How to Trace Titration Errors to the Final Concentration

A student titrates a 25.00 mL sample of hydrochloric acid with standard sodium hydroxide. Their three titres agree to within 0.05 mL, so they calculate the acid concentration and feel confident in the result.

But there is a problem: the burette was rinsed with water instead of sodium hydroxide. The titrant was diluted before the experiment even began.

Will the calculated acid concentration be too high or too low?

Now consider a second student who leaves distilled water in the conical flask before adding the acid sample. Does that cause an error too?

The important distinction is that the first mistake changes the concentration of the titrant, while the second changes only the volume of liquid in the flask. They do not have the same effect.

The method for tracing a titration error is to follow it through to the measured titre, then through the mole calculation to the final concentration.

For a direct titration with the unknown solution in the conical flask and the standard solution in the burette:

• A titre that is too large generally produces a calculated unknown concentration that is too high.
• A titre that is too small generally produces a calculated unknown concentration that is too low.
• An error that does not change the number of moles delivered or the measured titre may have no effect on the calculated concentration.

There is one further complication: some errors affect the actual concentration of the titrant or the amount of analyte transferred. You must account for those as well.

01Start with the relationship between titre and concentration

Before investigating individual mistakes, establish the calculation that connects them.

Imagine a standard sodium hydroxide solution in a burette and an unknown hydrochloric acid solution in a conical flask.

The reaction is:

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

The balanced equation gives a mole ratio of \(1:1\).

At the equivalence point, the moles of hydrochloric acid equal the moles of sodium hydroxide delivered.

Therefore:

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

Using \(n=cV\):

\[ c_{\ce{HCl}}V_{\ce{HCl}} = c_{\ce{NaOH}}V_{\ce{NaOH}} \]

Rearranging:

\[ \boxed{ c_{\ce{HCl}} = \frac{c_{\ce{NaOH}}V_{\ce{NaOH}}} {V_{\ce{HCl}}} } \]

Here:

• \(c_{\ce{HCl}}\) is the unknown acid concentration in \(\text{mol L}^{-1}\).
• \(c_{\ce{NaOH}}\) is the known sodium hydroxide concentration in \(\text{mol L}^{-1}\).
• \(V_{\ce{HCl}}\) is the fixed volume of acid transferred into the flask.
• \(V_{\ce{NaOH}}\) is the titre, or volume of sodium hydroxide delivered from the burette.

All volumes must be expressed in litres when calculating moles using \(n=cV\).

Notice something important: if the other quantities remain correct, the calculated acid concentration is directly proportional to the titre.

\[ c_{\ce{HCl}}\propto V_{\ce{NaOH}} \]

This gives us our first decision rule.

This relationship is the foundation of error tracing.

For reactions that do not have a \(1:1\) mole ratio, use the balanced equation rather than assuming this simplified expression applies. The same reasoning still works once the stoichiometric coefficients are included.

02Follow the solution through the apparatus

A titration involves more than adding one solution to another. Each piece of apparatus has a specific job, and the way it is prepared affects the quantity of substance transferred.

The easiest way to investigate an experimental error is to follow the solutions in the order they enter the experiment.

Conditioning the burette

The burette contains the standard solution, also called the titrant.

Suppose you have a burette that has just been washed with distilled water. There is still a thin film of water coating its internal walls.

What happens if you fill it directly with 0.1000 mol L\(^{-1}\) sodium hydroxide?

The remaining water mixes with the sodium hydroxide, reducing its concentration.

The solution in the burette is now more dilute than the standard solution you intended to use.

This is why the burette must be conditioned before use.

The correct procedure is:

1. Rinse the burette with distilled water to remove contaminants.
2. Drain the water through the tap and delivery tip.
3. Rinse the burette with a small portion of the sodium hydroxide solution.
4. Rotate the burette so the solution contacts the internal walls.
5. Drain this rinse through the tip.
6. Fill the burette with the standard sodium hydroxide solution.
7. Check that the delivery tip is completely filled and contains no air bubbles.

