How to Calculate an Average Titre from Concordant Results

Imagine your accurate titration runs give titres of 24.62 mL, 24.65 mL, 24.63 mL, and 24.98 mL. A straight average of all four is 24.72 mL, but that includes a result that does not agree with the others.

To calculate an average titre from concordant results, calculate each titre by subtracting the initial burette reading from the final reading, identify the accurate results that agree within the accepted tolerance, and calculate their arithmetic mean. Exclude the rough titration and any results that are not concordant.

For the examples in this article, we will use a maximum difference of 0.10 mL between concordant titres.

01What is a titre?

A titre is the volume of solution delivered from a burette during one titration.

In a typical acid-base titration:

• The burette contains the titrant, usually a solution of known concentration.
• The volumetric pipette transfers a fixed volume of the analyte, the solution being analysed.
• The conical flask holds the analyte and a few drops of indicator.
• The indicator changes colour near the equivalence point, allowing you to identify the endpoint.

The burette has a graduated scale that starts at zero near the top and increases downwards. You record its reading before and after the titration.

The titre is the difference between these readings:

\[ \boxed{\text{Titre}=\text{Final reading}-\text{Initial reading}} \]

For example, suppose the initial reading is 0.35 mL and the final reading is 24.98 mL.

\[ \begin{aligned} \text{Titre}&=24.98-0.35\\ &=24.63\text{ mL} \end{aligned} \]

The titre is 24.63 mL.

Notice that the initial reading does not have to be zero. You must subtract it every time.

Also, \(1\text{ mL}=1\text{ cm}^3\), so either unit is acceptable if you use it consistently.

Reading the burette accurately

A small reading error can affect every calculation that follows.

For a typical school burette:

• Read the bottom of the concave meniscus for a clear aqueous solution.
• Position your eyes level with the meniscus to avoid parallax error.
• Record readings to two decimal places, estimating the second decimal place.
• Use the same reading method for both the initial and final measurements.

For example, record 24.65 mL rather than 24.6 mL.

Do not calculate an average from final burette readings alone. Calculate each individual titre first.

02Condition your apparatus before titrating

Before calculating an average titre, you need to understand how the apparatus affects the measurements.

The most important principle is that different pieces of glassware must preserve different things.

A burette and pipette must preserve the concentration of the solution they contain or transfer. A conical flask must preserve the number of moles of analyte.

Conditioning the burette

First, clean the burette with deionised water.

Then rinse it with small portions of the titrant that will go inside it. Rotate the burette so the solution contacts the internal walls, and allow the rinse liquid to drain.

Why?

Imagine filling a burette with 0.1000 mol L⁻¹ sodium hydroxide after rinsing it with water only. Some water remains inside and dilutes the sodium hydroxide.

The actual concentration is now lower than 0.1000 mol L⁻¹.

You need a greater volume of this diluted solution to react with the same amount of analyte. Your measured titre will therefore be too high.

Rinsing with the titrant prevents this dilution.

After filling the burette:

1. Open the tap briefly to fill the delivery tip and remove any air bubbles.
2. Check that no bubbles remain in the tip.
3. Remove the filling funnel.
4. Record the initial burette reading.

The funnel must be removed because liquid dripping from it could change the burette reading after you have recorded it.

Conditioning the pipette

The pipette must also be rinsed with the solution it will transfer.

For example, if you are transferring hydrochloric acid, rinse the pipette with hydrochloric acid before measuring the final aliquot.

If you leave water inside the pipette, the acid becomes diluted.

The pipette still delivers its calibrated volume, but that volume now contains fewer moles of acid.

Consequently, less sodium hydroxide is needed to neutralise it, and the measured titre becomes too low.

When transferring the solution:

• Use a pipette filler.
• Draw the solution above the calibration mark.
• Adjust the bottom of the meniscus to the mark at eye level.
• Allow the pipette to drain vertically into the conical flask.
• Touch the pipette tip against the inside wall of the flask.

A standard volumetric pipette is designed to deliver its calibrated volume with a small amount of liquid remaining in the tip. Do not blow this residue out unless the pipette is specifically designed as a blow-out pipette.

Why is the conical flask rinsed with water?

This is a common source of confusion.

Suppose you transfer 25.00 mL of hydrochloric acid into a conical flask. The flask also contains 10 mL of deionised water left over from cleaning.

Has the number of moles of hydrochloric acid changed?

No.

The additional water dilutes the acid, but it does not remove or add any hydrochloric acid.

The same number of moles still needs to react with the sodium hydroxide. Therefore, the theoretical titre remains unchanged.

