RP1: Volumetric Solution & Acid-Base Titration

Revise the theory

This practical links to these notes pages:

Amount of Substance & Solution Calculations (Unit 1 · Spec 3.1.2.1 to 3.1.2.4)
Balanced Equations and Calculations (Unit 1 · Spec 3.1.2.5)

1. Practical Aim & Theoretical Background

The objective of Required Practical 1 is twofold:

To prepare 250.0 cm³ of a standard volumetric solution of known concentration from a primary solid standard (such as anhydrous sodium carbonate Na₂CO₃, or hydrated ethanedioic acid H₂C₂O₄ • 2H₂O).
To perform an accurate acid-base titration using this standard solution to find the unknown concentration of an acid (e.g. hydrochloric acid HCl) or base (e.g. sodium hydroxide NaOH).

Primary Standard Definition A substance of known high purity, high stability, known molar mass, and non-hygroscopic character (does not absorb moisture from the air), which can be weighed out directly to make a standard solution with precise known concentration.

2. Step-by-Step Standard Solution Protocol

To prepare 250.0 cm³ of standard solution with minimal transfer losses, follow this sequential laboratory procedure:

Weighing by Difference: Place a clean, dry weighing boat on a 3-decimal place balance (±0.001 g). Weigh out the required mass of solid primary standard (e.g. approximately 1.325 g of anhydrous Na₂CO₃). Record the exact mass: m1 = mass of boat + solid.
Transfer to Beaker: Tip the solid carefully into a clean 250 cm³ glass beaker. Reweigh the empty boat containing any residual powder: m2 = mass of boat after transfer. Calculate the exact mass of solid transferred: m_transferred = m1 - m2.
Dissolve Solid: Add approximately 100 cm³ of deionised (or distilled) water to the beaker. Stir thoroughly with a clean glass rod until all crystals have dissolved completely.
Quantitative Transfer to Volumetric Flask: Insert a clean filter funnel into the neck of a 250.0 cm³ volumetric flask. Pour the solution from the beaker into the flask using the glass rod to guide the stream.
Rinsing Protocols: Rinse the beaker, glass rod, and funnel multiple times with small volumes of deionised water from a wash bottle. Pour all washings into the volumetric flask to ensure 100% quantitative transfer of solute.
Making Up to the Mark: Add deionised water until the water level approaches ~1 cm below the graduation line. Using a dropping pipette, add water drop-by-drop until the bottom of the curved meniscus sits exactly on the 250.0 cm³ graduation mark at eye level.
Inversion: Insert the ground-glass or plastic stopper firmly into the flask. Invert the volumetric flask 10 to 15 times to ensure thorough mixing and a uniform concentration throughout the entire solution.

Examiner Warning: Why Inversion is Mandatory

A standard volumetric flask has a very long, narrow neck. When water is added to the mark, the lighter distilled water in the neck does not mix spontaneously with the denser dissolved solution in the bulb. Failure to invert the flask results in an unmixed solution that is more dilute at the top and more concentrated at the bottom, producing wildly erratic titres.

3. Titration Protocol & Glassware Rinsing Rules

Incorrect glassware rinsing is the most common cause of systematic error in volumetric analysis. Follow these strict laboratory rinsing rules:

Glassware	Rinse With	What NOT to Rinse With	Consequence of Incorrect Rinsing
Burette	The solution it will contain (acid or base) after initial distilled water rinse.	Distilled water immediately before filling.	Residual water droplets inside the burette dilute the titrant, artificially increasing the titre volume required.
Volumetric Pipette	The solution it will transfer after initial distilled water rinse.	Distilled water immediately before filling.	Residual water droplets dilute the sample aliquot, artificially lowering the number of moles delivered.
Conical Flask	Deionised (distilled) water only.	The acid or base solution.	Rinsing with solution leaves extra unmeasured moles of reactant in the flask, ruining calculation accuracy. Adding distilled water to the flask during titration (e.g. washing down splashes) does NOT change moles of reactant present.

Try it: Put the titration in order. Read the steps, then check the order below.

