Exercise 4 | Buffers and Titration of Acids and Bases
Calculate the pH of a solution that is 0.25 M in HNO2 (aq) and 0.10 M in NaNO2 (aq).
Data: Ka (HNO2/NO2-) = 5.62 x 10-4
HNO2 is a weak acid so it does not ionize very much.
NaNO2 completely ionizes to Na+ a spectator ion, and NO2- the conjugate base of HNO2.
So [Na+] = [NO2-] = 0.10 M
Henderson-Hasselbalch equation can be used:
pH = pKa + log
pH = -log (5.62 x 10-4) + log
pH = 2.85