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Conjugate Pairs and Amphoteric Species

Learn and apply Conjugate Pairs and Amphoteric Species in the published Chemistry course sequence.

  • GCE A-Level H2 Chemistry 9476-2027
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Conjugate Pairs and Amphoteric Species: Orientation

Conjugate-pair questions are bookkeeping questions: track one H⁺ and one unit of charge change. Amphiprotic examples extend the same reasoning by showing one species acting as a proton donor in one reaction and a proton acceptor in another.

Use this together with the Acids and Bases (Theories) hub and the Chemistry of Aqueous Solutions hub so definitions and calculations stay connected.

Definitions (Must Know)

A. Conjugate acid–base pair

A conjugate acid–base pair differs by exactly one proton, H⁺.

  • acid → conjugate base (after donating H⁺)
  • base → conjugate acid (after accepting H⁺)

B. Conjugate acid / conjugate base (what changes)

  • An acid loses one H⁺ to form its conjugate base; the charge becomes one unit more negative.
  • A base gains one H⁺ to form its conjugate acid; the charge becomes one unit more positive.
One-proton bookkeeping for conjugate pairs: removing a proton makes the charge one unit more negative; adding a proton makes it one unit more positive.

C. Amphiprotic and amphoteric species

An amphiprotic species can both donate and accept a proton, so it can act as both a Brønsted–Lowry acid and a Brønsted–Lowry base. It is therefore amphoteric in this proton-transfer context. Amphoteric is the broader term for a species that can act as an acid and as a base; the two words are not universal synonyms.

Detailed Explanations

A. Finding conjugate pairs (workflow)

  1. Write the full equation (or use the one given).
  2. Identify the acid/base by seeing who loses/gains H⁺.
  3. Pair:
    • acid with the product after losing H⁺ (conjugate base)
    • base with the product after gaining H⁺ (conjugate acid)

Because removing H⁺ removes a + 1 charge, the conjugate base is always one charge unit more negative than its acid; therefore conjugate pairs always differ by exactly one H⁺ and a charge difference of 1.

Mini example: H₂SO₄ + H₂O → H₃O⁺ + HSO₄⁻

  • Conjugate pair: H₂SO₄ / HSO₄⁻ (lost H⁺).
  • Conjugate pair: H₂O / H₃O⁺ (gained H⁺).

B. Proving amphiprotic behaviour (workflow)

  1. Show the species accepting H⁺ from an acid, so it acts as a base.
  2. Show the species donating H⁺ to a base, so it acts as an acid.
  3. Write both equations and state the role explicitly.

Worked Examples

Modelled example 1

Follow both sides of a proton transfer

Core

Problem

Identify the conjugate acid–base pairs in HCl + H₂O → H₃O⁺ + Cl⁻.
Study the worked solution

  1. Follow the proton donor

    Method

    Pair hydrogen chloride with chloride.

    Reason

    HCl loses exactly one H⁺ to form Cl⁻.

    Working

    HCl/Cl⁻ is an acid/conjugate-base pair.

  2. Follow the proton acceptor

    Method

    Pair water with hydronium.

    Reason

    H₂O gains exactly one H⁺ to form H₃O⁺.

    Working

    H₃O⁺/H₂O is a conjugate-acid/base pair.

Guided practice 2

Assign roles and conjugates together

About 7 min

Problem

For NH₃ + H₂O ⇌ NH₄⁺ + OH⁻, identify the Brønsted–Lowry acid and base on the left and both conjugate pairs.

Try this before viewing the solution

Hints

Hint 1: assign roles
Water loses H⁺; ammonia gains it.
Hint 2: pair across the equation
Match each left-hand species to the product differing by exactly one proton.
View solution step by step

  1. Assign donor and acceptor

    Method

    Name water as acid and ammonia as base.

    Reason

    Water donates the proton accepted by ammonia.

    Working

    Acid: H₂O; base: NH₃.

  2. Pair the acid

    Method

    Match water with hydroxide.

    Reason

    Removing one proton from water gives OH⁻.

    Working

    H₂O/OH⁻.

