Question

A beaker with 1.90×10² 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 buffer is 0.100 mol L−1. A student adds 5.00 mL of a 0.490 mol L−1 HCl solution to the beaker. How much will the pH change? The pKa of acetic acid is 4.760.

Express your answer numerically to two decimal places. Use a minus ( − ) sign if the pH has decreased.

Answer 1

First, we need to figure out how many moles of each species there are before adding the acid.
Buffer problems use the henderson-Hasselbalch equation:

pH = pKa + log ([base]/[acid]) . . .where [base] = [acetate] and [acid] = [acetic acid].

5.000 = 4.760 + log ([acetate]/[acetic acid])
0.240 = log ([acetate]/[acetic acid])
([acetate]/[acetic acid]) = 10^0.240 = 1.74
[acetate] = 1.74[acetic acid]

We also know that

[acetate] + [acetic acid] = 0.100 . . .substituting 1.74[acetic acid] for [acetate] gives us
1.74[acetic acid] + [avcetic acid] = 0.100
2.74[acetic acid] = 0.100
[acetic acid] = 0.0365 M . .and [acetate] = 0.100 - 0.0365 = 0.0635 M. In 190 mL of solution,

initial moles acetate = M acetate x L of solution = (0.0635)(0.190) = 0.0121 moles acetate
initial moles acetic acid = M acetic acid x L of solution = (0.0365)(0.190) = 0.00694 moles acetic acid

moles HCl to be added = M HCl x L HCl = (0.490)(0.0050) = 0.00245 moles HCl

HCl will react with the base part of the buffer (acetate) to produce acetic acid:

Moles . . . . . . . . . .acetate + HCl ==> acetic acid + Cl-
initial . . . . . . . . . .0.0121 . . .0.00245 . . . .0.00694 . . . .0
change . . . . . . .-0.00245 . .-0.00245 . . . .+0.00245 . .+0.00245
final . . . . . . . . . . .0.00965 . . . .0 . . . . . . . .0.00939 . . .0.00245

So after adding the HCl we now have

[acetate] = 0.00965 moles / 0.198 L = 0.0487 M
[acetic acid] = 0.00939 moles / 0.198 L = 0.0474 M

pH = 4.760 + log (0.00965 / 0.00939) = 4.760 +0.0118 = 4.7718

So the pH raised by 0.0118 units.