How to Read a Burette Meniscus and Calculate a Titre

Your burette reads 1.42 cm³ before a titration and 24.86 cm³ afterwards. A student records 24.86 cm³ as the titre because that is where the liquid finished.

Is that correct?

No. The final reading tells you the position of the liquid, not the volume delivered. The actual titre is:

\[ 24.86-1.42=23.44\text{ cm}^3 \]

Getting this measurement right starts with knowing exactly where to read the curved surface of the liquid.

01How to read a burette meniscus correctly

For a clear aqueous solution, always read the bottom of the concave meniscus at eye level.

A meniscus is the curved surface that forms where a liquid meets the walls of a container. Water is attracted to glass, so the liquid rises slightly at the edges and dips in the middle.

Think of it as a shallow bowl. You want the lowest point at the centre, not either of the raised edges.

There are four things to remember when taking a reading.

1. Read the bottom, not the top.

For ordinary clear aqueous titrants, the lowest point of the meniscus is your reference.

2. Keep your eye level with the meniscus.

Looking down or up changes the apparent position of the liquid against the scale. This is called parallax error.

The reading may look reasonable, but it won’t represent the actual position of the meniscus.

3. Remember that the burette scale increases downwards.

Unlike most measuring cylinders, a burette has its zero mark at the top. As liquid leaves the burette, the meniscus moves down and the numerical reading increases.

4. Record two decimal places.

A typical school burette has graduations every \(0.1\text{ cm}^3\). You should estimate one additional decimal place, giving readings to \(0.01\text{ cm}^3\).

For example:

• \(12.30\text{ cm}^3\) is acceptable.
• \(12.3\text{ cm}^3\) does not show the expected precision.
• \(12.304\text{ cm}^3\) claims more precision than the instrument supports.

The final decimal place is estimated between the marked graduations. It isn’t another directly marked division.

Remember that \(1\text{ cm}^3=1\text{ mL}\), so the numerical readings are identical in either unit.

02Why you need an initial and final reading

A titration presents a measurement problem: you don’t know beforehand exactly how much titrant will be needed to reach the endpoint.

A measuring cylinder can measure liquid volume, but it isn’t designed to deliver small, controlled amounts accurately near the endpoint.

A burette solves this problem using three main parts:

• A long, graduated glass tube that holds the titrant.
• A stopcock (tap) near the bottom that controls the flow.
• A narrow delivery tip that allows small quantities to enter the conical flask.

The titrant enters through the top, travels down the tube, passes through the stopcock, and exits through the tip into the flask.

As the titrant leaves, the meniscus moves down the graduated scale.

The scale tells you how much liquid would have been delivered if the burette had started at zero.

But what if it didn’t?

That’s why you record both readings.

The volume delivered, also called the titre, is:

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

Where:

• \(V_f\) = final burette reading, in \(\text{cm}^3\).
• \(V_i\) = initial burette reading, in \(\text{cm}^3\).

The sign convention is important: subtract the initial reading from the final reading.

The final reading should normally be larger because the scale increases downwards as liquid is delivered.

There is no requirement for your initial reading to be exactly \(0.00\text{ cm}^3\). Starting at \(1.35\text{ cm}^3\), for example, is perfectly acceptable if you record it correctly.

03Condition your apparatus before taking measurements

Reading the meniscus correctly won’t save a titration if the apparatus contains leftover water that changes the concentration of your solutions.

This is where conditioning comes in.

Conditioning means rinsing volumetric equipment with the solution it will contain or transfer immediately before use. It removes residual water that could dilute the solution.

The burette, pipette, and conical flask have different preparation requirements.

Preparing the burette

Follow this sequence:

Step 1: Clean the burette.

Rinse it with deionised water to remove leftover chemicals or contaminants.

Step 2: Rinse with the titrant.

Add a small amount of the solution that will go into the burette.

Rotate the burette so the solution contacts the internal walls, then discard the rinse.

Repeat with fresh small portions.

This matters because a thin film of water can remain on the glass even after draining. If you fill the burette without conditioning it, that water dilutes the titrant.

