Why Conductivity Falls and Then Rises During Some Titrations

You add sodium hydroxide to a sample of hydrochloric acid while measuring its electrical conductivity. The reading starts high, drops steadily, reaches a minimum, and then rises again.

There’s an apparent puzzle here. Sodium hydroxide adds ions to the solution at every step. So why doesn’t conductivity increase from the beginning?

The reason is that conductivity depends on the identity and mobility of the ions, not just how many ions are present. Before equivalence, highly mobile hydrogen ions are removed by neutralisation and replaced mainly by less mobile sodium ions. After equivalence, excess hydroxide ions accumulate, causing conductivity to rise.

This produces the familiar V-shaped conductivity graph seen in some strong acid-strong base titrations. The important detail is that this pattern does not occur in every acid-base titration.

01First, establish the equivalence point

Before interpreting a conductivity graph, work out the reaction stoichiometry. You need to know how much titrant is required to reach equivalence before deciding what the conductivity readings mean.

Consider a sample of hydrochloric acid, \(\ce{HCl}\), in a beaker. Sodium hydroxide, \(\ce{NaOH}\), is added gradually from a burette.

The balanced molecular equation is:

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

Both reactants dissociate essentially completely in water.

The net ionic equation is:

\[ \ce{H3O+(aq) + OH-(aq) -> 2H2O(l)} \]

This tells us that one mole of hydroxide ions reacts with one mole of hydronium ions.

Therefore, at equivalence:

\[ n(\ce{H3O+})_{\text{initial}} = n(\ce{OH-})_{\text{added}} \]

Here, \(n\) represents the amount of substance in moles.

02What does the conductivity experiment actually measure?

Picture the experimental arrangement.

A burette containing sodium hydroxide is positioned above a beaker containing the hydrochloric acid. The beaker sits on a magnetic stirrer, which keeps the solution mixed as titrant enters.

A conductivity probe is immersed in the solution and connected to a conductivity meter. Its electrodes detect how readily electrical current passes through the solution.

The probe does not measure pH. It measures electrical conductivity, usually represented by \(\kappa\).

Conductivity depends on two main factors:

• The concentration of dissolved ions.
• How effectively those ions move through the solution and carry electrical charge.

A useful model for dilute solutions is:

\[ \kappa\approx\sum_i\lambda_i c_i \]

Here:

• \(\kappa\) is electrical conductivity, measured in \(\text{S m}^{-1}\).
• \(c_i\) is the concentration of ion \(i\), measured in \(\text{mol m}^{-3}\).
• \(\lambda_i\) is the ionic molar conductivity, measured in \(\text{S m}^2\text{ mol}^{-1}\).

You do not need to calculate conductivity using this equation in a typical HSC-level question. Its purpose is to show why counting ions alone is insufficient.

Some ions carry electrical charge much more effectively than others.

In particular, \(\ce{H3O+}\) and \(\ce{OH-}\) have unusually high mobility in water. Proton transfer through the hydrogen-bonded water network contributes to their effective movement.

Sodium ions, \(\mathrm{Na}^{+}\), are considerably less mobile.

This difference explains the falling part of the graph.

03Why conductivity falls before equivalence

Return to the reaction:

\[ \ce{H3O+ + OH- -> 2H2O} \]

Initially, the hydrochloric acid has dissociated into hydronium and chloride ions:

\[ \ce{HCl + H2O -> H3O+ + Cl-} \]

The solution therefore contains:

• A high concentration of \(\ce{H3O+}\).
• A high concentration of \(\mathrm{Cl}^{-}\).

The hydronium ions contribute strongly to conductivity because of their high mobility.

Now add sodium hydroxide.

It dissociates into:

\[ \ce{NaOH -> Na+ + OH-} \]

The hydroxide ions react immediately with hydronium ions:

\[ \ce{H3O+ + OH- -> 2H2O} \]

The hydroxide ions do not accumulate while excess strong acid remains.

Instead, hydronium ions are progressively removed, while sodium ions remain in solution.

The important change is therefore:

\[ \ce{H3O+}\quad\longrightarrow\quad\mathrm{Na}^{+} \]

The solution is losing a highly mobile ion and replacing it with a less mobile ion.

Chloride ions also remain in solution. They are spectator ions, meaning they do not participate in the net ionic reaction. However, spectator ions still conduct electricity.

There is another contribution: dilution.

Every addition of sodium hydroxide increases the total solution volume. This reduces the concentrations of ions already present, including chloride.

