Question

A)

A buffer solution contains 0.336 M KHSO₃ and 0.447 M Na₂SO₃.

If 0.0144 moles of hydroiodic acid are added to 125 mL of this buffer, what is the pH of the resulting solution ?
(Assume that the volume change does not change upon adding hydroiodic acid)

pH =

B)  
If 0.0335 moles of perchloric acid are added to 250 mL of this buffer, what is the pH of the resulting solution ?
(Assume that the volume change does not change upon adding perchloric acid)

pH =

Answer 1

The pH of the resulting buffer solution can be calculated using Henderson-Hasselbalch equation-

pH = pKa + log([salt]/[acid])

where, salt = [SO₃^-]

acid = [HSO₃^-]

pKa of acid = 7.19

Molairty of Na₂SO₃ = 0.447 M

Therefore, no. of moles of SO₃^2- = 0.447 moles

Molairty of KHSO₃ = 0.336 M

Therefore, no. of moles of HSO₃^- = 0.336 moles

A)

Calculate the concentrations of SO₃^2-, and HSO₃^- after addition of 0.0144 moles of hydroiodic acid (HI). Since, 1 mole of HI reacts with 1 mole of SO₃^2-. Therefore, 0.0144 moles of hydroiodic acid (HI) react with 0.0144 moles of SO₃^2- and 0.0144 moles of HSO₃^- formed.

No. of moles of SO₃^2- left = (0.447 - 0.0144) moles

= 0.4326 moles

Total No. of moles of HSO₃^- = (0.336 + 0.0144) moles

= 0.3504 moles

Calculate resulting pH

pH = 7.19 + log(0.4326/0.3504)

= 7.28

pH of resulting solution = 7.28 (Ans)

B)

Calculate the concentrations of SO₃^2-, and HSO₃^- after addition of 0.0335 moles of perchloric acid. Since, 1 mole of perchloric acid reacts with 1 mole of SO₃^2-. Therefore, 0.0335 moles of perchloric acid (HI) react with 0.0335 moles of SO₃^2- and 0.0335 moles of HSO₃^- formed.

No. of moles of SO₃^2- left = (0.447 - 0.0335) moles

= 0.4135 moles

Total No. of moles of HSO₃^- = (0.336 + 0.0335) moles

= 0.3695 moles

Calculate resulting pH

pH = 7.19 + log(0.4135/0.3695)

= 7.24

pH of resulting solution = 7.24 (Ans)