Step 3: Fill the burette.

Clamp it vertically and use a funnel to add the titrant.

Step 4: Fill the tip.

Open the stopcock briefly so that titrant flows through the tap and fills the narrow delivery tip.

This removes air bubbles from the tip and stopcock.

Check that no bubbles remain. If there is an air bubble, the measured volume may include liquid that is used to fill the tip rather than delivered into the flask.

Step 5: Remove the funnel.

This step is easy to forget.

A drop falling from the funnel after the initial reading can enter the burette and change its liquid level. Your initial reading would no longer represent the actual starting point.

Step 6: Record the initial reading.

Read the bottom of the meniscus at eye level and record to two decimal places.

For safety, keep the top of the burette below eye level. Adjust your position to read the meniscus rather than raising the apparatus above your face.

Transferring the sample with a pipette

The burette delivers a variable volume of titrant. The volumetric pipette transfers a fixed volume of the solution being analysed.

The pipette also needs conditioning.

Rinse it with deionised water first, then rinse it with small portions of the solution you intend to transfer.

Use a pipette filler to draw the solution above the calibration mark.

Lower the liquid until the bottom of the meniscus aligns exactly with the calibration line, with your eye at the same height.

Allow the pipette to drain into the conical flask under gravity. Touch the tip against the inside wall of the flask and allow the normal drainage time.

Don’t blow out the final drop unless the pipette is specifically calibrated as a blow-out pipette. A standard volumetric pipette is designed to deliver its stated volume while retaining a small amount of liquid inside the tip.

Why can you rinse the conical flask with water?

This is a common point of confusion.

You can rinse the conical flask with deionised water before transferring your sample.

Why?

Because adding water changes the volume and concentration of the solution in the flask, but it does not change the number of moles of analyte transferred by the pipette.

The number of moles determines how much titrant is required for the reaction.

However, rinsing the flask with the analyte solution would introduce additional, unmeasured moles of analyte. That would increase the titre.

The distinction is:

• Water in the conical flask: no change to the number of moles of analyte.
• Water in the pipette: dilutes the measured aliquot, reducing the moles delivered.
• Water in the burette: dilutes the titrant, reducing its concentration.

These three situations produce different experimental results.

04Making the correct endpoint decision

You have prepared the apparatus, transferred the sample, and recorded the initial burette reading.

Now you need to decide when to stop adding titrant.

The endpoint is the observed point at which the indicator changes colour. It is used as an experimental estimate of the equivalence point, where the reactants have combined in the correct stoichiometric proportions.

These two terms are related, but they aren’t identical.

The equivalence point is defined by the reaction stoichiometry. The endpoint is the visible indicator change. A suitable indicator is chosen so that its colour change occurs close to the equivalence point.

What should you actually look for?

Consider an example where hydrochloric acid is in the conical flask and sodium hydroxide is in the burette.

Suppose you’re using phenolphthalein as the indicator.

The initial solution is colourless. As sodium hydroxide enters, the pH increases.

Near the endpoint, a small amount of sodium hydroxide can produce a noticeable colour change.

Your aim is to stop at the first faint pink colour that remains after swirling.

A good endpoint decision follows this process:

1. Add titrant relatively quickly at the beginning.

Swirl the conical flask continuously to mix the reactants.

2. Slow down as you approach the endpoint.

If the indicator colour starts to linger after swirling, stop adding large amounts.

3. Add titrant drop by drop.

Swirl after each addition. Near the endpoint, even a fraction of a drop can matter.

4. Wash down the flask walls if necessary.

Use a wash bottle containing deionised water to rinse any droplets back into the main solution.

This doesn’t change the moles of reactants in the flask.

5. Stop at the first permanent indicator colour change.

Don’t continue adding titrant just to make the colour more obvious.

A deep pink colour with phenolphthalein generally indicates that too much sodium hydroxide has been added.

What happens if you overshoot?

Suppose the true endpoint occurs after \(24.50\text{ cm}^3\) of sodium hydroxide has been delivered.