Both ion replacement and dilution contribute to the decrease in conductivity.

Follow the ions through the titration

Use the original example:

• Initial acid: \(25.00\ \text{mL}\) of \(0.0800\ \text{mol L}^{-1}\) \(\ce{HCl}\).
• Titrant: \(0.1000\ \text{mol L}^{-1}\) \(\ce{NaOH}\).
• Equivalence volume: \(20.00\ \text{mL}\).

The initial amount of hydrochloric acid is \(2.00\ \text{mmol}\).

The following table tracks the major ions at four stages.

All amounts are in millimoles. Small concentrations of ions from water autoionisation are ignored.

Notice what happens.

At \(10.00\ \text{mL}\), half the original hydronium ions have reacted. Sodium ions have replaced them, but there is still excess acid.

At \(20.00\ \text{mL}\), all the original acid has reacted stoichiometrically. The solution contains mainly sodium and chloride ions.

The most important observation is that conductivity can fall even though the solution still contains plenty of ions.

The ions have changed, and their concentrations have changed.

04Why conductivity rises after equivalence

At \(20.00\ \text{mL}\), the acid has been completely consumed in the stoichiometric reaction.

What happens when you add another \(1.00\ \text{mL}\) of sodium hydroxide?

There is no longer excess hydronium available to neutralise the incoming hydroxide ions.

Therefore, the additional hydroxide remains in solution.

The solution now contains:

• \(\mathrm{Na}^{+}\) from the added sodium hydroxide.
• \(\mathrm{Cl}^{-}\) from the original hydrochloric acid.
• Excess \(\ce{OH-}\) from the additional sodium hydroxide.

Hydroxide ions are highly mobile in water. Their accumulation increases the solution’s ability to carry electrical current.

Consequently, conductivity rises.

There are now two contributions from the additional titrant:

1. More sodium ions enter the solution.
2. Excess hydroxide ions begin to accumulate.

The second contribution is particularly important because hydroxide ions have much higher mobility than sodium ions.

At equivalence, conductivity is generally close to its minimum for this simple strong acid-strong base system. It does not fall to zero because sodium and chloride ions remain dissolved.

What would the experimental readings look like?

The following is an illustrative data set, not a record of an actual experiment.

The pattern is clear:

• From 0 to 20 mL, conductivity decreases.
• Near 20 mL, conductivity reaches its minimum.
• Beyond 20 mL, conductivity increases.

The equivalence volume calculated from the reaction is \(20.00\ \text{mL}\), matching the approximate minimum in the illustrative data.

The graph is not necessarily made of two perfectly straight lines. Real conductivity measurements depend on temperature, dilution, ion interactions, and the concentration-dependent mobility of ions.

Also, do not confuse the conductivity minimum with a pH measurement. For a strong acid-strong base titration at \(25^\circ\text{C}\), the equivalence solution has a pH close to 7, but the conductivity probe does not measure that pH.

05A contrasting case: When conductivity rises throughout

Now replace hydrochloric acid with ethanoic acid, \(\ce{CH3COOH}\).

Keep the sodium hydroxide concentration and the initial acid amount the same:

• \(25.00\ \text{mL}\) of \(0.0800\ \text{mol L}^{-1}\) ethanoic acid.
• \(0.1000\ \text{mol L}^{-1}\) sodium hydroxide.

The reaction is:

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

The mole ratio is still \(1:1\).

The initial amount of ethanoic acid is:

\[ n=0.0800\times0.02500 =0.00200\ \text{mol} \]

Therefore, the equivalence volume is still:

\[ V_{\text{eq}} =\frac{0.00200}{0.1000} =0.02000\ \text{L} =20.00\ \text{mL} \]

The stoichiometry has not changed.

However, the conductivity graph looks different.

Why?

Ethanoic acid is a weak acid. It only partially ionises in water:

\[ \ce{CH3COOH + H2O <=> H3O+ + CH3COO-} \]

Initially, most of the acid exists as neutral molecules rather than ions. Its conductivity is therefore relatively low.

When sodium hydroxide is added, it reacts with ethanoic acid to produce acetate ions:

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

The sodium ions from the titrant also remain in solution.

In other words, adding sodium hydroxide converts relatively un-ionised acid into dissolved ionic salt.

The number of dissolved ions increases, so conductivity generally rises before equivalence.

After equivalence, excess hydroxide ions accumulate, causing conductivity to rise more steeply.

This comparison exposes an important misconception:

A conductivity minimum is not a feature of every acid-base titration.

