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 hydrogencarbonate NaHCO3, 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).
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 2-decimal place balance (+/- 0.01 g) and tare it. Weigh out the required mass of solid primary standard (e.g. approximately 2.10 g of NaHCO3). 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.
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. |
4. Concordant Titres & Mean Calculation Rules
In OxfordAQA chemistry examinations, calculating the mean titre follows strict mathematical rules:
- 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).
5. Exact Percentage Uncertainty Mathematics
In Paper 5, percentage uncertainty calculations for RP1 follow these exact relationships:
% 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%
- 2-Decimal Balance (Weighing by difference, tolerance +/- 0.01 g):
% Uncertainty = [ (0.01 * 2) / Mass Transferred ] * 100 = [ 0.02 / Mass Transferred ] * 100
6. Worked Calculation Problem: Finding Unknown Acid Concentration
Problem: A student prepares 250.0 cm3 of a standard solution by dissolving 1.325 g of anhydrous sodium carbonate (Na2CO3, Mr = 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 / Mr = 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?
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 cm^3 mark?
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.