The titrant rinse removes the remaining water film by replacing it with the solution that will actually be used.

What if the student forgets this step?

The sodium hydroxide becomes diluted. A larger volume is needed to deliver the same number of moles.

If the student still uses the original labelled concentration in their calculation, they will calculate too many moles of sodium hydroxide and overestimate the unknown acid concentration.

There is an important qualification: if the diluted sodium hydroxide is independently standardised and its new concentration is used correctly, the dilution itself does not cause this concentration error. The problem is using the original concentration when it is no longer correct.

Conditioning the pipette

The pipette has a different job.

It transfers a fixed volume of the unknown solution into the conical flask. For example, a 25.00 mL volumetric pipette should deliver a 25.00 mL aliquot.

The pipette must be rinsed with the unknown solution before use.

The procedure is:

1. Rinse the pipette with distilled water.
2. Allow the water to drain.
3. Rinse the pipette with a small portion of the unknown acid solution.
4. Rotate it so the internal walls are coated.
5. Discard the rinse.
6. Fill the pipette and adjust the bottom of the meniscus to the calibration mark at eye level.
7. Allow the solution to drain naturally into the conical flask.

Why not simply use distilled water?

Because residual water inside the pipette dilutes the unknown solution. The 25.00 mL transferred will contain fewer moles of acid than expected.

The titre will consequently be smaller, and the calculated acid concentration will be too low if the student assumes the full 25.00 mL contains undiluted acid.

Transferring the aliquot correctly

There is another small detail that can change the number of moles transferred.

A standard volumetric pipette is generally calibrated to deliver its stated volume by gravity. It is not designed to have every last drop forced out.

Once the solution has drained:

• Allow the specified drainage time.
• Touch the pipette tip against the inside wall of the conical flask.
• Do not blow out the remaining drop unless the pipette is specifically marked as a blow-out type.

If you force out the residual drop from an ordinary volumetric pipette, you deliver more analyte than the calibrated volume.

More analyte means more titrant is required.

The measured titre increases, so the calculated unknown concentration becomes too high.

If you fail to allow the pipette to drain properly, the opposite can happen: less analyte enters the flask, producing a smaller titre and a lower calculated concentration.

Why the conical flask can contain water

Here is a common point of confusion.

The conical flask is normally rinsed with distilled water before the analyte is transferred into it.

Unlike the burette and pipette, it does not need to be rinsed with the solution it will contain.

Why?

Because the water already in the flask does not change the number of moles of acid transferred by the pipette.

Suppose the pipette delivers:

\[ n(\ce{HCl})=0.00250\text{ mol} \]

Adding another 20 mL of distilled water does not change this amount.

The acid is more dilute, but it still contains 0.00250 mol.

The equivalence point depends on the number of moles that react, not simply the total volume of liquid in the flask.

This gives us a useful contrast:

The distinction is between changing a solution’s concentration before measuring or transferring it and simply adding more solvent after the correct amount of analyte has entered the flask.

03Trace the direction of common apparatus errors

Now that the roles of the three pieces of apparatus are clear, we can predict the effect of individual mistakes.

Assume throughout this section that the unknown solution is in the conical flask and the standard titrant is in the burette.

Let’s examine two errors that are particularly easy to overlook.

An air bubble in the burette tip

Imagine that the burette is filled with sodium hydroxide, but there is an air bubble trapped in the delivery tip.

The student records an initial reading and begins the titration.

As the tap is opened, some of the liquid leaving the burette fills the space previously occupied by air. This volume is recorded as having left the burette, even though it has not yet entered the conical flask.

The student therefore records a larger titre than the volume that actually reacted.

The calculated acid concentration is too high.

This error only occurs if the bubble is present at the initial reading and is displaced during the titration. A bubble that is removed before the initial reading does not create this particular error.

