How to Find Equivalence Volume on a Conductivity Graph

Suppose the conductivity of an acid solution decreases as NaOH is added, then starts increasing. Somewhere between those two trends is the equivalence volume. But should you choose the lowest point on the graph, or the point where the two trends intersect?

The equivalence volume is the x-coordinate where the extrapolated linear trends before and after a change in conductivity intersect. This is the standard graphical method for estimating the equivalence volume from a conductometric titration.

There is one important complication: not every titration produces a V-shaped graph, and some acids have more than one equivalence point. Before reading the volume, you need to understand the reaction stoichiometry and what the conductivity meter is actually measuring.

01Finding the equivalence volume: the basic method

The equivalence point occurs when the amount of titrant added is exactly the amount required to react with the analyte according to the balanced chemical equation.

The equivalence volume, \(V_{\mathrm{eq}}\), is the volume of titrant required to reach that point.

On a conductivity graph:

• The horizontal axis represents the cumulative volume of titrant added, usually in mL.
• The vertical axis represents the electrical conductivity of the solution, commonly in mS cm\(^{-1}\).
• A change in the graph’s gradient indicates a change in the ions contributing to conductivity.
• The intersection of the extrapolated linear trends estimates the equivalence volume.

The important distinction is that you read the equivalence volume from the horizontal axis, not the vertical axis.

Before using this method, check the balanced chemical equation. For example:

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

One mole of HCl reacts with one mole of NaOH.

Therefore, at equivalence:

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

The graph tells you the volume of NaOH required. The balanced equation tells you what that volume means in terms of the reaction.

These are two separate steps. Don’t skip the stoichiometry and assume every titration has a 1:1 mole ratio.

02Why does conductivity change during a titration?

A conductivity meter measures how effectively a solution conducts electricity.

In an aqueous solution, electrical current is carried by mobile ions. Conductivity depends on both the concentration of those ions and how easily they move through the solution.

This second factor is particularly important for acid-base titrations.

Hydrogen ions, \(\mathrm{H}^{+}\), and hydroxide ions, \(\mathrm{OH}^{-}\), have unusually high ionic mobility compared with many other common ions.

Think of replacing a group of fast runners with a group of slower walkers. Even if both groups contain the same number of people, they won’t move through a space at the same rate.

Similarly, replacing highly mobile \(\mathrm{H}^{+}\) ions with less mobile \(\mathrm{Na}^{+}\) ions changes the conductivity.

The analogy illustrates ionic mobility, but actual conductivity also depends on ion concentration, dilution, and temperature.

What does the experiment look like?

In a conductometric titration, a measured volume of the acid solution is placed in a beaker.

A burette containing a standard solution of the titrant is positioned above the beaker. A conductivity probe is immersed in the solution and connected to a conductivity meter. The beaker is stirred continuously, usually with a magnetic stirrer.

The experiment follows this sequence:

1. Measure and record the initial conductivity.
2. Add a known volume of titrant from the burette.
3. Stir the solution and allow the conductivity reading to stabilise.
4. Record the new conductivity and the cumulative volume of titrant added.
5. Repeat until the titration has passed equivalence.

The recorded measurements are then plotted as conductivity against volume of titrant added.

What should the experiment show?

The answer depends on which ions are present before, during, and after neutralisation.

Let’s start with the simplest case.

03Strong acid-strong base: the V-shaped graph

Consider hydrochloric acid being titrated with sodium hydroxide.

The reaction is:

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

Initially, the solution contains \(\mathrm{H}^{+}\) and \(\mathrm{Cl}^{-}\) ions.

As NaOH is added, the hydroxide ions react with hydrogen ions:

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

Sodium ions enter the solution, but hydrogen ions are removed through neutralisation.

This produces two important changes.

Before equivalence: conductivity decreases.

Hydrogen ions have very high mobility. As they are neutralised and effectively replaced by less mobile sodium ions, the conductivity decreases.

At equivalence: conductivity reaches a minimum or approaches one.

