Chemistry 5070/31 — October/November 2013

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

Q117MObservations and MeasurementsExperimental ContextsAnalysis, Conclusions and EvaluationFree sample

P is an aqueous solution which contains a mixture of sodium carbonate, $\text{Na}_2\text{CO}_3$ , and sodium hydroxide, $\text{NaOH}$ .

The concentration of sodium carbonate in P is $0.0200\text{ mol / dm}^3$ .

You are to determine by titration the volume of dilute hydrochloric acid, Q, needed to neutralise a volume of P, and then calculate the concentration of sodium hydroxide present.

Q is $0.200\text{ mol / dm}^3$ hydrochloric acid, $\text{HCl}$ .

(a)

Put Q into the burette.

Pipette a $25.0\text{ cm}^3$ (or $20.0\text{ cm}^3$ ) portion of P into a flask and titrate with Q, 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 Q used / $\text{cm}^3$
best titration results (✓)

Summary

Tick (✓) the best titration results.

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

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

12M

DifficultyMedium

Worked solution

Working

Fill the results table with initial and final burette readings recorded to $1$ or $2$ decimal places (e.g. $0.0\text{ cm}^3$ or $0.00\text{ cm}^3$ ).
Calculate each titre by subtracting the initial reading from the final reading:

\text{volume used} = \text{final reading} - \text{initial reading}

Repeat titrations until at least two titres are concordant (within $0.20\text{ cm}^3$ of each other).
Tick (✓) the concordant titres and calculate the mean average of only the ticked titres.
Record the pipette volume of P used ( $25.0\text{ cm}^3$ or $20.0\text{ cm}^3$ ).

Answer

Candidate-dependent experimental titration table, ticked concordant titres, calculated average titre of Q, and recorded pipette volume of P ( $25.0\text{ cm}^3$ or $20.0\text{ cm}^3$ ).

Final answer

Candidate-dependent titration readings and calculated average volume of Q

Detailed explanation

Walkthrough

In this practical titration task:

The burette is rinsed with and filled with solution Q ( $0.200\text{ mol / dm}^3\text{ HCl}$ ).
A pipette is used to accurately measure a known volume of solution P (usually $25.0\text{ cm}^3$ or $20.0\text{ cm}^3$ ) into a conical flask.
An appropriate indicator (such as methyl orange or screened methyl orange) is added to detect the end-point of neutralisation.
Initial and final burette readings must be recorded to a consistent precision ( $0.1\text{ cm}^3$ or $0.05\text{ cm}^3$ ).
Titrations are repeated until consistent, concordant results (within $0.2\text{ cm}^3$ ) are achieved.
The candidate ticks the concordant results and calculates the average titre using only the selected values.

Key Takeaways

All burette readings must be recorded systematically with units.
Concordant titres are values that are very close to each other (usually within $0.20\text{ cm}^3$ ).
Rough or non-concordant titres should not be included in the calculation of the average titre.

Common Mistakes

Including an initial non-concordant rough titre when calculating the average.
Recording burette readings to whole numbers only instead of appropriate decimal places (e.g. writing $24$ instead of $24.0$ ).
Inverting initial and final readings.

Things to Be Careful About

Ensure the average calculation is shown clearly and rounded correctly to $1$ or $2$ decimal places.

Techniques used

record initial and final burette readingscalculate titre volume by subtractionidentify concordant titrescalculate average titre volume

(b)

Q is $0.200\text{ mol / dm}^3$ hydrochloric acid, $\text{HCl}$ .
Calculate the number of moles of hydrochloric acid present in your average volume of Q.

moles of hydrochloric acid present = ______

DifficultyMedium-Easy

Worked solution

Working

\text{moles of HCl} = \frac{\text{average volume of } \mathbf{Q} \text{ in } \text{cm}^3}{1000} \times 0.200

For an average titre of $25.2\text{ cm}^3$ :

\text{moles of HCl} = \frac{25.2 \times 0.200}{1000} = 0.00504\text{ mol}

Answer

\frac{\text{average volume of } \mathbf{Q}}{1000} \times 0.200

Final answer

(average volume of Q / 1000) * 0.200 mol

Detailed explanation

Walkthrough

To find the number of moles of solute in a solution when volume and concentration are known, use the formula:

\text{moles} = \text{concentration (in } \text{mol / dm}^3\text{)} \times \text{volume (in } \text{dm}^3\text{)}

Since the average titre volume of Q is measured in $\text{cm}^3$ , divide by $1000$ to convert it to $\text{dm}^3$ , then multiply by the concentration of $\text{HCl}$ ( $0.200\text{ mol / dm}^3$ ).

