Question #37c75

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Question #37c75

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Answer 1 · 2015-08-18T17:54:51

$p H = 0.4$ $pH=0.4$

Explanation:

Since $H_{2} S O_{4}$ $H_2SO_4$ is a strong diprotic acid with virtual 100% dissociation: a $0.2 M$ $0.2"M"$ solution will be $0.4 M$ $0.4"M"$ with respect to $H^{+}$ $H^+$ ions.

$p H = - log [H^{+}]$ $pH=-log[H^+]$

$= - log [0.4]$ $=-log[0.4]$

= $0.4$ $0.4$

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Question #37c75

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