Chemistry 9701/52 — October/November 2016

Cambridge A-Level · Planning, Analysis and Evaluation · worked solutions for every part, with the mark scheme

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Topics Planning · Analysis, Conclusions and Evaluation

Q1PlanningAnalysis, Conclusions and EvaluationFree sample

Titrations using ethylenediaminetetraacetic acid (EDTA) can be used to determine the concentration of metal ions in solution, such as $\text{Zn}^{2+}(\text{aq})$ .

A solution of EDTA is usually prepared from the hydrated disodium salt, $\text{Na}_2\text{H}_2\text{Y}\cdot2\text{H}_2\text{O}$ . The anion of EDTA is $\text{H}_2\text{Y}^{2-}$ , where $\text{Y}$ represents the organic part of the ion.

The equation for the reaction between $\text{Zn}^{2+}(\text{aq})$ and EDTA is shown.

\text{Zn}^{2+}(\text{aq}) + \text{H}_2\text{Y}^{2-}(\text{aq}) \rightarrow \text{ZnY}^{2-}(\text{aq}) + 2\text{H}^+(\text{aq})

The indicator for the reaction is Solochrome Black, which changes colour at the endpoint from purple to blue. The indicator only works at pH 10, so a buffer solution is added to the metal ion solution to maintain the pH.

(a)

Explain why the pH would change during the titration if the buffer were not present.

DifficultyEasy

Worked solution

Answer

The reaction produces $\text{H}^+$ ions, so the $\text{H}^+$ concentration increases and the pH decreases.

Final answer

The reaction produces H+ ions, so pH decreases.

Detailed explanation

Background Concept

In an EDTA titration, the metal ion forms a stable complex with the EDTA anion. The equation given shows that each $\text{Zn}^{2+}$ ion reacts with one $\text{H}_2\text{Y}^{2-}$ ion and releases two $\text{H}^+$ ions. pH is defined as $-\log_{10}[\text{H}^+]$ , so an increase in $[\text{H}^+]$ lowers the pH. A buffer resists pH change by absorbing added $\text{H}^+$ or $\text{OH}^-$ ions.

Understanding the Question

The question asks why the pH would change during the titration if the buffer were not present. The key is the reaction equation: the titration itself produces hydrogen ions. Without a buffer, these ions would accumulate and change the pH.

Approach

Look at the stoichiometry of the EDTA reaction, identify $\text{H}^+$ as a product, and apply the relationship between $[\text{H}^+]$ and pH.

Step-by-Step Reasoning

The equation $\text{Zn}^{2+}(\text{aq}) + \text{H}_2\text{Y}^{2-}(\text{aq}) \rightarrow \text{ZnY}^{2-}(\text{aq}) + 2\text{H}^+(\text{aq})$ shows that two $\text{H}^+$ ions are produced for every $\text{Zn}^{2+}$ ion that reacts.
If no buffer is present, these $\text{H}^+$ ions are not removed, so the hydrogen ion concentration of the solution increases.
Since $\text{pH} = -\log_{10}[\text{H}^+]$ , an increase in $[\text{H}^+]$ means the pH decreases.
The buffer is needed because Solochrome Black only works at pH 10; a falling pH would change the indicator behaviour and ruin the endpoint.

Key Takeaways

EDTA titrations release $\text{H}^+$ ions, so a buffer is essential to keep the pH constant. More $\text{H}^+$ means lower pH.

Common Mistakes

Saying the pH increases: this is wrong because the reaction produces acid, not alkali.
Failing to mention that $\text{H}^+$ ions are produced.
Confusing pH with $[\text{OH}^-]$ .

Things to Be Careful About

State clearly that the reaction produces $\text{H}^+$ ions.
pH decreases when hydrogen ion concentration increases.
The buffer maintains the pH at 10 so the indicator works correctly.

Techniques used

identify the H+ ions produced by the reactionrelate increasing H+ concentration to decreasing pH

(b)

You are to plan a titration experiment to determine the concentration of zinc ions in a solution of zinc sulfate of concentration approximately $0.1\text{ mol dm}^{-3}$ .

You are provided with the following materials.

$20.0\text{ g}$ of hydrated disodium EDTA, $\text{Na}_2\text{H}_2\text{Y}\cdot2\text{H}_2\text{O}$ ( $M_r = 372.2$ )
aqueous zinc sulfate of approximate concentration $0.1\text{ mol dm}^{-3}$
buffer solution, pH 10
Solochrome Black indicator solution

(i)

Name three pieces of volumetric apparatus you would use, with their capacities in $\text{cm}^3$ .

