Why Diprotic Acids Need a 1:2 Mole Ratio in Titrations

Imagine two 25.00 mL samples, each containing 2.00 mmol of acid molecules. One contains a monoprotic acid, HA; the other contains a diprotic acid, H₂A. Both are titrated with 0.1000 mol L⁻¹ sodium hydroxide (NaOH).

Before calculating, predict which sample needs more NaOH for complete neutralisation.

The diprotic acid needs twice as much: 40.0 mL rather than 20.0 mL. Each H₂A molecule can transfer two acidic protons, so complete neutralisation requires 2 moles of OH⁻ for every 1 mole of H₂A.

There is an important catch, though. At the first equivalence point, the mole ratio is still 1:1. The 1:2 ratio applies at the second equivalence point, when both acidic protons have been neutralised.

Understanding this difference is the key to getting diprotic acid titration calculations right.

01Why a diprotic acid needs two moles of NaOH

Start with the general formula H₂A.

• H represents an acidic hydrogen that can be donated as a proton, H⁺.
• A represents the remaining part of the acid molecule.
• The subscript 2 tells us that each molecule can donate two acidic protons.

When NaOH dissolves in water, it provides hydroxide ions, OH⁻. Each hydroxide ion can accept one proton and form water.

This means neutralisation happens in two steps.

First neutralisation:

\[
\ce{H2A + OH^- -> HA^- + H2O}
\]

One hydroxide ion removes one acidic proton. The diprotic acid becomes HA⁻.

But notice something important: HA⁻ still contains one acidic proton.

Second neutralisation:

\[
\ce{HA^- + OH^- -> A^2- + H2O}
\]

A second hydroxide ion removes the remaining proton, producing A²⁻.

Combining both reactions gives the overall equation:

\[
\ce{H2A + 2OH^- -> A^2- + 2H2O}
\]

Read the coefficients carefully.

One mole of H₂A reacts with two moles of OH⁻. Therefore, the mole ratio for complete neutralisation is:

\[
\boxed{n(\ce{H2A}):n(\mathrm{OH}^{-})=1:2}
\]

This is the fundamental reason a diprotic acid needs a different mole ratio from a monoprotic acid.

For comparison, a monoprotic acid reacts according to:

\[
\ce{HA + OH^- -> A^- + H2O}
\]

Its mole ratio is 1:1 because each acid molecule can donate only one proton.

02The first and second equivalence points

This is where students often make mistakes.

A diprotic acid can undergo two separate neutralisation stages. Each stage has its own equivalence point.

An equivalence point occurs when the amount of titrant added matches the amount required by the balanced reaction stoichiometry.

For a diprotic acid, there are two relevant equivalence points.

First equivalence point

At the first equivalence point, one mole of OH⁻ has reacted for every mole of H₂A.

\[
n(\mathrm{OH}^{-})=n(\ce{H2A})
\]

The formal product of the first neutralisation is HA⁻.

Only one of the two acidic protons has been neutralised.

Second equivalence point

At the second equivalence point, two moles of OH⁻ have reacted for every mole of H₂A.

\[
n(\mathrm{OH}^{-})=2n(\ce{H2A})
\]

Both acidic protons have now been neutralised, giving A²⁻ as the formal product.

The difference is summarised below.

These are stoichiometric stages. Actual solutions contain acid-base equilibrium mixtures, so the species are not necessarily present in perfectly pure forms.

How the equivalence volumes are related

Suppose a diprotic acid is titrated using NaOH of constant concentration.

Let:

• \(V_1\) = total NaOH volume at the first equivalence point.
• \(V_2\) = total NaOH volume at the second equivalence point.

The first equivalence point requires one mole of NaOH per mole of acid.

The second requires two moles of NaOH per mole of acid.

Therefore:

\[
\boxed{V_2=2V_1}
\]

This relationship assumes the same acid sample and the same NaOH concentration throughout the titration.

For example, if the first equivalence point occurs at 15.00 mL, the second occurs at 30.00 mL.

Notice that the second equivalence volume is 30.00 mL in total. It is not another 30.00 mL added after the first equivalence point.

The additional volume between the two points is:

\[
30.00-15.00=15.00\text{ mL}
\]

The second proton requires the same number of moles of OH⁻ as the first.

