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

1 Answer
anor277
Jun 23, 2016

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

Explanation:

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.

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