Exercise 4 | Buffers and Acid-Base Titrations

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