The pH of a completely dissociated 0.005 M | vedclass

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(C) $H_2SO_4$ is a strong diprotic acid that dissociates completely in water as follows: $H_2SO_4 \rightarrow 2H^+ + SO_4^{2-}$ Since the concentratio

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$2$

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(C) $H_2SO_4$ is a strong diprotic acid that dissociates completely in water as follows: $H_2SO_4 ightarrow 2H^+ + SO_4^{2-}$ Since the concentration of $H_2SO_4$ is $0.005 \ M$,the concentration of $H^+$ ions is: $[H^+] = 2 	imes [H_2SO_4] = 2 	imes 0.005 \ M = 0.01 \ M = 10^{-2} \ M$ The $pH$ is calculated as: $pH = -\log[H^+] = -\log(10^{-2}) = 2$ Therefore,the correct option is $C$.

The $pH$ of a completely dissociated $0.005 \ M \ H_2SO_4$ solution is:

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