The shape depends on which ions are initially present and which ions are produced or consumed during the reaction.

There is another useful variation. If a strong base such as \(\ce{NaOH}\) is titrated with a strong acid such as \(\ce{HCl}\), conductivity also generally falls and then rises. Before equivalence, hydroxide ions are removed and replaced by chloride ions. After equivalence, highly mobile hydronium ions accumulate.

The direction of titration changes which ion appears in excess, but the underlying principle remains the same.

06Getting reliable conductivity measurements

A conductivity graph is only useful if the measurements are reliable.

A simple first trial might involve adding \(2.00\ \text{mL}\) portions of sodium hydroxide and recording the conductivity after each addition.

This would reveal the general downward and upward trend, but it might miss the precise location of the minimum.

A better experimental design uses the stoichiometric calculation to predict the equivalence volume first.

For the example above, equivalence is expected at \(20.00\ \text{mL}\).

You could then use:

• Larger additions well before equivalence.
• Smaller additions, perhaps \(0.20\)-\(0.50\ \text{mL}\), near equivalence.
• Larger additions again after the minimum has been established.

Keep the following conditions consistent:

1. Temperature: Conductivity changes with temperature, so avoid warming the solution during measurement.
2. Mixing: Stir after every addition and allow the reading to stabilise.
3. Probe position: Keep the electrodes submerged without touching the beaker walls or stir bar.
4. Calibration: Use a properly calibrated conductivity meter.
5. Contamination: Rinse the probe appropriately between solutions.

If the observed minimum does not match the calculated equivalence volume, do not immediately change your stoichiometric calculation.

Check the measurement conditions, the size of the titrant additions, the presence of other dissolved ions, and whether the reaction is actually a strong acid-strong base titration.

The conductivity minimum is useful experimental evidence, but it is not the definition of equivalence.

For a broader comparison of conductivity graphs and pH titration curves, see Titration Curves and Conductivity Graphs for HSC Chemistry.

07Worked examples

08Questions and solutions

Question 1

A \(25.00\ \text{mL}\) sample of \(0.0400\ \text{mol L}^{-1}\) hydrochloric acid is titrated with \(0.1000\ \text{mol L}^{-1}\) sodium hydroxide.

Calculate the equivalence volume.

Explain why conductivity decreases before equivalence and increases afterwards.

Solution 1

The equivalence volume is \(10.00\ \text{mL}\). Conductivity decreases because highly mobile hydronium ions are consumed and replaced by less mobile sodium ions. It increases after equivalence because excess hydroxide ions accumulate.

First, calculate the original amount of hydrochloric acid:

\[ \begin{aligned} n(\ce{HCl}) &=cV\\ &=0.0400\times0.02500\\ &=0.00100\ \text{mol} \end{aligned} \]

The balanced equation is:

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

The mole ratio is \(1:1\).

Therefore:

\[ \begin{aligned} V_{\text{eq}} &=\frac{0.00100}{0.1000}\\ &=0.01000\ \text{L}\\ &=10.00\ \text{mL} \end{aligned} \]

Before \(10.00\ \text{mL}\), there is excess acid. Added hydroxide reacts with hydronium, reducing the contribution of this highly mobile ion to conductivity.

At equivalence, sodium and chloride ions remain in solution.

Beyond \(10.00\ \text{mL}\), hydroxide is in excess. Its accumulation causes conductivity to rise.

The solution does not become non-conductive at equivalence because spectator ions still carry electrical current.

Question 2

A \(30.00\ \text{mL}\) sample of \(0.120\ \text{mol L}^{-1}\) nitric acid is titrated with \(0.0800\ \text{mol L}^{-1}\) potassium hydroxide.

At the point when \(30.00\ \text{mL}\) of potassium hydroxide has been added:

(a) Calculate the equivalence volume.

(b) Calculate the concentrations of the remaining hydronium ions, potassium ions, and nitrate ions.

(c) Explain why conductivity has decreased compared with the initial solution.

Solution 2

The equivalence volume is \(45.00\ \text{mL}\). After \(30.00\ \text{mL}\) of potassium hydroxide has been added, the major ion concentrations are \(0.0200\ \text{mol L}^{-1}\) hydronium, \(0.0400\ \text{mol L}^{-1}\) potassium, and \(0.0600\ \text{mol L}^{-1}\) nitrate.

Part (a): Equivalence volume

Calculate the original amount of nitric acid:

\[ \begin{aligned} n(\ce{HNO3}) &=0.120\times0.03000\\ &=0.00360\ \text{mol} \end{aligned} \]

The reaction is:

\[ \ce{HNO3 + KOH -> KNO3 + H2O} \]

The mole ratio is \(1:1\).

