pH of a mixture of 1 M benzoic acid pKa = 4.20 and 1 M C6H5COONa is 4.5. In 300 ml buffer, benzoic acid is [log 2 = 0.3]
200 ml
150 ml
100 ml
50 ml
pH = pKa + log SaltAcid ∴ 4.5 = 4.2 + log SaltAcid = log SaltAcid = 0.3 since anti log 0.3= 2 ∴ SaltAcid = 2
Let V ml of 1 M C6H5 COOH solution and (300 - V) ml of 1 M C6H5COONa solution be mixed together
Acid = V × 11000 × 1000300 ; Salt = 300 - V1000 × 1000300 Acid = V300 ; Salt = 300 - V300 ∴ 300 - V/300V/300 = 2=300-V = 2V ∴ V=100 ml