However, rinsing the flask with hydrochloric acid would introduce extra, unmeasured moles of acid. More sodium hydroxide would then be required, increasing the titre.

Remember:

• Water in the conical flask: no change to the number of analyte moles.
• Water in the pipette: can dilute the measured aliquot.
• Water in the burette: can dilute the titrant.

03Make a consistent endpoint decision

Getting concordant results depends on more than reading the burette correctly.

You must also stop each titration at the same chemical endpoint.

The equivalence point is when the reactants have combined in the exact stoichiometric proportions given by the balanced chemical equation.

The endpoint is the observable signal, usually an indicator colour change, that tells you the titration is complete.

These points are not automatically identical. A suitable indicator is chosen so that its colour change occurs very close to the equivalence point.

What should you look for?

Consider a titration where sodium hydroxide is added from the burette to hydrochloric acid containing phenolphthalein.

Phenolphthalein is colourless in acidic solution and pink in alkaline solution.

As sodium hydroxide enters the flask, the indicator may briefly turn pink where the drop enters. Swirling distributes the sodium hydroxide throughout the solution, and the pink colour disappears while excess acid remains.

As you approach the endpoint, the colour takes longer to disappear.

You should then:

1. Add sodium hydroxide one drop at a time.
2. Swirl the flask after each addition.
3. Rinse any solution clinging to the flask walls into the mixture using deionised water.
4. Stop when the first faint pink colour persists after mixing.

A dark pink solution usually indicates that too much sodium hydroxide has been added.

Why does overshooting matter?

Suppose the correct endpoint requires 24.60 mL of sodium hydroxide.

You accidentally add another 0.20 mL before stopping.

Your recorded titre becomes:

\[ 24.60+0.20=24.80\text{ mL} \]

Your measured titre is now 0.20 mL too high.

If you overshoot every trial by approximately the same amount, you may still obtain concordant results. However, your average will also be too high.

This is the difference between precision and accuracy:

• Precision describes how closely repeated measurements agree.
• Accuracy describes how close a measurement is to the correct value.

Concordant results improve confidence in precision. They do not automatically guarantee accuracy.

The role of the rough titration

Your first titration is usually a rough titration.

Its purpose is to estimate the volume needed to reach the endpoint.

For example, if your rough titre is approximately 24.50 mL, you know to slow down as you approach that volume during subsequent trials.

The rough titre is not included in your average, even if it happens to be close to your accurate results.

Accurate titrations should use fresh aliquots, consistent indicator amounts, and the same endpoint technique.

04Identify concordant titres

Now you can decide which results belong in your average.

Concordant titres are repeated titre measurements that agree within an accepted tolerance.

For this article, we will use the common classroom criterion:

\[ \text{Maximum spread}\leq 0.10\text{ mL} \]

The maximum spread is calculated by subtracting the smallest titre from the largest titre in the selected group.

For example, consider these accurate titres:

The maximum spread is:

\[ 24.65-24.62=0.03\text{ mL} \]

Since 0.03 mL is less than 0.10 mL, all three titres are concordant.

Now consider another result of 24.98 mL.

Its difference from the other measurements is too large for it to belong to this concordant group.

You should not include it in the average.

A note about the tolerance: A tolerance of 0.10 mL is used in many school and university laboratory procedures. Other protocols specify different tolerances. For example, the University of Sydney Chemistry Laboratory Manual uses 0.10 mL, while the Royal Australian Chemical Institute’s Guide to Titration discusses agreement of approximately 1%. Always follow the tolerance specified in your practical instructions.

The decision process

Use this sequence whenever you are given a table of titration results.

1. Calculate the titre for every trial using final reading minus initial reading.
2. Identify and exclude the rough titration.
3. Compare the accurate titres.
4. Find a group whose maximum spread satisfies the stated concordance tolerance.
5. Exclude non-concordant results from the mean.
6. Calculate the arithmetic mean of the selected titres.
7. Round the final answer to the same decimal precision as the recorded measurements.

If your results do not form a concordant group, perform additional accurate titrations rather than forcing an average.

05Worked examples

06How experimental errors affect your average titre

When analysing titration results, you should be able to explain not just that an error occurs, but also which direction it moves the measured titre.

For the following table, assume that:

• The known titrant is in the burette.
• The unknown analyte is transferred using a volumetric pipette.
• The calculated analyte concentration uses the stated titrant concentration and fixed pipette volume.

A positive titre error means the measured titre is too high. A negative titre error means it is too low.

Why doesn’t averaging remove these errors?

Averaging is useful for reducing the effect of random errors.

