1. Consider one litre of buffer solution containing 0.8 m CH3COOH and O.8 m CH3COONa. Assume that the volume change due to the addition of 0.01 mol of solid NaOH is negligible. Ka for CH3COOH is 1.8 x 10-5.
2.
3. The dissociation constant for CH3COOH is given by
The above expression shows that the concentration of H+ is directly proportional to
degree of dissociation of CH3COOH = α
5. Given that Ka for CH3COOH is 1.8 x 10-5
[H+] = 1.8 x 10-5
pH = – log [H+]
= – log [1.8 x 10-5]
= 5 – log 1.8
= 5 – 0.26 pH
= 4.74
6. After adding 0.01 moI NaOH to I litre of buffer. Given that volume change due to the addition of NaOH is negligible. [OH-] = 0.01 M. The consumption of OH- are expressed by the following equation.
7. The addition of a strong base (0.01 M NaOH) increased the pH only slightly i.e., from 4.74 to 4.75. So the buffer action is verified.