What will be the H^{+} ion concentration when | vedclass

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(B) First,calculate the number of moles of $NaOH$: $	ext{moles} = \frac{	ext{mass}}{	ext{molar mass}} = \frac{4 \ g}{40 \ g/mol} = 0.1 \ mol$.Next,

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$10^{-13}$

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(B) First,calculate the number of moles of $NaOH$: $	ext{moles} = \frac{	ext{mass}}{	ext{molar mass}} = \frac{4 \ g}{40 \ g/mol} = 0.1 \ mol$.Next,calculate the molarity of the $NaOH$ solution: $[OH^{-}] = \frac{0.1 \ mol}{1 \ L} = 0.1 \ M = 10^{-1} \ M$.Calculate $pOH$: $pOH = -\log[OH^{-}] = -\log(10^{-1}) = 1$.Calculate $pH$: $pH = 14 - pOH = 14 - 1 = 13$.Finally,calculate the $H^{+}$ ion concentration: $[H^{+}] = 10^{-pH} = 10^{-13} \ M$.

What will be the $H^{+}$ ion concentration when $4 \ g$ $NaOH$ is dissolved in $1000 \ mL$ solution?

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