Number of equivalents of H_2SO_4 present in 1 | vedclass

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(B) Given $pH = 5$. Since $pH = -\log[H^+]$,we have $[H^+] = 10^{-5} \ M$. For $H_2SO_4$,which is a strong diprotic acid,the normality $N$ is equal to

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$10^{-6}$

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(B) Given $pH = 5$. Since $pH = -\log[H^+]$,we have $[H^+] = 10^{-5} \ M$. For $H_2SO_4$,which is a strong diprotic acid,the normality $N$ is equal to the molarity $M$ multiplied by the $n$-factor $(n=2)$. However,$[H^+] = N$ for a strong acid solution. Thus,$N = [H^+] = 10^{-5} \ N$. Volume of solution $V = 100 \ mL = 0.1 \ L$. Number of equivalents = $N 	imes V(L) = 10^{-5} 	imes 0.1 = 10^{-6} \ 	ext{equivalents}$.

Number of equivalents of $H_2SO_4$ present in $100 \ mL$ of its solution whose $pH$ is $5$.

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