You accidentally add another \(0.20\text{ cm}^3\) before stopping.

Your recorded titre becomes:

\[ 24.50+0.20=24.70\text{ cm}^3 \]

The measured titre is \(0.20\text{ cm}^3\) too large.

If you’re using this titre to calculate the concentration of an unknown acid, you will calculate a concentration that is too high, assuming the titrant concentration and pipetted volume are correct.

The extra sodium hydroxide is a real addition to the flask. It isn’t a reading error.

06Common burette errors and their direction

When you make a measurement error, don’t just state that the result becomes inaccurate. Trace what physically happens to the liquid, then determine how that affects the titre.

Assume throughout this table that the titrant concentration is known and the volume of analyte transferred by the pipette is fixed.

Let’s examine the reasoning behind these errors.

Water remaining in the burette

The water dilutes the titrant.

Its actual concentration is lower than the labelled concentration. More volume is therefore needed to react with the same number of moles of analyte.

The titre increases.

If you still use the original labelled concentration in your calculation, you overestimate the number of moles of titrant delivered. This makes the calculated analyte concentration too high.

Water remaining in the pipette

The water dilutes the analyte inside the pipette.

The fixed volume transferred now contains fewer moles of analyte than it should.

Less titrant is needed to reach the endpoint, so the titre decreases.

If you use the original titrant concentration and the nominal pipette volume in your calculation, the calculated analyte concentration is too low.

Water in the conical flask

The water increases the total volume of the solution, but doesn’t add more analyte.

The same number of moles still needs to react.

Therefore, the stoichiometric titre remains unchanged, assuming the indicator still functions appropriately.

An air bubble in the burette tip

Imagine the tip contains an air bubble when you record the initial reading.

As you open the tap, some titrant leaves the main burette tube to fill the space occupied by the bubble.

That liquid doesn’t initially enter the conical flask.

The meniscus falls, so the burette records a larger volume than the amount of titrant actually delivered into the flask.

The measured titre is too large.

A note about reading errors

Not every reading error has a fixed direction.

Suppose the final reading is recorded \(0.04\text{ cm}^3\) too high.

The calculated titre is also \(0.04\text{ cm}^3\) too high.

But if the initial reading is recorded \(0.04\text{ cm}^3\) too high, the calculated titre is \(0.04\text{ cm}^3\) too low.

You can express this using:

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

Here, \(\delta V_f\) is the error in the final reading and \(\delta V_i\) is the error in the initial reading.

A positive reading error means the recorded numerical reading is larger than the true reading.

This explains why consistent reading technique matters. Even if two readings have the same numerical offset, their errors may cancel in the subtraction. You shouldn’t rely on this happening.

07Questions and solutions

Question 1

A student records an initial burette reading of \(2.38\text{ cm}^3\) and a final reading of \(26.71\text{ cm}^3\).

Calculate the titre.

Solution 1

The titre is \(24.33\text{ cm}^3\).

The volume delivered is the difference between the final and initial readings:

\[ \begin{aligned} V_{\text{titre}}&=V_f-V_i\\ &=26.71-2.38\\ &=24.33\text{ cm}^3 \end{aligned} \]

The common mistake is to record \(26.71\text{ cm}^3\) as the titre. That is only the final scale reading.

The burette measures the change in its reading, not simply the position where the liquid finishes.

Question 2

A student performs a titration using a \(25.00\text{ cm}^3\) aliquot of a monoprotic acid in a conical flask and \(0.1000\text{ mol L}^{-1}\) sodium hydroxide in the burette.

The student correctly conditions the pipette but rinses the burette with deionised water and fills it with sodium hydroxide without first conditioning it with the titrant.

A small amount of water remains in the burette and mixes with the sodium hydroxide.

The conical flask has also been rinsed with deionised water.

Assume the water in the burette dilutes the titrant uniformly and there are no other experimental errors.

Explain:

(a) How the titre changes compared with a correctly prepared titration.

