Choosing a Pipette and Burette for HSC Titrations

An unknown acid is sitting in a conical flask. You need to determine its concentration by titrating it with a solution of known concentration. You have a 25 mL volumetric pipette, a 50 mL burette, and a measuring cylinder. Which should you use, and why?

For a standard titration, use a volumetric pipette to transfer a fixed volume of the analyte into the conical flask, and a burette to deliver and measure the variable volume of titrant needed to reach the endpoint.

The distinction is simple, but choosing the wrong apparatus, conditioning it incorrectly, or misjudging the endpoint can introduce systematic errors into your calculated concentration.

01Choose the pipette and burette for different jobs

Think about what each measurement needs to achieve.

You need a precisely measured starting quantity of the solution being analysed. You also need to measure exactly how much of the second solution is required to react with it.

These are two different measurement tasks.

Choosing the volumetric pipette

A volumetric pipette is designed to transfer one specific, accurately measured volume of solution.

Common sizes include:

• 10.00 mL
• 20.00 mL
• 25.00 mL
• 50.00 mL

For example, if your experimental method requires a 25.00 mL aliquot of hydrochloric acid, choose a 25.00 mL volumetric pipette.

The pipette establishes the initial volume of analyte. You then use this known volume in your concentration calculation.

A measuring cylinder might also hold 25 mL, but it is generally less precise. Its wider graduations and larger measurement uncertainty make it a poorer choice for quantitative titration work.

Choosing the burette

A burette is designed to deliver a variable volume of solution.

Unlike a volumetric pipette, it has a graduated scale, allowing you to measure how much solution leaves the burette during a titration.

A common school burette has:

• A capacity of 50 mL.
• Graduations at 0.1 mL intervals.
• A stopcock to control the flow of solution.
• A narrow tip for controlled delivery.

The burette is particularly useful because you do not know the exact volume needed to reach the endpoint before performing the experiment.

You measure this volume by recording the initial and final burette readings.

\[ V_{\text{titre}}=V_{\text{final}}-V_{\text{initial}} \]

For example, if the initial reading is 0.35 mL and the final reading is 24.82 mL:

\[ V_{\text{titre}}=24.82-0.35 \]

\[ \boxed{V_{\text{titre}}=24.47\text{ mL}} \]

Notice that you do not need to start the burette at exactly 0.00 mL. You only need accurate initial and final readings.

How do you choose the capacity?

The pipette volume is determined by the aliquot required in the experimental method.

The burette capacity should be sufficient to accommodate the expected titre without running out of solution.

For many school titrations, a 50 mL burette is a practical choice, with titres often designed to fall around 20-30 mL. This is a useful experimental target, not a compulsory rule.

Consider these situations:

Why avoid extremely small titres?

The absolute uncertainty in measuring the burette readings becomes a larger percentage of a small titre. A larger titre, within the available capacity, generally gives a smaller relative contribution from reading uncertainty.

The important distinction

Do not choose the pipette or burette based on whether the solution is an acid or a base.

The choice depends on the measurement being made. In a typical analysis, the known-concentration solution goes into the burette, while the unknown solution is transferred into the flask using the pipette.

02Condition the pipette and burette correctly

Choosing the correct equipment is only half the job.

Before using either instrument, you must condition it. Conditioning means rinsing the inside with a small amount of the solution that will actually be placed in that instrument.

Why bother?

Imagine that your pipette contains a small amount of distilled water before you draw up the acid sample. That water mixes with the acid and dilutes the aliquot.

You still transfer 25.00 mL, but the transferred solution contains fewer moles of acid than expected.

The same problem occurs if water remains inside the burette. It dilutes the titrant, changing its actual concentration.

Conditioning a volumetric pipette

Suppose you are using a 25.00 mL pipette to transfer an unknown hydrochloric acid solution.