Failing to wash down the flask walls

During a titration, small droplets of the solution can remain on the inside walls of the conical flask.

If these droplets contain analyte and are not washed back into the bulk solution, some of the analyte may not react before the student judges the endpoint.

This can produce an endpoint that is too early and a titre that is too small.

The solution is simple: use a wash bottle containing distilled water to rinse the inside walls of the flask during the titration.

Again, this added water does not change the total moles of analyte. It simply ensures that the existing analyte enters the reacting mixture.

What about burette reading errors?

The titre is calculated from two readings:

\[ V_{\text{titre}}=V_f-V_i \]

where:

• \(V_f\) is the final burette reading.
• \(V_i\) is the initial burette reading.

Both readings must be taken at eye level, using the bottom of the meniscus for a clear solution.

Remember that the burette scale increases downwards.

A reading error does not automatically make the titre too high or too low. You must identify which reading is incorrect.

For example:

• If the final reading is recorded too high, the calculated titre is too high.
• If the initial reading is recorded too high, the calculated titre is too low.

This is why recording both readings is more useful than simply writing down the final volume.

04Endpoint decisions can change the final concentration

Even with perfectly conditioned apparatus, a titration can produce an incorrect concentration if the student judges the endpoint poorly.

First, distinguish two terms.

The equivalence point is the point at which the reactants have combined in their exact stoichiometric ratio.

The endpoint is the observed experimental signal, usually an indicator colour change, that tells the student to stop adding titrant.

Ideally, the endpoint occurs very close to the equivalence point.

However, they are not automatically identical.

Overshooting the endpoint

Suppose the student is titrating hydrochloric acid with sodium hydroxide using phenolphthalein.

The endpoint is the first faint pink colour that persists after swirling.

If the student adds several extra drops after this colour appears, more sodium hydroxide has entered the flask than is required for equivalence.

The recorded titre is too large.

Because the calculated acid concentration is proportional to the titre, the reported concentration is too high.

Stopping too early

Now imagine the student sees a slight colour change and immediately stops, even though the colour disappears when the flask is swirled.

The endpoint has not actually been reached.

The student records a smaller titre than required.

The calculated acid concentration is too low.

For phenolphthalein, the correct endpoint is a very pale pink that persists for approximately 30 seconds, rather than a strong pink colour.

Choosing the wrong indicator

This is slightly more subtle.

Different indicators change colour over different pH ranges. The correct indicator depends on the shape of the titration curve near the equivalence point.

Consider three common situations:

Phenolphthalein changes colour over approximately pH 8.2-10.0.

Methyl orange changes colour over approximately pH 3.1-4.4.

For a weak acid titrated with a strong base, the equivalence point is above pH 7. Methyl orange would change colour too early, before the equivalence point.

The student would record a titre that is too small and calculate an acid concentration that is too low.

For a strong acid titrated with a weak base, phenolphthalein would generally change colour too late, after equivalence. The titre would be too large, producing an overestimate of the acid concentration.

The important lesson is that choosing the wrong indicator does not always produce the same direction of error.

You need to compare its transition range with the equivalence region of the particular titration.

07A reliable decision process for any titration error

When an HSC question asks you to explain how an experimental error affects the final concentration, use the following sequence.

1. Identify the unknown solution and the standard solution.

Establish which one is in the conical flask and which one is in the burette.

2. Identify what the mistake changes.

Does it change:

• The concentration of the standard solution?
• The amount of analyte transferred?
• The volume of titrant actually delivered?
• The point at which the student stops the titration?

3. Predict the effect on the measured titre.

Ask whether more or less titrant is needed to reach equivalence, or whether the recorded volume differs from the amount actually delivered.

4. Trace the change through the mole calculation.

Use the balanced equation and the appropriate concentration equation.

Do not assume that every error affects the titre alone.