For example, slight differences in judging the exact endpoint colour may produce titres that vary slightly above and below the correct value.

Taking an average helps reduce the effect of this random variation.

However, consider a student who overshoots the endpoint by approximately 0.15 mL in every trial.

Their results might be:

• 24.75 mL
• 24.76 mL
• 24.74 mL

These titres are concordant because their maximum spread is only 0.02 mL.

Their average is:

\[ \frac{24.75+24.76+24.74}{3} =24.75\text{ mL} \]

If the correct titre was 24.60 mL, the average is still 0.15 mL too high.

Repeating the same mistake does not eliminate it.

This is a systematic error, and it must be addressed by correcting the experimental technique rather than simply collecting more results.

07Questions and solutions

Question 1

A student obtains four accurate titres:

The accepted maximum difference between concordant titres is 0.10 mL.

Calculate the average titre and identify which result should be excluded.

Solution 1

The average titre is 23.46 mL, using Trials 1, 2, and 3 only.

First, calculate the spread of the first three results:

\[ 23.49-23.44=0.05\text{ mL} \]

They are concordant because their spread is below 0.10 mL.

Trial 4 is excluded because it is too far from the other results.

Now calculate the mean:

\[ \begin{aligned} \text{Average titre} &=\frac{23.44+23.49+23.45}{3}\\ &=\frac{70.38}{3}\\ &=\boxed{23.46\text{ mL}} \end{aligned} \]

The important point is that you select the concordant results before calculating the mean.

Question 2

A student titrates 25.00 mL of a monoprotic acid using 0.1000 mol L⁻¹ sodium hydroxide.

The three accurate titres are:

• 18.42 mL
• 18.45 mL
• 18.43 mL

The student discovers two things:

1. The volumetric pipette was rinsed with deionised water only, and a film of water remained inside.
2. The conical flask contained 12.0 mL of deionised water from cleaning.

Using a concordance tolerance of 0.10 mL:

(a) Calculate the average titre.

(b) Explain which experimental mistake changes the titre and predict the direction of the error.

(c) Explain whether the additional water in the conical flask affects the titre.

Solution 2

The average titre is 18.43 mL. The wet pipette introduces a systematic error that makes the titre too low, while the extra water in the conical flask has no theoretical effect on the titre.

Step 1: Check concordance.

\[ 18.45-18.42=0.03\text{ mL} \]

All three titres are concordant.

Step 2: Calculate the mean.

\[ \begin{aligned} \text{Average titre} &=\frac{18.42+18.45+18.43}{3}\\ &=\frac{55.30}{3}\\ &=\boxed{18.43\text{ mL}} \end{aligned} \]

Step 3: Analyse the pipette error.

The water remaining inside the pipette dilutes the acid.

Although the pipette still delivers 25.00 mL, that volume contains fewer moles of acid than it should.

Less sodium hydroxide is therefore needed to reach the equivalence point.

The measured titre is too low, and the calculated acid concentration will be underestimated if the student assumes the original acid concentration was delivered.

Step 4: Analyse the conical flask water.

The additional 12.0 mL of water dilutes the acid but does not change the number of acid moles.

The same amount of sodium hydroxide is still required for complete neutralisation.

Therefore, the extra water in the conical flask does not change the theoretical titre.

The distinction is between diluting a measured sample before it enters the flask and simply increasing the volume of liquid already in the flask.

Question 3

A student uses a burette to deliver sodium hydroxide into a conical flask.

The initial burette reading is 0.20 mL, and the final reading is 24.90 mL.

However, an air bubble with a volume of 0.10 mL was present in the burette tip before titration. During the titration, this bubble was expelled. Assume the 0.10 mL of liquid used to fill the tip did not enter the conical flask.

(a) Calculate the recorded titre.

(b) Calculate the actual volume of sodium hydroxide delivered into the conical flask.

(c) Explain whether repeating this procedure and averaging three concordant titres would eliminate the error.

Solution 3

The recorded titre is 24.70 mL, but the actual volume entering the conical flask is only 24.60 mL. Repeating the same procedure will not eliminate this systematic error.

Step 1: Calculate the recorded titre.

\[ \begin{aligned} \text{Titre} &=24.90-0.20\\ &=24.70\text{ mL} \end{aligned} \]

Step 2: Account for the air bubble.

The burette reading measures the volume that has left the main burette section.

However, 0.10 mL of that volume was used to fill the previously empty part of the delivery tip.

It did not enter the conical flask.