Why the equivalence points do not necessarily have pH 7

It is tempting to assume that neutralisation always produces a pH of 7.

That is not correct.

At the first equivalence point, HA⁻ is amphiprotic. This means it can either donate a proton or accept one.

Its pH depends on the relative strengths of the acid dissociation steps, represented by their acid dissociation constants, \(K_{a1}\) and \(K_{a2}\).

At the second equivalence point, A²⁻ is present as the formal product. It can react with water to produce hydroxide ions:

\[
\ce{A^2- + H2O <=> HA^- + OH^-}
\]

This can make the solution basic.

The equivalence point is determined by stoichiometry, not by whether the pH reaches exactly 7.

There is another complication: not every diprotic acid produces two clearly separated jumps on a titration curve. If the two acid dissociation steps overlap, the curve may show only one obvious steep region.

A single visible jump does not automatically mean the acid is monoprotic.

03What does the titration experiment actually measure?

To understand the measurements, picture the apparatus.

A measured volume of diprotic acid is transferred into a conical flask using a pipette. An indicator is added, or a pH probe is placed in the solution.

The burette above the flask contains a standardised NaOH solution.

When the tap is opened, NaOH flows into the flask. The flask is swirled to mix the reactants. The volume delivered is calculated by subtracting the initial burette reading from the final burette reading.

The experiment measures how much base is required to reach a particular reaction stage.

It does not directly count the number of acidic protons in each molecule.

What should the titration show?

If both neutralisation stages are sufficiently separated, a pH probe should detect two distinct changes in the titration curve.

The first occurs when one proton per acid molecule has been neutralised.

The second occurs when both protons have been neutralised.

For an illustrative titration, the first equivalence point might occur at 16.40 mL, with the second at 32.80 mL.

The second volume is twice the first, consistent with the two-stage neutralisation.

However, an experiment using only one indicator may fail to identify both stages. The indicator changes colour within a particular pH range, and that range might capture the final equivalence point without clearly identifying the first.

A better approach is to use a calibrated pH probe and add small, measured amounts of NaOH. Record the pH after each addition and plot pH against the cumulative volume of NaOH.

This can reveal two separate inflection regions if the chemistry allows them to be distinguished.

If the two neutralisation stages overlap, even a carefully measured pH curve may not show two clear equivalence points. That is a limitation of the chemical system, not necessarily an experimental mistake.

It is also useful to distinguish the observed indicator endpoint from the theoretical equivalence point. They are related, but they are not identical. See Endpoint vs Equivalence Point: The Difference in Titrations.

04How to calculate the mole ratio from titration data

Once you know which equivalence point the question refers to, the calculation becomes straightforward.

Start with the standard relationship:

\[
n=cV
\]

where:

• \(n\) is the amount of substance in mol.
• \(c\) is the concentration in mol L⁻¹.
• \(V\) is the volume in L.

For the following calculations, assume the acid is pure, NaOH is standardised, and the neutralisation reaction proceeds quantitatively to the specified equivalence point.

Let:

• \(c_b\) = concentration of NaOH.
• \(V_a\) = initial volume of acid.
• \(V_1\) = NaOH volume at the first equivalence point.
• \(V_2\) = total NaOH volume at the second equivalence point.

All volumes used in \(n=cV\) must be in litres.

At the first equivalence point

The mole ratio is 1:1.

Therefore:

\[
n(\ce{H2A})=n(\mathrm{OH}^{-})
\]

And:

\[
n(\ce{H2A})=c_bV_1
\]

At the second equivalence point

The mole ratio is 1:2.

Therefore:

\[
n(\mathrm{OH}^{-})=2n(\ce{H2A})
\]

Rearranging:

\[
\boxed{n(\ce{H2A})=\frac{c_bV_2}{2}}
\]

Once the moles of acid are known, calculate its concentration:

\[
\boxed{c_a=\frac{c_bV_2}{2V_a}}
\]

Here, \(c_a\) is the concentration of the diprotic acid.

The factor of 2 is essential. Leaving it out would double your calculated acid concentration.

A quick comparison of the mole ratios

The key decision is whether the question gives you the first equivalence point or the second.