Most of the original hydrogen ions have been neutralised. The solution mainly contains spectator ions and the ions from the salt formed.

After equivalence: conductivity increases.

Additional NaOH introduces excess hydroxide ions. These ions have high mobility, so conductivity increases.

The resulting graph has a general V-shaped appearance.

For this type of titration, the lowest region of the graph is close to the equivalence point.

However, don’t rely on the lowest individual experimental reading. Instead, extend the two linear trends and find their intersection.

04Weak acid-strong base: why the graph looks different

Here is where a common misconception causes problems.

A student learns that conductivity decreases before equivalence and increases afterwards. They then assume every conductometric titration must produce a V-shaped graph.

That isn’t correct.

Consider ethanoic acid titrated with sodium hydroxide.

The reaction is:

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

Ethanoic acid is a weak acid. It only partially ionises in water, so its initial solution contains relatively few ions compared with a similarly concentrated strong acid.

As NaOH is added, ethanoic acid reacts to form sodium ethanoate.

Sodium ethanoate dissociates into sodium and ethanoate ions:

\[
\ce{CH3COONa -> CH3COO^- + Na+}
\]

This increases the number of mobile ions in the solution.

Consequently, conductivity generally increases gradually before equivalence.

After equivalence, excess hydroxide ions enter the solution. Because hydroxide ions have high mobility, conductivity increases more steeply.

The graph therefore has two rising regions rather than a V shape.

The key is to identify the change in gradient, not to search for a minimum.

For a weak acid-strong base titration, the equivalence volume is still estimated from the intersection of the extrapolated linear trends.

05What if the graph has more than one intersection?

So far, we’ve considered acids that donate one proton per molecule.

Some acids can donate more than one proton. These are called polyprotic acids.

A diprotic acid, represented as \(\mathrm{H_2A}\), can donate two protons in separate neutralisation stages.

The reactions with hydroxide ions are:

First neutralisation:

\[
\mathrm{H_2A+OH^-\rightarrow HA^-+H_2O}
\]

Second neutralisation:

\[
\mathrm{HA^-+OH^-\rightarrow A^{2-}+H_2O}
\]

Each stage requires one mole of hydroxide ions per mole of acid molecules.

If the two stages are sufficiently separated and both are visible in the conductivity graph, two changes in gradient may appear.

For a pure diprotic acid with both equivalence stages resolved:

• The first equivalence volume corresponds to one mole of hydroxide per mole of acid.
• The second equivalence volume corresponds to two moles of hydroxide per mole of acid.
• Complete neutralisation occurs at the second equivalence volume.

Therefore:

\[
V_2=2V_1
\]

This relationship assumes the same titrant concentration throughout and that both protons are fully neutralised at the respective equivalence points.

Be careful with the wording of a question.

If the first intersection occurs at 8.0 mL and the second occurs at 16.0 mL, the second equivalence volume is 16.0 mL, not another 8.0 mL.

The second value is the cumulative volume added from the beginning of the titration.

There is another limitation worth remembering: not every diprotic acid produces two clearly separated conductivity changes. If its two ionisation stages overlap significantly, the graph may not resolve them cleanly.

The balanced equation tells you how many neutralisation stages are possible. The graph tells you which changes can actually be identified experimentally.

06A reliable method for HSC Chemistry questions

When you’re given an unfamiliar conductivity graph, follow this sequence.

Step 1: Identify the reaction.

Write the balanced chemical equation before interpreting the graph.

Determine whether the acid is monoprotic, diprotic, or polyprotic, and establish the mole ratio between the reactants.

Step 2: Read the axes.

Confirm that the horizontal axis represents volume of titrant added.

Check the units. You might be given mL or L, and conductivity might be expressed in mS cm\(^{-1}\) or S m\(^{-1}\).

Step 3: Identify the linear regions.

Look for the approximately straight sections before and after a change in gradient.

Don’t assume that conductivity must decrease before equivalence.

Step 4: Extend the trend lines.

