How to Calculate an Unknown Acid Concentration from Titration Data

A 25.00 mL sample of an unknown acid is titrated with sodium hydroxide of known concentration. The pH rises slowly, then suddenly jumps. You record the volume of sodium hydroxide needed to reach this point.

Can you now calculate the acid concentration?

Yes, but there is one decision to make before touching your calculator: what is the mole ratio between the acid and the base?

A monoprotic acid reacts with hydroxide ions in a 1:1 ratio. A diprotic acid can require twice as many moles of hydroxide for complete neutralisation. Using the wrong ratio gives you the wrong concentration, even if every other calculation is perfect.

The method is straightforward:

1. Write the balanced neutralisation equation.
2. Identify the equivalence point and its corresponding titrant volume.
3. Calculate the moles of the known solution using \(n=cV\).
4. Apply the mole ratio to find the moles of acid.
5. Divide by the original acid sample volume to find its concentration.

For a monoprotic acid, the calculation simplifies to:

\[
c_{\mathrm{acid}}=
\frac{c_{\mathrm{base}}V_{\mathrm{eq}}}
{V_{\mathrm{acid}}}
\]

For a diprotic acid at its second equivalence point:

\[
c_{\mathrm{acid}}=
\frac{c_{\mathrm{base}}V_{\mathrm{eq}}}
{2V_{\mathrm{acid}}}
\]

The difference is the factor of two. We’ll work through why it appears, how to identify the correct volume from experimental data, and what happens when the pH curve doesn’t behave as expected.

01Understand what the titration is measuring

In an acid-base titration, you use a solution of accurately known concentration to determine the concentration of an unknown solution.

There are two important solutions:

• Titrant: the solution of known concentration, usually delivered from a burette.
• Analyte: the solution being analysed, usually measured into a conical flask using a pipette.

The measured portion of the analyte is called an aliquot.

Imagine adding sodium hydroxide to an unknown hydrochloric acid solution.

The reaction is:

\[
\ce{HCl(aq) + NaOH(aq) -> NaCl(aq) + H2O(l)}
\]

Every mole of hydrochloric acid reacts with one mole of sodium hydroxide.

If you know the concentration and volume of sodium hydroxide required to reach equivalence, you can calculate how many moles of hydrochloric acid were originally present.

That is the central measurement: the titration determines the amount of acid in your aliquot by measuring how much base it takes to neutralise it.

What is the equivalence point?

The equivalence point is the stage at which the acid and base have reacted in exactly the mole ratio specified by their balanced chemical equation.

For hydrochloric acid and sodium hydroxide:

\[
n(\ce{HCl})=n(\ce{NaOH})
\]

For a different reaction, the relationship may be different.

The equivalence point is a stoichiometric condition. It is not simply a particular pH.

There is also an important distinction between two experimental terms:

A suitable indicator changes colour close to the equivalence point. The two points are intended to coincide as closely as possible, but they are not identical by definition.

02Find the correct volume from your titration data

Before calculating concentration, you need the volume of titrant delivered at equivalence.

There are two common ways this information appears in HSC-style questions.

Method A: Using burette readings

A burette records the volume of solution that has passed through its tap.

The volume delivered is:

\[
V_{\mathrm{titre}}=V_{\mathrm{final}}-V_{\mathrm{initial}}
\]

Notice the subtraction order.

Burette readings increase as the liquid level falls, so the final reading is usually larger than the initial reading.

Consider the following illustrative results.

Trials 1, 2, and 3 are closely grouped. They are concordant titres.

Calculate their mean:

\[
\begin{aligned}
V_{\mathrm{mean}}
&=\frac{24.98+25.02+25.00}{3}\\
&=25.00\ \mathrm{mL}
\end{aligned}
\]

The rough titre is excluded because it is used to estimate where the endpoint occurs, not to provide another precise measurement.

Use the mean of your concordant titres as the experimental titre.

Common mistake: Using the final burette reading as the titre.

If the initial reading is 0.15 mL and the final reading is 25.13 mL, the volume delivered is 24.98 mL, not 25.13 mL.

Method B: Using a pH titration curve

Sometimes the experiment uses a pH probe instead of an indicator.

The probe records the pH as increasing volumes of base are added. You can then plot pH against the volume of titrant.

The curve usually has three recognisable regions:

1. Before equivalence, the original acid is in excess.
2. Near equivalence, the pH changes rapidly.
3. After equivalence, the added base is in excess.