DifficultyEasy

Worked solution

Answer

Volumetric flask: $250\text{ cm}^3$
Pipette: $25\text{ cm}^3$
Burette: $50\text{ cm}^3$

Final answer

Volumetric flask (250 cm3), pipette (25 cm3), burette (50 cm3)

Detailed explanation

Background Concept

A standard solution is prepared in a volumetric flask, which has a single accurate mark. A pipette delivers a fixed volume of solution, and a burette delivers variable volumes during a titration. All are volumetric apparatus because they measure volumes accurately.

Understanding the Question

You need to name three pieces of volumetric apparatus and give their capacities in $\text{cm}^3$ . The apparatus must be suitable for preparing a standard EDTA solution and carrying out the titration.

Approach

Think about the steps: preparing the standard solution needs a volumetric flask; transferring a sample of zinc sulfate needs a pipette; delivering the EDTA titrant needs a burette.

Step-by-Step Reasoning

Volumetric flask: used to make the standard EDTA solution up to an exact volume, commonly $250\text{ cm}^3$ .
Pipette: used to transfer a fixed volume, usually $25\text{ cm}^3$ , of the zinc sulfate solution into the conical flask.
Burette: used to add the EDTA solution gradually during the titration, typically $50\text{ cm}^3$ capacity.

Key Takeaways

Standard solution preparation and titration require a volumetric flask, pipette and burette. Capacities must be stated in $\text{cm}^3$ .

Common Mistakes

Choosing a measuring cylinder, which is not sufficiently accurate for volumetric work.
Giving capacities in $\text{dm}^3$ or forgetting the capacity.
Naming a conical flask as volumetric apparatus.

Things to Be Careful About

The pipette is used for the solution being analysed, not for the EDTA titrant.
The burette is rinsed with the titrant before use.
Use the same capacities in later parts of the question.

Techniques used

identify volumetric apparatusstate capacities in cm3

(ii)

Calculate the mass of hydrated disodium EDTA that would be required for the preparation of a standard solution of concentration $0.100\text{ mol dm}^{-3}$ , using the apparatus you have specified in (i).

DifficultyMedium-Easy

Worked solution

Working

Volume of volumetric flask = $250\text{ cm}^3 = 0.250\text{ dm}^3$

n = cV = 0.100 \times 0.250 = 0.0250\text{ mol}

\text{mass} = n \times M_r = 0.0250 \times 372.2 = 9.305\text{ g}

Answer

$9.31\text{ g}$ (using a $250\text{ cm}^3$ volumetric flask)

Final answer

9.31 g

Detailed explanation

Background Concept

Concentration $c = n/V$ , so moles $n = cV$ . The volume must be in $\text{dm}^3$ . Mass is found from $m = n \times M_r$ . The hydrated salt has $M_r = 372.2$ .

Understanding the Question

You need the mass of $\text{Na}_2\text{H}_2\text{Y}\cdot2\text{H}_2\text{O}$ needed to make $0.100\text{ mol dm}^{-3}$ solution using the volumetric flask chosen in (i), which we take as $250\text{ cm}^3$ .

Approach

Convert the flask volume to $\text{dm}^3$ , calculate the moles required, then multiply by $M_r$ .

Step-by-Step Reasoning

$250\text{ cm}^3 = 0.250\text{ dm}^3$ .
$n = cV = 0.100 \times 0.250 = 0.0250\text{ mol}$ .
$m = n \times M_r = 0.0250 \times 372.2 = 9.305\text{ g}$ .
To three significant figures, this is $9.31\text{ g}$ .

Key Takeaways

Always convert $\text{cm}^3$ to $\text{dm}^3$ by dividing by 1000 before using $n = cV$ .

Common Mistakes

Forgetting to divide the volume by 1000, giving a mass 1000 times too large.
Using the volume in $\text{cm}^3$ directly in $n = cV$ .
Quoting too many significant figures, e.g. $9.305\text{ g}$ , when the data support three.

Things to Be Careful About

Use the $M_r$ of the hydrated salt, $372.2$ , not the anhydrous salt.
The mass depends on the volumetric flask volume chosen in (i).
Include the unit $\text{g}$ .

Techniques used

calculate moles from concentration and volumeconvert moles to mass using Mr

(iii)

Describe how you would prepare this standard solution for use in your titration.

