Conjugate Acid–Base Pairs: Examples and Practice

H₂PO₄⁻ becomes HPO₄²⁻ when it donates one H⁺. The hydrogen count drops by one, and the charge goes from −1 to −2. That second change is easy to miss: removing something positive makes the remaining ion more negative.

Conjugate acid–base pairs differ by one H⁺. Keep the other atoms unchanged, adjust the hydrogen count and charge together, and you can check the answer without memorizing a long table.

Keep the proton and charge in the same calculation

In the Brønsted–Lowry model, an acid donates a proton and a base accepts it. The donor becomes its conjugate base; the acceptor becomes its conjugate acid. These definitions follow OpenStax's treatment of Brønsted–Lowry acids and bases.

To find a conjugate base, remove one H⁺. To find a conjugate acid, add one H⁺. Track its +1 charge along with the hydrogen count:

Requested partner	Hydrogen count	Total species charge
Conjugate base	Remove one H	Subtract 1
Conjugate acid	Add one H	Add 1

“Add 1” means ordinary signed arithmetic. A charge of −2 becomes −1; a charge of −1 becomes 0; a neutral species becomes +1. It doesn't mean every conjugate acid has a positive charge.

The transferred particle is H⁺, not a neutral hydrogen atom. That's why changing the hydrogen count alone leaves the answer incomplete.

For example, finding the conjugate acid of SO₄²⁻ gives HSO₄⁻: one added H and −2 + 1 = −1. Finding the conjugate base of HSO₄⁻ reverses those changes and gives SO₄²⁻.

The charge here belongs to the whole molecule or ion. Don't substitute an individual atom's formal charge or oxidation number. If those labels keep getting mixed together, work through the separate formal charge vs oxidation number examples.

This method checks proton bookkeeping for a specified acid–base step. With a more complicated structure, you may also need to identify which site gains or loses the proton; a molecular formula alone doesn't always locate it.

Match each reactant to its own product

Consider this proton-transfer equation, read from left to right:

H₂PO₄⁻ + NH₃ ⇌ NH₄⁺ + HPO₄²⁻

Start with one reactant and find the product that preserves its non-hydrogen atoms. H₂PO₄⁻ keeps its phosphorus and four oxygens in HPO₄²⁻. NH₃ keeps its nitrogen in NH₄⁺. The product order doesn't determine the pairs.

Reactant → its product	H count	Charge calculation	Role in the forward direction
H₂PO₄⁻ → HPO₄²⁻	2 → 1	−1 − 1 = −2	Acid → conjugate base
NH₃ → NH₄⁺	3 → 4	0 + 1 = +1	Base → conjugate acid

The two pairs are H₂PO₄⁻ / HPO₄²⁻ and NH₄⁺ / NH₃. Here the slash lists the acid member first, regardless of its position in the equation.

H₂PO₄⁻ and NH₃ react with each other, but they aren't conjugate partners. They don't have the same non-hydrogen atoms and don't differ by one proton. Pairing the two reactants would skip the actual change each species undergoes.

Now check the equation as a whole:

• Hydrogen: 2 + 3 = 4 + 1 = 5.
• Charge: −1 + 0 = +1 − 2 = −1.
• Phosphorus, oxygen, and nitrogen counts also match on both sides.

The proton lost by one species is gained by the other. Writing “−H⁺” beside the donor and “+H⁺” beside the acceptor is bookkeeping; it doesn't introduce an extra free proton into the balanced equation.

Read the same equation backward and the roles change: NH₄⁺ donates H⁺ to HPO₄²⁻. The pairs stay the same, but NH₄⁺ is now the reacting acid and HPO₄²⁻ the reacting base.

Stop after one proton, even when more hydrogens remain

Phosphate makes this mistake visible. Its successive proton-loss steps are:

H₃PO₄ → H₂PO₄⁻ → HPO₄²⁻ → PO₄³⁻

Each arrow here means removal of one H⁺ for identifying the neighboring partner; this is a bookkeeping sequence, not a claim that all steps happen completely in water.