Therefore:

\[ \begin{aligned} V_{\text{eq}} &=\frac{0.00360}{0.0800}\\ &=0.0450\ \text{L}\\ &=45.00\ \text{mL} \end{aligned} \]

Part (b): Ion concentrations after \(30.00\ \text{mL}\)

Calculate the potassium hydroxide amount added:

\[ \begin{aligned} n(\ce{OH-}) &=0.0800\times0.03000\\ &=0.00240\ \text{mol} \end{aligned} \]

The hydronium remaining is:

\[ \begin{aligned} n(\ce{H3O+}) &=0.00360-0.00240\\ &=0.00120\ \text{mol} \end{aligned} \]

The potassium amount is \(0.00240\ \text{mol}\), while the nitrate amount remains \(0.00360\ \text{mol}\).

Total volume:

\[ 30.00+30.00=60.00\ \text{mL} \]

\[ V_{\text{total}}=0.06000\ \text{L} \]

Calculate each concentration:

\[ [\ce{H3O+}] =\frac{0.00120}{0.06000} =0.0200\ \text{mol L}^{-1} \]

\[ [\mathrm{K}^{+}] =\frac{0.00240}{0.06000} =0.0400\ \text{mol L}^{-1} \]

\[ [\ce{NO3-}] =\frac{0.00360}{0.06000} =0.0600\ \text{mol L}^{-1} \]

Part (c): Explain the conductivity change

The solution is still before equivalence because only \(30.00\ \text{mL}\) has been added, compared with the required \(45.00\ \text{mL}\).

Hydronium ions are being removed through neutralisation and replaced by potassium ions.

Potassium ions are less mobile than hydronium ions. Dilution also reduces the concentration of ions already present.

Therefore, conductivity decreases.

Question 3

A student places \(25.00\ \text{mL}\) of \(0.0800\ \text{mol L}^{-1}\) sodium hydroxide in a beaker and titrates it with \(0.1000\ \text{mol L}^{-1}\) hydrochloric acid.

After \(24.00\ \text{mL}\) of hydrochloric acid has been added:

(a) Determine whether the solution is before or after equivalence.

(b) Calculate the concentration of excess hydronium ions.

(c) Explain the expected conductivity trend throughout the titration.

Solution 3

The solution is after equivalence, with an excess hydronium concentration of \(0.00816\ \text{mol L}^{-1}\). Conductivity initially decreases as hydroxide ions are removed, then increases after equivalence as highly mobile hydronium ions accumulate.

Part (a): Find the equivalence volume

Initial hydroxide amount:

\[ \begin{aligned} n(\ce{OH-}) &=0.0800\times0.02500\\ &=0.00200\ \text{mol} \end{aligned} \]

The reaction is:

\[ \ce{H3O+ + OH- -> 2H2O} \]

The mole ratio is \(1:1\).

Therefore:

\[ \begin{aligned} V_{\text{eq}} &=\frac{0.00200}{0.1000}\\ &=0.02000\ \text{L}\\ &=20.00\ \text{mL} \end{aligned} \]

Since \(24.00\ \text{mL}\) has been added, the solution is after equivalence.

Part (b): Calculate excess hydronium concentration

The amount of hydronium added is:

\[ \begin{aligned} n(\ce{H3O+}) &=0.1000\times0.02400\\ &=0.00240\ \text{mol} \end{aligned} \]

The excess is:

\[ \begin{aligned} n(\ce{H3O+})_{\text{excess}} &=0.00240-0.00200\\ &=0.00040\ \text{mol} \end{aligned} \]

Total volume:

\[ 25.00+24.00=49.00\ \text{mL} \]

\[ [\ce{H3O+}] =\frac{0.00040}{0.04900} \]

\[ =0.00816\ \text{mol L}^{-1} \]

Part (c): Explain the conductivity trend

Initially, sodium hydroxide dissociates into sodium and hydroxide ions.

As hydrochloric acid is added, hydronium reacts with hydroxide. The hydroxide concentration decreases, while chloride ions replace the consumed hydroxide ions.

Because hydroxide ions have higher mobility than chloride ions, conductivity decreases.

At equivalence, the solution contains mainly sodium and chloride ions.

After equivalence, hydronium ions accumulate. These ions have exceptionally high mobility, so conductivity increases.

The curve therefore falls and then rises, even though the direction of titration is opposite to the earlier examples.

