What is the pH of a solution of 0.300 M HNO₂ containing 0.120 M NaNO₂? (Ka of HNO₂ is 4.5 × 10⁻⁴)

Question

Note the presence of an acid and a salt of its conjugate base in the solution. To solve this problem, you will need to plug the appropriate pKa value into the Henderson-Hasselbalch equation.



Answer 1

The presence of a weak acid (HNO₂) and its conjugate base (NO₂^-) forms a buffer. For most buffer problems, we can use the Henderson Hasselbalch equation:

pH = pKa + log [conj.base] / [acid]

pH = ?

pKa = -log Ka = -log 4.5x10^-4 = 3.35

[conj.base] = 0.120 M

[acid] = 0.300 M

Solving for pH...

pH = 3.35 + log (0.120 / 0.300)

pH = 3.35 + log 0.4

pH = 3.35 -0.40

pH = 2.95