What is the normality of an aqueous solution | vedclass

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(A) The dissociation of $H_2SO_4$ is given by: $H_2SO_4 \rightarrow 2H^+ + SO_4^{2-}$.Given $pH = 1$,we know that $[H^+] = 10^{-pH} = 10^{-1} = 0.1 \

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$0.1$

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(A) The dissociation of $H_2SO_4$ is given by: $H_2SO_4 ightarrow 2H^+ + SO_4^{2-}$.Given $pH = 1$,we know that $[H^+] = 10^{-pH} = 10^{-1} = 0.1 \ M$.Since $1 \ mol$ of $H_2SO_4$ produces $2 \ mol$ of $H^+$,the molarity $(M)$ of $H_2SO_4$ is $[H^+] / 2 = 0.1 / 2 = 0.05 \ M$.Normality $(N)$ is calculated as $Molarity 	imes n$-factor.For $H_2SO_4$,the $n$-factor is $2$.Therefore,$N = 0.05 	imes 2 = 0.1 \ N$.

What is the normality of an aqueous solution of $H_2SO_4$ having $pH = 1$ (in $N$)?

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