You need to prepare 100.0 mL of a pH 4.00 buffer solution using 0.100 M benzoic acid (p𝐾a=4.20) and 0.200 M sodium benzoate.

Question

How many milliliters of each solution should be mixed to prepare this buffer?

Benzoic acid: _____ ml

Sodium Benzoate:_____ ml

Answer 1

Use the Henderson Hasselbalch equation:

pH = pKa + log [salt]/[acid]

4.00 = 4.20 + log [benzoate]/[benzoic acid]

log [benzoate]/[benzoic acid] = -0.2

[benzoate]/[benzoic acid] = 0.631

let x = ml of sodium benzoate. Then 100 ml - x = mls benzoic acid:

(x ml)(0.2 mmol / L) / (100 ml - x)(0.1 mmol/L) = 0.631

6.31 = 0.2631x

x = 23.98 mls of sodium benzoate = 24.0 mls sodium benzoate

100 - 23.98 = 76.02 mls benzoic acid = 76.0 mls benzoic acid