  3. Pair the base

    Method

    Match ammonia with ammonium.

    Reason

    Adding one proton to ammonia gives NH₄⁺.

    Working

    NH₄⁺/NH₃.

Common misconception 3

Reject a false conjugate pair

Find and correct the mistake

Learner pairing

For HCl + H₂O → H₃O⁺ + Cl⁻, a learner pairs HCl with H₂O because both are reactants. Diagnose the pairing.

Apply the one-proton test

The conjugate base of HCl is

View solution step by step

  1. Apply the structural test

    Method

    Compare formulas rather than equation positions.

    Reason

    Conjugates must differ by exactly one H⁺.

    Working

    HCl and H₂O are not related by removal of one proton.

  2. Build the valid pairs

    Method

    Pair donor with its product and acceptor with its product.

    Reason

    Those formula changes show the transferred proton.

    Working

    HCl/Cl⁻ and H₃O⁺/H₂O.

Examiner practice 4

Prove that hydrogencarbonate is amphiprotic

4 marks

Problem

Show that HCO₃⁻ is amphiprotic by writing two equations and stating its role in each. [4 marks]

Try this before viewing the solution

View solution step by step

  1. Show proton acceptance

    1 mark

    Method

    React hydrogencarbonate with water to form carbonic acid.

    Reason

    HCO₃⁻ gains one proton.

    Working

    HCO₃⁻ + H₂O ⇌ H₂CO₃ + OH⁻

  2. Name the base role

    1 mark

    Method

    Call hydrogencarbonate the Brønsted–Lowry base.

    Reason

    It accepts H⁺.

    Working

    Proton acceptor ⇒ base.

  3. Show proton donation

    1 mark

    Method

    React hydrogencarbonate with water to form carbonate and hydronium.

    Reason

    HCO₃⁻ loses one proton.

    Working

    HCO₃⁻ + H₂O ⇌ CO₃²⁻ + H₃O⁺

  4. Name the acid role

    1 mark

    Method

    Call hydrogencarbonate the Brønsted–Lowry acid.

    Reason

    It donates H⁺.

    Working

    Showing both roles proves it is amphiprotic.

Challenge 5

Deduce missing conjugates

Minimal support

Problem

Deduce (a) the conjugate acid of HPO₄²⁻ and (b) the conjugate base of NH₄⁺. Justify each formula and charge.

Try this before viewing the solution

Hints

Hint 1: conjugate acid
Add one H⁺, so both hydrogen count and charge change.
Hint 2: conjugate base
Remove one H⁺ and make the charge one unit more negative.
View solution step by step

  1. Form the conjugate acid

    Method

    Add one proton to hydrogenphosphate.

    Reason

    Adding H⁺ increases the charge by one.

    Working

    HPO₄²⁻ + H⁺ → H₂PO₄⁻.

  2. Form the conjugate base

    Method

    Remove one proton from ammonium.

    Reason

    Removing H⁺ makes the charge one unit more negative.

    Working

    NH₄⁺ → NH₃ + H⁺, so the conjugate base is NH₃.

Mind Stretchers

Mind stretcher 1Extension

In the equilibrium:

2H₂O ⇌ H₃O⁺ + OH⁻

identify the conjugate pairs and explain (using this equation) why water is amphiprotic.

Show Answer

Mark scheme:

  • Conjugate pairs: H₂O / OH⁻ (water donates H⁺) and H₂O / H₃O⁺ (water accepts H⁺).
  • In the same equilibrium, one water molecule acts as an acid and another acts as a base, so water is amphiprotic.

Mind stretcher 2Extension

Which of the following are conjugate acid–base pairs? Explain briefly using the “one H⁺ difference” test.

  • (a) H₂CO₃ and CO₃²⁻
  • (b) H₂CO₃ and HCO₃⁻
  • (c) HCO₃⁻ and CO₃²⁻
Show Answer

Mark scheme:

  • (a) Not a conjugate pair: differs by 2 H⁺ (and charge changes by 2).
  • (b) Conjugate pair: differs by exactly one H⁺.
  • (c) Conjugate pair: differs by exactly one H⁺.