Show the correct order

Rinse the burette with the hydrochloric acid, fill it, remove the funnel and record the initial reading to the nearest 0.05 cm³.
Rinse the pipette with the sodium carbonate solution, then use a pipette filler to transfer 25.0 cm³ of it into a clean conical flask.
Add 2 to 3 drops of methyl orange, which turns the solution yellow, and stand the flask on a white tile under the burette.
Rough titration: add the hydrochloric acid quickly with swirling until the first permanent orange colour, and record the rough titre.
Accurate titration: add quickly to about 2 cm³ before the rough titre, then drop by drop with swirling until the first permanent orange colour.
Repeat until two titres agree within 0.10 cm³, then calculate the mean of the concordant titres only.

Try it: Label the apparatus. Name each part in the diagram, then check the list below.

Show the parts

Retort stand
Clamp
Burette
Tap
Conical flask
White tile

Try it: Run the titration. This mini practical needs JavaScript. Open the full virtual lab.

4. Concordant Titres & Mean Calculation Rules

In OxfordAQA chemistry examinations, calculating the mean titre follows strict mathematical rules:

Concordant Titres Definition Titres that agree with one another within 0.10 cm³ (e.g. 24.30 cm³ and 24.35 cm³).

Three Mandatory Rules for Calculating Mean Titres

Rough Titre Must Be Discarded: The initial rough (trial) titration is carried out quickly to estimate the endpoint and must NEVER be included in the mean.
Use Concordant Titres Only: Only titres within 0.10 cm³ of each other may be averaged. Any non-concordant titres must be ignored.
State the Mean to 2 Decimal Places: Individual burette readings are recorded to 2 decimal places (ending in .00 or .05 cm³). The calculated mean titre must be reported to 2 decimal places (e.g. 24.33 cm³).

Try it: Which way does the titre go? Each mistake makes the titre too large, too small, or leaves it unchanged. Sort them. Then check the table below.

Show the answers

Mistake	Effect on the titre
Burette rinsed with distilled water only, not with the solution it will hold	Titre too large. Water left in the burette dilutes the solution, so a larger volume is needed to deliver the same number of moles.
Funnel left in the top of the burette during the titration	Titre too small. Drops fall from the funnel into the burette after the initial reading, so the final reading is lower than the volume really delivered.
End point overshot by adding too much from the burette	Titre too large. Extra solution is delivered after the end point, so the recorded titre is bigger than the volume that reacted.
Air bubble in the burette jet that fills with solution during the titration	Titre too large. Solution that fills the bubble leaves the burette but never reaches the flask, so the reading falls by more than the volume delivered.
Pipette rinsed with distilled water only, not with the solution it will measure	Titre too small. Water in the pipette dilutes the 25.0 cm³ sample, so fewer moles are in the flask and less solution is needed to react with them.
Conical flask rinsed with distilled water and still wet	No effect on the titre. Water changes the volume in the flask but not the number of moles delivered by the pipette, so the titre is unchanged.
Distilled water squirted down the inside of the flask during the titration	No effect on the titre. The number of moles in the flask is unchanged, so the same volume from the burette is needed.
Some of the weighed solid lost while making the standard solution	Titre too small. The standard solution is more dilute than calculated. Fewer moles of carbonate are pipetted into the flask, so less acid is needed and the calculated acid concentration is too high.

Now explain it in writing

[2 marks]

During a titration a student leaves the funnel in the top of the burette. Explain how this affects the titre.

Show Mark Scheme

Drops of solution can fall from the funnel into the burette after the initial reading has been taken [1]
So the final reading is lower than it should be and the titre is smaller than the volume actually delivered [1]

5. Exact Percentage Uncertainty Mathematics

In Unit 5, percentage uncertainty calculations for RP1 follow these exact relationships:

Glassware Percentage Uncertainty Formulas - Volumetric Pipette (25.00 cm³, tolerance ±0.06 cm³):
% Uncertainty = [ (0.06 × 1) / 25.00 ] × 100 = 0.24%

- Burette Titre (e.g. 24.20 cm³, tolerance ±0.05 cm³ per reading):
% Uncertainty = [ (0.05 × 2) / 24.20 ] × 100 = [ 0.10 / 24.20 ] × 100 = 0.41%