5. State the direction of the final error.

Use precise language:

• The calculated concentration is higher than the true concentration.
• The calculated concentration is lower than the true concentration.
• The error does not affect the theoretical titre or calculated concentration.

6. Explain the chemical or experimental reason.

A statement such as “the concentration is too high because of an experimental error” is incomplete.

Explain exactly what changed and why that change produces the final result.

One final caution: these direction rules assume the unknown solution is in the flask and the known standard solution is in the burette. If the arrangement is reversed, the unknown concentration may be calculated using a measured volume in the denominator. Re-derive the relationship before deciding the direction.

08Questions and solutions

Question 1

A student transfers 20.00 mL of an unknown hydrochloric acid solution into a conical flask.

The acid is titrated with 0.1000 mol L\(^{-1}\) sodium hydroxide. The average concordant titre is 18.40 mL.

Calculate the concentration of the hydrochloric acid.

Solution 1

The hydrochloric acid concentration is \(0.09200\text{ mol L}^{-1}\).

The balanced equation is:

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

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

First, calculate the moles of sodium hydroxide:

\[ \begin{aligned} n(\ce{NaOH}) &=cV\\ &=0.1000\times0.01840\\ &=0.001840\text{ mol} \end{aligned} \]

Therefore:

\[ n(\ce{HCl})=0.001840\text{ mol} \]

The acid volume is:

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

Calculate the concentration:

\[ \begin{aligned} c(\ce{HCl}) &=\frac{0.001840}{0.02000}\\ &=0.09200\text{ mol L}^{-1} \end{aligned} \]

The important point is to divide by the volume of the unknown acid in the flask, not the titre volume from the burette.

Question 2

A student titrates 25.00 mL of an unknown monoprotic acid using 0.1000 mol L\(^{-1}\) sodium hydroxide.

The burette contains an air bubble of volume 0.30 mL in its delivery tip at the beginning of the titration.

The student records a titre of 25.10 mL.

Assume the 0.30 mL required to fill the delivery tip is included in the burette reading but does not enter the conical flask.

Calculate:

a) The concentration the student reports.

b) The correct concentration.

c) The percentage by which the reported concentration differs from the correct concentration.

Solution 2

The student reports \(0.1004\text{ mol L}^{-1}\), but the correct acid concentration is \(0.09920\text{ mol L}^{-1}\). The reported result is approximately 1.21% too high.

Part a: Calculate the reported concentration.

The student uses the full recorded titre:

\[ V_{\text{recorded}}=25.10\text{ mL} \]

Using the \(1:1\) mole ratio:

\[ \begin{aligned} c_{\text{reported}} &=\frac{0.1000\times0.02510}{0.02500}\\ &=0.1004\text{ mol L}^{-1} \end{aligned} \]

Part b: Calculate the correct concentration.

The first 0.30 mL of the recorded volume filled the delivery tip rather than entering the flask.

Therefore, the volume that actually reacted was:

\[ \begin{aligned} V_{\text{delivered}} &=25.10-0.30\\ &=24.80\text{ mL} \end{aligned} \]

Convert to litres:

\[ V_{\text{delivered}}=0.02480\text{ L} \]

Now calculate the correct concentration:

\[ \begin{aligned} c_{\text{correct}} &=\frac{0.1000\times0.02480}{0.02500}\\ &=0.09920\text{ mol L}^{-1} \end{aligned} \]

Part c: Calculate the percentage error.

\[ \begin{aligned} \%\text{ error} &=\frac{|0.1004-0.09920|} {0.09920}\times100\\ &=1.21\% \end{aligned} \]

The error occurs because the burette measures the volume leaving its calibrated barrel, not necessarily the volume entering the conical flask.

The air bubble creates a difference between those two quantities.

Question 3

An unknown hydrochloric acid solution has a true concentration of 0.1200 mol L\(^{-1}\).

A student intends to transfer a 25.00 mL aliquot using a volumetric pipette. However, the pipette contains 0.50 mL of residual distilled water.

