Question

A beaker with 130 mL of an acetic acid buffer with a pH of 5.00 is sitting on a benchtop. The total molarity of acid and conjugate base in this buffer is 0.100 M. A student adds 6.20 mL of a 0.490 M HCl solution to the beaker. How much will the pH change? The pKa of acetic acid is 4.760.

Express answer numerically to two decimal places. Use a minus (-) sign if the pH has decreased.

Answer 1

According to Henderson-Hesselbalch equation,

5.00 = 4.76 + log [Acetate ] / [acetic acid]

5.00-4.76 = log [Acetate ] / [acetic acid]

0.24 = log [Acetate ] / [acetic acid]

10^-0.24 = [Acetate ] / [acetic acid]

[Acetate ] / [acetic acid] = 0.575

Given, total molarity of acid and conjugate base (i.e. Acetate + acetic acid ) = 0.100 M

Moles of acetate = 0.0362 M acid = 0.0635 M

Moles acetate = 0.0362 x 0.130 L = 0.00471 moles

Moles of acid = 0.0635 x 0.130 L = 0.00826 Moles

HCl added = 0.0062 x 0.490 = 0.00304 moles

Now,

H⁺ + CH₃COO^- ß-à CH₃COOH

Thus, Moles of acetate = 0.00471 - 0.00304 = 0.00167 moles

          Moles of acid = 0.00826 + 0.00304= 0.0113 moles

            Total volume = 130 + 6.20 = 136.20 mL = 0.1362 L

Molarity of acetate = no of moles / volume = 0.00167 moles / 0.1362 L

                               = 0.0127 M

Similarly, molarity of acid = 0.0113 moles / 0.1362 L

                                         = 0.083 M

New pH = 4.76 + log 0.0127M / 0.083 M

             = 4.76 + [-1.9 – (-1.081)]

             = 4.76 + (-1.9 + 1.081)

            = 4.76 – 0.819

      pH = 3.94