Salts and buffer solutions

IB Chemistry HL· Topic 18.1· 25 min read

1. Salt Hydrolysis and Salt Solution pH★★☆☆☆⏱ 8 min

📘 Definition

Salt hydrolysis

A reaction between a dissolved salt ion and water that produces $H^{+}$ or $O H^{-}$ ions, changing the solution pH from neutral.

Example:

Ammonium ions reacting with water to produce hydronium ions

Salt of strong acid + strong base: Neutral (no hydrolysis, pH ≈ 7)
Salt of weak acid + strong base: Basic (conjugate base of weak acid hydrolyzes to produce $O H^{-}$ , pH > 7)
Salt of strong acid + weak base: Acidic (conjugate acid of weak base hydrolyzes to produce $H^{+}$ , pH < 7)
Salt of weak acid + weak base: pH depends on relative $K_{a}$ of the acid and $K_{b}$ of the base

📐 Worked Example

Predict whether an aqueous solution of sodium ethanoate ( $C H_{3} C O O N a$ ) is acidic, basic, or neutral, and justify your answer.

1
Dissociate the salt into ions and identify parent acid/base:
2
$CH_3COONa(s) \rightarrow CH_3COO^-(aq) + Na^+(aq)$
3
Parent acid: ethanoic acid ( $C H_{3} C O O H$ , weak acid), parent base: sodium hydroxide ( $N a O H$ , strong base)
4
Identify the hydrolyzable ion: $N a^{+}$ does not react with water, $C H_{3} C O O^{-}$ is the conjugate base of a weak acid so it hydrolyzes:
5
$CH_3COO^-(aq) + H_2O(l) \rightleftharpoons CH_3COOH(aq) + OH^-(aq)$
6
Production of $O H^{-}$ increases pH above 7, so the solution is basic.

Exam tip:

Always check both the cation and anion for hydrolysis before predicting pH — never assume all salts are neutral.

2. Buffer Composition and Mechanism of Action★★★☆☆⏱ 7 min

📘 Definition

Buffer solution

Composed of a weak conjugate acid-base pair

A solution that resists changes in pH when small amounts of strong acid, strong base, or water are added.

Buffers are made in two common ways: mixing a weak acid with its conjugate base (as a soluble salt), or mixing a weak base with its conjugate acid (as a soluble salt). Strong acids and bases cannot form buffers, as they dissociate completely with no equilibrium to absorb added ions.

📐 Worked Example

Explain how an ammonia / ammonium chloride buffer resists an increase in pH when a small amount of $N a O H$ is added.

1
Main species in buffer: $N H_{3}$ (weak base, high concentration), $N H_{4}^{+}$ (conjugate acid from ammonium chloride, high concentration)
2
Added $N a O H$ dissociates completely to produce extra $O H^{-}$ ions.
3
Excess $O H^{-}$ reacts with the conjugate acid $N H_{4}^{+}$ :
4
$NH_4^+(aq) + OH^-(aq) \rightleftharpoons NH_3(aq) + H_2O(l)$
5
Almost all excess $O H^{-}$ is consumed, so pH increases only slightly.

3. Buffer pH Calculations: Henderson-Hasselbalch Equation★★★☆☆⏱ 8 min

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🔬 Derivation

Goal:

Derive the Henderson-Hasselbalch equation for an acidic buffer

Starting from:

Acid dissociation equilibrium of a weak acid $H A$

1
Write the acid dissociation equilibrium and $K_{a}$ expression:
2
$HA(aq) + H_2O(l) \rightleftharpoons H_3O^+(aq) + A^-(aq)$
3
$K_a = \frac{[H_3O^+][A^-]}{[HA]}$
4
Rearrange to isolate $[H_{3} O^{+}]$ :
5
$[H_3O^+] = K_a \times \frac{[HA]}{[A^-]}$
6
Take $- lo g_{10}$ of both sides and simplify using $p H = - lo g [H_{3} O^{+}]$ and $p K_{a} = - lo g K_{a}$ :
7
$-\log[H_3O^+] = -\log K_a - \log\left(\frac{[HA]}{[A^-]}\right)$

Result:

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The common ion effect suppresses dissociation of the weak acid, so we can approximate $[H A] \approx$ initial weak acid concentration and $[A^{-}] \approx$ initial salt concentration for calculations.

