Buffer solutions

Chemistry· 18.2 Buffer solutions· 20 min read

1. What is a Buffer Solution and How Does It Work?★★☆☆☆⏱ 5 min

📘 Definition

Buffer Solution

A solution that opposes changes in pH when small volumes of strong acid or strong base are added. All buffers consist of a conjugate acid-base pair: either a weak acid plus its conjugate base from a salt, or a weak base plus its conjugate acid from a salt.

Example:

A mixture of ethanoic acid ( $C H_{3} C O O H$ ) and sodium ethanoate ( $C H_{3} C O O N a$ ) is a common acidic buffer.

All buffers work by using the excess conjugate pair to neutralize any added $H^{+}$ (from acid) or $O H^{-}$ (from base), shifting equilibrium to maintain a roughly constant concentration of $H^{+}$ .

📐 Worked Example

Explain how an ethanoic acid / sodium ethanoate buffer resists an increase in pH when a small amount of NaOH is added.

1
Added $O H^{-}$ ions react with the excess weak acid in the buffer:
$OH^- + CH_3COOH \rightarrow CH_3COO^- + H_2O$
2
Most of the added $O H^{-}$ is consumed by the excess ethanoic acid, so the $O H^{-}$ concentration does not increase significantly.
3
The equilibrium $C H_{3} C O O H ⇌ H^{+} + C H_{3} C O O^{-}$ shifts right to replace any $H^{+}$ consumed, so $[H^{+}]$ (and thus pH) remains almost unchanged.

Exam tip:

Always mention both the excess acid/base and the conjugate partner in buffer action explanations, otherwise you will not get full marks.

2. Calculating Buffer pH: The Henderson-Hasselbalch Equation★★★☆☆⏱ 7 min

For any buffer made from a weak acid $H A$ and its conjugate base $A^{-}$ , we can derive the pH equation from the $K_{a}$ expression. Starting from the dissociation equilibrium: $H A ⇌ H^{+} + A^{-}$ , so $K_{a} = \frac{[ H ^{+} ] [ A ^{-} ]}{[ H A ]}$ . Rearranging and taking negative logs gives the standard equation:

pH = pK_a + \log_{10}\left( \frac{[A^-]}{[HA]} \right)

📐 Worked Example

Calculate the pH of a buffer solution containing 0.10 mol dm⁻³ ethanoic acid ( $p K_{a} = 4.76$ ) and 0.20 mol dm⁻³ sodium ethanoate.

1
Identify values: $p K_{a} = 4.76$ , $[C H_{3} C O O^{-}] = 0.20$ mol dm⁻³, $[C H_{3} C O O H] = 0.10$ mol dm⁻³
2
Substitute into the Henderson-Hasselbalch equation:
$pH = 4.76 + \log\left( \frac{0.20}{0.10} \right)$
3
Simplify to get the final pH:
$pH = 4.76 + 0.30 = 5.06$

✓ Quick check

Check your understanding of approximations:

Why do we approximate $[A^{-}] \approx [s a l t]$ in a buffer?
- A: All the weak acid dissociates
- B: The common ion effect suppresses dissociation of the weak acid, so very little extra $A^{-}$ is produced
- C: The salt does not dissociate in solution
Reveal answer
B —
Correct: The high concentration of common ion $A^{-}$ from the fully dissociated salt shifts the weak acid dissociation equilibrium left, so almost no extra $A^{-}$ comes from the weak acid.

3. Acidic and Basic Buffers: Differences and Calculations★★★☆☆⏱ 5 min

📘 Definition

Acidic Buffer

A buffer with a pH below 7, made from a weak acid and a soluble salt of the weak acid (which provides the conjugate base).

Example:

$C H_{3} C O O H + C H_{3} C O O N a$

📘 Definition

Basic Buffer

A buffer with a pH above 7, made from a weak base and a soluble salt of the weak base (which provides the conjugate acid).

Example:

$N H_{3} + N H_{4} C l$

For basic buffers, you can calculate pH two ways: use $K_{b}$ to find $pO H$ then convert to $p H$ , or use the $p K_{a}$ of the conjugate acid of the weak base and apply the Henderson-Hasselbalch equation directly.

📐 Worked Example

Calculate the pH of a buffer containing 0.05 mol dm⁻³ $N H_{3}$ ( $K_{b} = 1.8 \times 1 0^{- 5}$ ) and 0.10 mol dm⁻³ $N H_{4} C l$ .

1
For the weak base equilibrium $N H_{3} + H_{2} O ⇌ N H_{4}^{+} + O H^{-}$ , approximate $[N H_{4}^{+}] = 0.10$ , $[N H_{3}] = 0.05$ . Rearrange $K_{b}$ to find $[O H^{-}]$ :
$[OH^-] = K_b \times \frac{[NH_3]}{[NH_4^+]} = 1.8 \times 10^{-5} \times \frac{0.05}{0.10} = 9.0 \times 10^{-6}$
2
Calculate $pO H$ then convert to $p H$ :
$pOH = -\log(9.0 \times 10^{-6}) = 5.05, \quad pH = 14 - 5.05 = 8.95$
3
Check with Henderson-Hasselbalch: $p K_{a} (N H_{4}^{+}) = 14 - 4.74 = 9.26$
$pH = 9.26 + \log\left(\frac{0.05}{0.10}\right) = 8.96$, which matches (small difference from rounding)$

4. Preparing a Buffer of a Given pH★★★★☆⏱ 6 min

Exam questions often ask you to calculate how much acid/salt or base/salt is needed to make a buffer of a target pH. The method rearranges the Henderson-Hasselbalch equation to find the required ratio of the conjugate pair.