Follow this procedure:

1. Rinse the pipette with distilled or deionised water to remove contaminants.
2. Allow the water to drain.
3. Rinse the inside with a small volume of the hydrochloric acid solution.
4. Rotate the pipette so the solution contacts the internal walls.
5. Discard the rinsing solution.
6. Repeat with another small portion if necessary.

The pipette is now conditioned with the solution it will transfer.

To measure the aliquot:

1. Use a pipette filler to draw the solution above the calibration mark.
2. Adjust the liquid level until the bottom of the meniscus aligns with the mark, viewed at eye level.
3. Wipe any solution from the outside of the tip.
4. Allow the solution to drain into the conical flask under gravity.
5. Touch the pipette tip against the inside wall of the flask to transfer the hanging drop.

A standard volumetric pipette is calibrated to deliver its stated volume while retaining a small amount of liquid in the tip.

Do not blow out the remaining liquid unless the particular pipette is explicitly designed as a blow-out pipette.

Conditioning a burette

Now suppose your burette will contain sodium hydroxide solution of known concentration.

Follow this procedure:

1. Rinse the burette with distilled or deionised water.
2. Allow the water to drain.
3. Add a small volume of the sodium hydroxide solution.
4. Rotate the burette so the solution rinses its internal walls.
5. Open the stopcock and allow some solution to pass through the tip.
6. Discard the rinsing solution.
7. Repeat the conditioning rinse if necessary.
8. Fill the burette with the sodium hydroxide solution.

The final rinse through the tip is important.

If air remains trapped in the burette tip, some of the solution measured as delivered will actually be used to fill the tip rather than enter the conical flask.

This creates an error in the titre.

After filling, remove the funnel and check that the tip is completely filled with solution. Make sure there are no air bubbles and that the stopcock does not leak.

You can then record your initial burette reading.

What about the conical flask?

This is where students sometimes apply the conditioning rule too broadly.

A conical flask does not need to be conditioned with the analyte.

It should be clean, but it can be rinsed with distilled water and used while still wet.

Why?

The pipette has already delivered the fixed number of moles of analyte into the flask. Adding distilled water increases the total volume of liquid but does not change the number of moles of analyte.

The equivalence point therefore requires the same amount of titrant.

Compare the following:

This distinction is worth remembering:

Water inside the pipette or burette can change the concentration of the solution being measured. Water added to the conical flask does not change the number of moles already present.

03Make the correct endpoint decision

The burette measures how much titrant is delivered, but how do you know when to stop?

You need an indicator that changes colour at an appropriate point in the titration.

Two terms are important:

• The equivalence point is the point at which the reactants have combined in the exact mole ratio specified by the balanced chemical equation.
• The endpoint is the observed colour change that tells you to stop adding titrant.

These points should be very close, but they are not necessarily identical.

Choose an appropriate indicator

The indicator must have a colour-change range that falls within the steep pH change around the equivalence point.

For example:

These choices depend on the actual titration curve and the indicator’s transition range.

How to identify the endpoint during a titration

Use this sequence:

1. Add a few drops of the appropriate indicator to the analyte in the conical flask.
2. Perform a rough titration to estimate the endpoint volume.
3. Repeat the titration using a fresh aliquot.
4. Add titrant relatively quickly at first.
5. As you approach the rough endpoint, slow down to individual drops.
6. Swirl the flask continuously to mix the solutions.
7. Stop at the first permanent colour change.

A white tile beneath the conical flask makes subtle colour changes easier to see.

For a phenolphthalein titration involving a colourless acid and sodium hydroxide, the endpoint is typically the first faint pink colour that persists after swirling.

Do not keep adding sodium hydroxide until the solution becomes a strong pink. That would be an overshot endpoint.

Why repeat the titration?

A rough titre is useful for finding the approximate endpoint, but it should not normally be included in the final average.

Perform additional accurate titrations until you obtain concordant titres.

Concordant titres are results that agree within the accepted experimental tolerance.

