IB DP Chemistry - R3.1.17 The pH of a buffer solution - Study Notes - New Syllabus - 2026, 2027 & 2028
IB DP Chemistry -R3.1.17 The pH of a buffer solution – Study Notes – New Syllabus
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Reactivity 3.1.17 — Factors Affecting Buffer pH & Buffer Calculations
Reactivity 3.1.17 — Factors Affecting Buffer pH & Buffer Calculations
A buffer solution maintains a relatively constant pH upon the addition of small amounts of acid or base. The pH of a buffer solution depends on two main factors:
- The pKa or pKb of the weak acid or weak base involved: This represents the acid or base strength, and hence determines the central point (most stable pH) of the buffer.
- The ratio of the concentrations of the acid/base and its conjugate partner: This influences how far the pH shifts when a small amount of acid or base is added.
Mathematical Expression for Buffer pH — The Henderson–Hasselbalch Equation
For acidic buffers:
\( \text{pH} = \text{p}K_a + \log_{10} \left( \frac{[\text{A}^-]}{[\text{HA}]} \right) \)
For basic buffers:
\( \text{pOH} = \text{p}K_b + \log_{10} \left( \frac{[\text{BH}^+]}{[\text{B}]} \right) \) and then \( \text{pH} = 14 – \text{pOH} \)
Using Equilibrium Constants to Calculate Buffer pH
Sometimes, you’re not directly given the concentrations of conjugate acid/base forms. Instead, you may have to use the dissociation equilibrium and ICE tables to determine concentrations at equilibrium, and then substitute into the appropriate equation.
For a weak acid \( \text{HA} \):
\( \text{HA} \rightleftharpoons \text{H}^+ + \text{A}^- \) \( K_a = \frac{[\text{H}^+][\text{A}^-]}{[\text{HA}]} \)
For a weak base \( \text{B} \):
\( \text{B} + \text{H}_2\text{O} \rightleftharpoons \text{BH}^+ + \text{OH}^- \) \( K_b = \frac{[\text{OH}^-][\text{BH}^+]}{[\text{B}]} \)
Optimal Buffering and When pH = pKa
When the concentrations of acid and conjugate base are equal, the logarithmic term becomes 0, and:
\( \text{pH} = \text{p}K_a \)
This is the ideal buffering point — the buffer is most effective at resisting changes in pH here.
Effect of Dilution on Buffer pH
- Dilution reduces the concentration of both acid and conjugate base.
- However, the ratio \( \frac{[\text{A}^-]}{[\text{HA}]} \) remains constant, so the pH stays approximately the same.
- Still, dilution lowers the buffer’s capacity — it can neutralize less added acid or base before significant pH change occurs.
Buffer pH Relationships
| Condition | Impact on pH |
|---|---|
| \( [\text{A}^-] > [\text{HA}] \) | pH > pKa |
| \( [\text{A}^-] = [\text{HA}] \) | pH = pKa (optimum buffer action) |
| \( [\text{A}^-] < [\text{HA}] \) | pH < pKa |
Example
A buffer solution is prepared by mixing 0.200 mol of benzoic acid (\( \text{C}_6\text{H}_5\text{COOH} \), \( K_a = 6.3 \times 10^{-5} \)) and 0.100 mol of sodium benzoate in 500 mL of solution.
Calculate the pH of the buffer.
▶️Answer/Explanation
\( \text{p}K_a = -\log(6.3 \times 10^{-5}) = 4.20 \)
\( [\text{HA}] = \frac{0.200}{0.500} = 0.400 \, \text{mol·dm}^{-3} \)
\( [\text{A}^-] = \frac{0.100}{0.500} = 0.200 \, \text{mol·dm}^{-3} \)
\( \text{pH} = 4.20 + \log_{10} \left( \frac{0.200}{0.400} \right) = 4.20 + \log_{10}(0.5) = 4.20 – 0.301 = 3.90 \)
Example
A buffer is made using 0.300 mol of ammonia and 0.400 mol of ammonium chloride in 1.00 dm³.
Calculate the pH, given that \( K_b(\text{NH}_3) = 1.8 \times 10^{-5} \).
▶️Answer/Explanation
\( \text{p}K_b = -\log(1.8 \times 10^{-5}) = 4.74 \)
\( \text{pOH} = 4.74 + \log \left( \frac{0.400}{0.300} \right) = 4.74 + 0.124 = 4.864 \)
\( \text{pH} = 14 – 4.864 = 9.14 \)
Example
A buffer solution of 0.50 mol·dm⁻³ of ethanoic acid and 0.50 mol·dm⁻³ sodium ethanoate is diluted by adding 500 mL of water to 500 mL of the buffer.
Will the pH change?
▶️Answer/Explanation
Initial ratio of \( \frac{[\text{A}^-]}{[\text{HA}]} = 1 \), so \( \text{pH} = \text{p}K_a = 4.74 \)
After dilution, both concentrations halve (0.25 mol·dm⁻³), but the ratio remains 1,
so: \( \text{pH} = \text{p}K_a = 4.74 \) Thus, the pH does not change, though the buffer capacity decreases.