[noiseordinance]

Alright, so here's the details that I have:

KHP_aq + NaOH_(aq)  KP^-_(aq) + H₂O_l + Na⁺_(aq)

Starting with 25mL of KHP with an unknown molarity, 25mL of 0.1067M NaOH was titrated. My equivalence point was at 18.82mL with a pH of 9.40. So, cutting the volume in half, 9.41mL, my pH was 5.41. At this point, my pH = pKa, 5.41 = -log(Ka), Ka = 3.89 x 10^-6. This all seems right, since the Ka of KHP is 3.9 x 10^-6. I'm trying to figure out how to determine the molarity of the unknown acid.

Can anyone offer any pointers?

[noiseordinance]

As a side note, I think I figured it out, but maybe someone could verify what I did as correct. Supposedly, if I don't know the concentration of one of the solutions, I can solve for it by using the equivalence point data. So if my strong base was 0.1067 M NaOH, and I hit my equivalence point after adding 18.82mL of NaOH to 25mL KHP, then I could use the M_aV_a=M_bV_b, or (x)(0.025 L)=(1.067 mol)(0.01882 L), x = 0.0803. This would mean my molarity is 0.083 molar, yes?

[MrTeo]

I think you should consider that, unlike strong bases and strong acids, KPH doesn't dissociate completely. You'd better revise your calculations considering this fact... you could also try to write down the equilibrium reaction of K_a

[eunChae]

you can use ICE approach   
   reaction:                              KHP(aq) + NaOH(aq) <--> KP-(aq) + H2O(l) + Na+(aq)
   Initial concentrations,M           x            0.1067M                0                             0
   Changes,M                              ?                 ?                       ?                             ? 
   Equilib. conc,M                        ?                 ?                       ?                             ?

 and use equilibrium constant Ka to find out x. (question marks should be filled by you, I m too lazy to do that particular job )