A beaker with 165 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.

Question

A student adds 4.40 mL of a 0.300 M HCl solution to the beaker.

How much will the pH change? The pKa of acetic acid is 4.740.

Keep getting the wrong answer!

Answer 1

Henderson Hasselbalch equation:

pH = pKa + log [conj.base]/[acid]

Original buffer: 5.000 = 4.740 + log [conj.base]/[acid]

log [conj.base][acid] = 0.2600

[conj.base]/[acid] = 1.82

[conj.base] = 1.82[acid] ... equation (i)

conj.base + acid = 0.1 M ... equation (ii)

solve eq.(i) and substitute in eq. (ii) to obtain 2.82[acid] =0.1M

[acid] = 0.0355 M x 0.165 L = 0.00586 moles acid

[acetate] = 0.0645 M x 0.165 L = 0.0106 moles acetate (conj. base)

Adding 4.40 ml of 0.300 mol/L HCl = 0.00132 moles H+ and final volume = 165 + 4.4 = 169.4 ml = 0.1694

The H+ reacts with the conj.base to produce acid, so [conj.base] decreases and [acid] increases.

Final [HAc] = 0.00586 mol + 0.00132 mol = 0.00718 mol / 0.1694 L = 0.04238 M = [acid]

Final [Acetate] = 0.0106 mol - 0.00132 mol = 0.00928 mol / 0.1694 L = 0.05478 M = [conj.base]

pH = pKa + log [conj.base]/[acid]

pH = 4.740 + log (0.05478 / 0.04238) = 4.740 + 0.111

pH = 4.85