Assume the final 25.00 mL delivered contains 24.50 mL of the original acid solution and 0.50 mL of water.

The aliquot is titrated with 0.1000 mol L\(^{-1}\) sodium hydroxide.

a) Calculate the titre.

b) Calculate the concentration reported if the student assumes that the full 25.00 mL consists of undiluted acid.

c) Explain what would happen if the same 0.50 mL of water were instead present in the conical flask before the correctly transferred 25.00 mL acid aliquot was added.

Solution 3

The contaminated pipette produces a titre of \(29.40\text{ mL}\) and a reported concentration of \(0.1176\text{ mol L}^{-1}\), which is 2.00% below the true concentration. Water in the conical flask would not cause this error.

Part a: Calculate the titre.

First, calculate the actual moles of hydrochloric acid transferred.

Only 24.50 mL of the original acid solution is present.

\[ \begin{aligned} n(\ce{HCl}) &=cV\\ &=0.1200\times0.02450\\ &=0.002940\text{ mol} \end{aligned} \]

The reaction is:

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

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

Therefore:

\[ n(\ce{NaOH})=0.002940\text{ mol} \]

Calculate the sodium hydroxide volume:

\[ \begin{aligned} V(\ce{NaOH}) &=\frac{n}{c}\\ &=\frac{0.002940}{0.1000}\\ &=0.02940\text{ L} \end{aligned} \]

Therefore:

\[ \boxed{V_{\text{titre}}=29.40\text{ mL}} \]

Part b: Calculate the reported concentration.

The student assumes that the full 25.00 mL aliquot contains the original acid concentration.

Therefore:

\[ \begin{aligned} c_{\text{reported}} &=\frac{0.1000\times0.02940}{0.02500}\\ &=0.1176\text{ mol L}^{-1} \end{aligned} \]

The reported concentration is:

\[ 0.1176\text{ mol L}^{-1} \]

The percentage error is:

\[ \begin{aligned} \%\text{ error} &=\frac{0.1176-0.1200}{0.1200}\times100\\ &=-2.00\% \end{aligned} \]

The result is 2.00% too low.

The error occurs because the pipette transferred fewer moles of acid than expected.

Part c: Compare with water in the conical flask.

If the pipette is correctly conditioned, it transfers the full 25.00 mL of original acid.

The number of moles is then:

\[ \begin{aligned} n(\ce{HCl}) &=0.1200\times0.02500\\ &=0.003000\text{ mol} \end{aligned} \]

The additional 0.50 mL of water in the conical flask does not change these moles.

The required sodium hydroxide volume becomes:

\[ \begin{aligned} V(\ce{NaOH}) &=\frac{0.003000}{0.1000}\\ &=0.03000\text{ L}\\ &=30.00\text{ mL} \end{aligned} \]

The calculated acid concentration remains:

\[ 0.1200\text{ mol L}^{-1} \]

The difference is not the presence of water itself. It is where the water is introduced and whether it replaces some of the analyte in the calibrated aliquot.

Question 4

A student titrates 25.00 mL of ethanoic acid with 0.1000 mol L\(^{-1}\) sodium hydroxide.

A pH curve shows that the equivalence point occurs at a titre of 20.00 mL.

However, the student uses methyl orange, and the indicator changes colour at 18.20 mL.

a) Calculate the true concentration of ethanoic acid.

b) Calculate the concentration the student reports.

c) Calculate the percentage error.

d) Explain why the indicator produces this error and identify a more suitable indicator.

Solution 4

The true ethanoic acid concentration is \(0.0800\text{ mol L}^{-1}\), while the student reports \(0.0728\text{ mol L}^{-1}\). The result is 9.00% too low because methyl orange changes colour before the equivalence point.

Part a: Calculate the true concentration.