Question 4

A \(25.00\ \text{mL}\) sample of \(0.0800\ \text{mol L}^{-1}\) ethanoic acid is titrated with \(0.1000\ \text{mol L}^{-1}\) sodium hydroxide.

A student predicts a V-shaped conductivity graph because sodium hydroxide is being added.

(a) Calculate the equivalence volume.

(b) Explain why the student’s predicted graph is not generally correct.

(c) Compare the conductivity behaviour with that of an equal-volume, equal-concentration sample of hydrochloric acid titrated with the same sodium hydroxide.

Solution 4

The equivalence volume is \(20.00\ \text{mL}\), but the weak-acid titration generally produces an increasing conductivity curve rather than a V-shaped curve. The difference comes from the initial ion populations and the formation of dissolved ionic products.

Part (a): Calculate equivalence

The initial amount of ethanoic acid is:

\[ \begin{aligned} n(\ce{CH3COOH}) &=0.0800\times0.02500\\ &=0.00200\ \text{mol} \end{aligned} \]

The balanced reaction is:

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

The mole ratio is \(1:1\).

Therefore:

\[ \begin{aligned} V_{\text{eq}} &=\frac{0.00200}{0.1000}\\ &=0.02000\ \text{L}\\ &=20.00\ \text{mL} \end{aligned} \]

Part (b): Explain the conductivity behaviour

Ethanoic acid is weak and only partially ionised.

Initially, relatively few ions are present, so conductivity is comparatively low.

As sodium hydroxide is added, ethanoic acid reacts to produce acetate ions:

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

Sodium ions also enter the solution.

The neutral acid is being converted into an ionic salt. Consequently, conductivity generally increases before equivalence.

After equivalence, excess hydroxide ions accumulate, producing a steeper increase.

Part (c): Compare with hydrochloric acid

Hydrochloric acid is a strong acid. It is already essentially completely ionised before titration.

Initially, its solution contains substantial concentrations of highly mobile hydronium ions.

As sodium hydroxide is added, hydronium ions are removed and replaced by sodium ions. Conductivity decreases until it reaches a minimum near equivalence.

The two samples have the same initial acid amount:

\[ n=0.00200\ \text{mol} \]

Therefore, both require \(20.00\ \text{mL}\) of sodium hydroxide.

However, their conductivity graphs are different because acid strength determines the initial ion population and the way ions change during neutralisation.

This is why the equivalence volume and the shape of the conductivity graph must be considered separately.

Question 5

Three separate experiments are performed using identical hydrochloric acid samples.

Each sample contains:

\[ 25.00\ \text{mL of }0.0800\ \text{mol L}^{-1}\ \ce{HCl} \]

The experiments are:

• Experiment A: Titrate with \(0.1000\ \text{mol L}^{-1}\) \(\ce{NaOH}\).
• Experiment B: Titrate with \(0.1000\ \text{mol L}^{-1}\) \(\ce{KOH}\).
• Experiment C: Add \(25.00\ \text{mL}\) of distilled water to the original acid sample, then titrate with \(0.1000\ \text{mol L}^{-1}\) \(\ce{NaOH}\).

Assume complete dissociation of the strong electrolytes, additive solution volumes, and a constant temperature of \(25^\circ\text{C}\).

At equivalence, ignore the small concentrations of hydronium and hydroxide ions arising from water autoionisation.

For an approximate conductivity calculation, use:

\[ \kappa\approx\sum_i\lambda_i c_i \]

The following ionic molar conductivity values are supplied:

When these values are multiplied by concentrations in \(\text{mol L}^{-1}\), the numerical result is approximately in \(\text{mS cm}^{-1}\).

Answer the following:

(a) Calculate the equivalence volume for each experiment.

(b) Calculate the approximate conductivity at equivalence for experiments A, B, and C.

(c) Explain why experiment B has a higher conductivity at equivalence than experiment A.

(d) Explain why experiment C has a lower conductivity at equivalence than experiment A.

(e) A student argues that the experiment with the lowest conductivity minimum must have contained the fewest moles of acid. Evaluate this conclusion.

Solution 5

All three experiments have the same equivalence volume of \(20.00\ \text{mL}\), despite having different conductivity values at equivalence. Experiment B has the highest approximate conductivity because potassium ions are more mobile than sodium ions. Experiment C has the lowest because the additional water dilutes the ions without changing the amount of acid.

This question tests an important distinction: the horizontal position of the equivalence point depends on stoichiometry, while the vertical conductivity value depends on the ions present and their concentrations.