Draw or calculate the intersection of the relevant linear sections.

For a single-stage titration, use the intersection’s horizontal coordinate as the equivalence volume.

For a titration with multiple resolved stages, identify each intersection separately.

Step 5: Check the stoichiometry.

Ask whether the number and positions of the intersections make sense for the balanced reaction.

For a pure diprotic acid with two resolved stages, for example, the second equivalence volume should be twice the first.

Step 6: Only then calculate concentration or consider pH.

The conductivity graph identifies a volume. It does not directly give you the concentration of the original acid or the pH at equivalence.

You need additional information and the relevant chemical relationships for those calculations.

Common mistakes to avoid

Improving the experimental measurement

The intersection method depends on having enough reliable measurements around the change in gradient.

Imagine an initial experiment where NaOH is added in 2.0 mL increments. The graph might clearly show a change in gradient, but the exact intersection could lie between two measurements.

The experiment can be improved by repeating it with smaller additions near the suspected equivalence volume, perhaps 0.2-0.5 mL at a time.

Other important controls include:

• Maintaining a constant temperature, since conductivity changes with temperature.
• Keeping the conductivity probe at a consistent depth.
• Ensuring the probe does not touch the beaker.
• Stirring consistently and allowing readings to stabilise.
• Using a calibrated conductivity meter.
• Avoiding unnecessary dilution and contamination.

A conductivity graph is an experimental measurement, not a perfect theoretical curve. A good estimate should account for the quality and resolution of the measurements.

The intersection estimates the equivalence point. It is not the same thing as an indicator endpoint, which is an experimentally observed colour change. These distinctions are explained further in Endpoint vs Equivalence Point: The Difference in Titrations.

07Questions and solutions

Try these questions before reading their solutions. They progress from straightforward graph interpretation to problems involving stoichiometry, experimental limitations, and competing chemical models.

Question 1

A 20.0 mL sample of hydrochloric acid is titrated with sodium hydroxide.

The conductivity graph shows a decreasing linear region followed by an increasing linear region. Extrapolating these two sections gives an intersection at 19.6 mL.

The balanced reaction is:

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

Determine the equivalence volume and explain what is happening chemically at this point.

Solution 1

The equivalence volume is 19.6 mL of NaOH.

The intersection of the two linear regions gives the estimated volume of titrant required to reach equivalence.

The balanced equation shows that hydrochloric acid and sodium hydroxide react in a 1:1 mole ratio.

At 19.6 mL, the amount of NaOH added is stoichiometrically equivalent to the original amount of HCl.

Before this point, hydrogen ions are progressively neutralised, causing conductivity to decrease. After this point, additional hydroxide ions remain in solution, causing conductivity to increase.

The important point is that the horizontal coordinate gives the equivalence volume, while the balanced equation explains the chemical significance of that volume.

Question 2

A 20.0 mL sample of a diprotic acid, \(\mathrm{H_2A}\), is titrated with 0.100 mol L\(^{-1}\) NaOH.

The conductivity graph shows two distinct changes in gradient, at 8.00 mL and 16.0 mL.

Assume both neutralisation stages are clearly resolved and the sample contains only the diprotic acid.

Determine:

a) The amount of diprotic acid initially present.

b) The original concentration of the acid.

c) Which equivalence volume represents complete neutralisation.

Solution 2

The original amount of diprotic acid is \(8.00\times10^{-4}\) mol, its concentration is 0.0400 mol L\(^{-1}\), and complete neutralisation occurs at 16.0 mL.

The important idea is that each acid molecule can donate two protons, so two moles of hydroxide are required to completely neutralise one mole of acid.

First, consider the first equivalence point.

The first neutralisation stage is:

\[
\mathrm{H_2A+OH^-\rightarrow HA^-+H_2O}
\]

At the first equivalence volume of 8.00 mL:

\[
\begin{aligned}
n(\mathrm{OH}^{-})&=cV\\
&=0.100\times0.00800\\
&=8.00\times10^{-4}\text{ mol}
\end{aligned}
\]

The mole ratio is 1:1 at this stage.