The reaction is:

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

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

At equivalence:

\[ n(\ce{CH3COOH})=n(\ce{NaOH}) \]

Using the actual equivalence titre:

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

Therefore:

\[ \begin{aligned} c(\ce{CH3COOH}) &=\frac{0.002000}{0.02500}\\ &=0.0800\text{ mol L}^{-1} \end{aligned} \]

Part b: Calculate the reported concentration.

The student stops at 18.20 mL.

\[ 18.20\text{ mL}=0.01820\text{ L} \]

Therefore:

\[ \begin{aligned} c_{\text{reported}} &=\frac{0.1000\times0.01820}{0.02500}\\ &=0.0728\text{ mol L}^{-1} \end{aligned} \]

Part c: Calculate the percentage error.

\[ \begin{aligned} \%\text{ error} &=\frac{0.0728-0.0800}{0.0800}\times100\\ &=-9.00\% \end{aligned} \]

The calculated concentration is 9.00% too low.

Part d: Explain the indicator error.

Ethanoic acid is a weak acid, while sodium hydroxide is a strong base.

The equivalence point therefore occurs above pH 7 because the ethanoate ions remaining in solution react with water to produce hydroxide ions.

Methyl orange changes colour in an acidic pH range. Its transition occurs before the steep equivalence region for this titration.

The student consequently stops adding sodium hydroxide too early.

Phenolphthalein is more suitable because its transition range overlaps the steep pH change around the equivalence point.

The misconception is that any indicator colour change signals the correct endpoint. The indicator must be chosen for the particular acid-base combination.

Question 5

Two students independently determine the concentration of the same monoprotic acid.

For this investigation, the true acid concentration is 0.1000 mol L\(^{-1}\).

Both students intend to transfer 25.00 mL of acid into a conical flask and titrate it with sodium hydroxide labelled 0.1000 mol L\(^{-1}\).

Student A conditions the pipette and burette correctly.

Student B makes two mistakes:

• The volumetric pipette is not conditioned. Its 25.00 mL delivered volume contains 24.00 mL of the original acid solution and 1.00 mL of residual water.
• The burette is not conditioned. The sodium hydroxide is diluted to an actual concentration of 0.0960 mol L\(^{-1}\), but Student B still uses the labelled concentration of 0.1000 mol L\(^{-1}\) in the calculation.

Both students identify the true equivalence point accurately, and neither makes any other experimental errors.

a) Calculate the actual number of moles of acid transferred by Student B.

b) Calculate Student B’s actual titre.

c) Calculate the concentration Student B reports.

d) Explain why Student B can obtain the correct numerical concentration despite making two systematic errors.

e) If Student B overshoots the endpoint by an additional 0.20 mL, calculate the new reported concentration and explain why the previous cancellation no longer produces the correct result.

Solution 5

Student B transfers \(0.002400\text{ mol}\) of acid, obtains a titre of \(25.00\text{ mL}\), and reports the correct concentration of \(0.1000\text{ mol L}^{-1}\) by accidental cancellation. However, an additional 0.20 mL endpoint overshoot changes the reported result to \(0.1008\text{ mol L}^{-1}\).

This question demonstrates why two systematic errors can sometimes cancel rather than simply add together.

Part a: Calculate the actual moles of acid transferred.

Student B’s pipette delivers only 24.00 mL of the original acid solution.

Therefore:

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

The remaining 1.00 mL is water and contributes no additional acid.

Part b: Calculate the actual titre.

The sodium hydroxide concentration is actually:

\[ c(\ce{NaOH})=0.0960\text{ mol L}^{-1} \]

The reaction has a \(1:1\) mole ratio.

Therefore:

\[ n(\ce{NaOH})=0.002400\text{ mol} \]

Calculate the volume:

\[ \begin{aligned} V(\ce{NaOH}) &=\frac{n}{c}\\ &=\frac{0.002400}{0.0960}\\ &=0.02500\text{ L} \end{aligned} \]

Hence:

\[ \boxed{V_{\text{titre}}=25.00\text{ mL}} \]

Student A also obtains a titre of 25.00 mL.

