Edexcel International A Level (IAL) Chemistry (YCH11) - Unit 4 - 14.20 Buffer preparation calculations-Study Notes - New Syllabus
Edexcel International A Level (IAL) Chemistry (YCH11) -Unit 4 – 14.20 Buffer preparation calculations- Study Notes- New syllabus
Edexcel International A Level (IAL) Chemistry (YCH11) -Unit 4 – 14.20 Buffer preparation 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
14.20 Preparing a Buffer of a Given pH
To prepare a buffer solution with a specific pH, the required ratio of concentrations of a weak acid and its conjugate base must be calculated using the Henderson–Hasselbalch equation.
Key Equation
\( \mathrm{pH = pK_a + \log_{10}\left(\frac{[A^-]}{[HA]}\right)} \)
Rearranged Form
To find required ratio:
\( \mathrm{\frac{[A^-]}{[HA]} = 10^{(pH – pK_a)}} \)
Step-by-Step Method
- Identify weak acid and its \( \mathrm{pK_a} \).
- Substitute desired pH into equation.
- Calculate ratio \( \mathrm{\frac{[A^-]}{[HA]}} \).
- Choose suitable concentrations (or volumes) to match this ratio.
Important Notes
- Ratio can be applied to concentrations or moles.
- Absolute values can be chosen conveniently as long as ratio is correct.
- Best buffer when \( \mathrm{pH \approx pK_a} \).
Key Insight
- If \( \mathrm{pH = pK_a} \), then:
\( \mathrm{[A^-] = [HA]} \)
- Equal amounts give maximum buffering capacity.
Key Features
- Uses logarithmic relationship between pH and ratio.
- Only ratio matters, not absolute amounts.
- Central equation: Henderson–Hasselbalch.
Example 1:
Calculate the ratio \( \mathrm{\frac{[CH_3COO^-]}{[CH_3COOH]}} \) required to prepare a buffer of pH 5.76. Given \( \mathrm{pK_a = 4.76} \).
▶️ Answer/Explanation
\( \mathrm{\frac{[A^-]}{[HA]} = 10^{(pH – pK_a)}} \)
\( \mathrm{\frac{[A^-]}{[HA]} = 10^{(5.76 – 4.76)} = 10^1} \)
\( \mathrm{\frac{[A^-]}{[HA]} = 10} \)
Therefore, 10 times more conjugate base is required than acid.
Example 2:
A buffer of pH 4.00 is required using an acid with \( \mathrm{pK_a = 5.00} \). Calculate the required ratio \( \mathrm{\frac{[A^-]}{[HA]}} \).
▶️ Answer/Explanation
\( \mathrm{\frac{[A^-]}{[HA]} = 10^{(pH – pK_a)}} \)
\( \mathrm{\frac{[A^-]}{[HA]} = 10^{(4.00 – 5.00)} = 10^{-1}} \)
\( \mathrm{\frac{[A^-]}{[HA]} = 0.10} \)
Therefore, acid concentration must be 10 times greater than conjugate base.