Use your school’s specified tolerance rather than assuming there is one universal numerical requirement.

Repeating the experiment helps reduce the influence of random errors. However, repeated measurements cannot automatically eliminate systematic errors caused by incorrectly conditioned equipment or an unsuitable indicator.

04Trace errors through the concentration calculation

This is where choosing the correct equipment becomes especially important.

Assume that:

• The burette contains a solution of known concentration, called the titrant.
• The pipette transfers a fixed volume of an unknown solution, called the analyte.
• You are calculating the analyte concentration.
• The balanced equation gives the mole ratio between the two solutions.

For a general reaction:

\[ aA+bT\rightarrow \text{products} \]

where:

• \(A\) represents the analyte.
• \(T\) represents the titrant.
• \(a\) and \(b\) are their stoichiometric coefficients.

The mole relationship is:

\[ \frac{n_A}{a}=\frac{n_T}{b} \]

Since \(n=cV\), the analyte concentration is:

\[ \boxed{c_A=\frac{a}{b}\times \frac{c_TV_T}{V_A}} \]

where:

• \(c_A\) is the analyte concentration in mol L\(^{-1}\).
• \(c_T\) is the titrant concentration in mol L\(^{-1}\).
• \(V_T\) is the titre in litres.
• \(V_A\) is the pipetted analyte volume in litres.
• \(a\) and \(b\) are the balanced equation coefficients.

For a 1:1 reaction, this simplifies to:

\[ c_A=\frac{c_TV_T}{V_A} \]

Look carefully at the equation.

The calculated analyte concentration is directly proportional to the titre and inversely proportional to the aliquot volume.

This gives us a useful way to predict the direction of experimental errors.

Common titration errors and their effects

Assume that the titrant concentration and nominal pipette volume are used in the calculation.

The direction of error follows from the actual number of moles and volume delivered, not simply from whether a piece of equipment was wet.

A useful way to remember the two most important water errors

Consider two students who both forget to condition their glassware.

Student A leaves water in the burette.

The titrant becomes diluted. Its actual concentration decreases, so a larger volume must be delivered to provide the required number of moles.

If the student still uses the original labelled concentration in the calculation, the calculated analyte concentration is too high.

Student B leaves water in the pipette.

The analyte becomes diluted. Fewer moles are transferred into the flask, so a smaller titre is required.

The calculated analyte concentration is too low.

The location of the water matters.

05Worked examples: Choosing and using the apparatus

06Questions and solutions

All numerical values in the following questions are illustrative classroom data. Assume that temperature effects and other experimental errors are negligible unless stated otherwise.

Question 1

A student wants to determine the concentration of sodium carbonate solution by titrating it with 0.2000 mol L\(^{-1}\) hydrochloric acid.

The student needs to transfer 20.00 mL of sodium carbonate solution into a conical flask.

The initial burette reading is 0.15 mL, and the final reading is 15.77 mL.

(a) Identify the appropriate pipette and burette.

(b) Calculate the titre.

(c) Determine the concentration of sodium carbonate.

The balanced equation is:

\[ \ce{Na2CO3 + 2HCl -> 2NaCl + H2O + CO2} \]

Solution 1

The appropriate equipment is a 20.00 mL volumetric pipette and a suitable burette, such as a 50 mL burette. The sodium carbonate concentration is 0.07810 mol L\(^{-1}\).

(a) Selecting the apparatus

The volumetric pipette is used because the student needs to transfer a fixed 20.00 mL aliquot.

The burette is used because the volume of hydrochloric acid required to reach the endpoint is initially unknown.

(b) Calculating the titre

\[ V_T=V_f-V_i \]

\[ V_T=15.77-0.15 \]

\[ \boxed{V_T=15.62\text{ mL}} \]

Convert to litres:

\[ V_T=0.01562\text{ L} \]

(c) Calculating the sodium carbonate concentration

First, calculate the moles of hydrochloric acid:

\[ n_{\ce{HCl}}=cV \]

\[ n_{\ce{HCl}}=0.2000\times0.01562 \]

\[ n_{\ce{HCl}}=0.003124\text{ mol} \]

From the balanced equation, two moles of hydrochloric acid react with one mole of sodium carbonate.