05Questions and solutions

Question 1

A 20.00 mL sample of 0.0600 mol L⁻¹ diprotic acid, H₂A, is titrated with 0.1000 mol L⁻¹ NaOH.

Calculate the volume of NaOH required to reach the second equivalence point.

Solution 1

The required NaOH volume is 24.0 mL.

Step 1: Calculate the moles of acid.

\[
n(\ce{H2A})=cV
\]

\[
=0.0600\times0.02000
\]

\[
=0.001200\text{ mol}
\]

Step 2: Apply the 1:2 mole ratio.

\[
n(\mathrm{OH}^{-})=2\times0.001200
\]

\[
=0.002400\text{ mol}
\]

Step 3: Calculate the NaOH volume.

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

\[
=\frac{0.002400}{0.1000}
\]

\[
=0.02400\text{ L}
\]

\[
\boxed{V=24.0\text{ mL}}
\]

The acid requires twice as many moles of NaOH as there are moles of acid molecules because both acidic protons must be neutralised.

Question 2

A 25.00 mL aliquot of diprotic acid H₂A is titrated with 0.0800 mol L⁻¹ NaOH.

The titration curve shows two equivalence points:

• First equivalence point: 18.60 mL.
• Second equivalence point: 37.20 mL.

Calculate:

(a) The concentration of H₂A.

(b) The number of moles of H₂A in the original 250.0 mL stock solution from which the aliquot was taken.

Solution 2

The acid concentration is 0.0595 mol L⁻¹, and the original stock solution contains 0.0149 mol of H₂A.

Step 1: Calculate the moles of NaOH at the second equivalence point.

Use the total volume of 37.20 mL:

\[
n(\mathrm{OH}^{-})=cV
\]

\[
=0.0800\times0.03720
\]

\[
=0.002976\text{ mol}
\]

Step 2: Calculate the moles of H₂A.

At the second equivalence point, the ratio is 1:2.

\[
n(\ce{H2A})=\frac{0.002976}{2}
\]

\[
=0.001488\text{ mol}
\]

Step 3: Calculate the acid concentration.

\[
c_a=\frac{n}{V_a}
\]

\[
=\frac{0.001488}{0.02500}
\]

\[
\boxed{c_a=0.0595\text{ mol L}^{-1}}
\]

Step 4: Calculate the moles in the original stock solution.

The stock volume is 250.0 mL, or 0.2500 L.

\[
n=cV
\]

\[
=0.05952\times0.2500
\]

\[
\boxed{n=0.0149\text{ mol}}
\]

The original stock solution contains 0.0149 mol of H₂A.

Important check: The first equivalence point gives:

\[
n(\mathrm{OH}^{-})=0.0800\times0.01860
\]

\[
=0.001488\text{ mol}
\]

This is exactly the same number of moles of acid calculated using the second equivalence point.

Both equivalence points must give a consistent result.

Question 3

An illustrative titration curve for a diprotic acid shows:

• First equivalence point at 12.50 mL.
• Second equivalence point at 25.00 mL.

A student makes the following claim:

“At both equivalence points, the pH must be 7 because the acid has been neutralised.”

Explain why this statement is incorrect. In your answer, describe the chemical species present at each equivalence point and explain what determines the pH.

Solution 3

The student’s statement is incorrect. Both equivalence points represent specific stoichiometric amounts of NaOH, but neither is automatically at pH 7.

Step 1: Identify what happens at the first equivalence point.

At 12.50 mL, one mole of OH⁻ has reacted per mole of H₂A.

The formal reaction is:

\[
\ce{H2A + OH^- -> HA^- + H2O}
\]

The acid has lost one acidic proton, but HA⁻ still has another proton that can be donated.

HA⁻ is amphiprotic, meaning it can behave as either an acid or a base.

Its pH depends on the relative acid strengths of the two dissociation steps, not simply on the fact that the first equivalence point has been reached.

Step 2: Identify what happens at the second equivalence point.

At 25.00 mL, two moles of OH⁻ have reacted per mole of H₂A.

The formal product is A²⁻.

This ion can react with water to produce OH⁻:

\[
\ce{A^2- + H2O <=> HA^- + OH^-}
\]

As a result, the solution may be basic.

