What will be the concentration of H^{+} ions | vedclass

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(B) The molar mass of $NaOH$ is $40 \ g/mol$. Number of moles of $NaOH = \frac{4 \ g}{40 \ g/mol} = 0.1 \ mol$. Volume of solution = $100 \ mL = 0.1 \

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$10^{-14}$

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(B) The molar mass of $NaOH$ is $40 \ g/mol$. Number of moles of $NaOH = \frac{4 \ g}{40 \ g/mol} = 0.1 \ mol$. Volume of solution = $100 \ mL = 0.1 \ L$. Concentration of $NaOH$ (or $[OH^{-}]$) = $\frac{0.1 \ mol}{0.1 \ L} = 1 \ M$. Using the ionic product of water,$K_w = [H^{+}][OH^{-}] = 10^{-14}$ at $25 \ ^{\circ}C$. $[H^{+}] = \frac{10^{-14}}{[OH^{-}]} = \frac{10^{-14}}{1} = 10^{-14} \ M$.

What will be the concentration of $H^{+}$ ions in a solution prepared by dissolving $4 \ g$ of $NaOH$ in $100 \ mL$ of water at $25 \ ^{\circ}C$?

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