Therefore:

\[ n_{\ce{Na2CO3}}= \frac{0.003124}{2} \]

\[ n_{\ce{Na2CO3}}=0.001562\text{ mol} \]

Now calculate the concentration of sodium carbonate:

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

\[ c_{\ce{Na2CO3}}= \frac{0.001562}{0.02000} \]

\[ \boxed{c_{\ce{Na2CO3}}=0.07810\text{ mol L}^{-1}} \]

The factor of two in the balanced equation is essential. Using a 1:1 mole ratio would produce an incorrect concentration.

Question 2

A student titrates 25.00 mL of hydrochloric acid with 0.1200 mol L\(^{-1}\) sodium hydroxide.

The initial burette reading is 0.20 mL, and the final reading is 24.80 mL.

However, an air bubble initially occupies 0.30 mL of the burette tip. During the titration, the bubble disappears as sodium hydroxide fills the tip.

Assume that the full 0.30 mL is taken from the burette during the titration but does not enter the conical flask.

(a) Calculate the apparent titre recorded from the burette.

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

(c) Calculate the hydrochloric acid concentration that the student would report if they ignored the bubble.

(d) Calculate the actual hydrochloric acid concentration and explain the direction of error.

Solution 2

The apparent titre is 24.60 mL, but only 24.30 mL of sodium hydroxide actually reaches the flask. The student therefore overestimates the hydrochloric acid concentration.

(a) Calculating the apparent titre

\[ V_T=24.80-0.20 \]

\[ \boxed{V_T=24.60\text{ mL}} \]

(b) Calculating the actual delivered volume

The burette reading includes the sodium hydroxide used to fill the initially empty section of the tip.

Therefore:

\[ V_{\text{actual}}=24.60-0.30 \]

\[ \boxed{V_{\text{actual}}=24.30\text{ mL}} \]

(c) Calculating the reported acid concentration

The student incorrectly assumes that all 24.60 mL entered the flask.

The hydrochloric acid and sodium hydroxide react in a 1:1 mole ratio.

Using the apparent titre:

\[ c_{\ce{HCl}}= \frac{0.1200\times0.02460}{0.02500} \]

\[ \boxed{c_{\text{reported}}=0.11808\text{ mol L}^{-1}} \]

(d) Calculating the actual acid concentration

Only 24.30 mL entered the flask.

Therefore:

\[ c_{\ce{HCl}}= \frac{0.1200\times0.02430}{0.02500} \]

\[ \boxed{c_{\text{actual}}=0.11664\text{ mol L}^{-1}} \]

The reported concentration is too high because the burette records a larger volume decrease than the volume actually delivered to the flask.

This is why checking the burette tip for air bubbles before starting a titration is essential.

Question 3

Two students perform the same titration using a 25.00 mL volumetric pipette and 0.1000 mol L\(^{-1}\) sodium hydroxide.

The unknown monoprotic acid has an actual concentration of 0.1000 mol L\(^{-1}\).

Student A transfers 25.00 mL of acid into a conical flask and then accidentally adds 5.00 mL of distilled water.

Student B has 0.50 mL of distilled water remaining inside the pipette before filling it with acid to the calibration mark. Assume that the water mixes uniformly with the acid, and that the final delivered mixture has a total volume of 25.00 mL.

(a) Calculate the expected titre for Student A.

(b) Calculate the expected titre for Student B.

(c) Determine the concentration each student would calculate if both used the nominal 25.00 mL pipette volume.

(d) Explain why the two students obtain different results even though both have water in their apparatus.