(b) How the calculated acid concentration changes if the student still uses \(0.1000\text{ mol L}^{-1}\) as the sodium hydroxide concentration.

(c) Why the water in the conical flask does not produce the same effect.

Solution 2

The titre becomes larger, and the calculated acid concentration becomes too high. However, the deionised water in the conical flask does not change the stoichiometric titre.

(a) Effect on the titre

The residual water dilutes the sodium hydroxide.

Its actual concentration is now lower than \(0.1000\text{ mol L}^{-1}\).

A lower concentration means fewer moles of sodium hydroxide are delivered per unit volume.

Therefore, a larger volume is needed to neutralise the fixed number of moles of acid in the flask.

The measured titre increases.

(b) Effect on the calculated acid concentration

The student still uses the original concentration of \(0.1000\text{ mol L}^{-1}\).

However, the actual sodium hydroxide concentration is lower.

Because the measured titre is larger, the student calculates too many moles of sodium hydroxide:

\[ n=cV \]

Using an incorrect concentration with the increased volume produces an overestimate of the moles delivered.

For a 1:1 reaction, this also overestimates the moles of acid.

The calculated acid concentration is therefore too high.

(c) Effect of water in the conical flask

The deionised water does not add any acid.

The number of moles transferred by the pipette remains unchanged.

The acid is more dilute in the flask, but the same number of moles of sodium hydroxide is still required to reach the stoichiometric equivalence point.

Therefore, the water in the flask does not change the titre under the stated assumptions.

The important distinction is where the water remains: in the burette it changes the titrant concentration, while in the conical flask it only changes the solution volume.

Question 3

During a titration, a student has an air bubble trapped in the burette delivery tip.

The initial burette reading is \(0.25\text{ cm}^3\), and the final reading is \(25.03\text{ cm}^3\).

The bubble is expelled during the titration.

Assume that \(0.16\text{ cm}^3\) of titrant leaves the main burette tube to fill the tip before any titrant reaches the conical flask.

The titration involves a monoprotic acid and \(0.1000\text{ mol L}^{-1}\) sodium hydroxide, with a \(25.00\text{ cm}^3\) acid aliquot.

(a) Calculate the titre recorded by the student.

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

(c) Explain the direction of error in the calculated acid concentration.

Solution 3

The recorded titre is \(24.78\text{ cm}^3\), but only \(24.62\text{ cm}^3\) reaches the conical flask. The calculated acid concentration is therefore too high.

(a) Recorded titre

Apply the standard formula:

\[ \begin{aligned} V_{\text{titre}}&=V_f-V_i\\ &=25.03-0.25\\ &=24.78\text{ cm}^3 \end{aligned} \]

(b) Actual volume delivered

The \(0.16\text{ cm}^3\) used to fill the tip is included in the burette reading, but it hasn’t reached the flask.

Subtract this amount:

\[ \begin{aligned} V_{\text{actual}}&=24.78-0.16\\ &=24.62\text{ cm}^3 \end{aligned} \]

Only \(24.62\text{ cm}^3\) of sodium hydroxide actually reacts with the acid.

(c) Effect on the calculated concentration

The student uses \(24.78\text{ cm}^3\) instead of the correct \(24.62\text{ cm}^3\).

For a 1:1 reaction:

\[ c_{\text{acid}}=\frac{c_{\text{NaOH}}V_{\text{NaOH}}}{V_{\text{acid}}} \]

Using the recorded titre:

\[ \begin{aligned} c_{\text{calculated}}&=\frac{0.1000\times24.78}{25.00}\\ &=0.09912\text{ mol L}^{-1} \end{aligned} \]

Using the actual delivered volume:

\[ \begin{aligned} c_{\text{correct}}&=\frac{0.1000\times24.62}{25.00}\\ &=0.09848\text{ mol L}^{-1} \end{aligned} \]

The student overestimates the acid concentration because the recorded titre includes liquid that never entered the conical flask.

This is why you should always check the burette tip for air bubbles before recording the initial reading.

