Chemistry 5070/41 — May/June 2018

Walkthrough

The student must read the burette values for titration 4 from Fig. 1.2 and calculate the volume of sodium hydroxide used.

• Burette readings are always read to 1 decimal place (e.g., 0.6, 18.8). The meniscus is read at eye level from the bottom of the curve.
• Initial reading: The meniscus is between 0.0 and 1.0. There are 10 small divisions between 0.0 and 1.0, so each small division is 0.1 cm³. The meniscus is 6 small divisions below 0.0, so the reading is 0.6 cm³.
• Final reading: The meniscus is between 18.0 and 19.0. It is 8 small divisions below 18.0, so the reading is 18.8 cm³.
• Volume used (titre) = Final reading − Initial reading = 18.8 − 0.6 = 18.2 cm³.

Key Takeaways

Burette readings must be recorded to 1 decimal place. The volume of titrant used is always the final reading minus the initial reading. Always check the direction of the numbers on the burette (they increase downwards).

Common Mistakes

• Reading the burette to 2 decimal places (e.g., 0.60) — O Level burettes are read to 1 d.p.
• Subtracting in the wrong order (initial minus final), giving a negative volume.
• Misreading the meniscus level by ignoring the 0.1 cm³ graduations.

Things to Be Careful About

Ensure all readings in the table are to 1 decimal place, including 0.0. The volume used column must be calculated correctly for each row. Concordant results are those within 0.20 cm³ of each other.

Walkthrough

The volumes for titrations 1 and 3 are calculated by subtracting the initial reading from the final reading.

• Titration 1: 19.0 − 0.0 = 19.0 cm³
• Titration 3: 19.1 − 0.4 = 18.7 cm³
  These values are simply entered into the results table.

Key Takeaways

Calculating the titre is a straightforward subtraction. Ensure the units (cm³) are consistent and the arithmetic is correct.

Common Mistakes

• Arithmetic errors in subtraction (e.g., 19.1 − 0.4 = 18.3 instead of 18.7).
• Forgetting to include the unit in the final answer if required, though the table header provides the unit.

Things to Be Careful About

Keep the decimal places consistent. All burette readings and calculated volumes should be to 1 decimal place.

Walkthrough

To find the average titre, the student must first identify the concordant results. Concordant results in a titration are volumes that are within 0.20 cm³ of each other.

• Titration 1: 19.0 cm³
• Titration 2: 18.0 cm³
• Titration 3: 18.7 cm³
• Titration 4: 18.2 cm³

Comparing the values:

• 18.0 and 18.2 are within 0.20 cm³ (difference = 0.2 cm³). These are concordant and should be ticked.
• 19.0 and 18.7 are not within 0.20 cm³ of the concordant pair, nor of each other (difference = 0.3 cm³). These are not concordant and are discarded.

The average titre is calculated from the concordant results:

Key Takeaways

Always tick concordant results (within 0.20 cm³) before calculating the average. Do not include anomalous results in the average calculation.

Common Mistakes

• Averaging all four results: (19.0 + 18.0 + 18.7 + 18.2) / 4 = 18.475 cm³. This is incorrect because anomalous results must be excluded.
• Selecting the wrong pair as concordant.

Things to Be Careful About

The mark scheme for part (c) uses an average titre of 18.4 cm³ for a second student. Do not confuse this with the average titre calculated in part (b)(iii) for the first student, which is 18.1 cm³. Use the correct value for each part.

Walkthrough

The student used a measuring cylinder to transfer 25 cm³ of diluted vinegar into the conical flask. Measuring cylinders are not very accurate for measuring specific volumes (they have a large uncertainty, often ±0.5 cm³ or more).

Improvement: Use a pipette (e.g., a 25 cm³ volumetric pipette) to transfer the diluted vinegar.

Explanation: A pipette is designed to deliver a precise, fixed volume of liquid with much less uncertainty (or less apparatus error) than a measuring cylinder. This makes the volume of analyte more accurate, leading to more accurate final results.

Alternative improvement: Repeat the titration more times. Explanation: This provides more data points to identify anomalous results and calculate a more reliable average.

Key Takeaways

In quantitative analysis, the accuracy of the final result depends on the accuracy of all measurements. Volumetric pipettes are far more accurate than measuring cylinders for transferring fixed volumes.

Common Mistakes

• Suggesting "use a bigger measuring cylinder" — this does not necessarily improve accuracy.
• Saying "a pipette is more accurate" without explaining why it improves the result (i.e., less uncertainty/error in the volume measured).
• Suggesting "use a more accurate balance" — mass is not being measured here.

Things to Be Careful About

The explanation must link the improvement to a reduction in error or uncertainty. "Less uncertainty" or "less apparatus error" are acceptable phrases. Ensure the improvement is specific to the step mentioned (measuring 25 cm³ into the flask).

