Edexcel International A Level (IAL) Chemistry (YCH11) - Unit 4 - 14.23 Core Practical: Ka determination-Study Notes - New Syllabus

Edexcel International A Level (IAL) Chemistry (YCH11) -Unit 4 – 14.23 Core Practical: Ka determination- Study Notes- New syllabus

Edexcel International A Level (IAL) Chemistry (YCH11) -Unit 4 – 14.23 Core Practical: Ka determination- Study Notes -International A Level (IAL) Chemistry (YCH11) – per latest Syllabus.

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Edexcel International A Level (IAL) Chemistry (YCH11) -Concise Summary Notes- All Topics

14.23 CORE PRACTICAL 11: Finding the \( \mathrm{K_a} \) Value of a Weak Acid

This practical determines the acid dissociation constant (\( \mathrm{K_a} \)) of a weak acid using a titration curve. The key idea is that at the half-equivalence point, \( \mathrm{pH = pK_a} \).

Aim

To determine the value of \( \mathrm{K_a} \) for a weak acid using titration data.

Apparatus

  • Burette
  • Pipette
  • Conical flask
  • pH meter (or probe)
  • Magnetic stirrer (optional)

Chemicals

  • Weak acid solution (e.g. \( \mathrm{CH_3COOH} \))
  • Standard \( \mathrm{NaOH} \) solution (strong base)

Procedure

  1. Pipette a known volume of weak acid into a conical flask.
  2. Place pH probe in the solution.
  3. Add \( \mathrm{NaOH} \) from burette in small increments.
  4. Record pH after each addition.
  5. Continue until well past equivalence point.
  6. Plot pH vs volume of base added.

Processing Results

Step 1: Identify Equivalence Point

  • Point of steepest vertical increase in pH.

Step 2: Find Half-Equivalence Point

  • Half the volume at equivalence.

Step 3: Read pH at this point

  • \( \mathrm{pH = pK_a} \)

Step 4: Calculate \( \mathrm{K_a} \)

\( \mathrm{K_a = 10^{-pK_a}} \)

Theory

At half-neutralisation:

  • \( \mathrm{[HA] = [A^-]} \)

From:

\( \mathrm{K_a = \frac{[H^+][A^-]}{[HA]}} \)

Therefore:

\( \mathrm{K_a = [H^+]} \Rightarrow pH = pK_a \)

Key Features

  • Uses titration curve data.
  • Half-equivalence point is critical.
  • No need for complex equilibrium calculations.

Sources of Error

  • pH meter calibration errors.
  • Inaccurate volume readings.
  • Adding base in large increments near equivalence.

Improvements

Use smaller volume additions near equivalence.

  • Use a calibrated pH probe.
  • Repeat experiment for accuracy.

Example 1:

The equivalence point occurs at 25.0 cm³ of base added. The pH at 12.5 cm³ is 4.80. Calculate \( \mathrm{K_a} \).

▶️ Answer/Explanation

Half-equivalence = 12.5 cm³

\( \mathrm{pK_a = 4.80} \)

\( \mathrm{K_a = 10^{-4.80}} \)

\( \mathrm{K_a = 1.58 \times 10^{-5}} \)

Example 2:

Explain why the half-equivalence point is used to determine \( \mathrm{K_a} \).

▶️ Answer/Explanation

At half-equivalence, concentrations of acid and conjugate base are equal.

This simplifies the equilibrium expression.

Therefore, \( \mathrm{pH = pK_a} \), allowing direct determination of \( \mathrm{K_a} \).

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