How to Compare Titration Curves with Different Initial Concentrations

Two titration curves start at different pH values and reach equivalence at different volumes. Which acid solution was more concentrated?

Suppose both curves represent 25.00 mL samples of the same monoprotic strong acid, titrated with 0.1000 mol L⁻¹ NaOH. Curve A reaches equivalence after 20.00 mL of NaOH, while Curve B reaches equivalence after 32.00 mL.

Curve B represents the more concentrated acid. It contains more acid particles in the same starting volume, so it requires more NaOH to neutralise them.

But there’s an important condition: this conclusion only works directly because the acid sample volumes, titrant concentrations, and reaction stoichiometry are identical.

The method: identify the balanced reaction, locate the equivalence point on each curve, calculate the moles of acid present, and divide by the original acid sample volume. Only then use the initial pH and curve shape to interpret the difference.

01The method for comparing two titration curves

First, be clear about what the graph actually measures.

• The horizontal axis shows the volume of titrant added, usually in mL.
• The vertical axis shows the pH of the solution in the conical flask.
• The equivalence point is where the acid and base have reacted in their exact stoichiometric proportions.

A higher equivalence volume does not automatically mean a higher acid concentration. It means more titrant was required under the particular experimental conditions.

Step 1: Identify the reaction stoichiometry

Before comparing equivalence volumes, write the balanced neutralisation equation.

For hydrochloric acid reacting with sodium hydroxide:

\[ \ce{HCl + NaOH -> NaCl + H2O} \]

The mole ratio is 1:1.

This means one mole of HCl reacts with one mole of NaOH. At equivalence:

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

However, not every acid-base reaction has a 1:1 ratio.

Consider a diprotic acid, represented by \(\ce{H2A}\):

\[ \ce{H2A + 2NaOH -> Na2A + 2H2O} \]

One mole of \(\ce{H2A}\) requires two moles of NaOH for complete neutralisation.

Therefore:

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

If you assume a 1:1 ratio here, your calculated acid concentration will be twice its correct value.

Step 2: Locate the equivalence point on each curve

The equivalence point is generally identified from the inflection point, near the centre of the steepest section of the titration curve.

Record the volume of titrant added at this point.

For example:

Illustrative data for two hydrochloric acid samples.

Notice that Curve B requires more NaOH to reach equivalence.

For a 1:1 reaction, this means Curve B initially contained more moles of acid.

One important detail: equivalence does not always occur at pH 7.

At 25°C, a strong acid-strong base titration has an equivalence pH of approximately 7. A weak acid-strong base titration generally has an equivalence pH above 7.

The equivalence point is determined by stoichiometry, not by memorising a particular pH value.

For a fuller explanation of the characteristic curve shapes, see Strong Acid-Strong Base Titration Curves for HSC Chemistry.

Step 3: Calculate the initial acid concentration

Once you know the equivalence volume, calculate the moles of titrant added.

Use:

\[ n=cV \]

where:

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

Then use the balanced equation to determine the initial moles of acid.

Finally:

\[ c_\mathrm{acid}= \frac{n_\mathrm{acid}}{V_\mathrm{acid}} \]

The general expression for a reaction with acid and base stoichiometric coefficients is:

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

Here:

• \(\nu_\mathrm{acid}\) is the acid’s coefficient in the balanced equation.
• \(\nu_\mathrm{base}\) is the base’s coefficient.
• \(c_\mathrm{base}\) is the titrant concentration.
• \(V_\mathrm{eq}\) is the volume of titrant at equivalence.
• \(V_\mathrm{acid}\) is the original acid aliquot volume.

Use litres when calculating moles. The two volumes in the concentration ratio can also both be expressed in mL because their units cancel.

Step 4: Compare the concentrations, not just the volumes

There is a useful shortcut when both experiments have identical conditions.

If the following are the same:

• acid identity and reaction stoichiometry,
• original acid sample volume,
• titrant concentration,

then:

\[ \frac{c_A}{c_B} = \frac{V_{\mathrm{eq},A}}{V_{\mathrm{eq},B}} \]

For example, if Curve B has an equivalence volume twice that of Curve A, its initial acid concentration is twice as large.

But if any of these conditions change, you must calculate the concentrations separately.

Think of it this way: 20 mL of a concentrated NaOH solution contains more hydroxide ions than 20 mL of a more dilute solution. Comparing their volumes alone is like comparing spoonfuls of cordial without knowing how concentrated each mixture is.

The volume tells you how much solution was added. The concentration tells you how much reacting substance that volume contained.

