What is the pH of a centimolar solution of H_ | vedclass

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(A) $H_2SO_4$ is a strong diprotic acid,so it dissociates completely as $H_2SO_4 \rightarrow 2H^+ + SO_4^{2-}$.For a centimolar solution,concentration

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$1.7$

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(A) $H_2SO_4$ is a strong diprotic acid,so it dissociates completely as $H_2SO_4 ightarrow 2H^+ + SO_4^{2-}$.For a centimolar solution,concentration $C = 0.01 \ M = 10^{-2} \ M$.The concentration of $H^+$ ions is $[H^+] = 2 	imes C = 2 	imes 10^{-2} \ M$.$pH = -\log_{10} [H^+] = -\log_{10} (2 	imes 10^{-2})$.$pH = -(\log_{10} 2 + \log_{10} 10^{-2}) = -(\log_{10} 2 - 2) = 2 - \log_{10} 2$.Using $\log_{10} 2 \approx 0.3010$,we get $pH = 2 - 0.3010 = 1.699 \approx 1.7$.

What is the $pH$ of a centimolar solution of $H_2SO_4$?

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