The equivalence point is located at the inflection point of the curve, corresponding to its steepest region.

Here is the important detail: the equivalence point does not always occur at pH 7.

For a strong acid reacting with a strong base, the equivalence point is approximately pH 7 at 25°C.

For a weak acid reacting with a strong base, the equivalence point is above pH 7.

For a strong acid reacting with a weak base, it is below pH 7.

Why?

Consider a weak acid, represented by \(\ce{HA}\), reacting with sodium hydroxide.

At equivalence, most of the original acid has been converted into its conjugate base, \(\mathrm{A}^{-}\).

This conjugate base reacts with water:

\[
\ce{A^-(aq) + H2O(l) <=> HA(aq) + OH^-(aq)}
\]

The production of hydroxide ions makes the solution basic.

So a weak acid titrated with a strong base can reach equivalence at a pH greater than 7.

A student who automatically selects pH 7 as the equivalence point will identify the wrong volume.

What does pH actually tell you?

This is another important distinction.

pH measures the hydrogen ion concentration of a solution at a particular moment. It does not directly tell you the total amount of acid originally present.

A weak acid can have a relatively high concentration while producing a much smaller concentration of hydrogen ions because it only partially ionises.

The titration solves this problem by allowing the acid to react with a known amount of base.

You use the pH curve to locate equivalence. You use the balanced equation and the titrant concentration to calculate the original acid concentration.

03Use the balanced equation to establish the mole ratio

Once you have the equivalence volume, the next job is identifying the reaction stoichiometry.

This is where many otherwise correct calculations go wrong.

Consider two different acids.

Monoprotic acids: a 1:1 mole ratio

A monoprotic acid can donate one proton per formula unit in its complete neutralisation.

For example, ethanoic acid reacts with hydroxide ions:

\[
\ce{CH3COOH(aq) + OH^-(aq) -> CH3COO^-(aq) + H2O(l)}
\]

The mole ratio is:

\[
\ce{CH3COOH}:\mathrm{OH}^{-}=1:1
\]

One mole of ethanoic acid reacts with one mole of hydroxide ions.

The same stoichiometric ratio applies to other monoprotic acids, including hydrochloric acid and nitric acid, when completely neutralised by a strong base.

Diprotic acids: a 1:2 mole ratio

A diprotic acid can donate two protons per formula unit.

Consider a generic diprotic acid, \(\ce{H2A}\).

Its complete neutralisation is:

\[
\ce{H2A(aq) + 2OH^-(aq) -> A^2-(aq) + 2H2O(l)}
\]

The mole ratio is:

\[
\ce{H2A}:\mathrm{OH}^{-}=1:2
\]

One mole of diprotic acid requires two moles of hydroxide ions.

This is why you cannot automatically assume that the acid and base have equal numbers of moles at equivalence.

The balanced equation is your recipe. Its coefficients tell you how many moles of each substance are required, regardless of the volumes or concentrations of the solutions.

The general stoichiometric relationship

For a reaction:

\[
aA+bB\rightarrow\text{products}
\]

At equivalence:

\[
\frac{n(A)}{a}=\frac{n(B)}{b}
\]

The coefficients \(a\) and \(b\) are taken from the balanced equation.

Since the amount of substance is:

\[
n=cV
\]

you can substitute concentration and volume:

\[
\frac{c_AV_A}{a}
=
\frac{c_BV_B}{b}
\]

This relationship works for any suitable titration reaction, provided the measured endpoint corresponds to the reaction represented by the equation.

For HSC calculations, working through the moles explicitly is often safer than memorising a shortcut.

04Calculate the unknown acid concentration

Now we can connect the experimental volume to the unknown concentration.

The two equations you need are:

\[
n=cV
\]

and

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

where:

• \(n\) is the amount of substance, measured in mol.
• \(c\) is the molar concentration, measured in \(\mathrm{mol\,L^{-1}}\).
• \(V\) is the solution volume, measured in L.

Watch the units: if you are calculating moles using \(n=cV\), convert millilitres to litres by dividing by 1000.

For example:

\[
25.00\ \mathrm{mL}=0.02500\ \mathrm{L}
\]

You can also use both volumes in mL in a concentration-ratio equation, provided they are expressed in the same unit. However, converting to litres when calculating moles is the clearest approach.

