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 0.240 M sodium benzoate.

How much Benzoic Acid and Sodium Benzoate should be added in mL?

Answer 1

benzoic acid is the acid and sodium benzoate is the salt

so as we know

pH = pKa + log [salt]/[acid]

4 = 4.20 + log [salt] / [acid]

[salt]/[acid] = 0.631

no. of moles of salt / no. of moles of acid = 0.631

molarity = no. of moles of solute / volume of solutionin litres

so no. of moles of solute = molarity X volume of solution in litres

as the buffer solution is of 100 ml or 0.1 L

let acid be x L ...then salt will be ( 0.1 - x) L
no. of moles of salt = 0.24 X (0.1-x)

no. of moles of acid = 0.1 X x

therefore, we can wright as

no. of moles of salt / no. of moles of acid = 0.631
0.24(0.1-x) / 0.1x = 0.631

0.24(0.1-x) = 0.1x X 0.631 = 0.0631x

0.024 - 0.24x = 0.0631x
0.024 = 0.24x + 0.0631x = 0.3031x

x = 0.024/0.3031 = 0.079 L

x = volume of acid used = 0.079 L or 79 ml
0.1-x = volume of salt used = 0.1-0.079 = 0.021 L or 21 ml