Edexcel International A Level (IAL) Chemistry (YCH11) - Unit 4 - 14.19 Buffer pH calculations-Study Notes - New Syllabus
Edexcel International A Level (IAL) Chemistry (YCH11) -Unit 4 – 14.19 Buffer pH calculations- Study Notes- New syllabus
Edexcel International A Level (IAL) Chemistry (YCH11) -Unit 4 – 14.19 Buffer pH calculations- Study Notes -International A Level (IAL) Chemistry (YCH11) – per latest Syllabus.
Key Concepts:
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Edexcel International A Level (IAL) Chemistry (YCH11) -Concise Summary Notes- All Topics
14.19 Calculating the pH of a Buffer Solution
The pH of a buffer solution depends on the relative concentrations of a weak acid and its conjugate base. This is calculated using a rearranged form of the \( \mathrm{K_a} \) expression.
Key Equilibrium
\( \mathrm{HA \rightleftharpoons H^+ + A^-} \)
\( \mathrm{K_a = \frac{[H^+][A^-]}{[HA]}} \)
Rearranged Expression
Rearranging for \( \mathrm{[H^+]} \):
\( \mathrm{[H^+] = K_a \times \frac{[HA]}{[A^-]}} \)
Taking logs:
\( \mathrm{pH = pK_a + \log_{10}\left(\frac{[A^-]}{[HA]}\right)} \)
This is the Henderson–Hasselbalch equation.
Using the Equation
- Identify weak acid (\( \mathrm{HA} \)) and conjugate base (\( \mathrm{A^-} \)).
- Substitute concentrations (or moles if volume same).
- Use \( \mathrm{pK_a} \) or calculate it from \( \mathrm{K_a} \).
- Solve for pH.
Important Notes
- Can use mole ratio instead of concentration if volumes cancel.
- Works best when both components are present in significant amounts.
- If \( \mathrm{[A^-] = [HA]} \), then:
\( \mathrm{pH = pK_a} \)
Key Features
- pH depends on ratio of conjugate base to acid.
- Uses logarithmic relationship.
- Central equation: Henderson–Hasselbalch.
Example 1:
Calculate the pH of a buffer containing \( \mathrm{0.10\ mol\ dm^{-3}} \) \( \mathrm{CH_3COOH} \) and \( \mathrm{0.20\ mol\ dm^{-3}} \) \( \mathrm{CH_3COONa} \). Given \( \mathrm{pK_a = 4.76} \).
▶️ Answer/Explanation
\( \mathrm{pH = pK_a + \log\left(\frac{[A^-]}{[HA]}\right)} \)
\( \mathrm{pH = 4.76 + \log\left(\frac{0.20}{0.10}\right)} \)
\( \mathrm{pH = 4.76 + \log(2)} \)
\( \mathrm{pH = 4.76 + 0.301 = 5.06} \)
Example 2:
A buffer contains 0.050 mol of \( \mathrm{NH_3} \) and 0.100 mol of \( \mathrm{NH_4^+} \) in the same volume. Given \( \mathrm{pK_a = 9.25} \), calculate the pH.
▶️ Answer/Explanation
\( \mathrm{pH = pK_a + \log\left(\frac{[base]}{[acid]}\right)} \)
\( \mathrm{pH = 9.25 + \log\left(\frac{0.050}{0.100}\right)} \)
\( \mathrm{pH = 9.25 + \log(0.5)} \)
\( \mathrm{pH = 9.25 – 0.301 = 8.95} \)