05Common mistakes when calculating an unknown acid concentration

Before attempting exam questions, check these mistakes. They can turn a perfectly good experimental result into an incorrect answer.

One misconception deserves a closer look.

A weak acid is not necessarily only partly neutralised when titrated with a strong base.

For example, ethanoic acid is only partially ionised in water:

\[
\ce{CH3COOH(aq) <=> H^+(aq) + CH3COO^-(aq)}
\]

A student might think that only the ethanoic acid molecules that have already ionised can react with hydroxide ions.

But as hydroxide removes hydrogen ions, the equilibrium shifts to the right. More ethanoic acid ionises, allowing further reaction.

The overall neutralisation is:

\[
\ce{CH3COOH(aq) + OH^-(aq) -> CH3COO^-(aq) + H2O(l)}
\]

The 1:1 stoichiometric relationship still applies.

Weakness affects the equilibrium and the pH curve. It does not change the balanced reaction’s mole ratio.

06A reliable decision process for any titration calculation

When an exam gives you unfamiliar titration data, use this sequence.

First: identify the acid and base.

Determine whether the acid is monoprotic, diprotic, or has more ionisable protons.

Second: balance the reaction.

Do not assume a 1:1 mole ratio.

Third: identify the correct equivalence volume.

• For burette data, calculate the titre and use the mean of concordant results.
• For a pH curve, locate the inflection point.
• For a polyprotic acid, check whether more than one equivalence point is present.

Fourth: calculate the known reagent’s moles.

Use:

\[
n=cV
\]

Make sure the volume is in litres.

Fifth: convert using the mole ratio.

Use the coefficients from the balanced equation.

Finally: calculate the unknown concentration.

Divide the unknown acid’s moles by the original aliquot volume.

The final volume must be the volume of the original acid sample, not the combined volume of acid and titrant.

These assumptions should also hold: the titrant concentration is known accurately, the reaction proceeds sufficiently to completion, the endpoint is close to equivalence, and no other substances in the sample consume the titrant.

07Questions and solutions

The following questions build from routine concentration calculations to more demanding interpretations of titration data. Try each question before reading its solution.

Question 1

A 20.00 mL sample of nitric acid, \(\ce{HNO3}\), is titrated with \(0.1250\ \mathrm{mol\,L^{-1}}\) potassium hydroxide.

The initial burette reading is 0.35 mL, and the final burette reading is 20.45 mL.

The reaction is:

\[
\ce{HNO3(aq) + KOH(aq) -> KNO3(aq) + H2O(l)}
\]

Calculate the concentration of the nitric acid.

Solution 1

The nitric acid concentration is \(0.1256\ \mathrm{mol\,L^{-1}}\).

Start by calculating the actual volume of potassium hydroxide delivered.

\[
\begin{aligned}
V_{\mathrm{titre}}
&=V_{\mathrm{final}}-V_{\mathrm{initial}}\\
&=20.45-0.35\\
&=20.10\ \mathrm{mL}
\end{aligned}
\]

Convert to litres:

\[
V=0.02010\ \mathrm{L}
\]

Calculate the moles of potassium hydroxide:

\[
\begin{aligned}
n(\ce{KOH})
&=cV\\
&=0.1250\times0.02010\\
&=2.5125\times10^{-3}\ \mathrm{mol}
\end{aligned}
\]

The balanced equation gives a 1:1 mole ratio.

Therefore:

\[
n(\ce{HNO3})=2.5125\times10^{-3}\ \mathrm{mol}
\]

The original acid volume is:

\[
20.00\ \mathrm{mL}=0.02000\ \mathrm{L}
\]

Calculate the concentration:

\[
\begin{aligned}
c(\ce{HNO3})
&=\frac{n}{V}\\
&=\frac{2.5125\times10^{-3}}{0.02000}\\
&=0.125625\ \mathrm{mol\,L^{-1}}
\end{aligned}
\]

To four significant figures:

\[
\boxed{c(\ce{HNO3})=0.1256\ \mathrm{mol\,L^{-1}}}
\]

The important detail here is that the initial burette reading was not zero. Using the final reading alone would have overestimated the amount of potassium hydroxide delivered.

Question 2

A 25.00 mL sample of an unknown weak monoprotic acid, \(\ce{HA}\), is titrated with \(0.08000\ \mathrm{mol\,L^{-1}}\) sodium hydroxide.

The pH probe records the following results near the equivalence point.