Step 5: Use the initial pH as supporting evidence

The initial pH can help distinguish two solutions, but only if you account for acid strength.

For a strong monoprotic acid, such as HCl, complete ionisation means:

\[ [\ce{H3O+}]\approx c_\mathrm{acid} \]

Therefore:

\[ \mathrm{pH}=-\log_{10}(c_\mathrm{acid}) \]

A more concentrated solution of the same strong acid will have a lower initial pH.

For a weak acid, however, only a fraction of the acid molecules ionise.

For a sufficiently dilute weak acid where ionisation is small:

\[ [\ce{H3O+}]\approx\sqrt{K_a c_\mathrm{acid}} \]

The initial pH still generally decreases as concentration increases, but the relationship is different.

Most importantly, you cannot compare the concentrations of two different acids using their initial pH values alone. A strong acid can have a much lower pH than a weak acid at exactly the same analytical concentration.

02Worked examples

03A quick decision process for HSC questions

When you’re given two titration curves, work through these checks in order.

1. Identify the acid and base. Are they the same substances in both experiments?
2. Balance the reaction. What is the mole ratio at the relevant equivalence point?
3. Locate equivalence. Read the volume of titrant at the centre of the steep region.
4. Calculate reacting moles. Use the titrant concentration and equivalence volume.
5. Find the original acid moles. Apply the balanced equation’s mole ratio.
6. Calculate concentration. Divide the acid moles by the original acid aliquot volume.
7. Interpret the pH. Consider acid strength, initial concentration, and the species present at equivalence.

Keep these distinctions in mind:

04Questions and solutions

Question 1

Two 25.00 mL samples of the same monoprotic strong acid are titrated with 0.1000 mol L⁻¹ NaOH.

• Curve A reaches equivalence at 15.00 mL.
• Curve B reaches equivalence at 22.50 mL.

Calculate the concentration of each acid and determine the concentration ratio. Which curve should have the lower initial pH?

Solution 1

Sample B is 1.50 times more concentrated than Sample A. Its initial pH should therefore be lower.

The acid reacts with NaOH in a 1:1 ratio, so the acid moles equal the NaOH moles at equivalence.

For Sample A:

\[ \begin{aligned} n_A&=0.1000\times0.01500\\ &=0.001500\ \mathrm{mol} \end{aligned} \]

Therefore:

\[ \begin{aligned} c_A&=\frac{0.001500}{0.02500}\\ &=0.06000\ \mathrm{mol\,L^{-1}} \end{aligned} \]

For Sample B:

\[ \begin{aligned} n_B&=0.1000\times0.02250\\ &=0.002250\ \mathrm{mol} \end{aligned} \]

Therefore:

\[ \begin{aligned} c_B&=\frac{0.002250}{0.02500}\\ &=0.09000\ \mathrm{mol\,L^{-1}} \end{aligned} \]

Compare the concentrations:

\[ \frac{c_B}{c_A} = \frac{0.09000}{0.06000} =1.50 \]

Because both acids are the same strong acid, Sample B also has the higher initial hydronium ion concentration and lower initial pH.

For reference, their approximate initial pH values are:

\[ \mathrm{pH}_A=1.22 \]

\[ \mathrm{pH}_B=1.05 \]

The equivalence pH is approximately 7 for both at 25°C.

Question 2

Two samples of the same monoprotic strong acid, HX, are titrated using 0.1000 mol L⁻¹ NaOH.

Sample A has an initial volume of 20.00 mL and reaches equivalence after 24.00 mL of NaOH has been added.

Sample B reaches equivalence after 30.00 mL of NaOH has been added. A laboratory record confirms that both acid solutions have exactly the same initial concentration.

Calculate the original volume of Sample B.

Explain why the two curves have different equivalence volumes despite having the same initial acid concentration.

Solution 2

Sample B originally had a volume of 25.00 mL. Both solutions have the same concentration, but Sample B contains more total moles of acid because its aliquot is larger.

First, calculate the concentration of Sample A.

