Calculate the pH at the equivalence point when 50 mL of 0.098 M propionic acid is titrated with 0.340 M NaOH.
I tried solving for the volume of NaOH and them using the total volume and then plugging it in to [H+][A-]/[HA] and used the Ka value for propionic acid but it was not correct.
I think it isn't correct because it is a weak acid with a strong base but I don't know how to take that into account. Help please.
You started right but slipped half-way through.
Calculate volume NaOH to arrive at the e4quivalence point, then M of the salt at the equivalence point is mols propionic acid/total volume in L. Sould be about 0.2 M but you should be more exact than that.
The pH at the equivalence point is determined by the hydrolysis of the salt. The salt will be NaP and P^- is the propionate ion.
...........P^- + HOH ==> HP + OH^-
I.........0.2.............0....0
C..........-x.............x....x
E........0.2-x...........x......x
Kb for P^- = (Kw/Ka for propionic acid) = (x)(x)/(0.2-x). Solve for x = OH^- and convert to H^+ and pH.