Frequently Asked Question

Here's how to calculate the pH at each point in the titration:
a) Beginning

Initial Concentration of Acetic Acid: 1.0 x 10^-2 M
ICE Table:
| | CH3COOH | H+ | CH3COO- |
|-------------|----------|--------|----------|
| Initial | 1.0E-2 | 0 | 0 |
| Change | -x | +x | +x |
| Equilibrium | 1.0E-2-x | x | x |

Ka Expression: Ka = [H+][CH3COO-] / [CH3COOH] = 1.7 x 10^-2
Solving for x: 1.7 x 10^-2 = (x)(x) / (1.0 x 10^-2 - x)

Approximation: Since Ka is relatively small, we can assume x Solving for x: x ≈ 1.3 x 10^-2 = [H+]
pH Calculation: pH = -log[H+] ≈ -log(1.3 x 10^-2) ≈ 1.89
b) Half-way Point

At the half-way point, the moles of NaOH added are equal to half the moles of acetic acid initially present. This means the concentration of acetic acid and acetate are equal, and the pH is equal to the pKa.
pH = pKa: pH = -log(Ka) = -log(1.7 x 10^-2) ≈ 1.77
c) Equilibrium Point

At the equilibrium point, all the acetic acid has reacted with the NaOH to form acetate. The solution will now be a buffer solution containing acetate and its conjugate acid (acetic acid).
Calculate the volume of NaOH needed: (10.00 cm³)(1.0 x 10^-2 M) = (0.00500 M)(Volume of NaOH) => Volume of NaOH ≈ 20.00 cm³
Moles of Acetate formed: (10.00 cm³)(1.0 x 10^-2 M) = 1.0 x 10^-3 moles
Concentration of Acetate: 1.0 x 10^-3 moles / (10.00 cm³ + 20.00 cm³) = 3.33 x 10^-3 M
pH of the Buffer: Using the Henderson-Hasselbalch equation: pH = pKa + log([CH3COO-] / [CH3COOH])
Since at the equivalence point, [CH3COO-] >> [CH3COOH] (negligibly small), the pH will be slightly basic.
Approximate pH: The pH will be slightly higher than the pKa due to the presence of the excess acetate. We can estimate the pH to be slightly greater than 1.77.