Key Takeaways

$\text{moles} = \frac{V (\text{in } \text{cm}^3)}{1000} \times C (\text{in } \text{mol / dm}^3)$ .

Common Mistakes

Forgetting to convert $\text{cm}^3$ to $\text{dm}^3$ by dividing by $1000$ .

Things to Be Careful About

Ensure the value used for the volume matches the average titre recorded in part (a).

Techniques used

convert volume in cm3 to dm3calculate moles from concentration and volume

(c)

The concentration of sodium carbonate, $\text{Na}_2\text{CO}_3$ , in P is $0.0200\text{ mol / dm}^3$ .
Calculate the number of moles of sodium carbonate present in your volume of P.

moles of sodium carbonate present = ______

DifficultyMedium-Easy

Worked solution

Working

\text{moles of } \text{Na}_2\text{CO}_3 = \frac{\text{volume of } \mathbf{P} \text{ in } \text{cm}^3}{1000} \times 0.0200

For a $25.0\text{ cm}^3$ portion of P:

\text{moles of } \text{Na}_2\text{CO}_3 = \frac{25.0 \times 0.0200}{1000} = 0.000500\text{ mol}

(For a $20.0\text{ cm}^3$ portion: $\frac{20.0 \times 0.0200}{1000} = 0.000400\text{ mol}$ )

Answer

0.000500\text{ mol (for } 25.0\text{ cm}^3 \text{ of } \mathbf{P}\text{)}

Final answer

0.000500 mol (for 25.0 cm3 of P)

Detailed explanation

Walkthrough

The number of moles of sodium carbonate, $\text{Na}_2\text{CO}_3$ , present in the pipetted aliquot of solution P is calculated using its given concentration ( $0.0200\text{ mol / dm}^3$ ) and the volume of P pipetted ( $25.0\text{ cm}^3$ or $20.0\text{ cm}^3$ ):

\text{moles} = \frac{V}{1000} \times C

For $25.0\text{ cm}^3$ :

\text{moles} = \frac{25.0}{1000} \times 0.0200 = 0.000500\text{ mol}

Key Takeaways

Fixed concentration and known volume directly give the moles of that component in the mixture.

Common Mistakes

Using the average titre of Q instead of the volume of P pipetted.

Things to Be Careful About

Check whether a $25.0\text{ cm}^3$ or $20.0\text{ cm}^3$ pipette was used in part (a).

Techniques used

convert volume in cm3 to dm3calculate moles from concentration and volume

(d)

Using your answer to (c), deduce the number of moles of hydrochloric acid which react with the sodium carbonate present in your volume of P.

\text{Na}_2\text{CO}_3 + 2\text{HCl} \rightarrow 2\text{NaCl} + \text{H}_2\text{O} + \text{CO}_2

moles of hydrochloric acid which react with the sodium carbonate = ______

DifficultyMedium-Easy

Worked solution

Working

From the equation:

\text{Na}_2\text{CO}_3 + 2\text{HCl} \rightarrow 2\text{NaCl} + \text{H}_2\text{O} + \text{CO}_2

$1\text{ mole}$ of $\text{Na}_2\text{CO}_3$ reacts with $2\text{ moles}$ of $\text{HCl}$ .

\text{moles of HCl} = 2 \times (\text{moles of } \text{Na}_2\text{CO}_3 \text{ from } \mathbf{(c)})

For $0.000500\text{ mol}$ of $\text{Na}_2\text{CO}_3$ :

\text{moles of HCl} = 2 \times 0.000500 = 0.00100\text{ mol}

Answer

0.00100\text{ mol (for } 25.0\text{ cm}^3 \text{ of } \mathbf{P}\text{)}

Final answer

0.00100 mol (for 25.0 cm3 of P)

Detailed explanation

Walkthrough

The balanced chemical equation shows that $1\text{ mol}$ of $\text{Na}_2\text{CO}_3$ requires $2\text{ mol}$ of $\text{HCl}$ for complete reaction.
Therefore, the moles of $\text{HCl}$ reacting with $\text{Na}_2\text{CO}_3$ is simply twice the moles of $\text{Na}_2\text{CO}_3$ calculated in part (c):

\text{moles of HCl} = 2 \times 0.000500 = 0.00100\text{ mol}

Key Takeaways

Mole ratio from balanced chemical equations is used to relate amounts of different reactants.

Common Mistakes

Dividing by $2$ instead of multiplying by $2$ .

