# 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 $\frac{{\left[\mathrm{base}\right]}_{0}}{{\left[\mathrm{acid}\right]}_{0}}$

pH = -log (5.62 x 10-4) + log $\frac{0.10}{0.25}$

pH = 2.85