Question 4

A student records the following burette readings:

• Initial reading: \(0.42\text{ cm}^3\)
• Final reading: \(24.68\text{ cm}^3\)

A later check establishes that:

• The initial reading was \(0.04\text{ cm}^3\) too high.
• The final reading was \(0.03\text{ cm}^3\) too low.
• The student overshot the true indicator endpoint by \(0.12\text{ cm}^3\).

Calculate:

(a) The titre recorded by the student.

(b) The actual volume delivered by the time the student stopped.

(c) The correct titre at the true indicator endpoint.

(d) Whether the student’s recorded titre is too high or too low, and by how much.

Solution 4

The student’s recorded titre is \(24.26\text{ cm}^3\), but the correct endpoint titre is \(24.21\text{ cm}^3\).

The reported titre is therefore \(0.05\text{ cm}^3\) too high.

The important part is keeping track of the direction of each error.

(a) Calculate the recorded titre

\[ \begin{aligned} V_{\text{recorded}}&=24.68-0.42\\ &=24.26\text{ cm}^3 \end{aligned} \]

(b) Correct the initial and final readings

The initial reading was too high, so the true initial reading is:

\[ V_{i,\text{true}}=0.42-0.04 \]

\[ V_{i,\text{true}}=0.38\text{ cm}^3 \]

The final reading was too low, so the true final reading is:

\[ V_{f,\text{true}}=24.68+0.03 \]

\[ V_{f,\text{true}}=24.71\text{ cm}^3 \]

Therefore, the actual volume delivered by the time the student stopped is:

\[ \begin{aligned} V_{\text{delivered}}&=24.71-0.38\\ &=24.33\text{ cm}^3 \end{aligned} \]

(c) Correct for the overshoot

The student added \(0.12\text{ cm}^3\) more titrant than required.

Subtract this from the actual delivered volume:

\[ \begin{aligned} V_{\text{endpoint}}&=24.33-0.12\\ &=24.21\text{ cm}^3 \end{aligned} \]

(d) Compare the recorded and correct titres

\[ \begin{aligned} \text{Difference}&=24.26-24.21\\ &=0.05\text{ cm}^3 \end{aligned} \]

The reported titre is \(0.05\text{ cm}^3\) too high.

Notice that the two reading errors actually made the recorded titre smaller than the volume delivered by the time the student stopped.

However, the endpoint overshoot was large enough to reverse the overall direction of the error.

This is why you should trace each error separately before deciding whether a result is too high or too low.

Question 5

A student performs an acid-base titration using a monoprotic acid and sodium hydroxide.

The original acid solution has a concentration of \(0.1000\text{ mol L}^{-1}\). The sodium hydroxide stock solution also has a concentration of \(0.1000\text{ mol L}^{-1}\).

The reaction has a 1:1 mole ratio, so a correctly conditioned \(25.00\text{ cm}^3\) pipette and burette would produce a titre of \(25.00\text{ cm}^3\).

However, two preparation errors occur in the same titration:

• The burette contains \(5.00\text{ cm}^3\) of residual water mixed with \(45.00\text{ cm}^3\) of the original sodium hydroxide solution.
• The \(25.00\text{ cm}^3\) volumetric pipette contains \(2.50\text{ cm}^3\) of residual water before being filled to its calibration mark. Therefore, only \(22.50\text{ cm}^3\) of the original acid solution is transferred.

Assume:

• Both mixtures are uniform.
• Volumes are additive.
• The pipette delivers the entire mixture within its calibrated volume.
• There are no other experimental errors.

(a) Calculate the actual sodium hydroxide concentration in the burette.

(b) Calculate the moles of acid transferred into the flask.

(c) Determine the observed titre.

(d) A student uses the nominal sodium hydroxide concentration and the nominal pipette volume to calculate the acid concentration. Determine the value obtained.

(e) Explain how the result can appear correct despite both preparation errors. Would obtaining repeated concordant titres necessarily reveal the problem?

Solution 5

The observed titre is exactly \(25.00\text{ cm}^3\), and the student calculates an acid concentration of \(0.1000\text{ mol L}^{-1}\).