The balanced equation is:

\[ \ce{HX + NaOH -> NaX + H2O} \]

The mole ratio is 1:1.

\[ \begin{aligned} n(\ce{NaOH}) &=0.1000\times0.02400\\ &=0.002400\ \mathrm{mol} \end{aligned} \]

Therefore:

\[ \begin{aligned} c_A&=\frac{0.002400}{0.02000}\\ &=0.1200\ \mathrm{mol\,L^{-1}} \end{aligned} \]

The two solutions have equal concentrations, so:

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

Now calculate the initial moles of acid in Sample B.

\[ \begin{aligned} n_B&=c_\mathrm{NaOH}V_\mathrm{eq}\\ &=0.1000\times0.03000\\ &=0.003000\ \mathrm{mol} \end{aligned} \]

Finally, calculate the original acid volume:

\[ \begin{aligned} V_B&=\frac{n_B}{c_B}\\ &=\frac{0.003000}{0.1200}\\ &=0.02500\ \mathrm{L}\\ &=25.00\ \mathrm{mL} \end{aligned} \]

Sample B has a larger equivalence volume because it contains more total acid moles, not because its concentration is higher.

Since the acid identity and concentration are the same, both samples should also have approximately the same initial pH at the same temperature.

Question 3

Two 25.00 mL acid samples are titrated separately using 0.1000 mol L⁻¹ NaOH.

Sample P contains a monoprotic acid HX. Its equivalence point occurs at 20.00 mL.

Sample Q contains a diprotic acid \(\ce{H2Y}\). Its titration curve shows two clearly separated equivalence points:

• First equivalence point: 20.00 mL.
• Second equivalence point: 40.00 mL.

Assume both protons of \(\ce{H2Y}\) are neutralised at the second equivalence point.

Calculate the initial concentration of each acid.

A student argues that Sample Q must be twice as concentrated as Sample P because it requires twice as much NaOH. Explain the mistake.

Solution 3

Both acid samples have the same initial concentration of 0.0800 mol L⁻¹. The student’s mistake is assuming that the amount of NaOH required is directly proportional to the number of acid molecules without accounting for the number of acidic protons.

For Sample P, the reaction is:

\[ \ce{HX + NaOH -> NaX + H2O} \]

The mole ratio is 1:1.

At equivalence:

\[ \begin{aligned} n(\ce{HX}) &=0.1000\times0.02000\\ &=0.002000\ \mathrm{mol} \end{aligned} \]

Therefore:

\[ \begin{aligned} c_P&=\frac{0.002000}{0.02500}\\ &=0.0800\ \mathrm{mol\,L^{-1}} \end{aligned} \]

For Sample Q, the complete neutralisation reaction is:

\[ \ce{H2Y + 2NaOH -> Na2Y + 2H2O} \]

The mole ratio is 1:2.

At the second equivalence point:

\[ n(\ce{NaOH}) = 0.1000\times0.04000 = 0.004000\ \mathrm{mol} \]

One mole of \(\ce{H2Y}\) requires two moles of NaOH.

Therefore:

\[ n(\ce{H2Y}) = \frac{0.004000}{2} = 0.002000\ \mathrm{mol} \]

The initial concentration is:

\[ \begin{aligned} c_Q&=\frac{0.002000}{0.02500}\\ &=0.0800\ \mathrm{mol\,L^{-1}} \end{aligned} \]

The two acids have equal concentrations.

The reason Sample Q requires twice the total volume of NaOH is that each acid molecule can donate two protons, whereas each HX molecule can donate only one.

There is also a useful consistency check. At the first equivalence point of the diprotic acid, one mole of NaOH has reacted per mole of \(\ce{H2Y}\). Thus:

\[ n(\ce{H2Y})=0.1000\times0.02000 =0.002000\ \mathrm{mol} \]

This gives the same initial concentration.

Both equivalence points support the same result when their respective stoichiometric relationships are applied correctly.

Question 4

Two 25.00 mL acid samples are prepared:

• Sample A: 0.100 mol L⁻¹ hydrochloric acid.
• Sample B: 0.100 mol L⁻¹ ethanoic acid.

Both are titrated with 0.100 mol L⁻¹ NaOH.

The curves show:

The acid dissociation constant of ethanoic acid is:

\[ K_a=1.8\times10^{-5} \]

A student claims that Sample A must have a higher initial concentration because it starts at a lower pH.

Evaluate this claim. Explain why the curves have the same equivalence volume but different initial and equivalence pH values.

Solution 4

The student is incorrect. Both acid solutions have the same initial concentration of 0.100 mol L⁻¹. Their different initial pH values are caused by their different acid strengths.

First, consider the neutralisation reactions.

For hydrochloric acid:

\[ \ce{HCl + NaOH -> NaCl + H2O} \]

For ethanoic acid:

\[ \ce{CH3COOH + NaOH -> CH3COONa + H2O} \]

Both reactions have a 1:1 mole ratio.