Step 3: Explain the misconception.

The student has confused complete stoichiometric neutralisation with a neutral pH.

Equivalence describes the amount of titrant required according to the balanced chemical equation.

The pH depends on the acid-base properties of the species remaining in solution.

A pH of 7 is not a universal requirement for an equivalence point.

Question 4

A 25.00 mL sample contains a mixture of two acids:

• Monoprotic acid HA, containing 0.600 mmol of HA.
• Diprotic acid H₂B, with an unknown concentration.

The mixture is titrated with 0.1000 mol L⁻¹ NaOH until both acids are completely neutralised.

The total volume of NaOH required is 22.00 mL.

Assume both neutralisation reactions proceed quantitatively and no other substances consume OH⁻.

Calculate the concentration of H₂B.

Solution 4

The concentration of H₂B is 0.0320 mol L⁻¹.

The important point is that the two acids do not consume the same number of hydroxide ions per molecule.

HA consumes one OH⁻ per molecule, while H₂B consumes two.

Step 1: Calculate the total moles of NaOH added.

\[
n(\mathrm{OH}^{-})=cV
\]

\[
=0.1000\times0.02200
\]

\[
=0.002200\text{ mol}
\]

This is 2.200 mmol of OH⁻.

Step 2: Calculate the amount of OH⁻ consumed by HA.

The reaction is:

\[
\ce{HA + OH^- -> A^- + H2O}
\]

The ratio is 1:1.

Therefore, 0.600 mmol of HA consumes 0.600 mmol of OH⁻.

Step 3: Calculate the OH⁻ available for H₂B.

\[
n(\mathrm{OH}^{-})_{\ce{H2B}}=2.200-0.600
\]

\[
=1.600\text{ mmol}
\]

Step 4: Calculate the moles of H₂B.

H₂B requires two moles of OH⁻ per mole of acid.

\[
n(\ce{H2B})=\frac{1.600}{2}
\]

\[
=0.800\text{ mmol}
\]

Convert to moles:

\[
n(\ce{H2B})=0.000800\text{ mol}
\]

Step 5: Calculate the concentration of H₂B.

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

\[
=\frac{0.000800}{0.02500}
\]

\[
\boxed{c(\ce{H2B})=0.0320\text{ mol L}^{-1}}
\]

The trap is assuming that the total moles of NaOH equal the total moles of acid molecules.

They do not.

A titration of a mixture measures the total neutralisable proton equivalents. To calculate the individual amounts of different acids, you need additional information about their identities, amounts, or reaction stages.

Question 5

A 25.00 mL aliquot of a weak diprotic acid solution contains 2.00 mmol of H₂A, as established independently.

Before titration, a pH measurement gives:

\[
[\ce{H3O+}]=1.0\times10^{-3}\text{ mol L}^{-1}
\]

The aliquot is transferred into a conical flask and diluted with 75.00 mL of distilled water.

The diluted solution is then titrated with 0.1000 mol L⁻¹ NaOH.

An illustrative pH trace shows one clear, steep rise at 40.00 mL of added NaOH. No separate sharp rise is resolved at 20.00 mL.

A student makes two claims:

1. The initial pH tells us how many moles of H⁺ are available, so only that amount of NaOH is required.
2. There is only one clear equivalence-point jump, so the acid must be monoprotic.

Evaluate both claims.

Calculate the theoretical first and second equivalence volumes, and explain why the experimental observations do not contradict the acid being diprotic.

Solution 5

Both claims are incorrect.

The theoretical first equivalence volume is 20.0 mL, and the second equivalence volume is 40.0 mL. The initial pH measures the concentration of free hydronium ions, not the total number of acidic protons available for neutralisation. Dilution changes the concentration but does not change the number of acid molecules.

This question brings together stoichiometry, acid dissociation equilibrium, dilution, and the interpretation of titration curves.

Step 1: Identify the total amount of diprotic acid.

We are told that the original aliquot contains:

\[
n(\ce{H2A})=2.00\text{ mmol}
\]

Each molecule can donate two acidic protons.

Therefore, complete neutralisation requires:

\[
n(\mathrm{OH}^{-})=2\times2.00
\]

\[
=4.00\text{ mmol}
\]

This is the total amount of hydroxide needed at the second equivalence point.

