Question

A buffer solution contains 0.11mol of acetic acid and 0.14mol of sodium acetate in 1.00 L.

What is the pH of this buffer?

What is the pH of the buffer after the addition of 2

Answer 1

Use the Henderson-Hasselbalch equation, which I have shown below:

pH = pKa + log [CH₃COO^-] / [CH₃COOH]

The K_a of acetic acid = 1.8 x 10^-5 so the pKa = -log ( 1.8 x 10^-5) = 4.74. Thus:

pH = 4.74 + log ( 0.14 mol / 0.11 mol) = 4.74 + 0.105 = 4.85

---

If you add 0.02 mol of KOH, the OH^- coverts 0.02 mol of the acetic acid into acetate. So we have to re-calculate how much acetic acid and acetate we have:

0.11 mol acetic acid - 0.02 mol = 0.09 mol acetic acid

0.14 mol acetate + 0.02 = 0.16 mol acetate

Re-calculating pH using Henderson-Hasselbalch:

pH = 4.74 + log ( 0.16 / 0.09) = 4.74 + 0.25 = 4.99

----

Finally, if we had 0.02 mol of strong acid, it will react with acetate to make acetic acid. Re-calculating:

0.11 mol acetic acid + 0.02 mol = 0.13 mol acetic acid

0.14 mol acetate - 0.02 mol = 0.12 mol acetate

pH = 4.74 + log ( 0.12 / 0.13) = 4.74 - 0.035 = 4.71