Each original acid sample contains:

\[ \begin{aligned} n&=cV\\ &=0.100\times0.02500\\ &=0.002500\ \mathrm{mol} \end{aligned} \]

Therefore, both require:

\[ V_\mathrm{eq} = \frac{0.002500}{0.100} = 0.02500\ \mathrm{L} \]

or 25.00 mL of NaOH.

This explains why their equivalence volumes are identical.

Now consider their initial pH values.

Hydrochloric acid is a strong acid. It ionises essentially completely, so:

\[ [\ce{H3O+}]\approx0.100\ \mathrm{mol\,L^{-1}} \]

Therefore:

\[ \mathrm{pH}=-\log_{10}(0.100)=1.00 \]

Ethanoic acid is weak and only partially ionises.

Using the weak acid approximation:

\[ [\ce{H3O+}]\approx\sqrt{K_a c} \]

Substitute the given values:

\[ \begin{aligned} [\ce{H3O+}] &\approx\sqrt{1.8\times10^{-5}\times0.100}\\ &=1.34\times10^{-3}\ \mathrm{mol\,L^{-1}} \end{aligned} \]

Therefore:

\[ \mathrm{pH} = -\log_{10}(1.34\times10^{-3}) \approx2.87 \]

The ethanoic acid has a higher initial pH because a smaller fraction of its molecules ionises.

Finally, consider the equivalence pH.

For hydrochloric acid and NaOH, the remaining salt ions do not significantly react with water. At 25°C, the equivalence pH is approximately 7.

For ethanoic acid and NaOH, the solution contains ethanoate ions:

\[ \ce{CH3COO-} \]

These ions react with water:

\[ \ce{CH3COO- + H2O <=> CH3COOH + OH-} \]

The production of hydroxide ions makes the solution basic, so its equivalence pH is above 7.

The key distinction is that acid concentration determines the amount of acid present per unit volume, while acid strength determines how extensively that acid ionises.

A lower initial pH does not automatically mean a higher initial acid concentration.

Question 5

Two samples of the same monoprotic weak acid, HA, are titrated separately with sodium hydroxide at 25°C.

The acid dissociation constant is:

\[ K_a=1.8\times10^{-5} \]

The experimental conditions and illustrative curve readings are:

A student observes that both curves have the same equivalence volume and the same pH at half-equivalence.

They conclude that the two acid samples must have the same initial concentration.

Evaluate this conclusion.

Calculate:

(a) The initial concentration of each acid sample and the concentration ratio.

(b) Why both curves have the same pH at half-equivalence despite their different concentrations.

(c) The approximate equivalence pH of each sample, and explain why these values differ.

Assume ideal concentration behaviour, additive solution volumes, and that the weak acid approximations are valid.

Solution 5

The student’s conclusion is incorrect. Sample B is five times more concentrated than Sample A, despite both curves having the same equivalence volume and the same half-equivalence pH.

This question combines three ideas:

• Equivalence volume depends on the amount of titrant delivered, not just its volume.
• Half-equivalence pH depends on the acid’s pKa, not its initial concentration.
• Equivalence pH for a weak acid-strong base titration depends on the concentration of the conjugate base remaining in solution.

Let’s work through each part.

(a) Calculate the initial concentration of each acid.

The neutralisation reaction is:

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

The mole ratio is 1:1.

For Sample A:

\[ \begin{aligned} n(\ce{OH-}) &=0.0800\times0.02000\\ &=0.001600\ \mathrm{mol} \end{aligned} \]

Therefore, the original acid amount was:

\[ n(\ce{HA})=0.001600\ \mathrm{mol} \]

Calculate the initial concentration:

\[ \begin{aligned} c_A&=\frac{0.001600}{0.02500}\\ &=0.0640\ \mathrm{mol\,L^{-1}} \end{aligned} \]

For Sample B:

\[ \begin{aligned} n(\ce{OH-}) &=0.1600\times0.02000\\ &=0.003200\ \mathrm{mol} \end{aligned} \]

Therefore:

\[ n(\ce{HA})=0.003200\ \mathrm{mol} \]

Calculate its concentration:

\[ \begin{aligned} c_B&=\frac{0.003200}{0.01000}\\ &=0.320\ \mathrm{mol\,L^{-1}} \end{aligned} \]

The concentration ratio is:

\[ \frac{c_B}{c_A} = \frac{0.320}{0.0640} =5.00 \]

Sample B is five times more concentrated.

The tempting mistake is to assume that equal equivalence volumes mean equal initial moles.

That would only be true if the titrant concentrations were also equal. Here, Sample B uses twice the NaOH concentration, so the same equivalence volume delivers twice as many moles of hydroxide.

