[WHITECHIC9]

26.20 mL of .102M NaOH is needed to neutralize 25mL of an unknown acid, HA. The pH at the equivilance pt was measured to be 8.77. What's the Ka of the acid?

[Borek]

Please read forum rules. You just don't ask. You try, you tell us what you did, we push you in the right direction.

[WHITECHIC9]

well i think i need to use the hendersson equation but i'm not sure how to go about it.. should I change the mL to L? I'm not sure why it's giving me volume..

[Borek]

No need for HH. Do you know how to calculate equivalence point pH for a weak acid? Just reverse the process - take equation that describes pH as a function of Ka and solve it for Ka.

I'm not sure why it's giving me volume.

You have to calculate acid concentration somehow.

[WHITECHIC9]

i don't know how to calculate the equilivalence pt that's why I don't understand how to get the Ka if I don't have the name of the acid or if i don't use HH..

[Borek]

What is present in the solution at equivalence point?

[WHITECHIC9]

would there be excess NaOH because there's more mL left if you minus 26.20-25.00ml of acid?

[Borek]

No excess by definition. What is equivalence point then?

Looks like we will stop at multiplication table 

[WHITECHIC9]

IN my problem it states the pH is 8.77 at the equivalence pt. and it wants the Ka of the acid.

[Borek]

What is the DEFINITION of equivalence point. Not pH at, but DEFINITION of.

[WHITECHIC9]

the equilivalance pt is when enuogh reactant (NaOH) is added to neutralize the solution.(weak acid) Correct?

[Borek]

the equilivalance pt is when enuogh reactant (NaOH) is added to neutralize the solution.(weak acid) Correct?

Yes. So, if the equivalence point is when you have added just enough NaOH to neutralize the weak acid - is it possible that there is an excess of base at the equivalence point?