How to Identify a Conjugate Base After One Proton Is Removed

Take phosphoric acid, \(\ce{H3PO4}\). If it loses one proton, what is its conjugate base?

Is it \(\mathrm{H_{2}PO_{4}}^{-}\), \(\mathrm{HPO_{4}}^{2-}\), or \(\mathrm{PO_{4}}^{3-}\)?

The answer is \(\mathrm{H_{2}PO_{4}}^{-}\). To identify a conjugate base, remove exactly one \(\mathrm{H}^{+}\) from the acid species and reduce its charge by one unit. The other two formulas appear only after the acid loses its second and third protons.

This is a straightforward rule, but there is an important catch: you must track which species is donating the proton at each stage. That becomes particularly important when an acid can donate more than one proton.

01The one-proton test

A conjugate base is the species formed when an acid donates one proton.

For a simple neutral acid represented by \(\ce{HA}\), the process is:

\[ \ce{HA <=> H+ + A-} \]

The original acid is \(\ce{HA}\), and its conjugate base is \(\mathrm{A}^{-}\).

In water, we usually represent the proton transfer as:

\[ \ce{HA + H2O <=> H3O+ + A-} \]

Water accepts the proton, while the acid becomes its conjugate base.

There are three things to check when identifying that product.

1. Start with the correct acid species.

Look at the species before it donates the proton. Do not automatically start with the original molecule if it has already lost a proton in an earlier reaction.

2. Remove exactly one \(\mathrm{H}^{+}\).

A proton is a hydrogen nucleus without its electron. You are not removing a neutral hydrogen atom.

3. Adjust the charge.

Because the species loses a positive charge, its overall charge decreases by one.

For example:

\[ \ce{H3PO4 -> H2PO4- + H+} \]

Phosphoric acid starts with no overall charge. After losing \(\mathrm{H}^{+}\), its conjugate base has a charge of \(-1\).

The charge rule works for negatively charged and positively charged acids as well.

Notice that the conjugate base does not have to be negatively charged. Ammonia, \(\ce{NH3}\), is the conjugate base of ammonium, \(\ce{NH4+}\), even though it is electrically neutral.

The important feature is the loss of one proton, not whether the resulting species has a particular charge.

02Why polyprotic acids need step-by-step tracking

Some acids can donate more than one proton.

These are called polyprotic acids. Phosphoric acid is a useful example because it can donate three protons in successive reactions.

The important word is successive. The acid does not lose all three protons in one step.

Consider the three dissociation reactions.

First proton:

\[ \ce{H3PO4 + H2O <=> H3O+ + H2PO4-} \]

The conjugate base after the first proton is removed is:

\[ \boxed{\ce{H2PO4-}} \]

Second proton:

\[ \ce{H2PO4- + H2O <=> H3O+ + HPO4^2-} \]

The conjugate base after the second proton is removed is:

\[ \boxed{\mathrm{HPO_{4}}^{2-}} \]

Third proton:

\[ \ce{HPO4^2- + H2O <=> H3O+ + PO4^3-} \]

The conjugate base after the third proton is removed is:

\[ \boxed{\mathrm{PO_{4}}^{3-}} \]

Here is the distinction worth remembering:

• \(\ce{H2PO4-}\) is the conjugate base of \(\ce{H3PO4}\).
• \(\mathrm{HPO_{4}}^{2-}\) is the conjugate base of \(\ce{H2PO4-}\).
• \(\mathrm{PO_{4}}^{3-}\) is the conjugate base of \(\mathrm{HPO_{4}}^{2-}\).

Each conjugate pair differs by exactly one proton.

There is another interesting detail. \(\ce{H2PO4-}\) is a conjugate base in the first reaction, but it can donate another proton in the second reaction.

Its role depends on the reaction taking place. A species does not have to remain permanently an acid or a base.

This is closely connected to amphiprotic behaviour. For more practice with the broader conjugate-pair framework, see Conjugate Acid-Base Pairs and Proton Transfer | HSC Chemistry.

