Question

Using Henderson Hasselbach, calculate the pH of a buffer that has 0.65M acetic acid and 0.50 M sodium acetate before and after 0.06 M nitric acid is added. Write the equilibrium and buffering reactions. Ka of HAc = 1.8 x 10 -5 (show all reactions and work, with units)

Answer 1

Henderson-Hasselbalch Equation is given as:

pH = pK_a + log ([A^-]/[HA])

[A^-] = molar concentration of a conjugate base (salt)

[HA] = molar concentration of a undissociated weak acid

pH = pK_a + log ([A^-]/[HA])

pH = pK_a + log ([C₂H₃O₂^-] / [HC₂H₃O₂])

pH = -log (1.8 x 10^-5) + log (0.50 M / 0.65 M)

pH = -log (1.8 x 10^-5) + log (2.5)

pH = 4.745 - 0.114 = 4.631

When acid, H⁺, is added to this acetic acid-acetate buffer, the additional H⁺ can react with the CH₃COO^- which is now in excess:

H⁺ + CH₃COO^- → CH₃COOH

The concentration of undissociated CH₃COOH molecules increases.

Because acetic acid is a weak acid, it dissociates to a small extent to produce hydrogen ions:

CH₃COOH ↔ H⁺ + CH₃COO^-

So the concentration of hydrogen ions in solution, H⁺_(aq), will increase slightly.

Since pH = -log₁₀[H⁺] the pH will decrease slightly.
The pH actually falls from 4.74 ( ≈ 4.7) to 4.66 ( ≈ 4.7)