Question

A beaker with 115 mL of an acetic acid buffer with a pH of 5.000 is sitting on a benchtop. The total molarity of acid and conjugate base in this buller is 0.100 M. A student adds 6,30 mL of a 0.410 M HCl solution to the beaker. How much will the pH change? The pk of acetic acid is 4.740. Express your answer numerically to two decimal places. Use a minus ( - ) sign if the pH has decreased.

Answer 1

pH change = -0.40

Explanation

Total concentration = 0.100 M

[acetic acid] + [acetate] = 0.100 M ...(1)

According to Henderson - Hasselbalch equation,

pH = pK_a + log([conjugate base] / [weak acid])

pH = pK_a + log([acetate] / [acetic acid])

5.000 = 4.740 + log([acetate] / [acetic acid])

log([acetate] / [acetic acid]) = 5.000 - 4.740

log([acetate] / [acetic acid]) = 0.260

[acetate] / [acetic acid] = 10^0.26

[acetate] / [acetic acid] = 1.82 ...(2)

Solving equations (1) and (2),

[acetic acid] = 0.035465 M

[acetate] = 0.064535 M

moles acetic acid = (concentration acetic acid) * (volume of buffer)

moles acetic acid = (0.035465 M) * (115 mL)

moles acetic acid = 4.078475 mmol

Similarly, moles acetate = 7.421525 mmol

moles HCl added = (concentration HCl) * (volume HCl)

moles HCl added = (0.410 M) * (6.30 mL)

moles HCl added = 2.583 mmol

new moles acetic acid = (initial moles acetic acid) + (moles HCl)

new moles acetic acid = (4.078475 mmol) + (2.583 mmol)

new moles acetic acid = 6.661475 mmol

new moles acetate = (initial moles acetate) - (moles HCl)

new moles acetate = (7.421525 mmol) - (2.583 mmol)

new moles acetate = 4.838525 mmol

Again using Henderson - Hasselbalch equation,

pH = pK_a + log([acetate] / [acetic acid])

pH = 4.740 + log(4.838525 mmol / 6.661475 mmol)

pH = 4.60

pH change = (final pH) - (initial pH)

pH change = (4.60) - (5.000)

pH change = -0.40