What are conjugate acid-base pairs?

In the Bronsted-Lowry model, every acid has a conjugate base formed when it loses a proton, and every base has a conjugate acid formed when it gains one. In the reaction HCl + H_2O H_3O^+ + Cl^-, HCl and Cl^- are one conjugate pair and H_2O and H_3O^+ are the other. Identifying these pairs makes clear that an acid-base reaction is simply a proton transfer, and that water can act as either an acid or a base depending on its partner.

Identify the conjugate base of the acid HCl. [1 mark]

WJEC WJEC 2019-style practice: In a titration, 25.0\ cm^3 of sodium hydroxide solution is exactly neutralised by 22.40\ cm^3 of 0.100\ mol dm^-3 hydrochloric acid. Calculate the concentration of the sodium hydroxide solution.

The equation is NaOH + HCl → NaCl + H_2O, a 1:1 ratio. Moles of HCl = c × V = 0.100 × (22.40/1000) = 2.24 × 10^-3 mol. Moles of NaOH = 2.24 × 10^-3 mol (same ratio). Concentration of NaOH = n/V = 2.24 × 10^-3 / (25.0/1000) = 0.0896 mol dm-3. Markers reward the balanced ratio, correct moles of acid, and the concentration with units.

WalesChemistrySyllabus dot point

What is dynamic equilibrium and how do acids and bases behave?

Dynamic equilibrium, Le Chatelier's principle, the Bronsted-Lowry model of acids and bases, strong and weak acids, and acid-base titrations.

A focused answer to WJEC A-Level Chemistry Unit 1, covering dynamic equilibrium, Le Chatelier's principle, the Bronsted-Lowry definition of acids and bases, strong versus weak acids, and acid-base titration calculations.

Generated by Claude Opus 4.811 min answerUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking

WJEC wants you to describe dynamic equilibrium, predict the effect of changing conditions using Le Chatelier's principle, define acids and bases by the Bronsted-Lowry model, distinguish strong from weak acids, and carry out acid-base titration calculations.

The answer

Dynamic equilibrium

Le Chatelier's principle

Bronsted-Lowry acids and bases

A Bronsted-Lowry acid is a proton ( $\text{H}^+$ ) donor; a base is a proton acceptor. When an acid donates a proton, the species left is its conjugate base, so acids and bases come in conjugate pairs, for example $\text{HCl}/\text{Cl}^-$ .

A strong acid fully dissociates ( $\text{HCl} \rightarrow \text{H}^+ + \text{Cl}^-$ ); a weak acid only partially dissociates ( $\text{CH}_3\text{COOH} \rightleftharpoons \text{CH}_3\text{COO}^- + \text{H}^+$ ), establishing an equilibrium.

Acid-base titrations

Finding an unknown acid concentration

A $25.0$ cm $^3$ sample of sulfuric acid is neutralised by $30.0$ cm $^3$ of $0.200$ mol dm $^{-3}$ sodium hydroxide. Find the acid concentration.

step 1 Balanced equation

$\text{H}_2\text{SO}_4 + 2\text{NaOH} \rightarrow \text{Na}_2\text{SO}_4 + 2\text{H}_2\text{O}$ .

step 2 Moles of base

$n(\text{NaOH}) = 0.200 \times (30.0/1000) = 6.00 \times 10^{-3}$ mol.

step 3 Moles of acid using the ratio

The ratio is $1$ acid to $2$ base, so $n(\text{H}_2\text{SO}_4) = 6.00 \times 10^{-3} / 2 = 3.00 \times 10^{-3}$ mol.

step 4 Concentration of acid

$c = n/V = 3.00 \times 10^{-3} / (25.0/1000) = 0.120$ mol dm $^{-3}$ .

