[Beasty-98]

I have got a question: "calculate pH, pOH, [H+] and [OH-] in 2M H2SO4"
Solution:

I put up the reaction like:
H2SO4 + H2O --><-- (HSO4-) + H3O

Since it's a very strong acid which therefore protolysis almost completely the molar proportions are 1:1 for H2SO4 becoming HSO4 (I thought!). Therefore [H+]= 2M, according to me.

Then the pH=-lg(2)=-0,3
pH+pOH = 14 => pOH = 14- (-0,3)=14,3
[OH-] = 10^-pOH = 5 x 10^-15

BUT, according to the key [H+]=4M
The molar proportion is then 1:2. I guess it is because the HSO4- also can react with the water and become (SO4)2-. I thought the Ka generally decreased with every time it reacts with the water. And that the production of H3O from the (SO4)2- therefore was negligible?

How do I know that 1 M H2SO4 gives 2M H3O+?

I really hope one of you guys can help me with understanding this! Also I would like to apologize for eventual mishaps with the language, since I'm from Sweden my English isn't perfect, and especially not my scientific English hope you at least understand what I mean!

Best regards,
Bea

[AWK]

It is very difficult to calculate pH of 2 M sulfuric acid. You can safely treat it in calculation as diprotic acid at the concentration of order 10^-4 M.

Many textbooks (nearly all) except advanced textbooks of physical chemistry treat it as the strong acid on both dissociation steps.

When in Rome, do as the Romans do!

[Borek]

Very bad question. HSO₄^- is a relatively weak acid, and doesn't dissociate completely. In a 2M solution only a very small fraction of the HSO₄^- is dissociated (below 1%), so the answer given as a correct one is off.

I understand teachers need to simplify things to show concepts at work, but I prefer when they use examples that are not disconnected from reality.

For a method of solving this type of problems take a look here: http://www.chembuddy.com/?left=pH-calculation&right=pH-polyprotic-simplified - please note it requires knowledge about equilibrium calculations, chances are you are not there yet.

[Beasty-98]

Thanks a lot!