Question

A beaker with 115 mL of an acetic acid buffer with a pH of 5.000 is sitting on a bench top. The total molarity of acid and conjugate base in this buffer is 0.100 M. A student adds 4.80 mL of a 0.420 M HCl solution to the beaker. How much will the pH change? The pKa 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

change in pH = -0.31

Explanation

Given : total molarity of acid and conjugate base = 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] = 10^0.260

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

Solving equations (1) and (2) we get,

[acetic acid] = 0.0355 M

[acetate] = 0.0645 M

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

initial moles of acetic acid = (0.0355 M) * (115 mL)

initial moles of acetic acid = 4.08 mmol

Similarly, initial moles of acetate = 7.42 mmol

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

moles of HCl added = (0.420 M) * (4.80 mL)

moles of HCl added = 2.016 mmol

HCl is a strong acid which will convert acetate to acetic acid in neutralization reaction

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

new moles of acetic acid = (4.08 mmol) + (2.016 mmol)

new moles of acetic acid = 6.096 mmol

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

new moles of acetate = (7.42 mmol) - (2.016 mmol)

new moles of acetate = 5.404 mmol

According to Henderson - Hasselbalch equation,

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

pH = pK_a + log(new moles of acetate / new moles of acetic acid)

pH = 4.740 + log(5.404 mmol / 6.096 mmol)

pH = 4.688

change in pH = final pH - initial pH

change in pH = 4.688 - 5.000

change in pH = -0.31