Acid Base Titration

[hospitaldoctor1]

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In an acid base titration where H2SO4 is titrated with NaOH... What occurs at the first equivalence point?
I know that all of the H2SO4 is gone and the amount of base added equals the original amount of H2SO4... but which species starts to appear at the eq pt?: HSO4 or SO4? Why?
What else occurs at the equivalence point?
What occurs at the half equivalence point?
Thanks

 

[Isoprop]

H2SO4 is a much stronger acid than HSO4-. So as we add NaOH, H2SO4 will lose its proton before any HSO4- reacts. At the half equiv point, half of the H2SO4 would have reacted and we have half of the conjugate base. At the equiv point, all of the H2SO4 would have reacted and the solution only contains HSO4- and sodium ions. Any addition of NaOH after this point will then take off the second proton and form sulfate anions (SO4-2)

so at half equiv: 1/2 H2SO4 + 1/2 HSO4-
at equiv point: all HSO4-
after equiv point: SO4-2

In reality, what i described doesn't really happen. When sulfuric acid is placed in solution, being a strong acid, it will dissociated completely into H+ (H3O+) and its conjugate base. The NaOH will react to the H+ before it reacts with the HSO4-. Thus at half equiv point, we actually have 1/2 H+ and 1/2 HSO4-. If what I said confused you, then just ignore it.

 

[hospitaldoctor1]

which species starts to appear at the first eq pt?: HSO4 or SO4? Why?

 

[Isoprop]

exactly at equiv point, all you have in solution is HSO4-. any NaOH added after that will start to create SO4-2.

so UP TO equiv point, you are just "making" HSO4-. It doesn't "start" to appear; rather, it's been appearing since the start of the titration. Immediately after equiv point, SO4-2 starts to appear. But exactly at equiv point, there is no SO4-2.