What is meant by the conjugate acid-base pair? Find the conjugate acid/base for the following species: HNO2, C

Exercises · 6.35

Q. What is meant by the conjugate acid-base pair? Find the conjugate acid/base for the following species: HNO2, CN-, HClO4, F-, OH-, CO₃²-, and S2-.

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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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Concept: Brønsted–Lowry Conjugate Pairs

A conjugate acid-base pair consists of two species that differ by exactly one proton ( $\text{H}^+$ ). The acid donates a proton to become its conjugate base; the base accepts a proton to become its conjugate acid.

Reasoning steps:

For each given species, identify whether it is an acid (donates $\text{H}^+$ ) or a base (accepts $\text{H}^+$ ).
Remove one $\text{H}^+$ from an acid to get its conjugate base; add one $\text{H}^+$ to a base to get its conjugate acid.
Write the resulting conjugate species.

Species	Acts as	Conjugate
$\text{HNO}_2$	Acid	$\text{NO}_2^-$
$\text{CN}^-$	Base	$\text{HCN}$
$\text{HClO}_4$	Acid	$\text{ClO}_4^-$
$\text{F}^-$	Base	$\text{HF}$

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A conjugate acid-base pair differs by exactly one proton ( $\mathrm{H}^+$ ). For each given species, removing a proton gives its conjugate base; adding a proton gives its conjugate acid. The results are: $\mathrm{HNO_2}$ (conjugate base $\mathrm{NO_2^-}$ ), $\mathrm{CN^-}$ (conjugate acid $\mathrm{HCN}$ ), $\mathrm{HClO_4}$ (conjugate base $\mathrm{ClO_4^-}$ ), $\mathrm{F^-}$ (conjugate acid $\mathrm{HF}$ ), $\mathrm{OH^-}$ (conjugate acid $\mathrm{H_2O}$ and conjugate base $\mathrm{O^{2-}}$ ), $\mathrm{CO_3^{2-}}$ (conjugate acid $\mathrm{HCO_3^-}$ ), $\mathrm{S^{2-}}$ (conjugate acid $\mathrm{HS^-}$ ).

The Core Idea: One Proton, That’s All

The Brønsted–Lowry theory defines an acid as a proton ( $\mathrm{H}^+$ ) donor and a base as a proton acceptor. This leads to a beautifully simple relationship: when an acid donates a proton, what remains is its conjugate base. When a base accepts a proton, what forms is its conjugate acid. The original and the product together form a conjugate acid-base pair — two species that differ by exactly one proton.

Think of it as a chemical handshake: the acid passes a proton to the base. After the handshake, the acid becomes a base (it can now accept a proton back), and the base becomes an acid (it can now donate that proton). Every acid has a conjugate base; every base has a conjugate acid.

\mathrm{HA}

Step-by-Step: Finding Conjugate Partners

We’ll go through each species one by one. The rule is simple:

If the species can donate a proton (it has an $\mathrm{H}$ to give), remove $\mathrm{H}^+$ to get its conjugate base.
If the species can accept a proton (it has a lone pair or negative charge), add $\mathrm{H}^+$ to get its conjugate acid.
Some species (like $\mathrm{OH}^-$ ) can do both — we’ll handle that carefully.

1. $\mathrm{HNO_2}$ (Nitrous acid)

This is clearly an acid — it has a hydrogen that can be donated. Remove $\mathrm{H}^+$ :

\mathrm{HNO_2} \rightarrow \mathrm{H}^+ + \mathrm{NO_2^-}

The species left is $\mathrm{NO_2^-}$ (nitrite ion). So the conjugate base of $\mathrm{HNO_2}$ is $\mathrm{NO_2^-}$ .

Tip

Notice that the charge changes by $-1$ when you remove $\mathrm{H}^+$ (since $\mathrm{H}^+$ has a $+1$ charge). This is a quick check: if the original is neutral, the conjugate base will have a $-1$ charge.

2. $\mathrm{CN^-}$ (Cyanide ion)

This is a base — it has a negative charge and can accept a proton. Add $\mathrm{H}^+$ :

\mathrm{CN^-} + \mathrm{H}^+ \rightarrow \mathrm{HCN}

The product is $\mathrm{HCN}$ (hydrogen cyanide). So the conjugate acid of $\mathrm{CN^-}$ is $\mathrm{HCN}$ .

3. $\mathrm{HClO_4}$ (Perchloric acid)

A strong acid — it readily donates its proton. Remove $\mathrm{H}^+$ :

\mathrm{HClO_4} \rightarrow \mathrm{H}^+ + \mathrm{ClO_4^-}

The conjugate base is $\mathrm{ClO_4^-}$ (perchlorate ion).

4. $\mathrm{F^-}$ (Fluoride ion)

A base — it can accept a proton to form $\mathrm{HF}$ . Add $\mathrm{H}^+$ :

\mathrm{F^-} + \mathrm{H}^+ \rightarrow \mathrm{HF}

The conjugate acid is $\mathrm{HF}$ (hydrofluoric acid).

Watch out

A common mistake is to think $\mathrm{F^-}$ could also act as an acid. But fluorine is highly electronegative and holds its electrons tightly — it does not donate a proton because it has no hydrogen to give. So only the conjugate acid is relevant here.

5. $\mathrm{OH^-}$ (Hydroxide ion)

This is interesting. $\mathrm{OH^-}$ is famously a base — it accepts a proton to become water:

\mathrm{OH^-} + \mathrm{H}^+ \rightarrow \mathrm{H_2O}

So its conjugate acid is $\mathrm{H_2O}$ .

But can $\mathrm{OH^-}$ also act as an acid? Yes — in principle, it can donate a proton (though it’s a very weak acid). If we remove $\mathrm{H}^+$ from $\mathrm{OH^-}$ : …

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Q.What is meant by the conjugate acid-base pair? Find the conjugate acid/base for the following species: HNO2, CN–, HClO4, F–, OH–, CO₃²–, and S2–.

The Core Idea: One Proton, That’s All

Step-by-Step: Finding Conjugate Partners

1. HNO2\mathrm{HNO_2}HNO2​ (Nitrous acid)

2. CN−\mathrm{CN^-}CN− (Cyanide ion)

3. HClO4\mathrm{HClO_4}HClO4​ (Perchloric acid)

4. F−\mathrm{F^-}F− (Fluoride ion)

5. OH−\mathrm{OH^-}OH− (Hydroxide ion)

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1. $\mathrm{HNO_2}$ (Nitrous acid)

2. $\mathrm{CN^-}$ (Cyanide ion)

3. $\mathrm{HClO_4}$ (Perchloric acid)

4. $\mathrm{F^-}$ (Fluoride ion)

5. $\mathrm{OH^-}$ (Hydroxide ion)

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