Solution 3

Student A obtains a titre of 25.00 mL and calculates the correct concentration of 0.1000 mol L\(^{-1}\). Student B obtains a titre of 24.50 mL and calculates a concentration of 0.09800 mol L\(^{-1}\).

The difference is caused by where the water is located.

(a) Student A: Water in the conical flask

Student A transfers the full 25.00 mL of acid before adding water.

The moles of acid are:

\[ n_{\text{acid}}=0.1000\times0.02500 \]

\[ n_{\text{acid}}=0.002500\text{ mol} \]

Adding distilled water does not change these moles.

Since the reaction is 1:1, the same number of moles of sodium hydroxide is required.

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

\[ V_T=\frac{0.002500}{0.1000} \]

\[ \boxed{V_T=25.00\text{ mL}} \]

(b) Student B: Water in the pipette

Student B’s 25.00 mL aliquot contains:

\[ V_{\text{acid}}=25.00-0.50 \]

\[ V_{\text{acid}}=24.50\text{ mL} \]

The actual moles of acid transferred are:

\[ n_{\text{acid}}=0.1000\times0.02450 \]

\[ n_{\text{acid}}=0.002450\text{ mol} \]

The required sodium hydroxide volume is:

\[ V_T=\frac{0.002450}{0.1000} \]

\[ \boxed{V_T=24.50\text{ mL}} \]

(c) Calculating the reported concentrations

Student A calculates:

\[ c_A=\frac{0.1000\times0.02500}{0.02500} \]

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

Student B calculates using the nominal 25.00 mL aliquot:

\[ c_A=\frac{0.1000\times0.02450}{0.02500} \]

\[ \boxed{c_A=0.09800\text{ mol L}^{-1}} \]

Student B underestimates the actual concentration by 2.00%.

(d) Explaining the difference

Student A adds water after the pipette has delivered the correct number of acid moles.

The solution becomes more dilute, but the number of moles remains unchanged. The equivalence point therefore requires the same volume of sodium hydroxide.

Student B introduces water before transferring the aliquot.

The acid concentration inside the pipette is reduced, so fewer acid moles are transferred. Less sodium hydroxide is needed to reach equivalence.

This is why the conical flask can be wet, but the pipette must be conditioned with the analyte.

Question 4

A student titrates a weak monoprotic acid with 0.1000 mol L\(^{-1}\) sodium hydroxide.

Preliminary pH measurements show that the steep region of the titration curve occurs between pH 8.2 and 10.0, with the equivalence point at approximately 24.62 mL.

Two indicators are available:

• Methyl orange: transition range pH 3.1-4.4.
• Phenolphthalein: transition range pH 8.2-10.0.

Two students perform repeat titrations:

(a) Identify the more appropriate indicator.

(b) Calculate the mean titre for each student.

(c) Calculate the percentage by which Student A underestimates the equivalence volume.

(d) Explain why Student A’s concordant titres do not establish that the results are accurate.

Solution 4

Phenolphthalein is the more appropriate indicator. Student A’s titres are concordant but systematically too low because methyl orange changes colour before the equivalence point.

(a) Selecting the indicator

The correct indicator should change colour within the steep pH region surrounding the equivalence point.

Phenolphthalein changes colour between pH 8.2 and 10.0, which matches the steep section of the supplied titration curve.

Methyl orange changes colour between pH 3.1 and 4.4, well before the equivalence point.

Therefore, phenolphthalein is the appropriate choice.

(b) Calculating the mean titres

Student A:

\[ \overline V_A=\frac{23.08+23.10}{2} \]

\[ \boxed{\overline V_A=23.09\text{ mL}} \]

Student B:

\[ \overline V_B=\frac{24.61+24.63}{2} \]

\[ \boxed{\overline V_B=24.62\text{ mL}} \]

(c) Calculating the percentage underestimation

Use the equivalence volume of 24.62 mL as the reference.

\[ \%\text{ error}= \frac{24.62-23.09}{24.62}\times100 \]

\[ =\frac{1.53}{24.62}\times100 \]

\[ \boxed{6.21\%\text{ underestimation}} \]

Since the calculated analyte concentration is directly proportional to the titre, Student A would also underestimate the acid concentration by approximately 6.21%, assuming all other quantities are correct.