The steepest part of the titration curve is centred at 20.00 mL.

Calculate the concentration of the unknown acid.

Explain why using the volume corresponding to pH 7.0 would produce an incorrect equivalence volume.

Solution 2

The acid concentration is \(0.06400\ \mathrm{mol\,L^{-1}}\), and the equivalence volume is 20.00 mL rather than 19.95 mL.

The balanced neutralisation equation is:

\[
\ce{HA(aq) + OH^-(aq) -> A^-(aq) + H2O(l)}
\]

This is a 1:1 reaction.

The inflection point occurs at 20.00 mL, so:

\[
V_{\mathrm{eq}}=0.02000\ \mathrm{L}
\]

Calculate the moles of sodium hydroxide:

\[
\begin{aligned}
n(\mathrm{OH}^{-})
&=cV\\
&=0.08000\times0.02000\\
&=1.600\times10^{-3}\ \mathrm{mol}
\end{aligned}
\]

Because the mole ratio is 1:1:

\[
n(\ce{HA})=1.600\times10^{-3}\ \mathrm{mol}
\]

The original acid volume is 0.02500 L.

Therefore:

\[
\begin{aligned}
c(\ce{HA})
&=\frac{1.600\times10^{-3}}{0.02500}\\
&=0.06400\ \mathrm{mol\,L^{-1}}
\end{aligned}
\]

The pH reaches 7.0 at 19.95 mL, but that does not identify the equivalence point.

At equivalence, the conjugate base \(\mathrm{A}^{-}\) reacts with water and produces hydroxide ions. The resulting solution is basic.

This is why the equivalence point occurs above pH 7 for a weak acid titrated with a strong base.

The tempting approach is to use pH 7 as a universal neutralisation target. The correct approach is to identify the inflection point and use the balanced reaction.

Question 3

A 20.00 mL sample of an unknown diprotic acid, \(\ce{H2A}\), is titrated with \(0.08000\ \mathrm{mol\,L^{-1}}\) sodium hydroxide.

A pH probe identifies two equivalence points:

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

Calculate the original concentration of the diprotic acid using both equivalence points.

Explain why the two calculations should give the same result.

Solution 3

The acid concentration is \(0.05000\ \mathrm{mol\,L^{-1}}\), regardless of which equivalence point is used, provided the correct mole ratio is applied.

The overall neutralisation equation is:

\[
\ce{H2A(aq) + 2NaOH(aq) -> Na2A(aq) + 2H2O(l)}
\]

The first equivalence point represents neutralisation of the first proton.

The second equivalence point represents neutralisation of both protons.

Using the first equivalence point

At the first equivalence point, one mole of sodium hydroxide has reacted per mole of diprotic acid.

Therefore:

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

Calculate the moles of sodium hydroxide:

\[
\begin{aligned}
n(\ce{NaOH})
&=0.08000\times0.01250\\
&=1.000\times10^{-3}\ \mathrm{mol}
\end{aligned}
\]

Therefore:

\[
n(\ce{H2A})=1.000\times10^{-3}\ \mathrm{mol}
\]

Calculate the acid concentration:

\[
\begin{aligned}
c(\ce{H2A})
&=\frac{1.000\times10^{-3}}{0.02000}\\
&=0.05000\ \mathrm{mol\,L^{-1}}
\end{aligned}
\]

Using the second equivalence point

At the second equivalence point, two moles of sodium hydroxide have reacted per mole of diprotic acid.

First calculate the moles of sodium hydroxide:

\[
\begin{aligned}
n(\ce{NaOH})
&=0.08000\times0.02500\\
&=2.000\times10^{-3}\ \mathrm{mol}
\end{aligned}
\]

Now apply the 1:2 ratio:

\[
\begin{aligned}
n(\ce{H2A})
&=\frac{2.000\times10^{-3}}{2}\\
&=1.000\times10^{-3}\ \mathrm{mol}
\end{aligned}
\]

Calculate the concentration:

\[
\begin{aligned}
c(\ce{H2A})
&=\frac{1.000\times10^{-3}}{0.02000}\\
&=0.05000\ \mathrm{mol\,L^{-1}}
\end{aligned}
\]

Both methods give the same concentration.

The reason is that the first equivalence point measures the first proton, while the second measures both. The volume of base doubles, but the number of moles of acid in the original aliquot remains unchanged.

