Question

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. How much of each solution should be mixed to prepare this buffer?

Answer 1

volume of buffer solution = 100.0 mL

volume of acid + salt = 100 mL

let volume of acid = V mL

then volume of salt = (100 - V) mL

millimoles of acid = 0.1xV = 0.1V

millimles of salt = 0.2 x (100 - V) = 0.2(100 - V)

pH = pKa + log [salt /acid]

4.00 = 4.20 + log [0.2(100-V) / 0.1V]

[0.2(100-V) / 0.1V] = 0.631

20 - 0.2 V / 0.1 V = 0.631

V = 76 mL

volume of benzoic acid V = 76 mL

volume of salt (sodium benzoate )= 100 - V = 100 - 76

                        = 24 mL

volume of sodium benzoate = 24 mL