The ionisation of hydrochloric in water is given below: HCl(aq) + H2O (l) ⇌ H3O^+ (aq) + Cl^- (aq) Label two c

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Q. The ionisation of hydrochloric in water is given below: HCl(aq) + H2O (l) ⇌ H3O^+ (aq) + Cl^- (aq) Label two conjugate acid-base pairs in this ionisation.

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Concept understanding — Brønsted Lowry Conjugate Pairs

From Intuition to Precision: Brønsted–Lowry Conjugate Pairs

Imagine you're at a party where people are passing around a single, very special coin. The coin represents a proton ( $\text{H}^+$ ). The game is simple: someone can give the coin to someone else, but only if the other person is willing to take it. You can't just throw the coin; you need a willing receiver.

In chemistry, acids and bases play exactly this game. An acid is the person who gives away the proton. A base is the person who accepts it. But here's the twist: the moment the acid gives away its proton, it transforms into something new — something that can now accept a proton back. That transformed form is called the conjugate base. Similarly, the base, after accepting the proton, becomes something that can donate it back — the conjugate acid.

This is the core of the Brønsted–Lowry theory: every acid-base reaction involves a pair of substances that are linked by the gain or loss of a single proton.

The Precise Statement

\text{Acid} \rightleftharpoons \text{Conjugate Base} + \text{H}^+

A conjugate acid-base pair consists of two species that differ by exactly one proton ( $\text{H}^+$ ). The acid has the proton; the conjugate base does not.

When an acid donates a proton, it becomes its conjugate base.

When a base accepts a proton, it becomes its conjugate acid.

Important

In any Brønsted–Lowry reaction, there are two conjugate pairs: one on the reactant side and one on the product side. They always appear together.

Seeing It in Action

Take the classic reaction between hydrochloric acid and water:

\text{HCl} + \text{H}_2\text{O} \rightarrow \text{Cl}^- + \text{H}_3\text{O}^+

Let's identify the pairs:

Pair 1: $\text{HCl}$ (acid) and $\text{Cl}^-$ (conjugate base). They differ by one $\text{H}^+$ .
Pair 2: $\text{H}_2\text{O}$ (base) and $\text{H}_3\text{O}^+$ (conjugate acid). They also differ by one $\text{H}^+$ .

Notice: water acted as a base here — it accepted the proton from HCl. But water can also act as an acid in other reactions. That's the beauty of the Brønsted–Lowry theory: a substance's role depends on the reaction, not on a fixed label.

Tip

To find the conjugate base of any acid, simply remove one $\text{H}^+$ and reduce the charge by +1.

To find the conjugate acid of any base, add one $\text{H}^+$ and increase the charge by +1.

A Quick Reference Table

Acid	Conjugate Base	Base	Conjugate Acid
$\text{HCl}$	$\text{Cl}^-$	$\text{NH}_3$	$\text{NH}_4^+$
$\text{H}_2\text{SO}_4$	$\text{HSO}_4^-$	$\text{H}_2\text{O}$	$\text{H}_3\text{O}^+$
$\text{NH}_4^+$	$\text{NH}_3$	$\text{OH}^-$	$\text{H}_2\text{O}$
$\text{H}_2\text{O}$	$\text{OH}^-$	$\text{CO}_3^{2-}$	$\text{HCO}_3^-$

Notice how water appears in both columns — it's amphoteric, meaning it can act as either an acid or a base depending on its partner.

Why This Matters …

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A conjugate acid–base pair differs by exactly one proton ( $\ce{H+}$ ). In the ionisation of $\ce{HCl}$ , the two pairs are $\ce{HCl/Cl-}$ and $\ce{H3O+/H2O}$ .

The Brønsted–Lowry theory defines acids and bases in terms of proton transfer. An acid donates a proton; a base accepts one. When an acid gives up its proton, what remains is its conjugate base. When a base accepts a proton, it becomes its conjugate acid. The two species—differing by a single $\ce{H+}$ —form a conjugate pair.

Every proton-transfer reaction involves two such pairs: one on the reactant side (acid and its conjugate base) and one on the product side (the other base and its conjugate acid). The key is to track where the proton goes.

Identifying the pairs

Spot the proton donor (acid) on the left.

$\ce{HCl}$ donates a proton to water. It is the acid. After losing $\ce{H+}$ , it becomes $\ce{Cl-}$ , which is its conjugate base.

First conjugate pair: $\ce{HCl}$ (acid) and $\ce{Cl-}$ (conjugate base).
Spot the proton acceptor (base) on the left.

$\ce{H2O}$ accepts the proton from $\ce{HCl}$ . It is the base. After gaining $\ce{H+}$ , it becomes $\ce{H3O+}$ , which is its conjugate acid.

Second conjugate pair: $\ce{H2O}$ (base) and $\ce{H3O+}$ (conjugate acid).
Verify the relationship.

Each pair differs by exactly one proton:
- $\ce{HCl} \xrightarrow{-H^+} \ce{Cl-}$
- $\ce{H2O} \xrightarrow{+H^+} \ce{H3O+}$ …

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Q.The ionisation of hydrochloric in water is given below: HCl(aq) + H2O (l) ⇌ H3O^+ (aq) + Cl^- (aq) Label two conjugate acid-base pairs in this ionisation.

Identifying the pairs

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Q.The ionisation of hydrochloric in water is given below:
HCl(aq) + H2O (l) ⇌ H3O^+ (aq) + Cl^- (aq)
Label two conjugate acid-base pairs in this ionisation.

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