Equivalence point vs half-equivalence point in titrations

[Veni vidi vicii]

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I understand in the equivalence point Moles of Acid= Moles of Base. Then for half-equivalence point I have written Concentration of HA=Concentration of A-. Which makes pH=pKA because of the HH equation. My question is, how is this different, aren't they the same thing?

Thanks

 

[Veni vidi vicii]

I think I figured it out. At half-equiv we have half of the acid and half of the conjugate base thus pH=pKa. At equiv point, MolAcid=MolBase so we can set them equal to each other to solve for a variable. Am I right?

 

[jLaws]

At half-equivalence point, moles acid = moles base and ph = pka since ph = pka + log (A-/HA). If A- (base) and HA (acid) are equal, A-/HA = 1 and log(1) = 0. So ph = pka +0. At the equivalence point, moles base does not equal moles acid. The equivalence point is when you have added enough base/acid to convert all the acid/base that was initially present to its conjugate base/acid.

strong base + strong acid = neutral at equivalence
weak base + strong acid = acidic at equivalence (only conjugate acid present)
weak acid + strong base = basic at equivalence (only conjugate base present)

 

[jLaws]

sorry if you knew all that already, my test is tomorrow and it helped me to write it out 🙂

 

[Veni vidi vicii]

sorry if you knew all that already, my test is tomorrow and it helped me to write it out 🙂

Well good luck tomorrow! I thought at the equiv point moles of acid and base are equal so this allows us to do neutralization reactions because if molesA=molesB then MVacid=MVbase which allows us to calculate morlarity for either acid or base.

 

[jLaws]

Well good luck tomorrow! I thought at the equiv point moles of acid and base are equal so this allows us to do neutralization reactions because if molesA=molesB then MVacid=MVbase which allows us to calculate morlarity for either acid or base.

I

Thanks! I think we're kind of saying the same thing, except I am talking about what's present in the solution after the reaction. If you have 1L of 1M weak acid and add 1L of 1M strong base like NaOH, you now have only have the conj. base of the original weak acid in solution. The base you added would be gone ( Na+ and H2O)

 

[lwergod]

@jLaws, how did your test go?

 

[Veni vidi vicii]

Yeah how was the test? I'm sure you killed it

 

[jLaws]

Yeah how was the test? I'm sure you killed it

27AA! Thank god this is over with! I'll prob do a write up later this week.

 

[Veni vidi vicii]

27AA! Thank god this is over with! I'll prob do a write up later this week.

WOW! Congrats that's amazing. Write a breakdown ASAP my test is next week haha

 

[lwergod]

congrats!

a write-up would be nice.