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 M. A student adds 7.60 mL of a 0.330 M HCl solution to the beaker. How much will the pH change? The pKa of acetic acid is 4.740.

The change in pH is -0.047

Explanation:

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

pH = pKa + log (base/acid)

5.000 = 4.740 + log (base/acid)

0.260 = log (base/acid)

100.240 = 10log (base/acid)

base/acid = 1.73780

1.90 x10^2 mL buffer (0.100 M) = 190 mmols of Acid + Base        (A + B from now on)

B/A = 1.7378

B = 1.73780(A)

190 mmol = A + B

190 mmol = A + 1.7378(A)

190 mmol = 2.7378A

A = 69.39 mmol = acetic acid

Amount of B: 190 mmol - 69.39 mmol = 120.60 mmol B = acetate

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

7.60 mL HCl (0.330 M) = 2.508 mmol HCl added

                   Acetate         +           HCl        --->    Acetic acid     +          H2O

Before           120.60                   2.508                         69.39

Change          -2.508                  -2.508                     +2.508

Final             117.49                           0                      71.898

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

pH = 4.740 + log (117.49/71.898) = 4.9532

ΔpH: 4.9532 - 5.000 = -0.0467=-0.047

it takes on an electronical energy