Therefore:

\[
n(\mathrm{H_2A})=8.00\times10^{-4}\text{ mol}
\]

Next, calculate the original concentration.

The original acid volume was 20.0 mL, or 0.0200 L.

\[
\begin{aligned}
c&=\frac{n}{V}\\
&=\frac{8.00\times10^{-4}}{0.0200}\\
&=0.0400\text{ mol L}^{-1}
\end{aligned}
\]

Finally, identify complete neutralisation.

The second neutralisation stage is:

\[
\mathrm{HA^-+OH^-\rightarrow A^{2-}+H_2O}
\]

At the second equivalence point, 16.0 mL of NaOH has been added in total.

The total hydroxide amount is:

\[
\begin{aligned}
n(\mathrm{OH}^{-})&=0.100\times0.0160\\
&=1.60\times10^{-3}\text{ mol}
\end{aligned}
\]

This is twice the original amount of acid.

Therefore, the second intersection represents complete neutralisation.

A common mistake is to report 8.00 mL as the volume for complete neutralisation. That is only the first equivalence point.

Question 3

A 10.0 mL aliquot of a monoprotic weak acid, HA, is diluted by adding 40.0 mL of distilled water.

The diluted solution is titrated with 0.100 mol L\(^{-1}\) NaOH.

The conductivity graph shows a gradual increase followed by a steeper increase. The extrapolated linear regions intersect at 12.0 mL.

A student argues that because the acid was diluted fivefold, the equivalence volume should also be divided by five.

Explain why this reasoning is incorrect, and calculate the original concentration of HA.

Assume the added water contains negligible ions and no acid is lost during dilution.

Solution 3

The student’s reasoning is incorrect. The original acid concentration is 0.120 mol L\(^{-1}\), and the equivalence volume remains 12.0 mL despite dilution.

Dilution reduces concentration, but it does not change the number of moles of acid present.

The original 10.0 mL aliquot still contains exactly the same amount of HA after 40.0 mL of water is added.

Step 1: Identify the equivalence volume.

The graph’s two linear regions intersect at:

\[
V_{\mathrm{eq}}=12.0\text{ mL}
\]

Step 2: Write the reaction.

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

The acid is monoprotic, so the mole ratio is 1:1.

Step 3: Calculate the moles of NaOH added.

Convert the volume into litres:

\[
12.0\text{ mL}=0.0120\text{ L}
\]

Then:

\[
\begin{aligned}
n(\ce{NaOH})&=cV\\
&=0.100\times0.0120\\
&=1.20\times10^{-3}\text{ mol}
\end{aligned}
\]

At equivalence:

\[
n(\mathrm{HA})=1.20\times10^{-3}\text{ mol}
\]

Step 4: Calculate the original acid concentration.

The original aliquot was 10.0 mL, not 50.0 mL.

Therefore:

\[
\begin{aligned}
c(\mathrm{HA})&=\frac{1.20\times10^{-3}}{0.0100}\\
&=0.120\text{ mol L}^{-1}
\end{aligned}
\]

The original acid concentration is:

\[
\boxed{0.120\text{ mol L}^{-1}}
\]

The diluted solution would initially have a concentration of 0.0240 mol L\(^{-1}\), but that is not the original concentration being requested.

The distinction is between the number of moles, which remains constant during dilution, and concentration, which changes when the volume increases.

Question 4

A student performs a conductometric titration of hydrochloric acid with sodium hydroxide.

The conductivity readings are recorded in 1.0 mL increments.

The lowest recorded conductivity occurs at 18.0 mL. However, when the linear regions before and after the change are extrapolated, their intersection occurs at 17.5 mL.

The student reports 18.0 mL as the equivalence volume because it is the lowest measured point.

Evaluate this decision and suggest one experimental improvement that would produce a more precise estimate.

Assume the readings have small random variations and the temperature remains constant.

