Acid-base equilibria

Chemistry· Unit 18: Further chemical equilibria· 15 min read

1. Bronsted-Lowry Acid-Base Theory★★☆☆☆⏱ 4 min

📘 Definition

Bronsted-Lowry Acid

A substance that can donate a proton ( $H^{+}$ ion) to another substance

Example:

Hydrochloric acid ( $H C l$ ) donates a proton to water in aqueous solution

A Bronsted-Lowry base is defined conversely as a proton acceptor. All bases have at least one lone pair of electrons to form a bond with the donated proton. This theory is more general than the older Arrhenius theory, which only applies to aqueous solutions.

📐 Worked Example

Identify which species act as the acid and base in the forward reaction: $N H_{3} (a q) + H C l (g) \to N H_{4} C l (s)$

1
First, track proton movement between the reactants to identify donor and acceptor.
2
$H C l$ loses a proton to form $C l^{-}$ , so it donates a proton.
3
$N H_{3}$ gains a proton to form $N H_{4}^{+}$ , so it accepts a proton.
4
Final answer: $H C l$ = acid, $N H_{3}$ = base

Exam tip:

CIE examiners almost always ask for identification of acid/base species in reactions, always check for proton movement not just charge.

2. Conjugate Acid-Base Pairs★★☆☆☆⏱ 3 min

📘 Definition

Conjugate Acid-Base Pair

Two species that differ by exactly one proton, formed when an acid donates a proton or a base accepts a proton. The conjugate base of an acid is the species remaining after proton donation; the conjugate acid of a base forms after proton acceptance.

Example:

$H C l$ (acid) and $C l^{-}$ (conjugate base) form one conjugate pair

Acid strength is inversely related to conjugate base strength: strong acids have very weak conjugate bases, while weak acids have relatively strong conjugate bases. This is because a strong acid fully dissociates, so its conjugate base has almost no tendency to re-accept a proton.

📐 Worked Example

Write all conjugate acid-base pairs for the equilibrium: $H N O_{2} (a q) + H_{2} O (l) ⇌ N O_{2}^{-} (a q) + H_{3} O^{+} (a q)$

1
Group species by proton difference across the equilibrium.
2
$H N O_{2}$ loses one proton to become $N O_{2}^{-}$ , so this is the first pair: $H N O_{2}$ (acid) and $N O_{2}^{-}$ (conjugate base).
3
$H_{2} O$ gains one proton to become $H_{3} O^{+}$ , so this is the second pair: $H_{2} O$ (base) and $H_{3} O^{+}$ (conjugate acid).

3. pH Scale and Strong Acid pH Calculations★★★☆☆⏱ 5 min

📘 Definition

pH

pH = -\log_{10}[H^+_{(aq)}]

A logarithmic scale measuring hydrogen ion concentration in aqueous solution, ranging from ~0 (strongly acidic) to ~14 (strongly alkaline) at 25°C.

Strong acids fully dissociate in aqueous solution, so for monoprotic strong acids (with one acidic proton), $[H^{+}] = [a c i d]_{ini t ia l}$ . For diprotic strong acids like $H_{2} S O_{4}$ , CIE assumes full dissociation so $[H^{+}] = 2 \times [a c i d]_{ini t ia l}$ .

📐 Worked Example

Calculate the pH of 0.050 mol dm⁻³ hydrochloric acid at 25°C.

1
HCl is a strong monoprotic acid, so full dissociation occurs:
2
$HCl(aq) \rightarrow H^+(aq) + Cl^-(aq)$
3
Therefore, $[H^{+}] = [H C l] = 0.050$ mol dm⁻³
4
Substitute into the pH formula:
5
$pH = -\log_{10}(0.050) = 1.30$
6
Final pH = 1.30 (2 decimal places)

Exam tip:

Always give pH values to 2 decimal places unless the question explicitly states otherwise, this is the CIE marking requirement.

4. Weak Acids and the Acid Dissociation Constant Ka★★★☆☆⏱ 6 min

📘 Definition

Acid Dissociation Constant (Ka)

K_a = \frac{[H^+][A^-]}{[HA]}

The equilibrium constant for dissociation of a weak acid $H A$ in aqueous solution. $p K a = - lo g_{10} K a$ , so lower pKa corresponds to a stronger acid.