- 3-Decimal Balance (Weighing by difference, tolerance ±0.001 g):
% Uncertainty = [ (0.001 × 2) / Mass Transferred ] × 100 = [ 0.002 / Mass Transferred ] × 100

6. Worked Calculation Problem: Finding Unknown Acid Concentration

Worked Example: Standard Na₂CO₃ Titration with Unknown HCl

Problem: A student prepares 250.0 cm³ of a standard solution by dissolving 1.325 g of anhydrous sodium carbonate (Na₂CO₃, M_r = 106.0 g mol⁻¹) in deionised water. A 25.00 cm³ aliquot of this standard solution is pipetted into a conical flask and titrated against hydrochloric acid (HCl) of unknown concentration using methyl orange indicator.

The titration results obtained are:

Rough titre: 25.40 cm³
Titre 1: 24.30 cm³
Titre 2: 24.50 cm³
Titre 3: 24.35 cm³

1. Identify the concordant titres and calculate the mean titre.
2. Write the balanced chemical equation.
3. Calculate the concentration of the hydrochloric acid in mol dm⁻³.

Step 1: Concordant Titres & Mean

Rough (25.40) is ignored.
Titre 1 (24.30) and Titre 3 (24.35) are concordant (|24.35 - 24.30| = 0.05 cm³ ≤ 0.10 cm³).
Titre 2 (24.50) is non-concordant and excluded.
Mean Titre = (24.30 + 24.35) / 2 = 24.325 cm³ → 24.33 cm³

Step 2: Balanced Chemical Equation

Na₂CO₃(aq) + 2HCl(aq) → 2NaCl(aq) + H₂O(l) + CO₂(g)

Step 3: Moles of Na₂CO₃ in 250.0 cm³

Moles in 250 cm³ = mass / M_r = 1.325 / 106.0 = 0.01250 mol
Concentration of standard solution = 0.01250 / 0.2500 = 0.05000 mol dm⁻³

Step 4: Moles of Na₂CO₃ in 25.00 cm³ aliquot

Moles in 25 cm³ = 0.05000 × (25.00 / 1000) = 1.250 × 10⁻³ mol

Step 5: Moles of HCl reacted

Ratio Na₂CO₃ : HCl = 1 : 2
Moles of HCl = (1.250 × 10⁻³) × 2 = 2.500 × 10⁻³ mol

Step 6: Concentration of HCl

Volume of HCl = 24.325 cm³ = 0.024325 dm³ (carry unrounded in calculator)
Concentration = moles / volume = (2.500 × 10⁻³) / 0.024325 = 0.10277 mol dm⁻³
To 3 significant figures: 0.103 mol dm⁻³

Final Answer: [HCl] = 0.103 mol dm⁻³

7. Practice Exam Questions

Question 1: Why should a student rinse the conical flask with deionised water rather than the acid or base solution before performing a titration?

A. Distilled water neutralises trace base on the walls of the flask. B. Rinsing with solution leaves extra unmeasured moles of reactant in the flask, whereas water does not change the moles delivered by pipette. C. Distilled water prevents the indicator from decomposing. D. Rinsing with water ensures the pH of the flask remains exactly 7.00.

Show Answer & Explanation

Correct Answer: B

The aliquot transferred by pipette contains a precisely known number of moles. Residual water in the conical flask dilutes the concentration but leaves the total moles unchanged. Rinsing with solution would introduce unmetered moles and invalidate the titration.

Question 2: In preparing a standard solution, why must a student invert the stoppered volumetric flask 10 to 15 times after adding water to the 250 cm³ mark?

A. To dissolve any remaining undissolved solute crystals in the neck. B. To ensure uniform concentration throughout the flask because pure water in the neck does not spontaneously mix with dense solution below. C. To release dissolved atmospheric gases that would interfere with indicator color. D. To lower the temperature of the solution back to 20 °C.

Show Answer & Explanation

Correct Answer: B

The narrow neck restricts convection currents. The water added to bring the meniscus to the mark sits on top of the denser solution. Thorough inversion ensures homogeneous solute distribution.

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Exam questions, marked instantly

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[3 marks]

Describe how to take an accurate burette reading.