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 0.180 M sodium benzoate. How much of each solutionshould be mixed to prepare this buffer?

____ mL of benzoic acid

____ mL sodium benzoate

Answer 1

we have to use the the following equation

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

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

log [salt]/[acid] = -0.20

[salt] / [acid] = 10^-0.20

[salt] / [acid] = 0.631

no. of moles of salt / no. of moles of acid = 0.631 ----->1

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

from this we can write

no of moles of solute = molarity x volume of solution in litres

volume of the buffer solution = 100 mL = 0.1L

suppose acid will be a Liters

salt will be 0.1-a liters

no. of moles of acid = 0.1 xa

no. of moles of salt = 0.18 x (0.1-a) = 0.018 - 0.18a

put these salt and acid mole values in equation 1

0.018-0.18a / 0.1a = 0.631

0.018-0.18a = 0.0631a

0.2431a = 0.018

a = 0.074 L

a = volume of the benzoic acid = 0.074L = 74 mL

volume of the sodium benzoate (salt) 0.1-0.074 = 0.026 = 26 mL