Adjacent entries are conjugate pairs. H₃PO₄ and HPO₄²⁻ are two steps apart, so they aren't one conjugate pair. H₂PO₄⁻ and PO₄³⁻ are also two steps apart.

If the question asks for the conjugate base of H₂PO₄⁻, stop at HPO₄²⁻. Removing every remaining hydrogen answers a different question. The requested partner can still contain hydrogen and can participate in another proton transfer.

An amphiprotic ion can take either role

An amphiprotic species can both donate and accept a proton. Water and certain hydrogen-containing ions can do this; OpenStax explains the distinction using separate proton-transfer reactions.

Use HPO₄²⁻ to compare the two jobs. Read both equations from left to right. The ions are aqueous and H₂O is liquid water; state symbols are omitted to keep the proton changes visible.

Equation	What HPO₄²⁻ does	Its partner in this step
HPO₄²⁻ + H₃O⁺ ⇌ H₂PO₄⁻ + H₂O	Accepts H⁺; acts as a base	H₂PO₄⁻, its conjugate acid
HPO₄²⁻ + OH⁻ ⇌ PO₄³⁻ + H₂O	Donates H⁺; acts as an acid	PO₄³⁻, its conjugate base

In the first row, its hydrogen count goes 1 → 2 and charge goes −2 → −1. H₃O⁺ supplies the proton and becomes H₂O. In the second row, HPO₄²⁻ goes 1 → 0 and −2 → −3; OH⁻ receives the proton and becomes H₂O. Same starting phosphate ion, different partner and role.

Neither its negative charge nor the H in its formula settles the role by itself. Follow what it does in the stated reaction. These equations also don't say the two processes occur to the same extent.

What a correct pair still doesn't tell you

“Write the conjugate acid of SO₄²⁻” is a partner-identification task. It asks for HSO₄⁻. It doesn't ask whether SO₄²⁻ will accept a large amount of H⁺ from a particular substance under particular conditions.

Relative strength and equilibrium extent require more information than the one-proton relationship. In aqueous chemistry, acid and base ionization constants quantify strength; stronger acids have weaker conjugate bases. See OpenStax on relative acid and base strengths.

Likewise, an ion's charge doesn't give a solution's pH. That requires solution composition and acid–base equilibrium information; OpenStax's pH section relates pH to hydronium concentration at this course level. A phosphate sequence alone won't tell you which form predominates in a sample.

Keep the scope straight when reading a worksheet: identifying a formal conjugate partner, labeling roles in a written equation, and predicting the behavior of an actual solution are different tasks.

Conjugate acid and base practice

Try these on paper with the answers covered. For partner questions, write the complete formula and charge. For reaction questions, show the two reactant-to-product changes, label roles in the requested direction, and check atoms and total charge.

1. Write the conjugate acid of HPO₄²⁻ and show the charge arithmetic.
2. Write the conjugate base of C₂H₅NH₃⁺. Remove a proton from the NH₃ group; keep C₂H₅ unchanged.
3. For HClO₂ + NH₃ ⇌ NH₄⁺ + ClO₂⁻, identify both conjugate pairs and label each reactant's role from left to right.
4. Decide whether each proposed pair is a conjugate acid–base pair: NO₂⁻ / NO₃⁻, H₂PO₄⁻ / PO₄³⁻, and HSO₄⁻ / SO₄²⁻. Explain using the formula and charge changes.
5. Fill both blanks in HSO₄⁻ + OH⁻ ⇌ _____ + _____. Identify both conjugate pairs and label the reactants' roles from left to right.
6. Read HPO₄²⁻ + H₃O⁺ ⇌ H₂PO₄⁻ + H₂O from right to left. Identify the reacting acid and base, and give both conjugate pairs.
7. Someone writes “H₂SO₄ loses H⁺ to form HSO₄²⁻.” Correct the product and explain which part of their bookkeeping failed.
8. You correctly identify HPO₄²⁻ / PO₄³⁻ as a pair. Does that alone establish the pH of a phosphate solution or prove that PO₄³⁻ is its predominant phosphate species? Explain.

