IB DP Chemistry - R3.1.16 Buffer solutions - Study Notes - New Syllabus - 2026, 2027 & 2028
IB DP Chemistry – R3.1.16 Buffer solutions- Study Notes – New Syllabus
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Reactivity 3.1.16 — Buffer Solutions
Reactivity 3.1.16 — Buffer Solutions
A buffer solution is one that resists a change in pH upon the addition of small amounts of acid (H⁺) or base (OH⁻). This resistance occurs due to the presence of a conjugate acid–base pair in significant and comparable concentrations.
General Mechanism of Buffer Action:
Buffers work on the principle of equilibrium. They contain both an acid (to react with added base) and a base (to react with added acid). The equilibrium shifts to minimize pH changes:
For a weak acid buffer: \( \text{HA} \rightleftharpoons \text{H}^+ + \text{A}^- \)
For a weak base buffer: \( \text{B} + \text{H}_2\text{O} \rightleftharpoons \text{BH}^+ + \text{OH}^- \)
Types of Buffer Solutions:
Acidic Buffer: A solution of a weak acid and its conjugate base (usually a salt).
Example: Ethanoic acid (CH₃COOH) and sodium ethanoate (CH₃COONa)
Reaction:
- Added H⁺: \( \text{A}^- + \text{H}^+ \rightarrow \text{HA} \) (removes added acid)
- Added OH⁻: \( \text{HA} + \text{OH}^- \rightarrow \text{A}^- + \text{H}_2\text{O} \) (removes added base)
Basic Buffer: A solution of a weak base and its conjugate acid (often as a salt).
Example: Ammonia (NH₃) and ammonium chloride (NH₄Cl)
Reaction:
- Added H⁺: \( \text{NH}_3 + \text{H}^+ \rightarrow \text{NH}_4^+ \) (removes added acid)
- Added OH⁻: \( \text{NH}_4^+ + \text{OH}^- \rightarrow \text{NH}_3 + \text{H}_2\text{O} \) (removes added base)
Buffer Capacity:
The capacity of a buffer refers to how much acid or base can be added before a significant change in pH occurs. It is most effective when:
- The concentrations of acid and conjugate base are large and approximately equal
- The pH is close to the pKa of the acid (or pKb of the base)
Henderson–Hasselbalch Equation (for acidic buffers):
\( \text{pH} = \text{p}K_a + \log_{10} \left( \frac{[\text{A}^-]}{[\text{HA}]} \right) \)
This equation helps estimate the pH of a buffer and design buffer solutions of a desired pH.
Applications of Buffers:
- Biological systems: Blood contains a carbonic acid–hydrogencarbonate buffer system
- Industrial processes: Fermentation, dyeing, electroplating
- Pharmaceuticals: Ensuring drug stability and proper absorption
Example
A buffer solution is prepared by mixing 0.100 mol of ethanoic acid (\( \text{CH}_3\text{COOH} \)) and 0.100 mol of sodium ethanoate (\( \text{CH}_3\text{COONa} \)) in 1.00 dm³ of solution. The \( K_a \) of ethanoic acid is \( 1.8 \times 10^{-5} \).
What is the pH of this buffer solution?
▶️Answer/Explanation
Use the Henderson–Hasselbalch equation:
\( \text{pH} = \text{p}K_a + \log_{10} \left( \frac{[\text{A}^-]}{[\text{HA}]} \right) \)
\( \text{p}K_a = -\log_{10}(1.8 \times 10^{-5}) = 4.74 \)
\([ \text{A}^- ] = [ \text{CH}_3\text{COO}^- ] = 0.100\) mol·dm⁻³
\([ \text{HA} ] = [ \text{CH}_3\text{COOH} ] = 0.100\) mol·dm⁻³
\( \text{pH} = 4.74 + \log_{10}(1) = 4.74 \) The buffer has a pH of 4.74.
Example
A basic buffer is made by combining 0.150 mol of aqueous ammonia (\( \text{NH}_3 \)) with 0.120 mol of ammonium chloride (\( \text{NH}_4\text{Cl} \)) in 1.00 dm³ of solution. The \( K_b \) of ammonia is \( 1.8 \times 10^{-5} \).
Calculate the pH of the buffer solution.
▶️Answer/Explanation
First, convert \( K_b \) to \( pK_b \): \( \text{p}K_b = -\log_{10}(1.8 \times 10^{-5}) = 4.74 \)
Now, use the base form of the Henderson–Hasselbalch equation: \( \text{pOH} = \text{p}K_b + \log_{10} \left( \frac{[\text{BH}^+]}{[\text{B}]} \right) \)
\( \text{pOH} = 4.74 + \log_{10} \left( \frac{0.120}{0.150} \right) \approx 4.74 + (-0.097) = 4.643 \)
\( \text{pH} = 14 – 4.643 = 9.36 \) Final pH ≈ 9.36.
Example
The pH of blood is maintained at approximately 7.4 using the carbonic acid–hydrogencarbonate buffer system.
Write an equation showing how this buffer resists a drop in pH when a small amount of HCl is added.
▶️Answer/Explanation
The buffer system is: \( \text{H}_2\text{CO}_3 \rightleftharpoons \text{H}^+ + \text{HCO}_3^- \)
When HCl is added (source of H⁺), the hydrogencarbonate ion neutralizes the extra acid: \( \text{H}^+ + \text{HCO}_3^- \rightarrow \text{H}_2\text{CO}_3 \)
This removes excess H⁺ and prevents a significant drop in pH, stabilizing blood at ~7.4.