03An experiment: Following proton removal through a titration

How could you investigate the successive proton losses of phosphoric acid rather than simply writing them from a chemical formula?

A titration provides a useful way to investigate this.

Imagine placing 25.0 mL of phosphoric acid solution into a conical flask. A burette containing standardised sodium hydroxide is clamped vertically above the flask.

A magnetic stirrer sits underneath the flask, keeping the solution mixed as sodium hydroxide enters drop by drop. A pH electrode is immersed in the solution and connected to a meter that records the pH.

What should happen?

Each hydroxide ion can accept one proton:

\[ \ce{H+ + OH- -> H2O} \]

As sodium hydroxide is added, hydroxide ions remove protons from the acid.

For phosphoric acid, the successive reactions can be represented as:

\[ \ce{H3PO4 + OH- -> H2PO4- + H2O} \]

\[ \ce{H2PO4- + OH- -> HPO4^2- + H2O} \]

\[ \ce{HPO4^2- + OH- -> PO4^3- + H2O} \]

Predict what would happen if exactly one mole of hydroxide ions were added for every mole of phosphoric acid.

The first proton equivalent would have been neutralised. The predominant phosphate species would be \(\ce{H2PO4-}\).

Adding a second equivalent would produce predominantly \(\mathrm{HPO_{4}}^{2-}\).

This gives us a way to connect the chemical formula changes to measurable quantities.

Why the first experimental design needs improving

A simple experiment might use an indicator and record the colour change.

That can help locate a particular endpoint, but one colour change does not reveal every stage of a triprotic acid’s neutralisation.

A better design uses a pH probe to record pH after successive additions of sodium hydroxide. Plotting pH against the volume of base added produces a titration curve.

The curve can reveal separate neutralisation stages when the equilibrium changes are sufficiently distinct.

The expected stoichiometric relationships are:

The first two neutralisation stages are generally easier to distinguish than the third. The third dissociation of phosphoric acid is very weak, making its final neutralisation stage much less distinct in an ordinary aqueous titration.

There is also an important experimental limitation.

A pH probe measures pH, not the identity of individual phosphate ions. We infer the dominant species by combining the pH behaviour with the reaction equations, acid-base equilibria, and the amount of sodium hydroxide added.

The curve supports the stepwise model; it is not a direct picture of the particles.

04Strength, amount, and concentration are not interchangeable

Identifying a conjugate base is a question about chemical identity. However, HSC questions often combine this with acid strength, concentration, and the amount of substance present.

These are different ideas.

For phosphoric acid, the successive acid dissociation constants follow:

\[ K_{a1}>K_{a2}>K_{a3} \]

Here, \(K_{a1}\), \(K_{a2}\), and \(K_{a3}\) describe the first, second, and third proton losses respectively.

The first proton is donated more readily than the second, and the second more readily than the third.

This also affects conjugate base strength. The conjugate base of a stronger acid has a weaker tendency to accept a proton, relative to the conjugate base of a weaker acid.

However, remember that strength does not determine the number of moles present.

For example, consider 25.0 mL of \(0.100\ \mathrm{mol\,L^{-1}}\) phosphoric acid.

Its amount is:

\[ n=0.100\times0.0250 \]

\[ n=0.00250\ \mathrm{mol} \]

If you dilute this solution to 100.0 mL, its concentration becomes:

\[ c=\frac{0.00250}{0.100} \]

\[ c=0.0250\ \mathrm{mol\,L^{-1}} \]

The concentration has decreased, but the amount of phosphoric acid is still \(0.00250\) mol.

If both samples are titrated with \(0.100\ \mathrm{mol\,L^{-1}}\) sodium hydroxide, they require the same amount of sodium hydroxide to reach the first equivalence point.

Dilution changes concentration and the equilibrium composition. It does not change the identity of the conjugate base or the acid’s intrinsic \(K_a\) at a fixed temperature.

