Chemistry 5070/31 — October/November 2011

Cambridge O-Level · Practical Test · worked solutions for every part, with the mark scheme

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Topics Observations and Measurements · Experimental Contexts · Analysis, Conclusions and Evaluation · Qualitative Analysis

Q116MObservations and MeasurementsExperimental ContextsAnalysis, Conclusions and EvaluationFree sample

The volume of battery acid contained in a car battery is $4.50\text{ dm}^3$ . Battery acid is an aqueous solution of sulfuric acid. You are to determine by titration the concentration of the battery acid by titrating a diluted solution of the acid with aqueous sodium hydroxide. You will then calculate the mass of sulfuric acid in the battery.

P is dilute sulfuric acid. It has been made by adding water to $10.0\text{ cm}^3$ of battery acid until the volume was $1000\text{ cm}^3$ .
Q is $0.100\text{ mol/dm}^3$ sodium hydroxide.

(a)

Put P into the burette.

Pipette a $25.0\text{ cm}^3$ (or $20.0\text{ cm}^3$ ) portion of Q into a flask and titrate with P, using the indicator provided.

Record your results in the table, repeating the titration as many times as you consider necessary to achieve consistent results.

Results

Burette readings

titration number	1	2
final reading / $\text{cm}^3$
initial reading / $\text{cm}^3$
volume of P used / $\text{cm}^3$
best titration results (✓)

Summary

Tick (✓) the best titration results.

Using these results, the average volume of P required was ______ $\text{cm}^3$ .

Volume of Q used was ______ $\text{cm}^3$ .

12M

DifficultyMedium

Worked solution

Working

Record initial and final burette readings to 1 decimal place (or 2 decimal places ending in .00 or .05) for all titrations.
Complete table: $\text{volume of } \textbf{P} \text{ used} = \text{final reading} - \text{initial reading}$ .
Tick $(\checkmark)$ at least two concordant titres (within $0.20\text{ cm}^3$ of each other).
Calculate the average volume of $\textbf{P}$ using only the ticked titres.
Record the volume of $\textbf{Q}$ used (e.g. $25.0\text{ cm}^3$ ).

Answer

Average volume of P required = candidate's calculated mean of ticked titres (e.g. $24.80\text{ cm}^3$ )

Volume of Q used = $25.0\text{ cm}^3$

Final answer

Candidate-dependent titration readings; mean calculated from concordant titres (within 0.2 cm3)

Detailed explanation

Walkthrough

Performing the titration: Rinse and fill the burette with solution P (dilute sulfuric acid). Ensure there are no air bubbles trapped below the stopcock. Pipette $25.0\text{ cm}^3$ (or $20.0\text{ cm}^3$ ) of solution Q (aqueous sodium hydroxide) into a clean conical flask and add a few drops of indicator (e.g. methyl orange or phenolphthalein).
Recording data: Record initial and final burette readings in the table to at least 1 decimal place (e.g. $0.0\text{ cm}^3$ ). Subtract the initial reading from the final reading to determine the volume of P added (the titre).
Repeating for concordance: Repeat the titration until at least two results agree closely (concordant, within $0.20\text{ cm}^3$ ). Tick $(\checkmark)$ these best results in the table.
Calculating the mean: Sum the ticked titres and divide by the number of ticked values to find the average volume of P used.

Key Takeaways

Always record burette readings with a consistent precision (to 1 or 2 decimal places).
Only concordant results (within $0.20\text{ cm}^3$ ) should be averaged; rough or outlier titres must be excluded.

Common Mistakes

Averaging all titration runs, including the rough initial run.
Inverting the subtraction (subtracting final from initial).
Recording whole numbers without decimal places (e.g. writing 24 instead of 24.0).

Things to Be Careful About

Ensure the volume of Q recorded matches the pipette size used ( $25.0\text{ cm}^3$ or $20.0\text{ cm}^3$ ).
The calculated average must be mathematically correct to within $0.05\text{ cm}^3$ of the selected values.

Techniques used

record initial and final burette readingscalculate titre volume by subtractionselect concordant titrescalculate mean titre

(b)

Q is $0.100\text{ mol/dm}^3$ sodium hydroxide.

Using your results from (a), calculate the concentration, in $\text{mol/dm}^3$ , of sulfuric acid in P.

2\text{NaOH} + \text{H}_2\text{SO}_4 \rightarrow \text{Na}_2\text{SO}_4 + 2\text{H}_2\text{O}

concentration of sulfuric acid in P = ______ $\text{mol/dm}^3$

DifficultyMedium-Easy

Worked solution

Working

Assuming a $25.0\text{ cm}^3$ pipette and an average titre of $24.8\text{ cm}^3$ :

\text{moles of NaOH (Q)} = \frac{25.0}{1000} \times 0.100 = 0.00250\text{ mol}

From the balanced equation:

2\text{NaOH} + \text{H}_2\text{SO}_4 \rightarrow \text{Na}_2\text{SO}_4 + 2\text{H}_2\text{O}

\text{moles of } \text{H}_2\text{SO}_4 \text{ in P} = \frac{0.00250}{2} = 0.00125\text{ mol}

