Calculate the ph of a buffered solution prepared by dissolving 21.5 g benzoic acid and 37.7 g sodium benzoate

Data Given:

Mass of Benzoic Acid = 21.5 g

Moles of Benzoic Acid = 21.5 g / 122.12 g/mol

Moles of Benzoic Acid = 0.176 mol

Mass of Sodium Benzoate = 37.7 g

Moles of Sodium Benzoate = 37.7 g / 144.11 g/mol

Moles of Sodium Benzoate = 0.2616 mol

pKa of Benzoic Acid = 4.19

Solution:

Using the formula,

pH = pKa + log [Sodium Benzoate] / [Benzoic Acid]

pH = 4.19 + log [0.2616] / [0.176]

pH = 4.19 + 0.1721

pH = 4.362