Question

Question: A beaker with 1.40×10² 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 6.60 mL of a 0.430 M HClsolution to the beaker. How much will the pH change? The pKa of acetic acid is 4.740.

Please explain your steps. Thank you!

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 = 1.73780

140 mL buffer (0.100 M) = 14.0 mmols of Acid + Base        (A + B from now on)

B/A = 1.737800829

B = 1.73780(A)

14.0 mmol = A + B

14.0 mmol = A + 1.737800829(A)

14.0 mmol = 2.737800829 A

A = 5.113 mmol = acetic acid

Amount of B: 14.0 mmol - 5.113 mmol = 8.887 mmol B = acetate

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

6.60 mL HCl (0.430 M) = 2.838 mmol HCl added

                    Acetate         +           HCl        --->    Acetic acid            +             H2O

Before            8.887            2.838                5.113

Change          -2.838 -2.838                +2.838

Final              6.049              0                      7.951

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

pH = 4.760 + log (6.049 /7.951) = 4.6413

deltapH: 4.6413 - 5.000 = -0.3587