📐 Worked Example

What mass of ammonium chloride must be added to 250 cm³ of 0.10 mol dm⁻³ ammonia to make a buffer of pH 9.00? $p K_{b} (N H_{3}) = 4.75$ , $M_{r} (N H_{4} C l) = 53.5$ .

1
First find $p K_{a}$ of the conjugate acid $N H_{4}^{+}$ :
$pK_a = 14 - pK_b = 14 - 4.75 = 9.25$
2
Rearrange the Henderson-Hasselbalch equation:
$9.00 = 9.25 + \log\left( \frac{[NH_3]}{[NH_4^+]} \right)$
3
Solve for the ratio of concentrations:
$\log\left( \frac{[NH_3]}{[NH_4^+]} \right) = -0.25 \implies \frac{[NH_3]}{[NH_4^+]} = 0.562$
4
Calculate concentration and moles of $N H_{4} C l$ needed:
$[NH_4^+] = \frac{0.10}{0.562} = 0.178 \text{ mol dm}^{-3}, \quad n = 0.178 \times 0.25 = 0.0445 \text{ mol}$
5
Calculate final mass:
$m = 0.0445 \times 53.5 = 2.38 \text{ g}$

Exam tip:

The ratio of concentrations is equal to the ratio of moles for both species in the same total volume, so you can skip converting concentration to moles early to save calculation time.

5. Common Pitfalls

Wrong move:

Claiming buffers change pH by a negligible amount regardless of how much acid/base is added.

Why:

Buffers only resist pH change for small amounts of added acid/base. Adding large amounts exhausts the buffer's neutralizing capacity, leading to a large pH change.

Correct move:

Always specify that buffers resist pH change when small amounts of acid or base are added.

Wrong move:

Swapping the ratio, writing $lo g (\frac{[ H A ]}{[ A ^{-} ]})$ instead of $lo g (\frac{[ A ^{-} ]}{[ H A ]})$ in the Henderson-Hasselbalch equation.

Why:

This gives an incorrect pH that is systematically lower/higher than the true value depending on the ratio.

Correct move:

Remember: $p H = p K_{a} + lo g (\frac{co nj ug a t e ba se}{w e ak a c i d})$ , conjugate base is always the numerator.

Wrong move:

Using the total buffer concentration in calculations instead of individual conjugate pair concentrations.

Why:

pH depends on the ratio of the conjugate pair, not total buffer concentration. Two buffers with the same ratio have the same pH regardless of total concentration.

Correct move:

Always use the individual starting concentrations of the weak acid/base and the corresponding salt in the equation.

Wrong move:

Claiming only acidic buffers neutralize added base and only basic buffers neutralize added acid.

Why:

All buffers contain both an acidic and basic component to neutralize both added acid and added base.

Correct move:

Explain buffer action for both added acid and added base, regardless of whether the buffer is acidic or basic.

6. Quick Reference Cheatsheet

Concept	Key Formula / Rule
Buffer definition	Resists pH change on adding small amounts of acid/base
Acidic buffer composition	Weak acid + salt of the weak acid (pH < 7)
Basic buffer composition	Weak base + salt of the weak base (pH > 7)
Henderson-Hasselbalch equation	$p H = p K_{a} + lo g (\frac{[ A ^{-} ]}{[ H A ]})$
Basic buffer calculation	$pO H = p K_{b} + lo g (\frac{[ B H ^{+} ]}{[ B ]}), p H = 14 - pO H$
Core approximation	$[A^{-}] \approx [s a l t], [H A] \approx [w e ak a c i d]$ (common ion effect)

7. Frequently Asked

Can a buffer be made from a strong acid and strong base?

No. Strong acids and bases dissociate completely, so they cannot neutralize small amounts of added acid/base effectively, and do not maintain a steady pH.

Why is the common ion effect important for buffer calculations?

The common ion from the fully dissociated salt suppresses the dissociation of the weak acid/base, so the ratio of conjugate pair concentrations is approximately equal to the ratio of their starting concentrations, simplifying calculations.

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2022 · 22
Calculate pH of ethanoic acid buffer
2023 · 12
Explain buffer action on adding NaOH
2024 · 31
Prepare buffer of given pH calculation

Going deeper

syllabusCIE 9701 Official SyllabusCheck for latest syllabus updates

What's Next

Buffer solutions are a core part of acid-base equilibria, and this concept underpins many other topics in CIE A-Level chemistry, including acid-base titrations, pH indicators, and biological chemistry, where buffering is critical for maintaining a steady pH in living systems. Buffer calculations and explanations of buffer action are extremely common across both multiple choice and structured written questions in CIE 9701, so mastering this content is key to scoring high marks on all equilibria-related questions. Next, you can extend your knowledge of buffer action to understand the shape of titration curves for weak acids and bases, and how pH indicators exploit buffer properties to work effectively in different titration scenarios.