Question

Calculate the pH of a buffer which is 0.1 M in acetic acid and 0.15 M in sodium acetate. Given that the ionization constant of acetic acid is 1.75 x 10^–5 . Also calculate the change in pH of the buffer if to 1 L of the buffer

(i) 1 c.c. of 1 M NaOH are added

(ii) 1 c.c. of 1 M HCl are added.

Assume that the change in volume is negligible.

(iii) What will be the buffer index of the above buffer?

Answer 1

Given,

[CH₃COOH] = 0.1 M

[CH₃COONa] = 0.15 M

pK_a = 1.75 × 10^–15

For an acidic buffer

pH = pK_a + log\(\frac{[Salt]}{[acid]}\)

pK_a = – log [K_a]

= – log [1.75 × 10^–15]

= – log 1.75 + 5 log 10

= – 0.2430 + 5

= 4.757

From equation (i)

pH = 4.757 + log\(\frac{(0.15)}{(1.10)}\)

= 4.757 + 1761

= 4.933

(i) 1 cc of 1 M NaOH contains NaOH = 10^–3 mol. This will convert 10^–3 mol of acetic acid into the salt so the salt formed = 10^–3 mol

[Acid] = 0.10 – 0.001 = 0.099 M

[Salt] = 0.15 + 0.001 = 0.151 M

\(\therefore\) pH = pK_a + log\(\frac{[Salt]}{[acid]}\)

= 4.757 + log\(\frac{0.151}{0.099}\)

= 4.757 + 0.183

= 4.940

\(\therefore\) Increase in pH = 4.940 – 4.933

= 0.077 which is negligible

(ii) 1 cc of 1 M HCl contains HCl = 10^–3 M

This will convert 10^–3 mol CH₃COONa into CH₃COOH.

\(\therefore\) [Acid] = 0.10 + 0.001 = 0.101 M

[Salt] = 0.15 – 0.001 = 0.149 M

pH = pK_a + log\(\frac{[salt]}{[acid]}\)

= 4.757 + log\(\frac{0.149}{0.101}\)

= 4.747 + 0.169

= 4.925

\(\therefore\) Decrease in pH = 4.933 – 4.925

= 0.008 which is neglibile

(iii) No. of moles of HCl or NaOH added in buffer = 0.001 mol

Change in pH ≃ 0.007

\(\therefore\) Buffer index = \(\frac{No.\,of\,moles\,of\,acidor\,base\,added}{change\,in\,pH}\)

= \(\frac{0.001}{0.007}\)

=\(\frac{1}{7}\) = 0.143