} = 0.1 \ mol$. Step $2$: Calculate the molarity $(M)$ of the solution. $M = \frac{n}{V(L)} = \frac{0.1 \ mol}{0.5 \ L} = 0.2 \ M$. Step $3$: Since $NaOH$ is a strong base,$[OH^-] = [NaOH] = 0.2 \ M$. Step $4$: Calculate $pOH$. $pOH = -\log[OH^-] = -\log(0.2) = -(\log 2 - \log 10) = -(0.3010 - 1) = 0.6990$. Step $5$: Calculate $pH$. $pH = 14 - pOH = 14 - 0.6990 = 13.3010$.

$4 \ g$ of $NaOH$ is added in water to form $500 \ mL$ solution at $298 \ K$. What is the $pH$ of the solution? (Molar mass of $NaOH = 40 \ g \ mol^{-1}$)

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