DifficultyMedium-Easy

Worked solution

Answer

Weigh out the calculated mass of hydrated disodium EDTA. Dissolve it completely in a small volume of distilled water in a beaker. Transfer the solution quantitatively to the $250\text{ cm}^3$ volumetric flask, rinsing the beaker and funnel with distilled water and adding the washings. Make up to the mark with distilled water, stopper and invert to mix thoroughly.

Final answer

Dissolve the salt in distilled water, transfer to a 250 cm3 volumetric flask, make up to the mark with distilled water and mix.

Detailed explanation

Background Concept

A standard solution is one whose concentration is accurately known. It is prepared by dissolving an accurately weighed mass of solute in a volumetric flask and making the solution up to the calibration mark with distilled water.

Understanding the Question

Describe the practical steps to prepare the EDTA standard solution from the mass calculated in (ii). The mark scheme requires that the salt is dissolved and then transferred to the volumetric flask and made up to the mark with distilled water.

Approach

Follow the standard sequence: dissolve, transfer quantitatively, make up to the mark, mix.

Step-by-Step Reasoning

Weigh out $9.31\text{ g}$ of the hydrated salt.
Dissolve it completely in a small volume of distilled water in a beaker, stirring to ensure all solid dissolves.
Transfer the solution to the $250\text{ cm}^3$ volumetric flask using a funnel.
Rinse the beaker, stirring rod and funnel with distilled water and add the washings to the flask; this ensures all solute is transferred.
Add distilled water until the bottom of the meniscus sits exactly on the calibration mark.
Stopper the flask and invert it several times to mix the solution thoroughly.

Key Takeaways

Quantitative transfer and accurate making-up to the mark are essential for a standard solution. Distilled or deionised water must be used.

Common Mistakes

Not mentioning distilled/deionised water.
Adding too much water and overshooting the mark.
Failing to rinse the beaker, so some solute is lost.
Not mixing the solution after making up to the mark.

Things to Be Careful About

The mark scheme awards one mark for dissolving and one for transferring and making up to the mark.
Water must be mentioned at least once; distilled/deionised/purified water is needed for full marks.
Read the meniscus at eye level.

Techniques used

prepare a standard solutionuse a volumetric flaskdissolve and make up to the mark

(iv)

After you have performed a rough titration, how would you ensure that your next titration is accurate?

DifficultyEasy

Worked solution

Answer

Add the EDTA solution dropwise when close to the endpoint, swirling the flask after each addition.

Final answer

Add the solution dropwise near the endpoint.

Detailed explanation

Background Concept

A rough titration gives an approximate endpoint. In subsequent accurate titrations, the titrant is added quickly at first and then dropwise near the endpoint to avoid overshooting.

Understanding the Question

After a rough titration, how do you make the next titration accurate? The key is careful addition near the endpoint.

Approach

Add the EDTA solution slowly, drop by drop, when the colour is about to change.

Step-by-Step Reasoning

Run the burette quickly at first, but slow down as the endpoint approaches.
Add the solution dropwise, swirling the conical flask after each drop.
Stop immediately at the first permanent colour change from purple to blue.

Key Takeaways

Dropwise addition near the endpoint prevents overshooting and gives an accurate titre.

Common Mistakes

Adding the titrant too quickly near the endpoint.
Not swirling, so the colour change is not uniform.
Continuing to add after the endpoint has been reached.

Things to Be Careful About

The endpoint colour change is purple to blue.
Record the burette reading to the nearest $0.05\text{ cm}^3$ .

Techniques used

control addition near the endpointswirl the flask during titration

(v)

How would you ensure that your titration result is reliable?

DifficultyEasy

Worked solution

Answer

Repeat the titration until at least two concordant titres are obtained (within $0.10\text{ cm}^3$ of each other) and calculate the mean titre.

Final answer

Repeat the titration until concordant titres are obtained and take the mean.

Detailed explanation

Background Concept

A reliable result is one that is reproducible. Repeating the titration and obtaining concordant titres reduces the effect of random errors.

Understanding the Question

How do you ensure your titration result is reliable? The answer is to repeat until concordant titres are obtained and use the mean.

Approach

Perform several titrations, check that the titres agree closely, and calculate the mean of the concordant values.

Step-by-Step Reasoning

Carry out the titration at least twice more after the rough titration.
Concordant titres are usually within $0.10\text{ cm}^3$ of each other.
Discard any anomalous results.
Calculate the mean of the concordant titres and use it in the calculation.

