Question

To each of 100 mL of three (A, B & C) 0.1 M acetate buffer solutions (pH of A = 4.4, pH of B = 4.76 and pH of C = 5.2), l0 mL of 0.1 M NaOH solution was added. Calculate the resulting change in the pH and the buffering capacity (AOH- /ApH) of each solution. Calculate also the change in the pH if instead of NaOH, 0.1 M HCI was used.

Answer 1

for solution A :

acetate buffer pKa = 4.76

pH = 4.4 is given

millimoles of buffer = 100 x 0.1 = 10

millimoles of ( acid + salt ) = 10 --------------------->1

pH = pKa + log [salt /acid]

4.4 = 4.76 + log (salt /acid)

0.4365 = (salt /acid)

salt = 0.4365 acid ---------------> 2

from 1 and 2

acid + 0.4365 acid = 10

acid = 6.96

salt = 3.04

millimoles of acetic acid = 6.96

millimoles of acetate = 3.04

addition of C millimoles of NaOH = 10 x 0.1 = 1

C = 1

on addition of C millimoles of base to this buffer acid millimoles decreases and salt moles increases

new pH = pKa + log [salt + C / acid - C]

pH = 4.76 + log [3.04 + 1 / 6.96 -1]

pH = 4.59

change in pH = 4.59 - 4.4

                    = 0.19

buffer capacity = 0.19

if HCl is used instead of NaOH :

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

pH = 4.76 + log [3.04 -1 / 6.96 +1]

pH = 4.17

change in pH = 4.4 - 4.17 = 0.23

buffer capacity = 0.23