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

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