The trap is applying the 1:2 ratio at the first equivalence point or the 1:1 ratio at the second. Both would produce an incorrect result.

Question 4

A 20.00 mL sample of an unknown weak monoprotic acid, \(\ce{HA}\), is titrated with \(0.1000\ \mathrm{mol\,L^{-1}}\) sodium hydroxide.

The equivalence point occurs after 30.00 mL of sodium hydroxide has been added.

At exactly half the equivalence volume, the measured pH is 4.76.

Determine:

(a) The original concentration of the acid.

(b) The acid dissociation constant, \(K_a\).

Explain why the pH measurement at half-equivalence provides information about acid strength rather than directly determining the original acid concentration.

Solution 4

The acid concentration is \(0.1500\ \mathrm{mol\,L^{-1}}\), and its acid dissociation constant is approximately \(1.74\times10^{-5}\).

There are two separate pieces of information in this question.

The equivalence volume determines the acid concentration, while the half-equivalence pH provides information about its acid dissociation constant.

(a) Calculate the acid concentration.

The reaction is:

\[
\ce{HA(aq) + OH^-(aq) -> A^-(aq) + H2O(l)}
\]

The mole ratio is 1:1.

Convert the equivalence volume:

\[
30.00\ \mathrm{mL}=0.03000\ \mathrm{L}
\]

Calculate the moles of sodium hydroxide:

\[
\begin{aligned}
n(\ce{NaOH})
&=cV\\
&=0.1000\times0.03000\\
&=3.000\times10^{-3}\ \mathrm{mol}
\end{aligned}
\]

Therefore:

\[
n(\ce{HA})=3.000\times10^{-3}\ \mathrm{mol}
\]

The original acid volume is:

\[
20.00\ \mathrm{mL}=0.02000\ \mathrm{L}
\]

Calculate its concentration:

\[
\begin{aligned}
c(\ce{HA})
&=\frac{3.000\times10^{-3}}{0.02000}\\
&=0.1500\ \mathrm{mol\,L^{-1}}
\end{aligned}
\]

(b) Calculate the acid dissociation constant.

Half the equivalence volume is:

\[
V_{\mathrm{half}}=\frac{30.00}{2}=15.00\ \mathrm{mL}
\]

At half-equivalence, exactly half the original acid has been neutralised.

For a monoprotic acid, the amount of remaining \(\ce{HA}\) equals the amount of conjugate base \(\mathrm{A}^{-}\) produced.

The acid dissociation expression is:

\[
K_a=
\frac{[\mathrm{H_{3}O}^{+}][\mathrm{A}^{-}]}
{[\ce{HA}]}
\]

Since the concentrations of \(\ce{HA}\) and \(\mathrm{A}^{-}\) are equal at half-equivalence:

\[
K_a=[\mathrm{H_{3}O}^{+}]
\]

Therefore:

\[
\mathrm{pH}=\mathrm{p}K_a
\]

The measured pH is 4.76, so:

\[
\mathrm{p}K_a=4.76
\]

Using:

\[
K_a=10^{-\mathrm{p}K_a}
\]

\[
\begin{aligned}
K_a&=10^{-4.76}\\
&=1.74\times10^{-5}
\end{aligned}
\]

The acid dissociation constant is:

\[
\boxed{K_a=1.74\times10^{-5}}
\]

The important distinction is that the equivalence volume tells you how much acid was present, while the half-equivalence pH tells you about the acid’s tendency to ionise.

A pH of 4.76 alone cannot tell you the original acid concentration.

Question 5

A laboratory technician is given a pure, unknown acid. Its formula is not known, but it is established that the acid is either monoprotic, \(\ce{HA}\), or diprotic, \(\ce{H2A}\).

A 25.00 mL aliquot is titrated with \(0.1000\ \mathrm{mol\,L^{-1}}\) sodium hydroxide.

The pH probe records one clearly resolved steep rise, centred at 30.00 mL of sodium hydroxide added. The solution passes through pH 7.0 at 29.95 mL.

The probe cannot establish whether a first neutralisation stage was missed or whether two proton-transfer stages are insufficiently resolved in the recorded curve.

Two students make different claims:

• Student A claims the acid concentration is \(0.1200\ \mathrm{mol\,L^{-1}}\), because only one steep rise was observed.
• Student B claims the acid concentration is \(0.06000\ \mathrm{mol\,L^{-1}}\), because the acid could be diprotic.