For a more detailed explanation of how acid strength and pH differ, see How to Explain Acid Strength Using Equilibrium Rather Than pH Alone.

05A reliable method for HSC questions

When an exam asks you to identify a conjugate base, follow this sequence.

1. Identify the acid species immediately before proton transfer.
2. Determine which proton is being donated.
3. Remove exactly one \(\mathrm{H}^{+}\), not a neutral hydrogen atom.
4. Reduce the original charge by one unit.
5. Check that the acid and conjugate base differ by exactly one proton.
6. If the acid is polyprotic, check whether the question asks about the first, second, or third proton loss.

If the question also gives concentration or titration information, calculate the amount of acid and base separately. Do not use pH or concentration to guess the conjugate base’s formula.

06Questions and solutions

Question 1

The hydrogen sulfate ion, \(\ce{HSO4-}\), donates one proton to water.

Write the balanced proton-transfer equation and identify the conjugate base of hydrogen sulfate.

Solution 1

The conjugate base is \(\mathrm{SO_{4}}^{2-}\).

Hydrogen sulfate loses exactly one proton:

\[ \ce{HSO4- -> SO4^2- + H+} \]

Its charge changes from \(-1\) to \(-2\).

When water accepts the proton, it becomes hydronium:

\[ \ce{HSO4- + H2O <=> SO4^2- + H3O+} \]

The equation is balanced for both atoms and charge.

A common mistake is to write \(\ce{SO4-}\), changing the hydrogen count but not the charge. Losing a positive proton must decrease the charge by one unit.

Question 2

Phosphoric acid, \(\ce{H3PO4}\), can donate three protons successively.

a. Write the three successive proton-transfer equations with water.

b. Identify the conjugate base after exactly one proton has been removed.

c. Explain why \(\mathrm{PO_{4}}^{3-}\) is not the conjugate base formed after the first proton is removed.

Solution 2

The conjugate base after exactly one proton has been removed is \(\ce{H2PO4-}\).

The three reactions are:

First proton:

\[ \ce{H3PO4 + H2O <=> H3O+ + H2PO4-} \]

Second proton:

\[ \ce{H2PO4- + H2O <=> H3O+ + HPO4^2-} \]

Third proton:

\[ \ce{HPO4^2- + H2O <=> H3O+ + PO4^3-} \]

The corresponding conjugate bases are:

The tempting approach is to look at the original formula and remove all three acidic hydrogens immediately.

That gives \(\mathrm{PO_{4}}^{3-}\), but it skips two separate proton-transfer steps.

A conjugate base is the immediate product after one proton donation. It is not necessarily the final product after every possible proton has been removed.

Question 3

A student titrates 20.0 mL of \(0.150\ \mathrm{mol\,L^{-1}}\) phosphoric acid with \(0.100\ \mathrm{mol\,L^{-1}}\) sodium hydroxide.

Calculate:

a. the volume of sodium hydroxide required to reach the first equivalence point

b. the volume required to reach the second equivalence point

c. the predominant phosphate species at each point.

Solution 3

The first equivalence point requires 30.0 mL of sodium hydroxide and produces predominantly \(\ce{H2PO4-}\). The second requires 60.0 mL and produces predominantly \(\mathrm{HPO_{4}}^{2-}\).

Step 1: Calculate the initial amount of phosphoric acid.

\[ n=cV \]

Convert the volume:

\[ 20.0\ \mathrm{mL}=0.0200\ \mathrm{L} \]

Substitute:

\[ n=0.150\times0.0200 \]

\[ n=0.00300\ \mathrm{mol} \]

Step 2: Calculate the first equivalence volume.

At the first equivalence point, one mole of hydroxide is required per mole of phosphoric acid.

\[ n(\ce{OH-})=0.00300\ \mathrm{mol} \]

Therefore:

\[ V=\frac{n}{c} \]

\[ V=\frac{0.00300}{0.100} \]

\[ V=0.0300\ \mathrm{L} \]

\[ \boxed{V=30.0\ \mathrm{mL}} \]

The predominant phosphate species is \(\ce{H2PO4-}\), the conjugate base after the first proton is removed.