Both results appear correct, even though the burette and pipette have been incorrectly prepared.

This is an example of two systematic errors cancelling each other.

(a) Actual sodium hydroxide concentration

The burette contains:

• \(45.00\text{ cm}^3\) of \(0.1000\text{ mol L}^{-1}\) sodium hydroxide.
• \(5.00\text{ cm}^3\) of water.

The total volume is \(50.00\text{ cm}^3\).

First, calculate the original moles of sodium hydroxide:

\[ \begin{aligned} n&=cV\\ &=0.1000\times0.04500\\ &=0.004500\text{ mol} \end{aligned} \]

The water adds volume but no sodium hydroxide.

The actual concentration is therefore:

\[ \begin{aligned} c_{\text{actual}}&=\frac{0.004500}{0.05000}\\ &=0.09000\text{ mol L}^{-1} \end{aligned} \]

The burette solution is now 10% less concentrated than the original stock solution.

(b) Moles of acid transferred

The pipette has a total calibrated volume of \(25.00\text{ cm}^3\).

However, \(2.50\text{ cm}^3\) of this volume is occupied by water.

Only \(22.50\text{ cm}^3\) of the original acid solution is transferred.

Calculate the moles of acid:

\[ \begin{aligned} n(\text{acid})&=0.1000\times0.02250\\ &=0.002250\text{ mol} \end{aligned} \]

The flask contains \(0.002250\text{ mol}\) of acid.

(c) Determine the observed titre

The reaction has a 1:1 mole ratio.

Therefore:

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

The required volume of sodium hydroxide is:

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

Substitute the actual sodium hydroxide concentration:

\[ \begin{aligned} V&=\frac{0.002250}{0.09000}\\ &=0.02500\text{ L}\\ &=25.00\text{ cm}^3 \end{aligned} \]

The observed titre is therefore \(25.00\text{ cm}^3\).

(d) Calculate the concentration using the nominal values

The student incorrectly assumes:

• Sodium hydroxide concentration is \(0.1000\text{ mol L}^{-1}\).
• Acid volume transferred is \(25.00\text{ cm}^3\).

Using the nominal values:

\[ c_{\text{acid}}= \frac{c_{\text{NaOH}}V_{\text{NaOH}}} {V_{\text{acid}}} \]

Substitute:

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

The calculated concentration matches the original acid concentration.

(e) Explain the cancellation

Here’s the important interaction.

The water in the burette reduces the sodium hydroxide concentration by 10%.

The water in the pipette reduces the amount of acid transferred by 10%.

Both reactants are affected by the same dilution factor of \(0.900\).

The acid aliquot contains only 90% of the expected moles, but the titrant also contains only 90% of the expected moles per unit volume.

These two effects cancel in the volume ratio.

The titre therefore remains exactly the same as it would have been with correctly prepared equipment.

Repeated concordant titres would not necessarily reveal the errors either. If the same preparation errors were reproduced consistently, the results could be very close together while still being obtained using an incorrect procedure.

This is the distinction between precision and accuracy:

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

A result can appear precise, and even give the correct calculated concentration, without proving that the experimental technique was sound.

The practical lesson is that you cannot use a plausible final answer to justify skipping apparatus conditioning.

08What to do with your burette readings next

Once you can read the meniscus reliably, calculate titres correctly, and identify concordant results, you’re ready to use the mean titre in a stoichiometric calculation.

The sequence is:

1. Use the bottom of the meniscus at eye level.
2. Record the initial and final readings to two decimal places.
3. Calculate each titre using final minus initial.
4. Identify concordant accurate trials and calculate their mean.
5. Convert the mean titre into moles using the titrant concentration.
6. Apply the balanced equation’s mole ratio to calculate the unknown concentration.

The next important decision is choosing an appropriate indicator. Even a perfectly read burette cannot give an accurate equivalence volume if the indicator changes colour too far from the actual equivalence point.

Reliable titration results depend on both: measuring the volume correctly and recognising the chemical endpoint correctly.