Question

Calculate the pH at the equivalence point if 25.00 mL of 0.010 Mbarbituric acid (HC4H3N2O3) is titrated with 0.20 M NaOH. (Kabarbituric acid = 1.0 x 10^-5).

I calculated the pH to be 5.3. But I'm not sure which is why Iam asking.

25.0 mL x 0.010 M = 0.25 mmol Bacid

25.0 mL x 0.020 M = 0.50 mmol NaOH

Bacid NaOH

B 0.25 0

C 0.5-0.25 0.5

A 0.25 0.5

pKa = -log [Ka]

= -log (1.0 x 10^-5)

= 5

pH = pKa + log Base/Acid

pH = 5 + log 0.5/0.25

pH = 5 + 0.3010

pH = 5.3