📐 Worked Example

Calculate the pH of a buffer containing 0.15 mol dm⁻³ propanoic acid ( $p K_{a} = 4.87$ ) and 0.10 mol dm⁻³ sodium propanoate.

1
Identify components: weak acid = propanoic acid, $[a c i d] = 0.15$ , conjugate base = propanoate, $[ba se] = 0.10$
2
Substitute into the Henderson-Hasselbalch equation:
3
$pH = 4.87 + \log_{10}\left(\frac{0.10}{0.15}\right)$
4
Calculate the log term: $\frac{0.10}{0.15} \approx 0.667$ , $lo g_{10} (0.667) \approx - 0.18$
5
Compute final pH:
6
$pH = 4.87 - 0.18 = 4.69$

4. Buffer Capacity★★☆☆☆⏱ 4 min

📘 Definition

Buffer capacity

A measure of the amount of strong acid or base that can be added to a buffer before the pH changes significantly.

Total concentration of buffer components: Higher total concentration = higher buffer capacity, as more ions are available to react with added acid/base.
Ratio of buffer components: Buffer capacity is maximum when $\frac{[ co nj ug a t e ba se ]}{[ w e ak a c i d ]} = 1$ , so when $p H = p K_{a}$ of the weak acid.
Buffers are only effective within $p H = p K_{a} \pm 1$ of the weak acid's $p K_{a}$ .

5. Common Pitfalls

Wrong move:

Assuming all salts form neutral aqueous solutions

Why:

Most salts made from weak acids or bases have ions that hydrolyze to change pH

Correct move:

Always check both the cation and anion for hydrolysis, using the parent acid/base rules to predict pH

Wrong move:

Swapping the [acid] and [base] terms in the Henderson-Hasselbalch equation

Why:

The derivation gives the ratio of conjugate base to weak acid in the log term; swapping gives an incorrect pH

Correct move:

Memorize that $p H = p K_{a} + lo g (\frac{[ ba se ]}{[ a c i d ]})$ , or re-derive quickly from the $K_{a}$ expression if unsure

Wrong move:

Using a strong acid and strong base to prepare a buffer solution

Why:

Strong acids/bases dissociate completely, so no equilibrium exists to absorb added $H^{+}$ or $O H^{-}$

Correct move:

Buffers always require a weak conjugate acid-base pair

Wrong move:

Thinking buffer capacity only depends on the 1:1 ratio of components

Why:

Total concentration of components also impacts capacity: dilute buffers have lower capacity even with a 1:1 ratio

Correct move:

Remember buffer capacity depends on both the ratio of components (maximum at 1:1) and total concentration (higher = higher capacity)

Wrong move:

Expecting large pH changes when diluting a buffer

Why:

Dilution changes both [acid] and [base] equally, so their ratio stays the same

Correct move:

Buffer pH remains almost unchanged upon dilution, only extreme dilution causes significant change

6. Quick Reference Cheatsheet

Property	Rule
Strong acid + strong base salt	Neutral, pH ≈ 7
Weak acid + strong base salt	Basic, pH > 7
Strong acid + weak base salt	Acidic, pH < 7
Henderson-Hasselbalch (acid buffer)	$p H = p K_{a} + lo g (\frac{[ A ^{-} ]}{[ H A ]})$
Maximum buffer capacity	$p H = p K_{a}$ , 1:1 [base]:[acid]
Effective buffer pH range	$p K_{a} \pm 1$

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 · 2
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What's Next

Salts and buffer solutions are a core foundation for understanding acid-base titrations, a common extended response topic in IB Chemistry HL exams. Buffers also have important biological applications, such as the bicarbonate buffer that maintains human blood pH, which is frequently tested in exam questions. Next, you can explore acid-base titration curves and indicators, where you will apply your understanding of salt pH and buffer action to calculate pH at different points of a titration, including at the equivalence point.