Conjugate acid-base pairs

In the Bronsted-Lowry model, every acid has a conjugate base formed when it loses a proton, and every base has a conjugate acid formed when it gains one. In the reaction $\text{HCl} + \text{H}_2\text{O} \rightleftharpoons \text{H}_3\text{O}^+ + \text{Cl}^-$ , $\text{HCl}$ and $\text{Cl}^-$ are one conjugate pair and $\text{H}_2\text{O}$ and $\text{H}_3\text{O}^+$ are the other. Identifying these pairs makes clear that an acid-base reaction is simply a proton transfer, and that water can act as either an acid or a base depending on its partner.

Strong, weak, concentrated and dilute

These four terms are easily confused but mean different things. Strong and weak describe the extent of dissociation: a strong acid is fully dissociated, a weak acid only partly. Concentrated and dilute describe the amount of solute per volume. So a solution can be a concentrated weak acid (many moles of ethanoic acid per dm $^3$ , only slightly dissociated) or a dilute strong acid (few moles of hydrochloric acid per dm $^3$ , fully dissociated). Keeping these axes separate is essential for both equilibrium and pH reasoning.

Examples in context

The Haber process. Ammonia synthesis is run at a compromise of about $450$ degrees C and $200$ atm: high pressure favours the product side (fewer gas moles), while moderate temperature balances yield against rate, a direct application of Le Chatelier. Titration in the lab. The titration calculation above is the workhorse of standardising solutions and analysing unknown concentrations throughout the course.

Try this

Q1. Define a Bronsted-Lowry base. [1 mark]

Cue. A proton ( $\text{H}^+$ ) acceptor.

Q2. State the effect of increasing temperature on an exothermic equilibrium. [1 mark]

Cue. Shifts left (endothermic direction), reducing the yield of product.

Q3. Calculate the moles of hydrochloric acid in $20.0$ cm $^3$ of $0.150$ mol dm $^{-3}$ solution. [1 mark]

Cue. $n = 0.150 \times (20.0/1000) = 3.00 \times 10^{-3}$ mol.

Q4. Identify the conjugate base of the acid $\text{HCl}$ . [1 mark]

Cue. The chloride ion, $\text{Cl}^-$ .

Q5. Explain the difference between a strong acid and a concentrated acid. [1 mark]

Cue. Strong means fully dissociated; concentrated means many moles of acid per unit volume.

Exam-style practice questions

Practice questions written in the style of WJEC exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

WJEC 20213 marksThe reaction

\text{N}_2(\text{g}) + 3\text{H}_2(\text{g}) \rightleftharpoons 2\text{NH}_3(\text{g})

is exothermic in the forward direction. State and explain the effect of increasing the pressure on the position of equilibrium.

Show worked answer →

By Le Chatelier's principle, the system shifts to oppose the change imposed.

Increasing the pressure favours the side with fewer gas molecules. There are $4$ gas molecules on the left and $2$ on the right, so the equilibrium shifts to the right.

This increases the yield of ammonia.

Markers reward stating the shift to the right, counting gas moles on each side, and the conclusion of greater ammonia yield.

WJEC 20194 marksIn a titration,

25.0\ \text{cm}^3

of sodium hydroxide solution is exactly neutralised by

22.40\ \text{cm}^3

0.100\ \text{mol dm}^{-3}

hydrochloric acid. Calculate the concentration of the sodium hydroxide solution.

Show worked answer →

The equation is $\text{NaOH} + \text{HCl} \rightarrow \text{NaCl} + \text{H}_2\text{O}$ , a $1:1$ ratio.

Moles of HCl $= c \times V = 0.100 \times (22.40/1000) = 2.24 \times 10^{-3}$ mol.

Moles of NaOH $= 2.24 \times 10^{-3}$ mol (same ratio).

Concentration of NaOH $= n/V = 2.24 \times 10^{-3} / (25.0/1000) = 0.0896$ mol dm-3.

Markers reward the balanced ratio, correct moles of acid, and the concentration with units.

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Sources & how we know this

WJEC A-level Chemistry specification — WJEC (2015)