\text{concentration of } \text{H}_2\text{SO}_4 \text{ in P} = \frac{\text{moles}}{\text{volume (dm}^3\text{)}} = \frac{0.00125}{\frac{24.8}{1000}} = \frac{25.0 \times 0.100}{2 \times 24.8} = 0.0504\text{ mol / dm}^3

Answer

$0.0504\text{ mol / dm}^3$ (or calculated as $\frac{\text{volume of Q} \times 0.100}{2 \times \text{average titre of P}}$ )

Final answer

0.0504 mol / dm3

Detailed explanation

Walkthrough

Calculate the number of moles of $\text{NaOH}$ present in the pipetted volume of solution Q using $\text{moles} = \text{concentration} \times \text{volume in dm}^3$ .
Use the stoichiometric ratio from the balanced chemical equation ( $2\text{ mol NaOH} : 1\text{ mol H}_2\text{SO}_4$ ) to determine the moles of $\text{H}_2\text{SO}_4$ neutralised by dividing the moles of $\text{NaOH}$ by 2.
Divide the moles of $\text{H}_2\text{SO}_4$ by the average volume of P in $\text{dm}^3$ to obtain the concentration of $\text{H}_2\text{SO}_4$ in solution P.

Key Takeaways

In acid-base reactions, mole ratios from balanced equations must be applied before converting back to concentration.
For a diprotic acid like sulfuric acid reacting with a monoprotic base like sodium hydroxide, 1 mole of acid reacts with 2 moles of base.

Common Mistakes

Forgetting the $2:1$ stoichiometric ratio and assuming a $1:1$ reaction.
Forgetting to convert volumes from $\text{cm}^3$ to $\text{dm}^3$ (dividing by 1000).

Things to Be Careful About

Round the final concentration to 3 significant figures.
Follow through (ecf) is applied to the candidate's average titre from part (a).

Techniques used

calculate moles of solute from volume and concentrationapply stoichiometry from balanced equationcalculate concentration in mol / dm3

(c)

Using your answer from (b) and information given in the question, calculate the concentration of sulfuric acid in battery acid.

concentration of sulfuric acid in battery acid = ______ $\text{mol/dm}^3$

DifficultyMedium-Easy

Worked solution

Working

Dilution factor:

\text{dilution factor} = \frac{1000\text{ cm}^3}{10.0\text{ cm}^3} = 100

\text{concentration of battery acid} = \text{answer from (b)} \times 100 = 0.0504 \times 100 = 5.04\text{ mol / dm}^3

Answer

$5.04\text{ mol / dm}^3$

Final answer

5.04 mol / dm3

Detailed explanation

Walkthrough

From the question stem, solution P was prepared by diluting $10.0\text{ cm}^3$ of the original battery acid to a total volume of $1000\text{ cm}^3$ .
The dilution factor is therefore $\frac{1000\text{ cm}^3}{10.0\text{ cm}^3} = 100$ .
To find the concentration of the undiluted battery acid, multiply the concentration of solution P calculated in part (b) by 100.

Key Takeaways

Diluting a solution decreases its concentration proportionally: $\text{concentration}_{\text{original}} = \text{concentration}_{\text{dilute}} \times \text{dilution factor}$ .

Common Mistakes

Dividing by 100 instead of multiplying.
Using an incorrect dilution factor (e.g. 10 or 1000).

Things to Be Careful About

Ensure error carried forward (ecf) is applied if an incorrect value was obtained in part (b).

Techniques used

apply dilution factor to calculate original concentration

(d)

Using your answer from (c), calculate the mass of sulfuric acid present in $4.50\text{ dm}^3$ of battery acid.

The relative formula mass of sulfuric acid is 98.

mass of sulfuric acid present in $4.50\text{ dm}^3$ of battery acid = ______ $\text{g}$

DifficultyMedium-Easy

Worked solution

Working

\text{moles of } \text{H}_2\text{SO}_4 \text{ in } 4.50\text{ dm}^3 = \text{concentration} \times \text{volume} = 5.04 \times 4.50 = 22.68\text{ mol}

\text{mass of } \text{H}_2\text{SO}_4 = \text{moles} \times M_r = 22.68 \times 98 = 2220\text{ g}

Answer

$2220\text{ g}$

Final answer

2220 g

Detailed explanation

Walkthrough

Calculate the number of moles of sulfuric acid present in the full $4.50\text{ dm}^3$ battery volume by multiplying the concentration from part (c) by $4.50\text{ dm}^3$ .
Convert the number of moles to mass in grams by multiplying by the relative formula mass ( $M_r = 98$ ):

\text{mass} = \text{concentration from (c)} \times 4.50 \times 98

Round the final value appropriately (to 3 significant figures: $2220\text{ g}$ ).

Key Takeaways

Total moles in a solution: $\text{moles} = \text{concentration (mol / dm}^3\text{)} \times \text{volume (dm}^3\text{)}$ .
Mass of substance: $\text{mass (g)} = \text{moles} \times M_r$ .

Common Mistakes

Forgetting to multiply by the total volume ( $4.50\text{ dm}^3$ ) and calculating the mass per $\text{dm}^3$ instead.
Dividing by $M_r$ instead of multiplying.

Things to Be Careful About

Check that the unit given on the answer line is $\text{g}$ .
State the final answer to 3 significant figures.

Techniques used

calculate moles from volume and concentrationconvert moles to mass using Mr

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