Question

A beaker with 190 mL of an acetic acid buffer with a pH of 5.00 is sitting on a benchtop. The total molarity of acid and conjugate base in this buffer is 0.100 M. A student adds 8.00 mL of a 0.490 M HClsolution to the beaker. How much will the pH change? The pKa of acetic acid is 4.760.

Answer 1

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

[base] = [CH3COO- ] and [acid] = [CH3COOH].

First calculate ratio of the moles of ([base]/[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] ---1

given that;

Total moles of   [acetate] + [acetic acid] = 0.100 ----2

Then

1.74[acetic acid] + [acetic 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 = molarity of HCl* volume in L = (0.490)(0.008) = 0.00392 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 . . .0.00392 . . . .0.00694 . . . .0
change . . . . . . .-0.00392 . .-0.00392 . . . .+0.00392. .+0.00392
final . . . . . . . . . . .0.0081 . . . .0 . . . . . . . .0.011 . . .0.00392

Now calculate the new molarities of acetate and acetic acid;

total volume =190 +8 ml =198 ml = 0.198 L

[acetate] = 0.0081 moles / 0.198 L = 0.041 M
[acetic acid] = 0.011 moles / 0.198 L = 0.056 M

pH = 4.760 + log (0.041 / 0.056)

= 4.760 - 0.135= 4.625

Here initial pH = 5.000

After HCl adds pH =4.625

So the pH dropped = 5.00- 4.625= 0.375 units.