Question

Describe how would you prepare a benzoic acid-benzoate buffer with pH = 4.25. Available to you are 4.0 M sodium benzoate, solid benzoic acid (MW= 122.12 g/mol), water and a full laboratory of glassware. Ka = 6.3 x 10-5 for benzoic acid (C6H5COOH). Kw = 1.0 x 10-14.

Answer 1

We can use the Henderson-Hasselbalch equation equation

pH = pK_a + log [base / acid]

pH = 4.25

Ka =  6.3 x 10^-5

pKa = -Log Ka

pKa = - Log  6.3 x 10^-5 = 4.2

4.25 = 4.2 + log [base / acid]

log [base / acid] = 0.05

[base / acid] = 10^-0.05 = 1.122

numerator = 1.122 Moles

Volume of Sodium benzoate = 1.122 Moles x 1000 ml / 4.0 M = 280.50 ml

denominator (Acetic acid) = (1 mol) (122.12 g/mol) =122.12 g

Hence 280.50 ml of 4M NaOH and 122.12 gm of benzoic acid gives buffer with pH = 4.25