Key Takeaways

Repeats and concordance improve reliability. The mean titre is used for calculations.

Common Mistakes

Saying 'repeat once' without mentioning concordance.
Including the rough titre in the mean.
Not discarding anomalous results.

Things to Be Careful About

Concordant titres are the key phrase.
The mean should be calculated from at least two concordant values.

Techniques used

repeat the titrationcheck concordance of titrescalculate the mean titre

(c)

The term hard water is used to describe water containing the dissolved metal ions, $\text{Ca}^{2+}(\text{aq})$ and $\text{Mg}^{2+}(\text{aq})$ . Both of these metal ions react with EDTA anions, $\text{H}_2\text{Y}^{2-}$ .

\text{Ca}^{2+}(\text{aq}) + \text{Mg}^{2+}(\text{aq}) + 2\text{H}_2\text{Y}^{2-}(\text{aq}) \rightarrow \text{CaY}^{2-}(\text{aq}) + \text{MgY}^{2-}(\text{aq}) + 4\text{H}^+(\text{aq})

In an experiment to determine the concentration of each of these metal ions, two separate titrations with EDTA need to be performed.

For titration 1, a $25.0\text{ cm}^3$ sample of hard water is titrated with $0.0100\text{ mol dm}^{-3}$ EDTA solution using Solochrome Black solution as indicator.

For titration 2, another $25.0\text{ cm}^3$ sample of the same hard water is first treated with excess $2\text{ mol dm}^{-3}$ $\text{NaOH}(\text{aq})$ which precipitates all of the $\text{Mg}^{2+}(\text{aq})$ ions as $\text{Mg(OH)}_2(\text{s})$ . After this treatment, no $\text{Mg}^{2+}(\text{aq})$ ions remain in solution, leaving only dissolved $\text{Ca}^{2+}(\text{aq})$ ions in solution. This solution is then titrated with $0.0100\text{ mol dm}^{-3}$ EDTA solution using Solochrome Black solution as indicator.

The following information gives some of the hazards associated with the chemicals used in the procedure.

Chemical	Hazard
Sodium hydroxide	Solutions equal to or more concentrated than $0.5\text{ mol dm}^{-3}$ are classified as corrosive.
Solochrome Black	Solid Solochrome Black is classified as health hazard and is irritating to eyes, respiratory system and skin. All solutions are made up in ethanol and so are classified as flammable and health hazard.

(i)

Identify one hazard that must be considered when planning the experiment and describe a precaution, other than eye protection, that should be taken to keep risks from this hazard to a minimum.

DifficultyEasy

Worked solution

Answer

Sodium hydroxide solution is corrosive; wear gloves when handling it.

Final answer

NaOH(aq) is corrosive; wear gloves.

Detailed explanation

Background Concept

Risk assessment involves identifying hazards and taking precautions to minimise risk. The table lists hazards for sodium hydroxide and Solochrome Black solution.

Understanding the Question

Identify one hazard and describe a precaution, other than eye protection, to minimise the risk.

Approach

Choose one chemical from the table, state its hazard and pair it with a suitable precaution.

Step-by-Step Reasoning

Sodium hydroxide solution at $2\text{ mol dm}^{-3}$ is corrosive because it is more concentrated than $0.5\text{ mol dm}^{-3}$ .
Precaution: wear gloves to protect the skin.
Alternative: Solochrome Black solution is flammable because it contains ethanol; keep it away from naked flames.
Alternative: Solochrome Black is a health hazard irritating the respiratory system; use a fume cupboard or face mask.

Key Takeaways

A precaution must be specific to the hazard. Eye protection alone is not enough for the mark.

Common Mistakes

Giving only 'wear eye protection'.
Stating a hazard without a precaution.
Using a vague precaution such as 'be careful'.

Things to Be Careful About

The precaution must be other than eye protection.
Pair the correct precaution with the chosen hazard.

Techniques used

identify a hazardsuggest a precaution

(ii)

Results obtained from this experiment are shown.

\text{titre 1} = 22.70\text{ cm}^3 \quad \text{titre 2} = 16.60\text{ cm}^3

Use the results of the titrations to determine the concentrations of $\text{Ca}^{2+}(\text{aq})$ and $\text{Mg}^{2+}(\text{aq})$ in the hard water.

