Question

Calculate the change in pH when 0.28 mol H+ is added to 1.00 L of each of the following buffers:

(a) A 0.58 M solution of pyridine (py) containing 0.52 M pyH+

(b) A 0.60 M solution of aniline (an) containing 0.92 M anH+

Answer 1

0.58 M solution of pyridine (py) containing 0.52 M pyH⁺

We have, pKb of pyridine = 5.25

Henderson Hassalbalch's equation can be written as,

pOH = pKb + log [salt]/[base] = 5.25 + log [0.52/0.58] = 5.2

pH = 14 - pOH = 14-5.2 = 8.8

The added 0.28 moles of H+ will react with 0.28 moles of base to give 0.28 moles of salt

Initial moles of base = 0.58 M x 1 L = 0.58 moles

Initial moles of salt = 0.52 M x 1 L = 0.52 moles

After the addition of 0.28 moles of acid

Moles of the base left unreacted = 0.58 moles - 0.28 moles = 0.3 moles

Moles of the salt = 0.52 moles + 0.28 moles = 0.86 moles

Therefore, pOH = pKb + log [salt]/[base] = 5.25 + log [0.86/0.3] = 5.70

pH = 14 – 5.7 = 8.3

Therefore, Change in pH = 8.8- 8.3 = 0.5

0.60 M solution of aniline (an) containing 0.92 M anH⁺

We have, pKb of ammonia = 4.74

pOH = pKb + log [salt] / [base] = 9.13 + log [0.92/0.6] = 9.31

pH = 14 – 9.31 = 4.69

The added 0.28 moles of H+ will react with 0.28 moles of base to give 0.28 moles of salt

Initial moles of the base = 0.60 M x 1 L = 0.60 moles

Initial moles of the salt = 0.92 M x 1 L = 0.92 moles

After the addition of acid

Moles of base left = 0.60 - 0.28 = 0.32 moles

Moles of the salt = 0.92 + 0.28 = 1.2 moles

pOH = pKb + log [salt]/[base] = 9.13 + log [1.2/0.32] = 9.70

pH = 14 - 70 = 4.3

Change in pH = 4.69 - 4.3 = 0.39