Calculate the pH at the equivalence point if 25.00 mL of 0.010 M barbituric acid (HC4H3N2O3) is titrated with 0.20 M NaOH. (Ka barbituric acid = 1.0 x 10^-5).
I calculated the pH to be 5.3. But I'm not sure which is why I am 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