Step 3: Calculate the second equivalence volume.

Two moles of hydroxide are required per mole of phosphoric acid.

\[ n(\ce{OH-})=2(0.00300) \]

\[ n(\ce{OH-})=0.00600\ \mathrm{mol} \]

Therefore:

\[ V=\frac{0.00600}{0.100} \]

\[ V=0.0600\ \mathrm{L} \]

\[ \boxed{V=60.0\ \mathrm{mL}} \]

The predominant phosphate species is now \(\mathrm{HPO_{4}}^{2-}\).

The important point is that the first equivalence volume corresponds to one proton equivalent per original acid molecule, not complete removal of every acidic proton.

Question 4

Two monoprotic acids, HX and HY, are investigated at \(25^\circ\mathrm{C}\).

Each solution contains 20.0 mL of acid at a formal concentration of \(0.0500\ \mathrm{mol\,L^{-1}}\).

The following illustrative data are obtained:

Assume ideal behaviour, and that the measured hydronium ions originate from the acid dissociation.

a. Calculate the initial hydronium concentration of each solution.

b. Calculate the percentage ionisation of each acid.

c. Identify the stronger acid and predict which conjugate base is stronger.

d. Explain why both solutions require the same volume of sodium hydroxide despite having different pH values.

e. Calculate the initial amount of conjugate base in each solution.

Solution 4

HX is the stronger acid, while \(\mathrm{Y}^{-}\) is the stronger conjugate base. Both solutions contain the same total amount of acid, so they require the same amount of sodium hydroxide at equivalence.

Step 1: Calculate the hydronium concentration.

Use:

\[ [\ce{H3O+}]=10^{-\mathrm{pH}} \]

For HX:

\[ [\ce{H3O+}]=10^{-2.00} \]

\[ =0.0100\ \mathrm{mol\,L^{-1}} \]

For HY:

\[ [\ce{H3O+}]=10^{-3.00} \]

\[ =0.00100\ \mathrm{mol\,L^{-1}} \]

Step 2: Calculate the percentage ionisation.

For a monoprotic acid:

\[ \%\text{ ionisation} = \frac{[\ce{H3O+}]}{c_{\text{acid}}} \times100 \]

For HX:

\[ \frac{0.0100}{0.0500}\times100 =20.0\% \]

For HY:

\[ \frac{0.00100}{0.0500}\times100 =2.00\% \]

HX has a greater percentage ionisation at the same initial concentration, so HX is the stronger acid.

Step 3: Compare the conjugate bases.

The stronger the acid, the weaker its conjugate base.

Therefore:

• HX is the stronger acid.
• \(\mathrm{X}^{-}\) is the weaker conjugate base.
• HY is the weaker acid.
• \(\mathrm{Y}^{-}\) is the stronger conjugate base.

Remember that stronger here is a relative comparison. It does not mean that \(\mathrm{Y}^{-}\) is necessarily a strong base in absolute terms.

Step 4: Explain why the sodium hydroxide volumes are identical.

Both samples initially contain:

\[ n=cV \]

\[ n=0.0500\times0.0200 \]

\[ n=0.00100\ \mathrm{mol} \]

Each acid is monoprotic, so one mole of acid requires one mole of hydroxide for complete neutralisation.

The amount of sodium hydroxide required is therefore:

\[ n(\ce{OH-})=0.00100\ \mathrm{mol} \]

Its required volume is:

\[ V=\frac{0.00100}{0.0500} \]

\[ \boxed{V=0.0200\ \mathrm{L}=20.0\ \mathrm{mL}} \]

Both acids need the same amount of base because their total initial acid amounts are equal.

Their different strengths affect their initial equilibrium compositions, not the total number of monoprotic acid equivalents available for neutralisation.

Step 5: Calculate the initial amount of conjugate base.