Solution 4

The better estimate of the equivalence volume is 17.5 mL, rather than 18.0 mL.

The reason is that the intersection of the extrapolated linear trends uses information from both sides of the equivalence region.

The lowest measured point is only the lowest value among the particular volumes tested.

Because measurements were taken at 1.0 mL intervals, the actual equivalence volume could lie between two recorded measurements.

Small random variations can also make one measured point appear slightly lower than its neighbours.

The line intersection provides an estimate of where the underlying trends meet, rather than relying on a single experimental reading.

An appropriate experimental improvement would be to repeat the titration using smaller additions around the suspected equivalence volume.

For example, the student could use 0.2 mL increments between 16.5 mL and 18.5 mL.

This would provide more data points around the change in gradient and allow a more precise intersection to be estimated.

The student should also maintain consistent stirring, probe position, and temperature.

The important lesson is that experimental resolution matters. A graph made from widely spaced measurements cannot always identify the equivalence volume precisely.

Question 5

A 25.0 mL acid sample is titrated with 0.100 mol L\(^{-1}\) NaOH.

Its conductivity graph shows two clear, reproducible changes in gradient, at 10.0 mL and 20.0 mL.

The sample is known to contain one of two possible compositions:

Model A: The sample contains 1.00 mmol of a diprotic acid, \(\mathrm{H_2A}\), with:

\[
\mathrm{p}K_{a1}=2.00
\]

\[
\mathrm{p}K_{a2}=6.00
\]

Model B: The sample contains 1.00 mmol of hydrochloric acid and 1.00 mmol of ethanoic acid.

For ethanoic acid:

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

Assume the two models produce well-resolved neutralisation stages, volumes are additive, the temperature is 25°C, and activity effects can be ignored.

Answer the following:

a) Calculate the amount of hydroxide ions added at each conductivity intersection.

b) Does the existence of two intersections prove that the sample contains a diprotic acid? Explain.

c) A pH measurement is taken at the first intersection. Calculate the approximate pH predicted by each model and explain how this additional measurement could distinguish between them.

Solution 5

No. Two conductivity intersections do not prove that the sample contains a diprotic acid. They establish two distinguishable neutralisation stages, but those stages could arise from one diprotic acid or a mixture of two monoprotic acids.

The key is that a conductivity graph measures changes in the ionic composition of a solution. It does not directly identify the molecular structure of the original acid.

Part a: Calculate the hydroxide amounts at both intersections.

The titrant concentration is:

\[
c(\ce{NaOH})=0.100\text{ mol L}^{-1}
\]

At the first intersection, 10.0 mL:

\[
\begin{aligned}
n(\mathrm{OH}^{-})&=cV\\
&=0.100\times0.0100\\
&=0.00100\text{ mol}\\
&=1.00\text{ mmol}
\end{aligned}
\]

At the second intersection, 20.0 mL:

\[
\begin{aligned}
n(\mathrm{OH}^{-})&=0.100\times0.0200\\
&=0.00200\text{ mol}\\
&=2.00\text{ mmol}
\end{aligned}
\]

Therefore, the two intersections correspond to 1.00 mmol and 2.00 mmol of hydroxide ions added.

Part b: Compare the two possible chemical models.

For Model A, the sample contains 1.00 mmol of diprotic acid.

Its two neutralisation stages are:

\[
\mathrm{H_2A+OH^-\rightarrow HA^-+H_2O}
\]

\[
\mathrm{HA^-+OH^-\rightarrow A^{2-}+H_2O}
\]

At the first intersection, 1.00 mmol of hydroxide has neutralised the first proton from each acid molecule.

At the second intersection, 2.00 mmol of hydroxide has neutralised both protons from each acid molecule.

This is consistent with the graph.

However, consider Model B.

The sample contains:

• 1.00 mmol of HCl.
• 1.00 mmol of \(\ce{CH3COOH}\).

The strong acid is neutralised first, followed by the weak acid.