(d) Explaining concordance and accuracy

Student A’s titres differ by only 0.02 mL.

This suggests good repeatability: the student can obtain similar results across repeated measurements.

However, both results are affected by the same unsuitable indicator.

The indicator produces a colour change before the equivalence point, so the student consistently stops adding sodium hydroxide too early.

This is a systematic error.

Student B obtains concordant titres close to the equivalence volume because the selected indicator changes colour in the appropriate pH region.

The important distinction is that precision describes how closely repeated measurements agree, while accuracy describes how close a result is to the accepted or true value.

Concordant titres alone do not prove accuracy.

Question 5

A student performs a titration to determine the concentration of a monoprotic acid using 0.1000 mol L\(^{-1}\) sodium hydroxide.

The student uses a 25.00 mL volumetric pipette and a 50 mL burette.

The acid’s actual concentration is also 0.1000 mol L\(^{-1}\).

However, two experimental errors occur simultaneously:

• The pipette contains 0.40 mL of distilled water before the student draws the acid up to the 25.00 mL calibration mark. Assume the delivered aliquot contains 24.60 mL of acid solution and 0.40 mL of water.
• The burette contains sodium hydroxide diluted to an actual concentration of 0.09840 mol L\(^{-1}\), although the student still uses the labelled concentration of 0.1000 mol L\(^{-1}\) in calculations.

Assume that the acid and sodium hydroxide react in a 1:1 mole ratio, the pipette delivers exactly 25.00 mL of mixture, and the endpoint is identified correctly.

(a) Calculate the actual number of moles of acid transferred into the flask.

(b) Calculate the actual titre required to reach equivalence.

(c) Calculate the acid concentration reported by the student using the labelled sodium hydroxide concentration and nominal pipette volume.

(d) A second experiment uses a correctly conditioned pipette but the same diluted sodium hydroxide in the burette. Calculate the new titre and the concentration the student would report.

(e) Explain how the first experiment can produce the correct numerical concentration despite two incorrect experimental procedures.

(f) Explain why obtaining concordant titres would not prove that either instrument was correctly conditioned. Propose a way to investigate the two errors independently.

Solution 5

The student obtains an apparently correct concentration of 0.1000 mol L\(^{-1}\), even though both instruments are incorrectly conditioned. The two systematic errors cancel because they reduce the analyte amount and titrant concentration by exactly the same proportion.

This is the important conceptual trap: a correct numerical answer does not necessarily demonstrate a correct experimental method.

(a) Calculating the actual moles of acid transferred

The actual acid volume in the aliquot is 24.60 mL.

Convert to litres:

\[ V_A=0.02460\text{ L} \]

Calculate the moles of acid:

\[ n_A=c_AV_A \]

\[ n_A=0.1000\times0.02460 \]

\[ \boxed{n_A=0.002460\text{ mol}} \]

The student has transferred fewer moles than a correctly conditioned pipette would have delivered.

(b) Calculating the actual titre

The sodium hydroxide concentration is 0.09840 mol L\(^{-1}\).

Because the reaction is 1:1:

\[ n_T=n_A \]

Therefore:

\[ V_T=\frac{n_T}{c_T} \]

\[ V_T=\frac{0.002460}{0.09840} \]

\[ V_T=0.02500\text{ L} \]

\[ \boxed{V_T=25.00\text{ mL}} \]

The diluted sodium hydroxide requires a larger volume than correctly conditioned 0.1000 mol L\(^{-1}\) sodium hydroxide would require for the same number of moles.

However, the pipette has also delivered fewer acid moles, so the required volume has decreased.

