Calculate the pH of 0.01 M sulphuric acid. | vedclass

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Question

(A) $H_2SO_4$ is a strong diprotic acid that dissociates completely in water: $H_2SO_{4(aq)} + 2H_2O_{(l)} \longrightarrow 2H_3O^{+}_{(aq)} + SO_{4(aq

Vedclass · Accepted Answer

$1.699$

Vedclass · Answer

(A) $H_2SO_4$ is a strong diprotic acid that dissociates completely in water: $H_2SO_{4(aq)} + 2H_2O_{(l)} \longrightarrow 2H_3O^{+}_{(aq)} + SO_{4(aq)}^{2-}$ Since it is a strong acid,the concentration of hydronium ions is twice the concentration of the acid: $[H_3O^{+}] = 2 	imes c = 2 	imes 0.01 \ M = 0.02 \ M = 2 	imes 10^{-2} \ M$ The $pH$ is calculated as: $pH = -\log_{10}[H_3O^{+}]$ $pH = -\log_{10}(2 	imes 10^{-2})$ $pH = -(\log_{10}2 + \log_{10}10^{-2})$ $pH = -(0.3010 - 2)$ $pH = 2 - 0.3010 = 1.699$

Calculate the $pH$ of $0.01 \ M$ sulphuric acid.

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