What is the ph of 0.45 M of $H_2SO_4$ ?

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Answer 1 · 2016-06-23T07:49:01

$pH=-log_10[H_3O^+]$ $=$ $-log_10[0.90]$ $~=0.05$

Sulfuric acid is reasonably treated as a strong diprotic acid that gives stoichiometric $[H_3O^+]$ and $[SO_4^(2-)]$ on dissolution.

$H_2SO_4(aq) + 2H_2O(l) rarr2H_3O^+ + SO_4^(2-)$

Thus $[H_3O^+]=2xx0.45*mol*L^-1$ to a first approx.

And $pH=-log_10[0.90]$

For a better approximation, we need $pKa_2$ for bisulfate anion, $HSO_4^-$ , though this is still pretty low.

What is the ph of 0.45 M of H_2SO_4H_2SO_4?