A beaker with 1.00×102 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 5.80 mL of a 0.440 mol L−1 HCl solution to the beaker. How much will the pH change? The pKa of acetic acid is 4.760.
Given that pH = 5.000
Henderson-Hasselbalch equation:
pH = pKa + log[CH3COO-] / [HC2H3O2- ]
5.00 = 4.760 + log[C2H3O2-] / [HC2H3O2]
0.240 = log[CH3COO-] / [HC2H3O2]
[CH3COO-]/ [HC2H3O2] =10^0.240
[CH3COO-]/ [HC2H3O2] =1.74
1.00×102 mL mL buffer (0.100 M) = 10.0 mmols of Acid + Base (A + B from now on)
[CH3COO-]/ [HC2H3O2] =1.74
[CH3COO-] = [HC2H3O2] 1.74 ----1
10.0 mmol = [CH3COO-] + [HC2H3O2] ---2
10.0 mmol =[HC2H3O2] 1.74+ [HC2H3O2]
10.0 mmol =[HC2H3O2] 2.74
[HC2H3O2] = 3.65 mmol = acetic acid
Therefore
[CH3COO-] = [HC2H3O2] 1.74 ----1
[CH3COO-] = 3.65* 1.74 = 6.35 mmol
Now that we have our mmol of Acid and acetate, we can see what remains after HCl is added:
5.80 mL HCl (0.440 M) = 2.552 mmol HCl added
Acetate + HCl ---> Acetic acid + H2O
Before 6.35 2.552 3.65
Change -2.552 -2.552 +2.552
Final 3.798 0 6.202
We still have a buffer, so we can use pH = pKa + log[CH3COO-] / [HC2H3O2- ] again to find the pH:
pH = 4.760 + log (3.798/6.202)
pH = 4.760 + (0..213)
= 4.547
ΔpH: 4.547 - 5.000 = -0.453
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