For a monoprotic acid in this simplified equilibrium:

\[ [\mathrm{A}^{-}]\approx[\ce{H3O+}] \]

For HX:

\[ n(\mathrm{X}^{-})=0.0100\times0.0200 \]

\[ \boxed{n(\mathrm{X}^{-})=2.00\times10^{-4}\ \mathrm{mol}} \]

For HY:

\[ n(\mathrm{Y}^{-})=0.00100\times0.0200 \]

\[ \boxed{n(\mathrm{Y}^{-})=2.00\times10^{-5}\ \mathrm{mol}} \]

HX initially has ten times as many moles of conjugate base as HY.

This does not mean HX contains ten times as much total acid. Both samples contain the same total amount of acid. They simply have different proportions ionised at equilibrium.

That is the distinction between acid strength and amount.

Question 5

This final problem combines sequential proton removal, equilibrium, dilution, and titration.

Two flasks each contain \(2.50\times10^{-3}\) mol of phosphoric acid.

• Flask A contains 25.0 mL of solution.
• Flask B contains the same amount of phosphoric acid but has been diluted to 100.0 mL.

Both flasks are at \(25^\circ\mathrm{C}\).

Each flask receives 12.5 mL of \(0.100\ \mathrm{mol\,L^{-1}}\) sodium hydroxide. A further 12.5 mL of the same sodium hydroxide is then added to each flask.

For this question, assume:

• the first dissociation of phosphoric acid dominates over the region being considered
• the second and third dissociations can be neglected during the first neutralisation stage
• solution volumes are additive
• ideal-solution behaviour applies
• \(pK_{a1}=2.15\).

Answer the following.

a. Calculate the initial phosphoric acid concentration in each flask.

b. Calculate the amount of hydroxide added in the first 12.5 mL addition.

c. After the first addition, determine the approximate amounts of \(\ce{H3PO4}\) and \(\ce{H2PO4-}\) in each flask. Calculate their concentrations, accounting for the added titrant volume.

d. Predict the pH of each flask after the first addition. Explain why the pH values are approximately equal despite the different concentrations.

e. After the second 12.5 mL addition, identify the predominant phosphate species. Explain why both flasks require the same total amount of sodium hydroxide, despite their different initial concentrations.

Solution 5

Both flasks require 25.0 mL of sodium hydroxide to reach the first equivalence point, and both have approximately the same pH after the first 12.5 mL addition. Their concentrations are different, but their acid amounts and acid-to-conjugate-base ratios are the same.

Step 1: Calculate the initial acid concentrations.

Both flasks contain:

\[ n=0.00250\ \mathrm{mol} \]

For Flask A:

\[ c=\frac{0.00250}{0.0250} \]

\[ \boxed{c_A=0.100\ \mathrm{mol\,L^{-1}}} \]

For Flask B:

\[ c=\frac{0.00250}{0.100} \]

\[ \boxed{c_B=0.0250\ \mathrm{mol\,L^{-1}}} \]

Flask A is four times as concentrated as Flask B.

However, both contain exactly the same number of moles of phosphoric acid.

Step 2: Calculate the hydroxide added in the first addition.

Each flask receives 12.5 mL of \(0.100\ \mathrm{mol\,L^{-1}}\) sodium hydroxide.

Convert the volume:

\[ 12.5\ \mathrm{mL}=0.0125\ \mathrm{L} \]

Calculate the amount:

\[ n=cV \]

\[ n=0.100\times0.0125 \]

\[ \boxed{n(\ce{OH-})=0.00125\ \mathrm{mol}} \]

This is half the amount needed to reach the first equivalence point.

Step 3: Determine the phosphate species after the first addition.

The first neutralisation reaction is:

\[ \ce{H3PO4 + OH- -> H2PO4- + H2O} \]

The hydroxide converts phosphoric acid into its conjugate base.

Starting with 0.00250 mol of phosphoric acid and adding 0.00125 mol of hydroxide gives the approximate stoichiometric amounts:

\[ n(\ce{H3PO4})=0.00125\ \mathrm{mol} \]

\[ n(\ce{H2PO4-})=0.00125\ \mathrm{mol} \]

Both flasks therefore contain equal amounts of the acid and its conjugate base.

