Question

A beaker with 1.80×102 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 8.60 mL of a 0.390 mol L−1 HCl solution to the beaker. How much will the pH change? The pKa of acetic acid is 4.760.

Answer 1

Let's first determine how much acetic acid and acetate we have in the buffer:

pH = pKa + log (base/acid)

5.000 = 4.760 + log (base/acid)

0.240 = log (base/acid)

10^0.240 = 10^{log (base/acid)}

base/acid = 1.738

180 mL buffer (0.100 M) = 18.0 mmols of Acid + Base        (A + B from now on)

B/A = 1.738

B = 1.738(A)

18.0 mmol = A + B

18.0 mmol = A + 1.738(A)  

18.0 mmol = 2.738A

A = 6.574 mmol = acetic acid

Amount of B: 18.0 mmol – 6.574 mmol = 11.426 mmol B = acetate

Now that we have our mmol of A and B, we can see what remains after HCl is added:

Number of moles = moarity * volumes in L

8.60 mL HCl (0.390 M) = 3.354 mmol HCl added

                    Acetate         +           HCl        --->    Acetic acid            +             H2O

Before            11.426                3.354                 6.574

Change          -3.354                      -3.354                +3.354

Final              8.072                       0                   9.928

We still have a buffer, so we can use pH = pKa + log (base/acid) again to find the pH:

pH = 4.760 + log (8.072/9.928)

pH = 4.760 -0.0899

= 4.6701

ΔpH: 4.6701 - 5.000 = -0.3299 = -0.330