Question

Calculate the change in ph when 0.24 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 0.88 M anH+

Answer 1

Calculate the change in ph when 0.24 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+

The buffer is of weak base and its salt

pKb of pyridine = 5.25

For buffer the Henderson Hassalbalch's equation will be

pOH = pKb + log [salt] / [base]

pOH =5.25 + log [0.52 / 0.58] = 5.20

pH = 14 - pOH = 8.8

Now as we have added 0.24 moles of H+. it will react with the same moles of base to give same moles of salt

Initial moles of base = Molarity X volume = 0.58 X 1 = 0.58 moles

Initial moles of salt = Molarity X volume = 0.52 X 1 = 0.52 moles

After addition of acid

Moles of base left = 0.58 - 0.24 = 0.34 moles

Moles of salt = 0.52 + 0.24 = 0.76 moles

pOH = pKb + log [salt] / [base] = 5.25 + log [0.76 / 0.34] = 6

pH = 14 - 6 = 8

Change in pH = 8.8- 8 = 0.8

b) a 0.60 M solution of aniline (an) containing 0.88 M anH+

The buffer is of weak base and its salt

pKb of ammonia = 4.74

For buffer the Henderson Hassalbalch's equation will be

pOH = pKb + log [salt] / [base]

pOH =9.13+ log [0.88 / 0.60] = 9.29

pH = 14 - pOH = 4.71

Now as we have added 0.24 moles of H+. it will react with the same moles of base to give same moles of salt

Initial moles of base = Molarity X volume = 0.60 X 1 = 0.60 moles

Initial moles of salt = Molarity X volume = 0.88X 1 = 0.88 moles

After addition of acid

Moles of base left = 0.60 - 0.24 = 0.36 moles

Moles of salt = 0.88 + 0.24 = 1.12 moles

pOH = pKb + log [salt] / [base] =9.13 + log [1.12 / 0.34] = 9.65

pH = 14 - 6 = 4.35

Change in pH =4.71 - 4.35 = 0.36