Question

You need to prepare 100 ml of a ph=4 buffer solution using .100M benzoic acid (pka=4.20) and .240M sodium benzoate. How much of each solution should be mixed to prepare this buffer? Find ml of benzoic acid and ml of sodium benzoate

Answer 1

pH = pKa + log(base/acid)
4 = 4.2 + log(base/acid)
-0.2 = log(base/acid)
10^-0.2 = base/acid
base/acid = 0.63

0.63 x 0.01 moles base = 0.0063 moles base.

0.0063 moles of a 0.24 M salt solution

0.24 moles = 1000 mL
1 mole = 1000/0.24 mL

0.0063 moles = 1000*0.0063/0.24 mL = 26.25 mL

Hence mixture of 100 mL 0.1 M bencoic acid + 26.25 mL of 0.24 Sodium benzoate has pH4.

Scale down the volume to 100 ml

100 mL total volume:
benzoic acid: 100 x 100/(126.25.) = 79.21 mL
sodium benzoate: 26.25 x 100/(126.25) = 20.79 mL