Determine what can be established from the data.

Explain whether either student can confidently claim the molecular concentration of the acid, and identify what additional information would resolve the uncertainty.

Solution 5

Neither student can establish the molecular concentration uniquely. The data establish an acid-equivalent concentration of \(0.1200\ \mathrm{mol\,L^{-1}}\), but the molecular concentration depends on whether the acid is monoprotic or diprotic.

This question is testing a subtle distinction between the number of acidic protons and the number of acid molecules.

Start with what the experiment actually measures.

The sodium hydroxide concentration and equivalence volume are known:

\[
c(\ce{NaOH})=0.1000\ \mathrm{mol\,L^{-1}}
\]

\[
V_{\mathrm{eq}}=0.03000\ \mathrm{L}
\]

Calculate the moles of hydroxide added:

\[
\begin{aligned}
n(\mathrm{OH}^{-})
&=cV\\
&=0.1000\times0.03000\\
&=3.000\times10^{-3}\ \mathrm{mol}
\end{aligned}
\]

These hydroxide ions have neutralised the acidic protons in the original 25.00 mL sample.

Therefore, the original sample contained \(3.000\times10^{-3}\) mol of neutralisable proton equivalents.

Calculate the acid-equivalent concentration:

\[
\begin{aligned}
c_{\mathrm{eq}}
&=\frac{3.000\times10^{-3}}{0.02500}\\
&=0.1200\ \mathrm{mol\,L^{-1}}
\end{aligned}
\]

This value is established regardless of whether the acid is monoprotic or diprotic.

Now consider the two possible models.

Possibility 1: The acid is monoprotic.

The reaction is:

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

The mole ratio is 1:1.

Therefore:

\[
n(\ce{HA})=3.000\times10^{-3}\ \mathrm{mol}
\]

Its molecular concentration would be:

\[
\begin{aligned}
c(\ce{HA})
&=\frac{3.000\times10^{-3}}{0.02500}\\
&=0.1200\ \mathrm{mol\,L^{-1}}
\end{aligned}
\]

Student A’s result is possible.

Possibility 2: The acid is diprotic.

The complete neutralisation is:

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

Now two moles of hydroxide react with every mole of acid.

Therefore:

\[
\begin{aligned}
n(\ce{H2A})
&=\frac{3.000\times10^{-3}}{2}\\
&=1.500\times10^{-3}\ \mathrm{mol}
\end{aligned}
\]

Its molecular concentration would be:

\[
\begin{aligned}
c(\ce{H2A})
&=\frac{1.500\times10^{-3}}{0.02500}\\
&=0.06000\ \mathrm{mol\,L^{-1}}
\end{aligned}
\]

Student B’s result is also possible.

Why doesn’t the single pH jump settle the question?

It is tempting to assume that one visible equivalence region proves the acid is monoprotic.

That conclusion goes beyond the evidence.

A diprotic acid has two neutralisation stages, but the pH changes associated with those stages do not always produce two clearly distinguishable steep rises. Depending on the acid’s dissociation behaviour and the experimental resolution, an intermediate stage may not be identified clearly.

The fact that the probe records one resolved jump is not proof that only one acidic proton is present per molecule.

The pH 7 crossing at 29.95 mL does not resolve the uncertainty either. Equivalence is determined by stoichiometry, not by reaching pH 7.

To establish the molecular concentration, you need additional evidence, such as:

• The chemical formula or known protonicity of the acid.
• A sufficiently resolved titration curve that identifies the neutralisation stages.
• An independent chemical analysis that establishes the acid’s identity.

The key result is:

\[
\boxed{c_{\mathrm{eq}}=0.1200\ \mathrm{mol\,L^{-1}}}
\]

But the molecular concentration could be either:

\[
\boxed{0.1200\ \mathrm{mol\,L^{-1}}}
\]

for a monoprotic acid, or:

\[
\boxed{0.06000\ \mathrm{mol\,L^{-1}}}
\]

for a diprotic acid.

This is the limitation of titration data: the measured amount of base tells you how many acidic proton equivalents were present, but you still need the correct reaction stoichiometry to determine the number of acid molecules.

Once you have established the unknown acid concentration, the next useful step is working out the pH during the titration itself. That requires identifying which reactant remains in excess before, at, or after equivalence. See Calculating pH Before and After Equivalence in Strong-Acid Titrations for that calculation.