Part (a): Calculate the equivalence volume

First, calculate the initial amount of hydrochloric acid:

\[ \begin{aligned} n(\ce{HCl}) &=cV\\ &=0.0800\times0.02500\\ &=0.002000\ \text{mol} \end{aligned} \]

The neutralisation reaction has a \(1:1\) ratio between acid and hydroxide.

Both sodium hydroxide and potassium hydroxide supply one mole of hydroxide per mole of base.

Therefore, all three experiments require:

\[ \begin{aligned} V_{\text{eq}} &=\frac{0.002000}{0.1000}\\ &=0.02000\ \text{L}\\ &=20.00\ \text{mL} \end{aligned} \]

Adding water in experiment C changes the concentration of the acid but not its original amount.

It therefore does not change the equivalence volume.

Part (b): Calculate the approximate conductivity

For experiment A, the total volume at equivalence is:

\[ 25.00+20.00=45.00\ \text{mL} \]

The major ions are \(\mathrm{Na}^{+}\) and \(\mathrm{Cl}^{-}\).

Each has an amount of \(0.002000\ \text{mol}\).

Their concentrations are:

\[ \begin{aligned} [\mathrm{Na}^{+}] &=\frac{0.002000}{0.04500}\\ &=0.04444\ \text{mol L}^{-1} \end{aligned} \]

The chloride concentration is identical.

Therefore:

\[ \begin{aligned} \kappa_A &\approx(50.1+76.3)(0.04444)\\ &\approx5.62\ \text{mS cm}^{-1} \end{aligned} \]

For experiment B, the major ions are \(\mathrm{K}^{+}\) and \(\mathrm{Cl}^{-}\).

Their concentrations are also \(0.04444\ \text{mol L}^{-1}\).

Therefore:

\[ \begin{aligned} \kappa_B &\approx(73.5+76.3)(0.04444)\\ &\approx6.66\ \text{mS cm}^{-1} \end{aligned} \]

For experiment C, the initial acid has been diluted with an additional \(25.00\ \text{mL}\) of water.

The total volume at equivalence is:

\[ 25.00+25.00+20.00 =70.00\ \text{mL} \]

The amounts of sodium and chloride ions remain \(0.002000\ \text{mol}\) each.

Their concentrations are:

\[ \begin{aligned} [\mathrm{Na}^{+}] &=\frac{0.002000}{0.07000}\\ &=0.02857\ \text{mol L}^{-1} \end{aligned} \]

The chloride concentration is also \(0.02857\ \text{mol L}^{-1}\).

Therefore:

\[ \begin{aligned} \kappa_C &\approx(50.1+76.3)(0.02857)\\ &\approx3.61\ \text{mS cm}^{-1} \end{aligned} \]

The approximate results are:

These are dilute-solution estimates. Actual readings can differ because ionic mobility is concentration-dependent and real solutions are not perfectly ideal.

Part (c): Why is experiment B more conductive?

Experiment B produces potassium chloride rather than sodium chloride.

Both solutions contain the same concentration of chloride ions and the same concentration of positive ions.

However, potassium ions have a higher ionic molar conductivity than sodium ions:

\[ 73.5>50.1 \]

Therefore, potassium ions carry electrical charge more effectively.

Experiment B has higher conductivity even though exactly the same amount of acid has been neutralised.

Part (d): Why is experiment C less conductive?

Adding distilled water does not change the original number of moles of hydrochloric acid.

However, it increases the total volume.

At equivalence, experiment C has the same amounts of sodium and chloride ions as experiment A, but these ions are spread through a larger volume.

Their concentrations are therefore lower.

Lower ion concentrations produce a lower conductivity.

Part (e): Evaluate the student’s conclusion

The student’s conclusion is incorrect.

The minimum conductivity does not directly measure the number of moles of acid originally present.

Experiment B has a higher minimum because potassium ions are more mobile than sodium ions.

Experiment C has a lower minimum because the solution is more dilute.

Yet all three experiments contain exactly \(0.002000\ \text{mol}\) of hydrochloric acid and require exactly \(20.00\ \text{mL}\) of the specified hydroxide titrant.

The correct approach is to use the reaction equation and titrant concentration to determine the equivalence volume. Conductivity then provides information about the changing ionic composition of the solution.

This is the distinction to carry into unfamiliar titration questions: stoichiometry tells you when equivalence occurs, while ion identity, concentration, and mobility explain the shape and height of the conductivity graph.