A buffer is prepared by mixing 25.2 mL of 0.183 M NaOH with 144.3 mL of 0.231 M acetic acid. What is the pH of this buffer?

Question

An acid–base buffer is able to resist changes in pH due to the addition of small amounts of strong acid or base to the system. An acid–base buffer typically contains a weak acid and its conjugate base.

A buffer is prepared by mixing 25.2 mL of 0.183  M NaOH with 144.3 mL of 0.231 M acetic acid. What is the pH of this buffer? (The pK_a for acetic acid is 4.75.)

Calculate the pH of this buffer after the addition of 0.192 g of solid NaOH.

Answer 1

For most buffer problems, we can use the Henderson Hasselbalch equation:

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

Initial mols NaOH =25.2 ml x 1 L / 1000 ml x 0.183 mol / L = 4.61x10^-3 mols

Initial mols CH3COOH = 144.3 ml x 1 L / 1000 ml x 0.231 mol / L = 0.0333 mols

NaOH + CH3COOH ==> CH3COONa + H2O

4.61x10^-3.....0.0333....................0....................Initial

-4.61x10^-3...-4.61x10^-3..........+4.61x10^-3.........Change

0................0.0287.................4.61x10^-3.......Equilibrium

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

pH = 4.75 + log (4.61x10^-3 / 0.0287)

pH = 4.75 - 0.794

pH = 3.96

If you add 0.192 g NaOH to the above buffer, and assuming no change in volume:

moles NaOH added = 0.192 g x 1 mol NaOH / 40 g = 0.00480 mols

NaOH + CH3COOH ==> CH3COONa + H2O

0.00480.........0.0287..............0.00461...............Initial

-0.00480.......-0.00480...........+0.00480............Change

0................0.0239..............0.00941..............Equilibrium

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

pH = 4.75 + log (0.00941 / 0.0239)

pH = 4.75 -0.405

pH = 4.35