For Student A:

\[ \begin{aligned} n(\ce{acid}) &=0.1000\times0.02500\\ &=0.002500\text{ mol} \end{aligned} \]

Using 0.1000 mol L\(^{-1}\) sodium hydroxide requires 25.00 mL.

The two students obtain identical titres, even though the actual amounts of acid transferred are different.

Part c: Calculate Student B’s reported concentration.

Student B uses the labelled sodium hydroxide concentration:

\[ c_{\text{used}}=0.1000\text{ mol L}^{-1} \]

They also assume that the full 25.00 mL pipette volume contains the original acid solution.

Therefore:

\[ \begin{aligned} c_{\text{reported}} &=\frac{0.1000\times0.02500}{0.02500}\\ &=0.1000\text{ mol L}^{-1} \end{aligned} \]

The reported concentration is:

\[ \boxed{0.1000\text{ mol L}^{-1}} \]

This happens to match the true concentration.

Part d: Explain the cancellation.

There are two systematic errors.

The pipette error reduces the amount of acid transferred.

The fraction of the intended acid volume actually delivered is:

\[ \frac{24.00}{25.00}=0.960 \]

So Student B transfers 96.0% of the expected moles of acid.

The burette error also reduces the concentration of the sodium hydroxide.

The fraction of the labelled titrant concentration remaining is:

\[ \frac{0.0960}{0.1000}=0.960 \]

So the titrant is also at 96.0% of its intended concentration.

These errors have opposite effects on the titre:

• Less acid in the flask requires a smaller volume of titrant.
• More dilute titrant requires a larger volume to supply the required moles.

Because both changes are exactly 4.0%, they cancel in this particular experiment.

The recorded titre remains 25.00 mL.

Student B’s calculation therefore gives the correct numerical concentration despite the incorrect procedure.

This is an important distinction: an accurate numerical result does not necessarily demonstrate an accurate experimental method.

Part e: Calculate the result after overshooting by 0.20 mL.

The new recorded titre is:

\[ \begin{aligned} V_{\text{recorded}} &=25.00+0.20\\ &=25.20\text{ mL} \end{aligned} \]

Student B still uses the labelled concentration and nominal pipette volume.

Therefore:

\[ \begin{aligned} c_{\text{reported}} &=\frac{0.1000\times0.02520}{0.02500}\\ &=0.1008\text{ mol L}^{-1} \end{aligned} \]

The new reported concentration is:

\[ \boxed{0.1008\text{ mol L}^{-1}} \]

This is 0.80% higher than the true concentration.

The extra 0.20 mL introduces a further error into the recorded titre. The original cancellation no longer balances all the errors.

The deeper lesson is that systematic errors can interact. You cannot simply count how many mistakes occurred and predict whether the final answer will be high or low.

You must consider the size and direction of each error in the actual calculation.

08What to remember when tracing titration errors

The most reliable approach is to think about what physically happens before reaching for the concentration equation.

There are two final distinctions worth keeping in mind.

First, precision and accuracy are not the same. Concordant titres indicate that repeated measurements agree. They do not eliminate systematic errors such as a diluted standard solution or an incorrectly conditioned pipette.

Second, the direction rules in this article apply to direct titrations where the unknown analyte is in the flask and the known titrant is in the burette. In a back titration, the unknown amount is often calculated by subtracting the measured excess reagent from the original amount added. An error in the second titration can therefore travel through the subtraction in a different direction.

The next useful step is understanding how experimental uncertainty affects the reliability of a calculated concentration. Once you can distinguish random variation from systematic bias, you can explain not only whether a titration result is wrong, but also why repeating the experiment may or may not fix it.

For another application of the mole-ratio method, see Precipitation Titrations: Calculate an Unknown Concentration. The same stoichiometric reasoning applies, but the endpoint chemistry and experimental considerations are different.