(c) Calculating the reported acid concentration

The student incorrectly uses:

• Labelled sodium hydroxide concentration: 0.1000 mol L\(^{-1}\).
• Nominal pipette volume: 25.00 mL.
• Measured titre: 25.00 mL.

Using the 1:1 relationship:

\[ c_A=\frac{c_TV_T}{V_A} \]

\[ c_A= \frac{0.1000\times0.02500}{0.02500} \]

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

This is numerically identical to the actual acid concentration.

The result is correct by coincidence, not because the method was correct.

(d) Repeating the experiment with a correctly conditioned pipette

Now the pipette transfers the full 25.00 mL of acid solution.

The actual moles transferred are:

\[ n_A=0.1000\times0.02500 \]

\[ n_A=0.002500\text{ mol} \]

The burette still contains diluted sodium hydroxide at 0.09840 mol L\(^{-1}\).

The new titre is:

\[ V_T=\frac{0.002500}{0.09840} \]

\[ V_T=0.0254065\text{ L} \]

\[ \boxed{V_T=25.41\text{ mL}} \]

The student now calculates the concentration using the incorrect labelled titrant concentration:

\[ c_A= \frac{0.1000\times0.0254065}{0.02500} \]

\[ \boxed{c_A=0.1016\text{ mol L}^{-1}} \]

The calculated concentration is now too high because the pipette delivers the correct amount of acid, but the diluted titrant still requires an excessive volume.

(e) Explaining the cancellation of errors

The two errors affect different parts of the titration.

The pipette error reduces the acid volume from 25.00 mL to 24.60 mL.

The fraction of the intended acid volume transferred is:

\[ \frac{24.60}{25.00}=0.984 \]

The burette error reduces the sodium hydroxide concentration from 0.1000 mol L\(^{-1}\) to 0.09840 mol L\(^{-1}\).

The fraction of the labelled titrant concentration remaining is:

\[ \frac{0.09840}{0.1000}=0.984 \]

Both errors therefore introduce the same proportional reduction: 1.6%.

The amount of acid decreases by a factor of 0.984, but the amount of sodium hydroxide available per litre also decreases by a factor of 0.984.

These two changes cancel when calculating the titre.

The student therefore obtains 25.00 mL, even though both pieces of equipment were incorrectly conditioned.

(f) Investigating the errors independently

Concordant titres would not expose these errors because both are systematic. They affect every repeat in the same way.

To investigate the burette independently, standardise the sodium hydroxide against a known amount of primary standard weighed directly into a flask. This avoids relying on the potentially incorrectly conditioned pipette.

To investigate the pipette independently, check its delivered volume using a separate calibrated reference or a gravimetric water-delivery test under controlled temperature conditions.

These checks separate the two measurement problems instead of allowing them to cancel within the same titration.

The lesson is that repeatability alone is not enough. A reliable titration requires appropriate apparatus, correct conditioning, accurate volume transfer, and a suitable endpoint.

07What to remember for your HSC Chemistry practicals

When choosing a pipette and burette, work through the following decisions:

1. Identify the fixed aliquot required. Choose a volumetric pipette of the appropriate volume.
2. Estimate the expected titre. Choose a burette with sufficient capacity and suitable graduations.
3. Condition the pipette with the analyte and the burette with the titrant.
4. Rinse the conical flask with distilled water, not the analyte.
5. Check that the burette tip contains no air bubbles.
6. Choose an indicator whose transition range matches the steep region of the titration curve.
7. Add titrant dropwise near the endpoint and record accurate initial and final readings.
8. Obtain concordant titres and investigate possible systematic errors before calculating the final concentration.

Once the apparatus and endpoint are reliable, the next step is applying the balanced chemical equation to calculate the unknown concentration and assessing how measurement uncertainty affects the final result.

If you’re also revising how experimental methods are selected in other areas of Chemistry, see How to Choose Separation Techniques in HSC Chemistry.