Therefore:

\[ \begin{aligned} V_{\text{delivered}} &=24.70-0.10\\ &=\boxed{24.60\text{ mL}} \end{aligned} \]

The recorded titre overestimates the volume that actually reacted.

Step 3: Consider repeated trials.

If the same air bubble error occurs during every titration, each recorded titre will be approximately 0.10 mL too high.

The results may still be concordant because they can agree closely with one another.

However, their average will also be too high.

The correct solution is to remove air bubbles from the burette tip before recording the initial reading, not to rely on averaging.

Question 4

A student titrates a weak monoprotic acid with sodium hydroxide.

Two sets of accurate titrations are performed using different indicators.

Set A: Phenolphthalein

• 22.34 mL
• 22.36 mL
• 22.35 mL

Set B: Methyl orange

• 21.90 mL
• 21.91 mL
• 21.92 mL

Both sets satisfy the 0.10 mL concordance criterion.

(a) Calculate the average titre for each set.

(b) Identify which set should be used to calculate the acid concentration.

(c) Explain why both sets can be concordant even though their average titres are different.

(d) Explain why averaging all six titres together would be inappropriate.

Solution 4

Set A should be used, giving an average titre of 22.35 mL. Set B gives a lower but concordant result because methyl orange changes colour before the equivalence point in this weak acid-strong base titration.

Step 1: Calculate the average for Set A.

\[ \begin{aligned} \bar V_A &=\frac{22.34+22.36+22.35}{3}\\ &=\boxed{22.35\text{ mL}} \end{aligned} \]

Step 2: Calculate the average for Set B.

\[ \begin{aligned} \bar V_B &=\frac{21.90+21.91+21.92}{3}\\ &=\boxed{21.91\text{ mL}} \end{aligned} \]

Step 3: Explain the difference between the indicators.

The equivalence point of a weak acid-strong base titration occurs in the basic pH range.

Phenolphthalein changes colour within the steep pH change near this equivalence point, making it a suitable indicator.

Methyl orange changes colour in a much more acidic pH range.

Its endpoint is therefore reached before the equivalence point, meaning less sodium hydroxide has been added than is chemically required for complete neutralisation.

This produces a titre that is systematically too low.

Step 4: Explain why both sets are concordant.

Concordance measures agreement between repeated results.

The student can use the same indicator consistently and obtain very similar readings each time, even if that indicator is unsuitable.

Therefore, concordance does not establish that an indicator is appropriate.

Step 5: Explain why all six titres should not be averaged together.

The two sets represent different endpoint conditions.

Their combined average would be:

\[ \begin{aligned} \bar V &=\frac{67.05+65.73}{6}\\ &=\boxed{22.13\text{ mL}} \end{aligned} \]

Although this calculation is mathematically correct, the result is not a valid average titre for the intended titration.

It combines measurements made using two different endpoint criteria. You must select the suitable indicator before deciding which results are chemically meaningful.

Question 5

A student is titrating a monoprotic acid with sodium hydroxide. The reaction has a 1:1 mole ratio.

Under correctly prepared conditions, a 25.00 mL aliquot of 0.1000 mol L⁻¹ acid requires 25.00 mL of 0.1000 mol L⁻¹ sodium hydroxide.

However, the student makes several experimental mistakes:

• The 50.00 mL working solution of sodium hydroxide contains 48.00 mL of 0.1000 mol L⁻¹ sodium hydroxide and 2.00 mL of residual water.
• The 25.00 mL pipette delivers a mixture containing 24.50 mL of the original acid solution and 0.50 mL of residual water.
• The conical flask already contains 10.0 mL of deionised water.

The student obtains three accurate titres:

• 25.51 mL
• 25.52 mL
• 25.53 mL

Assume complete mixing, additive solution volumes, and no other experimental errors.

(a) Calculate the average titre and determine whether the results are concordant.

(b) Determine how each of the two conditioning errors affects the titre. Do they cancel each other?

(c) Calculate the titre predicted by the actual concentrations and amounts delivered.

(d) Calculate the acid concentration the student would report if they incorrectly assumed the sodium hydroxide concentration was still 0.1000 mol L⁻¹ and the pipette had delivered 25.00 mL of undiluted acid.

(e) Explain why the additional water in the conical flask does not change the result, and why concordant readings do not prove that the experiment was accurate.

Solution 5

The average titre is 25.52 mL, and the three results are concordant. However, the student overestimates the acid concentration by approximately 2.08% because the two conditioning errors do not cancel.

This question is difficult because two errors affect the titre in opposite directions. You must compare their magnitudes rather than assume they cancel.

