A 25.0 mL sample of the acid is titrated against 2 M NaOH.?

Question

What is the pH when 30.0 mL of base have been added?

0.85 mol acid
Ka = 2.24×10^-6?

Answer 1

30 ml of 2 M NaOH = 0.06 moles NaOH

If you started with 0.85 moles acid, then you have 0.79 moles left, and 0.06 moles of the conjugate base has formed. Since they're in the same volume, you can use Henderson-Hasselbach with a ratio of the moles vs. the molarities.

pH = pKa + log [base]/[acid]
pH = 5.65 + log (0.06/0.79) = 5.65 - 1.12 = 4.53