Question

A beaker with 1.30×10² 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 M. A student adds 5.40 mL of a 0.440 MHCl 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

Given that pH = 5.00

Henderson-Hasselbalch equation:
pH = pKa + log[CH3COO-] / [HC2H3O2- ]
5.00 = 4.74 + log[C2H3O2-] / [HC2H3O2]
0.26 = log[CH3COO-] / [HC2H3O2]
[CH3COO-]/ [HC2H3O2] = 1.82

[HC2H3O2] + [CH3COO-] = 0.100
Let x = [CH3COO-]
Then 0.100 –x = [CH3COOH]
x/ 0.100 –x = 1.82
x = 0.182 – 1.82x
2.82x – 0.182
x = 0.0645 = [CH3COO-]
0.1-x = 0.10 – 0.0645 = 0.0355 = [CH3COOH]

0.440 moles HCl/ liter x 5.40 mL x 1 liter/1000 mL = 0.00238 moles of HC l

You wrote 1.30 x 102 which I assumed was 1.30 x 10^2
moles of CH3COOH initially present:
0.0355 moles / liter x 130 mL x 1liter /1000 mL = 0.00461
moles of CH3COO- initially present:
0.0645 moles/ liter x 130 mL x 1 liter/1000 mL = 0.00839
HCl + CH3COO- ===? CH3COOH + Cl-
0.00167 moles HCl will consume 0.00167 moles of CH3COO- and produce an additional 0.00167 moles of CH3COOH

Moles of CH3COOH after addition:
0.00461 + 0.00238 = 0.00699
Moles CH3COO- = 0.00839 – 0.00238 = 0.00601
Final Volume = 130 mL + 5.40 mL = 135.40 mL
[CH3COOH] = 0.00699 moles/0.1354 liters = 0.0516

[CH3COO-] = 0.00601/ 0.1354 liters = 0.0444

pH = 4.74 + log(0.0444)/0.0516)
pH = 4.740 -0 .0653 = 4.6747
Change in pH = 5.000 - 4.6747 = 0.3253 will decrease