Weak acids only partially dissociate, so we use Ka to quantify their strength. For most weak acids, dissociation is very small, so two simplifying approximations are accepted by CIE: $[H A]_{e q u i l ib r i u m} \approx [H A]_{ini t ia l}$ and $[H^{+}] \approx [A^{-}]$ .

📐 Worked Example

A 0.10 mol dm⁻³ solution of weak monoprotic acid HA has Ka = 1.8 × 10⁻⁵ mol dm⁻³ at 25°C. Calculate its pH.

1
Write the dissociation equilibrium for HA:
2
$HA(aq) \rightleftharpoons H^+(aq) + A^-(aq)$
3
Write the Ka expression:
4
$K_a = \frac{[H^+][A^-]}{[HA]}$
5
Substitute the accepted approximations: $[H^{+}] = [A^{-}]$ , $[H A] \approx 0.10$ :
6
$1.8 \times 10^{-5} = \frac{[H^+]^2}{0.10}$
7
Rearrange to solve for $[H^{+}]$ :
8
$[H^+]^2 = 1.8 \times 10^{-6} \rightarrow [H^+] = 1.34 \times 10^{-3} \text{ mol dm}^{-3}$
9
Calculate pH:
10
$pH = -\log_{10}(1.34 \times 10^{-3}) = 2.87$

5. Common Pitfalls

Wrong move:

Treating diprotic $H_{2} S O_{4}$ as monoprotic for pH calculations

Why:

CIE assumes full dissociation of sulfuric acid, so $[H^{+}]$ is twice the acid concentration

Correct move:

Always check the number of acidic protons before calculating $[H^{+}]$ for strong acids

Wrong move:

Forgetting the negative sign in the pH formula, leading to negative pH values

Why:

The negative sign reverses the scale so higher $[H^{+}]$ gives lower pH, matching standard convention

Correct move:

Always double-check your calculator input to confirm the negative sign is included

Wrong move:

Using Arrhenius definitions to answer acid-base definition questions

Why:

CIE expects Bronsted-Lowry definitions for all A-Level acid-base questions, as it is more general

Correct move:

Always define acids as proton donors and bases as proton acceptors

Wrong move:

Confusing conjugate acid and conjugate base in a pair

Why:

Students often mix up which species gained or lost the proton

Correct move:

Remember: +1 proton = conjugate acid, -1 proton = conjugate base

Wrong move:

Not using the weak acid approximation, leading to overly complex quadratic calculations

Why:

For all weak acids commonly tested in CIE, the approximation is valid and accepted

Correct move:

Use the approximation unless the question explicitly tells you not to, to save exam time

6. Quick Reference Cheatsheet

Concept	Formula / Rule
Bronsted-Lowry Acid	Proton donor
Bronsted-Lowry Base	Proton acceptor
pH	$p H = - lo g_{10} [H^{+}]$
Strong monoprotic acid	$[H^{+}] = [a c i d]_{ini t ia l}$
Strong diprotic acid	$[H^{+}] = 2 [a c i d]_{ini t ia l}$
Ka for weak acid HA	$K_{a} = \frac{[ H ^{+} ] [ A ^{-} ]}{[ H A ]}$
pKa	$p K a = - lo g_{10} K a$
Conjugate acid-base pair	Differs by exactly 1 proton

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.

2023 · Paper 1
Identify conjugate acid-base pairs
2022 · Paper 2
Calculate pH of weak acid
2021 · Paper 4
Bronsted-Lowry acid base definition

Going deeper

referenceIUPAC Gold Book: Acid-Base DefinitionsFor advanced reading on acid-base theory

What's Next

Acid-base equilibria is the foundation for all further acid-base topics in CIE A-Level Chemistry, including buffer solutions, pH curves, titrations and the ionic product of water. The calculation principles you learned here for Ka and pH will be extended to weak bases and salt hydrolysis in later topics, and underpin practical titration calculations commonly assessed in both Papers 2 and 4. Mastery of core definitions and basic pH calculations here is essential to avoid losing easy marks in extended response questions. Build on this knowledge by exploring the related topics below.