Step 4: Calculate the concentrations after titrant addition.

Remember that the sodium hydroxide adds volume to each flask.

For Flask A:

\[ V_{\text{final}}=25.0+12.5 \]

\[ V_{\text{final}}=37.5\ \mathrm{mL} \]

Therefore:

\[ [\ce{H3PO4}] = \frac{0.00125}{0.0375} \]

\[ =0.0333\ \mathrm{mol\,L^{-1}} \]

The conjugate base has the same concentration:

\[ [\ce{H2PO4-}] =0.0333\ \mathrm{mol\,L^{-1}} \]

For Flask B:

\[ V_{\text{final}}=100.0+12.5 \]

\[ V_{\text{final}}=112.5\ \mathrm{mL} \]

Therefore:

\[ [\ce{H3PO4}] = \frac{0.00125}{0.1125} \]

\[ =0.0111\ \mathrm{mol\,L^{-1}} \]

And:

\[ [\ce{H2PO4-}] =0.0111\ \mathrm{mol\,L^{-1}} \]

The results are:

Flask A has three times the concentration of each phosphate species after the first addition.

However, the ratio of conjugate base to acid is identical in both flasks.

Step 5: Predict the pH.

For the first dissociation:

\[ K_{a1} = \frac{[\ce{H3O+}][\ce{H2PO4-}]} {[\ce{H3PO4}]} \]

Rearrange:

\[ [\ce{H3O+}] = K_{a1} \frac{[\ce{H3PO4}]}{[\ce{H2PO4-}]} \]

Both flasks have equal concentrations of acid and conjugate base.

Therefore:

\[ [\ce{H3O+}]\approx K_{a1} \]

Taking the negative logarithm:

\[ \mathrm{pH}\approx pK_{a1} \]

\[ \boxed{\mathrm{pH}\approx2.15} \]

The pH values are approximately equal, despite the different concentrations.

Why? Because the equilibrium depends on the ratio of the acid and its conjugate base in this buffer region. Both flasks have the same ratio.

This is the important conceptual hinge: a difference in concentration does not necessarily produce a difference in pH when the acid-to-conjugate-base ratio remains the same.

Step 6: Determine what happens after the second addition.

Another 12.5 mL of sodium hydroxide supplies another:

\[ 0.00125\ \mathrm{mol\ OH-} \]

The total hydroxide added to each flask is now:

\[ 0.00125+0.00125 =0.00250\ \mathrm{mol} \]

This is one mole of hydroxide per mole of original phosphoric acid.

Both flasks have reached the first equivalence point.

The predominant phosphate species is:

\[ \boxed{\ce{H2PO4-}} \]

This is the conjugate base formed after the first proton is removed.

The second addition has completed the first proton-neutralisation stage. It has not removed a second proton from every original phosphoric acid molecule.

Step 7: Explain the difference between amount and concentration.

The tempting argument is that Flask B is more dilute, so it must require less sodium hydroxide.

But concentration is not amount.

Both flasks initially contain \(0.00250\) mol of phosphoric acid. Both therefore require \(0.00250\) mol of hydroxide to reach their first equivalence points.

The total volume of sodium hydroxide required is:

\[ V=\frac{0.00250}{0.100} \]

\[ \boxed{V=25.0\ \mathrm{mL}} \]

Dilution changes the concentration, but it does not remove any phosphoric acid from the flask.

It also does not change the intrinsic acid strength, represented by \(K_{a1}\), at the same temperature.

The distinction between the two flasks is their concentration, not their total acid amount or the identity of their conjugate base.

The most reliable habit is to identify the acid species first, remove one proton, and check the charge before thinking about strength or concentration.

Once you can do that confidently, the next useful step is understanding amphiprotic species such as \(\ce{H2PO4-}\), which can act as either an acid or a base depending on what they react with.