Edexcel International A Level (IAL) Chemistry (YCH11) - Unit 4 - 14.14 Ka from experimental data-Study Notes - New Syllabus
Edexcel International A Level (IAL) Chemistry (YCH11) -Unit 4 – 14.14 Ka from experimental data- Study Notes- New syllabus
Edexcel International A Level (IAL) Chemistry (YCH11) -Unit 4 – 14.14 Ka from experimental data- 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
14.14 Calculating \( \mathrm{K_a} \) from Experimental Data
The acid dissociation constant (\( \mathrm{K_a} \)) of a weak acid can be determined experimentally using the pH of a solution prepared from a known mass of the acid. This combines equilibrium concepts with quantitative analysis.
Step-by-Step Method
Step 1: Calculate Initial Concentration
- Convert mass → moles using molar mass.
- Divide by volume to get concentration \( \mathrm{c} \).
\( \mathrm{c = \frac{n}{V}} \)
Step 2: Calculate \( \mathrm{[H^+]} \) from pH
\( \mathrm{[H^+] = 10^{-pH}} \)
Step 3: Use Equilibrium Expression
For a weak acid:
\( \mathrm{HA \rightleftharpoons H^+ + A^-} \)
At equilibrium:
- \( \mathrm{[H^+] = [A^-] = x} \)
- \( \mathrm{[HA] \approx c} \) (assumption: weak acid dissociates slightly)
\( \mathrm{K_a = \frac{[H^+]^2}{c}} \)
Key Assumption
- \( \mathrm{[HA]_{equilibrium} \approx [HA]_{initial}} \)
- Valid when dissociation is small.
Final Expression
\( \mathrm{K_a = \frac{(10^{-pH})^2}{c}} \)
Key Features
- Requires converting experimental data into concentrations.
- Uses pH to determine \( \mathrm{[H^+]} \).
- Assumes weak acid dissociation is small.
- No quadratic equation required.
Example 1:
\( \mathrm{1.22\ g} \) of a weak acid (Mr = 122) is dissolved in \( \mathrm{1.00\ dm^3} \) of water. The pH is 3.00. Calculate \( \mathrm{K_a} \).
▶️ Answer/Explanation
Step 1:
\( \mathrm{n = \frac{1.22}{122} = 0.0100\ mol} \)
\( \mathrm{c = 0.0100\ mol\ dm^{-3}} \)
Step 2:
\( \mathrm{[H^+] = 10^{-3.00} = 1.0 \times 10^{-3}} \)
Step 3:
\( \mathrm{K_a = \frac{(1.0 \times 10^{-3})^2}{0.0100}} \)
\( \mathrm{K_a = \frac{1.0 \times 10^{-6}}{0.0100} = 1.0 \times 10^{-4}} \)
Example 2:
A \( \mathrm{0.050\ mol\ dm^{-3}} \) weak acid has pH = 2.72. Calculate \( \mathrm{K_a} \).
▶️ Answer/Explanation
Step 1:
\( \mathrm{[H^+] = 10^{-2.72} = 1.91 \times 10^{-3}} \)
Step 2:
\( \mathrm{K_a = \frac{(1.91 \times 10^{-3})^2}{0.050}} \)
\( \mathrm{K_a = \frac{3.65 \times 10^{-6}}{0.050} = 7.30 \times 10^{-5}} \)