A beaker with 185 mL of an acetic acid buffer with a pH of 5

A beaker with 185 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 8.40 mL of a 0.320 M HCl solution to the beaker. How much will the pH change? The p K a of acetic acid is 4.740.

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

Solution

pKa = 4.740

pH = 5.000

total mmoles of buffer = 185 x 0.1 = 18.5

pH = pKa + log [salt / acid]

5.000 = 4.740 + log [salt / acid]

[salt / acid] = 1.8197

[salt + acid] = 18.5

1.8197 acid + acid = 18.5

mmoles of acid = 6.56

mmoles of salt = 11.94

mmoles of HCl added = 8.40 x 0.320 = 2.688

pH = pKa + log [salt - C / acid + C]

    = 4.740 + log [11.94 - 2.688 / 6.56 + 2.688]

pH = 4.740

pH change = 4.740 - 5.000

pH change = - 0.26