Things to Be Careful About

Ensure error carried forward (ecf) is applied correctly if the value in (c) was calculated for a $20.0\text{ cm}^3$ pipette.

Techniques used

apply stoichiometry from balanced chemical equation

(e)

Using your answers to (b) and (d), calculate the number of moles of hydrochloric acid which react with the sodium hydroxide in your volume of P.

moles of hydrochloric acid which react with the sodium hydroxide = ______

DifficultyMedium-Easy

Worked solution

Working

\text{moles of HCl reacting with NaOH} = (\text{moles of HCl in } \mathbf{(b)}) - (\text{moles of HCl in } \mathbf{(d)})

For an average titre of $25.2\text{ cm}^3$ and a $25.0\text{ cm}^3$ pipette:

\text{moles of HCl} = 0.00504 - 0.00100 = 0.00404\text{ mol}

Answer

(\text{answer to } \mathbf{(b)}) - (\text{answer to } \mathbf{(d)})

Final answer

0.00404 mol (for example titre of 25.2 cm3 and 25.0 cm3 pipette)

Detailed explanation

Walkthrough

The solution P is a mixture of two bases: $\text{Na}_2\text{CO}_3$ and $\text{NaOH}$ . Both react with the added $\text{HCl}$ :

\text{total moles of HCl used (from } \mathbf{b}\text{)} = (\text{moles of HCl reacting with } \text{Na}_2\text{CO}_3) + (\text{moles of HCl reacting with } \text{NaOH})

Rearranging gives:

\text{moles of HCl reacting with NaOH} = \mathbf{(b)} - \mathbf{(d)}

Key Takeaways

In a mixture of two neutralising agents, total acid consumed is the sum of acid consumed by each individual component.

Common Mistakes

Adding the values instead of subtracting.

Things to Be Careful About

Carry forward the candidate's actual calculated values from parts (b) and (d).

Techniques used

subtract moles to find remaining amount in mixture

(f)

Using your answer to (e), calculate the concentration, in $\text{mol / dm}^3$ , of sodium hydroxide in P.

\text{NaOH} + \text{HCl} \rightarrow \text{NaCl} + \text{H}_2\text{O}

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

DifficultyMedium

Worked solution

Working

From the equation:

\text{NaOH} + \text{HCl} \rightarrow \text{NaCl} + \text{H}_2\text{O}

$1\text{ mole}$ of $\text{HCl}$ reacts with $1\text{ mole}$ of $\text{NaOH}$ .

\text{moles of NaOH in } \mathbf{P} = \text{moles of HCl in } \mathbf{(e)}

\text{concentration of NaOH} = \frac{\text{moles of NaOH} \times 1000}{\text{volume of } \mathbf{P} \text{ in } \text{cm}^3}

For $0.00404\text{ mol}$ and a $25.0\text{ cm}^3$ volume of P:

\text{concentration of NaOH} = \frac{0.00404 \times 1000}{25.0} = 0.162\text{ mol / dm}^3

Answer

0.162\text{ mol / dm}^3\text{ (based on example data)}

Final answer

0.162 mol / dm3 (based on example data)

Detailed explanation

Walkthrough

From the equation $\text{NaOH} + \text{HCl} \rightarrow \text{NaCl} + \text{H}_2\text{O}$ , the mole ratio between $\text{NaOH}$ and $\text{HCl}$ is $1:1$ . Therefore:

\text{moles of NaOH} = \text{moles of HCl from part } \mathbf{(e)}

To determine concentration in $\text{mol / dm}^3$ :

\text{concentration} = \frac{\text{moles}}{\text{volume in } \text{dm}^3} = \frac{\text{moles} \times 1000}{\text{volume in } \text{cm}^3}

Using the example values ( $0.00404\text{ mol}$ in $25.0\text{ cm}^3$ ):

\text{concentration} = \frac{0.00404 \times 1000}{25.0} = 0.162\text{ mol / dm}^3

Key Takeaways

$\text{concentration (mol / dm}^3\text{)} = \frac{\text{moles} \times 1000}{\text{volume (cm}^3\text{)}}$ .
Follow error-carried-forward values accurately from previous parts.

Common Mistakes

Using the average titre of acid instead of the pipetted volume of solution P in the denominator.
Forgetting to multiply by $1000$ when converting volume from $\text{cm}^3$ to $\text{dm}^3$ .

Things to Be Careful About

Quote the final concentration to an appropriate number of significant figures (typically 3 sig. fig.).

Techniques used

apply 1:1 stoichiometrycalculate solution concentration in mol / dm3

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