Step 2: Calculate the first equivalence volume.

At the first equivalence point, one mole of OH⁻ is required per mole of H₂A.

Therefore:

\[
n(\mathrm{OH}^{-})=2.00\times10^{-3}\text{ mol}
\]

Using \(n=cV\):

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

\[
=\frac{2.00\times10^{-3}}{0.1000}
\]

\[
=0.0200\text{ L}
\]

\[
\boxed{V_1=20.0\text{ mL}}
\]

Step 3: Calculate the second equivalence volume.

At the second equivalence point, two moles of OH⁻ are required per mole of H₂A.

\[
n(\mathrm{OH}^{-})=4.00\times10^{-3}\text{ mol}
\]

Therefore:

\[
V_2=\frac{4.00\times10^{-3}}{0.1000}
\]

\[
=0.0400\text{ L}
\]

\[
\boxed{V_2=40.0\text{ mL}}
\]

The second equivalence point occurs at 40.0 mL, matching the location of the observed steep rise.

Step 4: Examine the initial pH measurement.

The student uses the initial hydronium concentration to calculate the amount of free hydronium ions.

Before dilution, the original acid volume is:

\[
V=0.02500\text{ L}
\]

Therefore:

\[
n(\ce{H3O+})=[\ce{H3O+}]V
\]

\[
=1.0\times10^{-3}\times0.02500
\]

\[
=2.50\times10^{-5}\text{ mol}
\]

Converting to mmol:

\[
n(\ce{H3O+})=0.0250\text{ mmol}
\]

If the student neutralised only this amount, the predicted NaOH volume would be:

\[
V=\frac{2.50\times10^{-5}}{0.1000}
\]

\[
=0.000250\text{ L}
\]

\[
=0.250\text{ mL}
\]

That is nowhere near the 40.0 mL required for complete neutralisation.

Why?

Because the initial pH measures the hydronium ions already present in solution. It does not measure all the acidic protons still attached to H₂A and HA⁻.

A weak acid only partially dissociates in water.

When OH⁻ reacts with hydronium ions, it removes them from solution. This causes the acid dissociation equilibria to shift, allowing more acid molecules to release protons.

The process continues as NaOH is added.

Consequently, the total amount of base required is determined by the total number of neutralisable acidic protons, not just the initial concentration of free hydronium ions.

In this example, complete neutralisation requires 4.00 mmol of OH⁻, which is 160 times the initial amount of free hydronium.

Step 5: Explain why dilution does not change the equivalence volumes.

The original 25.00 mL acid sample contains 2.00 mmol of H₂A.

Adding 75.00 mL of water increases the total solution volume to:

\[
25.00+75.00=100.00\text{ mL}
\]

The acid concentration changes.

Before dilution:

\[
c=\frac{0.00200}{0.02500}
\]

\[
=0.0800\text{ mol L}^{-1}
\]

After dilution:

\[
c=\frac{0.00200}{0.10000}
\]

\[
=0.0200\text{ mol L}^{-1}
\]

However, the amount of H₂A remains 2.00 mmol.

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

The same number of moles of OH⁻ is therefore required to reach either equivalence point. The theoretical equivalence volumes remain 20.0 mL and 40.0 mL.

Step 6: Explain why only one clear pH jump can appear.

The absence of a separately resolved jump at 20.0 mL does not prove that the acid is monoprotic.

The two acid dissociation steps may overlap, making the first equivalence point difficult to identify on the pH curve.

The single clear rise at 40.0 mL is consistent with the second equivalence point, where both acidic protons have been neutralised.

The independently established amount of H₂A confirms the 1:2 stoichiometry.

The student has confused the number of visible pH changes with the number of acidic protons in each molecule. They are not necessarily the same.

The essential decision in diprotic acid titrations is to identify the reaction stage before using the volume data.

At the first equivalence point, use a 1:1 acid-to-base ratio. At the second equivalence point, use a 1:2 ratio.

Once the correct mole ratio is established, you can identify which species are present in the flask and move on to calculating pH. Stoichiometry tells you how much NaOH has reacted; acid-base equilibrium determines the pH at that stage.