DifficultyMedium

Worked solution

Working

Titration 2 (Ca $^{2+}$ only):

n(\text{EDTA}) = 0.0100 \times \frac{16.60}{1000} = 1.66 \times 10^{-4}\text{ mol} = n(\text{Ca}^{2+})

[\text{Ca}^{2+}] = \frac{1.66 \times 10^{-4}}{0.0250} = 6.64 \times 10^{-3}\text{ mol dm}^{-3}

Titration 1 (Ca $^{2+}$ + Mg $^{2+}$ ):

n(\text{EDTA}) = 0.0100 \times \frac{22.70}{1000} = 2.27 \times 10^{-4}\text{ mol}

n(\text{Mg}^{2+}) = 2.27 \times 10^{-4} - 1.66 \times 10^{-4} = 6.10 \times 10^{-5}\text{ mol}

[\text{Mg}^{2+}] = \frac{6.10 \times 10^{-5}}{0.0250} = 2.44 \times 10^{-3}\text{ mol dm}^{-3}

Answer

$[\text{Ca}^{2+}] = 6.64 \times 10^{-3}\text{ mol dm}^{-3}$ ; $[\text{Mg}^{2+}] = 2.44 \times 10^{-3}\text{ mol dm}^{-3}$

Final answer

[Ca2+] = 6.64 × 10^-3 mol dm^-3; [Mg2+] = 2.44 × 10^-3 mol dm^-3

Detailed explanation

Background Concept

In EDTA titrations, each $\text{M}^{2+}$ ion reacts with one $\text{H}_2\text{Y}^{2-}$ ion in a 1:1 ratio. Titration 1 measures the total amount of $\text{Ca}^{2+}$ and $\text{Mg}^{2+}$ because both react with EDTA. Titration 2, after removing $\text{Mg}^{2+}$ as $\text{Mg(OH)}_2$ , measures only $\text{Ca}^{2+}$ . The amount of $\text{Mg}^{2+}$ is found by subtraction.

Understanding the Question

We are given two titres for the same $25.0\text{ cm}^3$ sample of hard water. Titre 1 is for total $\text{Ca}^{2+}+\text{Mg}^{2+}$ ; titre 2 is for $\text{Ca}^{2+}$ only. We need concentrations in $\text{mol dm}^{-3}$ .

Approach

Calculate moles of EDTA used in each titration, apply 1:1 stoichiometry, then convert moles to concentrations using the sample volume. For $\text{Mg}^{2+}$ , subtract the $\text{Ca}^{2+}$ moles from the total moles.

Step-by-Step Reasoning

Titration 2: $n(\text{EDTA}) = 0.0100 \times 16.60/1000 = 1.66 \times 10^{-4}\text{ mol}$ .
This equals $n(\text{Ca}^{2+})$ because the ratio is 1:1.
$[\text{Ca}^{2+}] = 1.66 \times 10^{-4} / 0.0250 = 6.64 \times 10^{-3}\text{ mol dm}^{-3}$ .
Titration 1: $n(\text{EDTA}) = 0.0100 \times 22.70/1000 = 2.27 \times 10^{-4}\text{ mol}$ .
This equals $n(\text{Ca}^{2+}) + n(\text{Mg}^{2+})$ .
$n(\text{Mg}^{2+}) = 2.27 \times 10^{-4} - 1.66 \times 10^{-4} = 6.10 \times 10^{-5}\text{ mol}$ .
$[\text{Mg}^{2+}] = 6.10 \times 10^{-5} / 0.0250 = 2.44 \times 10^{-3}\text{ mol dm}^{-3}$ .

Key Takeaways

Two titrations allow determination of two ions when one can be selectively removed. Always use 1:1 stoichiometry and convert volumes to $\text{dm}^3$ .

Common Mistakes

Using the titre volume in $\text{cm}^3$ without dividing by 1000.
Forgetting to divide by the $25.0\text{ cm}^3$ sample volume.
Not subtracting to find $\text{Mg}^{2+}$ .
Confusing which titre corresponds to which ion.

Things to Be Careful About

The sample volume is $25.0\text{ cm}^3 = 0.0250\text{ dm}^3$ .
EDTA reacts 1:1 with each $\text{M}^{2+}$ ion.
Quote concentrations in $\text{mol dm}^{-3}$ with appropriate significant figures.

Techniques used

calculate moles of EDTA from titreapply 1:1 stoichiometrysubtract to find Mg2+ molesconvert moles to concentration

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