The corresponding reactions are:

\[
\ce{HCl + OH^- -> Cl^- + H2O}
\]

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

At the first intersection, 1.00 mmol of hydroxide has been added, corresponding to the first acid component.

At the second intersection, 2.00 mmol has been added in total, corresponding to neutralisation of both acid components.

This model is also consistent with the two observed intersections, assuming the two stages are sufficiently separated to be resolved.

Therefore, the two breakpoints alone cannot distinguish Model A from Model B.

This is the important conceptual distinction: a graph can reveal the number of observable neutralisation stages without uniquely revealing the number of acidic protons on each original molecule.

Part c: Use pH to distinguish the models.

An independent pH measurement at the first intersection can provide additional information.

For Model A, the solution at the first equivalence point contains the amphiprotic species \(\mathrm{HA^-}\).

An amphiprotic species can act as either an acid or a base.

For a solution dominated by an amphiprotic species, an approximate pH relationship is:

\[
\mathrm{pH}\approx
\frac{\mathrm{p}K_{a1}+\mathrm{p}K_{a2}}{2}
\]

Substitute the supplied values:

\[
\begin{aligned}
\mathrm{pH}&\approx\frac{2.00+6.00}{2}\\
&=4.00
\end{aligned}
\]

Therefore, Model A predicts a pH of approximately 4.00 at the first equivalence point.

Now consider Model B.

At the first intersection, the hydrochloric acid has been neutralised, leaving the ethanoic acid as the remaining weak acid.

The original solution volume was 25.0 mL. A further 10.0 mL of NaOH has been added.

The total solution volume is therefore:

\[
V_{\mathrm{total}}=25.0+10.0=35.0\text{ mL}
\]

The remaining ethanoic acid concentration is:

\[
\begin{aligned}
c&=\frac{n}{V}\\
&=\frac{0.00100}{0.0350}\\
&=0.0286\text{ mol L}^{-1}
\end{aligned}
\]

We can use the weak acid approximation:

\[
[\mathrm{H}^{+}]\approx\sqrt{K_a c}
\]

First, calculate the acid dissociation constant:

\[
K_a=10^{-4.76}
\]

\[
K_a\approx1.74\times10^{-5}
\]

Substitute:

\[
\begin{aligned}
[\mathrm{H}^{+}]&\approx
\sqrt{(1.74\times10^{-5})(0.0286)}\\
&\approx7.05\times10^{-4}\text{ mol L}^{-1}
\end{aligned}
\]

Calculate pH:

\[
\begin{aligned}
\mathrm{pH}&=-\log_{10}[\mathrm{H}^{+}]\\
&=-\log_{10}(7.05\times10^{-4})\\
&\approx3.15
\end{aligned}
\]

Therefore, Model B predicts a pH of approximately 3.15 at the first intersection.

The predictions are:

A pH measurement close to 4.00 would support Model A, while a measurement close to 3.15 would support Model B, under the stated assumptions.

The important point is that the conductivity graph alone cannot settle the question. The additional pH measurement provides independent evidence about the chemical species present.

This is a much stronger interpretation than simply counting the number of intersections and assuming each one corresponds to another proton on the same molecule.

08What to remember

When identifying equivalence volume from a conductivity graph, the most reliable method is to:

1. Balance the chemical equation and establish the mole ratio.
2. Identify the linear conductivity trends before and after each change.
3. Extrapolate the trends to their intersection.
4. Read the corresponding volume from the horizontal axis.
5. Check that the identified equivalence stages agree with the reaction stoichiometry.

For a strong acid-strong base titration, conductivity generally decreases and then increases.

For a weak acid-strong base titration, conductivity generally increases gradually and then more steeply.

For polyprotic acids, multiple equivalence stages may appear.

Once you have identified the correct equivalence volume, you can move on to calculating unknown concentrations using the balanced equation and \(n=cV\).

For further practice with the relationship between conductivity and pH curves, see Titration Curves and Conductivity Graphs for HSC Chemistry.

For the next calculation step, see How to Calculate an Unknown Acid Concentration from Titration Data.