Walkthrough

Calculate the moles of sodium hydroxide used in the titration.

• Concentration of NaOH = $0.0250 \text{ mol / dm}^3$
• Volume of NaOH used = $18.4 \text{ cm}^3$ (average titre given in the question)
• Convert volume to $\text{dm}^3$: $18.4 \text{ cm}^3 \div 1000 = 0.0184 \text{ dm}^3$
• Calculate moles: $\text{moles} = \text{concentration} \times \text{volume} = 0.0250 \times 0.0184 = 0.00046 \text{ mol}$

Key Takeaways

Always convert volume from $\text{cm}^3$ to $\text{dm}^3$ before using the concentration formula $n = c \times V$. The molar gas volume at r.t.p. is $24 \text{ dm}^3\text{/mol}$, but here we are dealing with solutions, so use $n = cV$.

Common Mistakes

• Forgetting to divide the volume by 1000, calculating $0.0250 \times 18.4 = 0.46$ moles (wrong by a factor of 1000).
• Using the wrong average titre (e.g., 18.1 from part b(iii) instead of 18.4 given in part c).

Things to Be Careful About

The question states "This student obtains an average titre of $18.4 \text{ cm}^3$". Do not use the 18.1 cm³ calculated in part (b)(iii); that was for a different student/method. Use 18.4 cm³ for all calculations in part (c).

Walkthrough

The balanced chemical equation shows that 1 mole of ethanoic acid ($\text{CH}_3\text{COOH}$) reacts with 1 mole of sodium hydroxide ($\text{NaOH}$).

Since the ratio is 1:1, the number of moles of ethanoic acid that reacted is equal to the number of moles of sodium hydroxide used.

Moles of ethanoic acid = 0.00046 mol

Key Takeaways

Always use the mole ratio from the balanced equation to relate the moles of reactants. For a 1:1 ratio, the moles are equal.

Common Mistakes

• Using a 1:2 or 2:1 ratio incorrectly.
• Copying the answer from part (c)(i) without stating the reasoning (though for a fill-in-the-blank, just the number is needed).

Things to Be Careful About

The equation is already balanced and provided. Do not alter it. The ratio is clearly 1:1.

Walkthrough

The titration was performed on a diluted sample of vinegar. We need to find the moles in the original $5.0 \text{ cm}^3$ sample.

• Dilution process: $5.0 \text{ cm}^3$ of vinegar was made up to $250 \text{ cm}^3$ with distilled water.
• Aliquot taken for titration: $25 \text{ cm}^3$ of this diluted solution was used.
• Scaling factor: The total diluted volume ($250 \text{ cm}^3$) is 10 times the volume used in the titration ($25 \text{ cm}^3$). So, $\frac{250}{25} = 10$.

Moles of ethanoic acid in the original $5.0 \text{ cm}^3$ = Moles in $25 \text{ cm}^3$ × 10

Key Takeaways

When a solution is diluted and an aliquot is taken for titration, you must scale up the moles found in the aliquot to find the moles in the original sample. The scaling factor is $\frac{\text{Total diluted volume}}{\text{Volume of aliquot used}}$.

Common Mistakes

• Forgetting to multiply by the dilution factor and leaving the answer as 0.00046 mol.
• Using the wrong dilution factor (e.g., $\frac{250}{5} = 50$, which is incorrect because the 25 cm³ aliquot already represents a fraction of the 250 cm³ total).

Things to Be Careful About

Ensure you are scaling up correctly. The 0.00046 mol is in the 25 cm³ aliquot. The 250 cm³ flask contains 10 times that amount. That total amount (0.0046 mol) came from the original 5.0 cm³ of vinegar.

Walkthrough

Calculate the concentration of ethanoic acid in the original vinegar.

• Moles of ethanoic acid = 0.0046 mol (this is the amount in the original 5.0 cm³ sample)
• Volume of original sample = 5.0 cm³

Convert volume to $\text{dm}^3$:

Calculate concentration:

Key Takeaways

Concentration is always calculated using volume in $\text{dm}^3$. Remember to use the volume and moles from the original sample, not the diluted aliquot.

Common Mistakes

• Using the volume of the diluted solution (250 cm³ or 25 cm³) instead of the original volume (5.0 cm³).
• Forgetting to convert cm³ to dm³, calculating $\frac{0.0046}{5.0} = 0.00092$.
• Using the moles from the aliquot (0.00046) instead of the scaled-up moles (0.0046).

Things to Be Careful About

The question asks for the concentration in the original sample of vinegar. Ensure you are using the moles and volume that correspond to that original sample. Units must be $\text{mol / dm}^3$.