Step 1: Calculate the average titre.

\[ \begin{aligned} \bar V &=\frac{25.51+25.52+25.53}{3}\\ &=\frac{76.56}{3}\\ &=\boxed{25.52\text{ mL}} \end{aligned} \]

The maximum spread is:

\[ 25.53-25.51=0.02\text{ mL} \]

The results are concordant.

Step 2: Calculate the actual sodium hydroxide concentration.

The working solution contains 48.00 mL of the original sodium hydroxide solution in a total volume of 50.00 mL.

Therefore:

\[ \begin{aligned} c_{\mathrm{NaOH}} &=0.1000\times\frac{48.00}{50.00}\\ &=\boxed{0.09600\text{ mol L}^{-1}} \end{aligned} \]

The residual water has reduced the titrant concentration by 4%.

This means a greater volume of sodium hydroxide is needed to deliver the required number of moles.

The titre is pushed upwards.

Step 3: Calculate the actual moles of acid delivered.

The pipette delivers 24.50 mL of the original acid solution, rather than 25.00 mL.

Convert the volume into litres:

\[ 24.50\text{ mL}=0.02450\text{ L} \]

Calculate the moles:

\[ \begin{aligned} n_{\mathrm{acid}} &=cV\\ &=0.1000\times0.02450\\ &=\boxed{0.002450\text{ mol}} \end{aligned} \]

The original acid aliquot would have contained 0.002500 mol.

The pipette error has therefore reduced the delivered acid moles by 2%.

Fewer moles of acid require less sodium hydroxide, pushing the titre downwards.

Step 4: Calculate the actual titre.

The reaction has a 1:1 mole ratio, so the sodium hydroxide must supply 0.002450 mol.

Using the actual sodium hydroxide concentration:

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

\[ \begin{aligned} V_{\mathrm{NaOH}} &=\frac{0.002450}{0.09600}\\ &=0.0255208\text{ L}\\ &=\boxed{25.52\text{ mL}} \end{aligned} \]

This agrees with the student’s measured average.

Notice how the two errors interact:

• The pipette error reduces the acid moles by 2%, lowering the titre.
• The burette error reduces the titrant concentration by 4%, increasing the titre.

Their combined effect is:

\[ \frac{V_{\mathrm{measured}}}{V_{\mathrm{correct}}} =\frac{0.98}{0.96} =1.02083 \]

The titre is therefore approximately 2.08% higher than the correct value.

The errors do not cancel because their relative magnitudes are different.

Step 5: Calculate the concentration the student would report.

The student incorrectly assumes:

• Sodium hydroxide concentration = 0.1000 mol L⁻¹.
• Acid volume = 25.00 mL.
• Reaction mole ratio = 1:1.

Using the measured average titre:

\[ n_{\mathrm{NaOH}}=cV \]

\[ n_{\mathrm{NaOH}} =0.1000\times0.02552 =0.002552\text{ mol} \]

The student assumes these are also the moles of acid.

Therefore:

\[ \begin{aligned} c_{\mathrm{acid}} &=\frac{n}{V}\\ &=\frac{0.002552}{0.02500}\\ &=\boxed{0.1021\text{ mol L}^{-1}} \end{aligned} \]

The actual acid concentration was 0.1000 mol L⁻¹.

The percentage error is:

\[ \begin{aligned} \%\text{ error} &=\frac{0.10208-0.1000}{0.1000}\times100\\ &=\boxed{2.08\%\text{ high}} \end{aligned} \]

Step 6: Explain the effect of water in the conical flask.

The 10.0 mL of deionised water increases the total volume of liquid in the flask.

However, it does not change the number of moles of acid being titrated.

Therefore, it has no effect on the theoretical titre.

Finally, the three measurements are concordant because they agree closely with one another. They are not necessarily accurate because the same systematic errors were present in every trial.

This is the key distinction: a precise average can still produce an inaccurate concentration if the experimental method is flawed.

07What comes after the average titre?

Once you have calculated a reliable average titre, the next step is usually to determine the unknown concentration using reaction stoichiometry.

The average titre gives you the volume of titrant required to react with the measured analyte aliquot.

You can then use:

\[ n=cV \]

where:

• \(n\) is the number of moles in mol.
• \(c\) is the concentration in mol L⁻¹.
• \(V\) is the volume in litres.

Use the balanced chemical equation to establish the mole ratio between the titrant and analyte. Then calculate the analyte concentration using the original pipetted volume.

Keep the two stages separate:

1. Use concordant titres to obtain the average titre.
2. Use the average titre and balanced equation to calculate the unknown concentration.

Getting the first stage right gives you a sound starting point for every calculation that follows.