(b) Explain why the half-equivalence pH values are identical.

For a weak acid reacting with a strong base, the neutralisation reaction produces the conjugate base:

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

Before equivalence, both \(\ce{HA}\) and \(\mathrm{A}^{-}\) are present.

At half-equivalence, exactly half of the original acid has been converted into its conjugate base.

Therefore:

\[ [\ce{HA}]=[\mathrm{A}^{-}] \]

The Henderson-Hasselbalch equation is:

\[ \mathrm{pH} = \mathrm{p}K_a+ \log_{10} \left( \frac{[\mathrm{A}^{-}]}{[\ce{HA}]} \right) \]

Since the concentration ratio is 1:

\[ \mathrm{pH} = \mathrm{p}K_a+\log_{10}(1) \]

And because \(\log_{10}(1)=0\):

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

Calculate the pKa:

\[ \begin{aligned} \mathrm{p}K_a &=-\log_{10}(1.8\times10^{-5})\\ &=4.74 \end{aligned} \]

Both samples contain the same weak acid, so they have the same pKa.

Their half-equivalence pH values are therefore both approximately 4.74.

This is the important distinction: half-equivalence pH identifies the acid’s strength, not its initial concentration.

(c) Calculate and compare the equivalence pH values.

At equivalence, essentially all the original HA has been converted into its conjugate base, \(\mathrm{A}^{-}\).

The conjugate base reacts with water:

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

This produces hydroxide ions, making the equivalence solution basic.

The base dissociation constant is:

\[ K_b=\frac{K_w}{K_a} \]

At 25°C:

\[ \begin{aligned} K_b&=\frac{1.00\times10^{-14}}{1.8\times10^{-5}}\\ &=5.56\times10^{-10} \end{aligned} \]

First, calculate the concentration of \(\mathrm{A}^{-}\) at equivalence for each sample.

For Sample A, the total solution volume is:

\[ V_\mathrm{total}=25.00+20.00=45.00\ \mathrm{mL} \]

The original acid amount was 0.001600 mol, so:

\[ \begin{aligned} [\mathrm{A}^{-}]_A &=\frac{0.001600}{0.04500}\\ &=0.0356\ \mathrm{mol\,L^{-1}} \end{aligned} \]

For Sample B:

\[ V_\mathrm{total}=10.00+20.00=30.00\ \mathrm{mL} \]

Therefore:

\[ \begin{aligned} [\mathrm{A}^{-}]_B &=\frac{0.003200}{0.03000}\\ &=0.1067\ \mathrm{mol\,L^{-1}} \end{aligned} \]

For a weak base where hydrolysis is small:

\[ [\ce{OH-}]\approx\sqrt{K_b[\mathrm{A}^{-}]} \]

For Sample A:

\[ \begin{aligned} [\ce{OH-}]_A &=\sqrt{5.56\times10^{-10}\times0.0356}\\ &=4.44\times10^{-6}\ \mathrm{mol\,L^{-1}} \end{aligned} \]

Calculate pOH:

\[ \mathrm{pOH}_A = -\log_{10}(4.44\times10^{-6}) =5.35 \]

At 25°C:

\[ \mathrm{pH}+\mathrm{pOH}=14.00 \]

Therefore:

\[ \mathrm{pH}_A=14.00-5.35=8.65 \]

For Sample B:

\[ \begin{aligned} [\ce{OH-}]_B &=\sqrt{5.56\times10^{-10}\times0.1067}\\ &=7.70\times10^{-6}\ \mathrm{mol\,L^{-1}} \end{aligned} \]

Therefore:

\[ \mathrm{pOH}_B = -\log_{10}(7.70\times10^{-6}) =5.11 \]

And:

\[ \mathrm{pH}_B=14.00-5.11=8.89 \]

The approximate equivalence pH values are:

Sample B has the higher equivalence pH because its equivalence solution contains a greater concentration of the conjugate base, producing more hydroxide ions through hydrolysis.

This is why two curves for the same weak acid can share the same half-equivalence pH and equivalence volume while still representing very different initial acid concentrations.

The important distinction is between the information carried by each part of the curve: equivalence calculations establish the reacting amounts, half-equivalence identifies acid strength, and the pH at equivalence reflects the behaviour of the conjugate base.

Once you can separate those three ideas, comparing titration curves becomes much more reliable. The next useful step is learning to use the buffer region and half-equivalence point to determine an unknown weak acid’s pKa, rather than trying to extract every chemical property from the equivalence volume alone.