Answers with the reasoning left in

1. H₂PO₄⁻. Add one H and one positive charge: −2 + 1 = −1. A conjugate acid can still be an anion.
2. C₂H₅NH₂. Remove one nitrogen-bound H⁺, and +1 − 1 = 0. The conjugate base is neutral; the five hydrogens in C₂H₅ remain in place.
3. HClO₂ / ClO₂⁻ and NH₄⁺ / NH₃. HClO₂ loses one H⁺ and changes charge from 0 to −1, so it is the reacting acid. NH₃ gains that H⁺ and changes from 0 to +1, so it is the reacting base. ClO₂⁻ is the first reactant's conjugate base, and NH₄⁺ is the second's conjugate acid. Each side contains four H, one Cl, two O, and one N. Total charge is 0 on each side: 0 + 0 = +1 − 1.
4. The first two fail; the third works. NO₂⁻ / NO₃⁻ changes the oxygen count and changes no hydrogen count. H₂PO₄⁻ / PO₄³⁻ preserves P and O but loses two H and changes charge by −2. Its one-step partner would be HPO₄²⁻. HSO₄⁻ / SO₄²⁻ preserves S and O, loses exactly one H, and changes charge from −1 to −2: a valid pair.
5. SO₄²⁻ + H₂O. HSO₄⁻ is the acid: it donates one H⁺ and becomes SO₄²⁻. OH⁻ is the base: it accepts H⁺ and becomes H₂O. The pairs are HSO₄⁻ / SO₄²⁻ and H₂O / OH⁻. Each side contains two H, one S, and five O. Charge is −1 − 1 = −2 on the left and −2 + 0 = −2 on the right.
6. H₂PO₄⁻ is the acid; H₂O is the base. In the requested reverse direction, H₂PO₄⁻ donates H⁺ and becomes HPO₄²⁻ (−1 → −2). H₂O accepts it and becomes H₃O⁺ (0 → +1). The pairs are H₂PO₄⁻ / HPO₄²⁻ and H₃O⁺ / H₂O. Each side contains four H, one P, and five O; total charge is −1 on each side. The reacting species are those on the right of the printed equation.
7. The product is HSO₄⁻. The hydrogen count was changed correctly, but a neutral starting molecule losing one H⁺ has charge 0 − 1 = −1. HSO₄²⁻ fails that check.
8. No to both. The pair identifies neighboring protonation states. Predicting pH and predominant species needs the relevant equilibrium data and solution conditions. Correct formula bookkeeping doesn't supply those missing facts.

For another set after these, the University of Texas at Austin's Preparation for Buffer Problems worksheet begins with conjugate-partner practice and then moves into equilibrium and salts. Treat those later questions as additional chemistry, beyond the pair checks practiced here.

Make a card for the mistake you want to stop repeating

Keep the full reactions in your worked practice. If the same error returns, save a short prompt that makes you perform the missed step. These can be paper cards or ordinary front/back cards in Nibomo.

Front	Back
HPO₄²⁻ gains one H⁺. Formula and charge arithmetic?	H₂PO₄⁻; −2 + 1 = −1.
C₂H₅NH₃⁺ loses one H⁺. Must its conjugate base be negative?	No. C₂H₅NH₂ is neutral: +1 − 1 = 0.
Why aren't H₂PO₄⁻ and PO₄³⁻ one conjugate pair?	They differ by two H⁺. The one-step partner is HPO₄²⁻.
In HClO₂ + NH₃ ⇌ NH₄⁺ + ClO₂⁻, which product belongs to HClO₂?	ClO₂⁻. Preserve Cl and both O atoms; remove one H⁺.
What decides whether HPO₄²⁻ is the acid or base in an equation?	Donating H⁺ makes it the acid; accepting H⁺ makes it the base. Read the stated direction.

Choose the card that matches your own miss rather than adding the whole table automatically. The advanced chemistry flashcard guide develops that mistake-repair workflow, and the getting-started guide covers creating cards in Nibomo.

After reviewing, return to a fresh reaction and write both changes without the answers visible. Check one proton, its charge, and the atoms that stayed put.