What is the #pH# of a solution of #"sulfuric acid"# of #2xx10^-2*mol*L^-1# concentration?

1 Answer
anor277
Aug 9, 2017

#pH=1.70...........#

Explanation:

By definition, #pH=-log_10[H_3O^+]#

Now for #H_2SO_4# with a concentration of #10^-2*mol*L^-1#, clearly, #[H_3O^+]=2xx10^-2*mol*L^-1#.

That is to a first approx. #H_2SO_4# is considered to be a diacid in aqueous solution and gives 2 equiv of #H_3O^+#. (In the absence of a value for #pK_(a2)# I am perfectly justified in making this assumption!)

And so #pH=-log_10(2xx10^-2)=-log_10(0.02)=1.70#

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