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
- Pipette a known volume of weak acid into a conical flask.
- Place pH probe in the solution.
- Add \( \mathrm{NaOH} \) from burette in small increments.
- Record pH after each addition.
- Continue until well past equivalence point.
- 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 \( \mathrm{K_a} \).
