Chemistry 9701/52 — May/June 2013

Background Concept

Le Chatelier's principle states that when a system at equilibrium is subjected to a change in conditions, the system responds in a way that partially counteracts the change. For temperature changes specifically: increasing temperature favours the endothermic direction, while decreasing temperature favours the exothermic direction. This is because adding heat is equivalent to adding a 'reactant' to the endothermic side of the equilibrium.

For a solubility equilibrium, the solid is in dynamic equilibrium with its dissolved ions. The enthalpy of solution ($\Delta H_{\text{soln}}$) tells us whether the dissolution process releases or absorbs heat. A negative value means the process is exothermic — heat is evolved when the solid dissolves.

Understanding the Question

This part asks you to predict the effect of increasing temperature on the solubility of $\text{Ca(OH)}_2$ and to justify your prediction using Le Chatelier's principle applied to the given equilibrium equation. The key piece of information is $\Delta H_{\text{soln}} = -16.7\text{ kJ mol}^{-1}$, which tells you the forward reaction (dissolution) is exothermic.

The command word is 'predict' followed by 'explain', so you must state the direction of change AND give the reasoning.

Approach

1. Identify the sign of $\Delta H$ for the forward reaction (exothermic, negative).
2. Apply Le Chatelier's principle: increasing temperature favours the endothermic direction (reverse reaction here).
3. Conclude that the equilibrium shifts left, meaning less solid dissolves, so solubility decreases.

Step-by-Step Reasoning

Prediction (B1): The solubility decreases. This is the direct consequence of the equilibrium shifting left.

Explanation (B1): The dissolution of $\text{Ca(OH)}_2$ is exothermic ($\Delta H_{\text{soln}} = -16.7\text{ kJ mol}^{-1}$, a negative value). When temperature is increased, the system responds by favouring the direction that absorbs heat — the reverse (endothermic) reaction. This shifts the equilibrium to the left, converting dissolved $\text{Ca}^{2+}$ and $\text{OH}^-$ ions back into solid $\text{Ca(OH)}_2$, thereby reducing solubility.

Note: The mark scheme explicitly states that saying 'increase' negates both marks, so the direction must be correct. The word 'exothermic' (or 'heat evolved') must appear in the explanation.

Key Takeaways

• A negative $\Delta H$ means the forward process is exothermic.
• For exothermic dissolutions, solubility decreases with increasing temperature (unusual — most solids have increasing solubility with temperature because their dissolution is endothermic).
• Le Chatelier's principle applied to temperature: the system shifts to oppose the change by favouring the endothermic direction when heated.

Common Mistakes

• Saying solubility increases (this negates both marks according to the mark scheme).
• Failing to mention 'exothermic' or 'heat evolved/absorbed' — the mark scheme requires this terminology.
• Confusing the direction of the shift: increasing temperature shifts the equilibrium in the endothermic direction, which here is the reverse (leftward) direction.
• Saying 'the equilibrium shifts to reduce the temperature' without specifying left/right or connecting to solubility.

Things to Be Careful About

• The sign of $\Delta H$ must be read correctly: negative = exothermic forward reaction.
• The explanation must connect the shift direction to the observable consequence (solubility decreases).
• Use precise language: 'exothermic' or 'heat evolved' for the forward direction, or 'endothermic' or 'heat absorbed' for the reverse direction.

Background Concept

When representing a relationship between two variables graphically, the independent variable goes on the x-axis and the dependent variable on the y-axis. Both axes must be labelled with the quantity name and its unit. Numerical values on the axes allow the reader to interpret the magnitude of the relationship.

For solubility-temperature graphs, the shape of the curve reflects how solubility changes with temperature. A decreasing solubility with increasing temperature (as for $\text{Ca(OH)}_2$) produces a downward-sloping curve or line.

Understanding the Question

You are given a blank set of axes (Fig. 1.1) and must sketch a graph showing how solubility of $\text{Ca(OH)}_2$ varies with temperature from $0\text{ }^\circ\text{C}$ to $100\text{ }^\circ\text{C}$. The question requires:

• Correct axis labels with units
• A decreasing trend (consistent with your prediction in (a)(i))
• The line must pass through the specific point $(25\text{ }^\circ\text{C}, 1\text{ g dm}^{-3})$
• The x-axis must span from 0 to 100

Approach

1. Label the x-axis as 'Temperature / $^\circ\text{C}$' with values from 0 to 100.
2. Label the y-axis as 'Solubility / $\text{g dm}^{-3}$' with a value of 1 marked at the appropriate height.
3. Draw a decreasing curve or straight line that passes through the point $(25, 1)$.
4. Ensure the line spans the full range 0–100 $^\circ\text{C}$.

Step-by-Step Reasoning

B1 — Axes and trend: Both axes must be labelled with the quantity and unit. The graph must show a decrease in solubility as temperature increases. The mark scheme says 'ignore units' for this mark, meaning the units on the axes are not strictly required for this particular mark, but the labels (Temperature, Solubility) and the decreasing trend are essential.

B1 — Specific point and range: The graph must pass through the point $(25\text{ }^\circ\text{C}, 1\text{ g dm}^{-3})$ and extend from $0\text{ }^\circ\text{C}$ to $100\text{ }^\circ\text{C}$. This means you need numerical values on both axes to show that the curve intersects the point where temperature = 25 and solubility = 1.

The mark scheme allows error carried forward from (a)(i), so if you incorrectly predicted an increase in (a)(i), you can still get the graph marks by drawing an increasing curve through (25, 1).

Key Takeaways

• Always label axes with both the quantity and its unit.
• Mark specific data points mentioned in the question.
• Ensure the graph spans the full range requested.
• The shape of the graph must be consistent with your prediction.

Common Mistakes

• Drawing an increasing curve (contradicts the exothermic dissolution).
• Forgetting to mark the point $(25, 1)$ explicitly with numerical values on the axes.
• Not extending the line to $100\text{ }^\circ\text{C}$.
• Labelling axes without units (though the mark scheme says 'ignore units' for the first B1, the second B1 requires numerical values which implicitly need units to be meaningful).

Things to Be Careful About

• The graph must be a curve or straight line showing decrease — a horizontal line or increase scores zero.
• The point $(25, 1)$ must be clearly identifiable on the graph with axis values marked.
• The x-axis must go from 0 to 100 (not just 25 or 50).

Background Concept

In any scientific investigation, the independent variable is the one factor that the experimenter deliberately changes or controls. It is the 'cause' in a cause-and-effect relationship. The dependent variable is what is measured in response, and control variables are everything else kept constant.

Understanding the Question

The investigation described is: 'how the solubility of calcium hydroxide varies as the temperature increases'. The question asks you to name the independent variable — the thing you are deliberately changing.

Approach

Identify what is being varied by the experimenter: temperature is being increased, so it is the independent variable.

Step-by-Step Reasoning

The phrase 'as the temperature increases' tells us that temperature is the variable being manipulated by the experimenter. This is the definition of the independent variable. Solubility is what changes in response — that is the dependent variable.

Key Takeaways

• The independent variable is what you change; the dependent variable is what you measure.
• Look for the phrase 'effect of X on Y' or 'as X changes' — X is the independent variable.

Common Mistakes

• Confusing independent and dependent variables.
• Writing 'solubility' for the independent variable.

Things to Be Careful About

• This is a simple identification question; just name the variable clearly.

Background Concept

The dependent variable is the quantity that is measured or observed in response to changes in the independent variable. It 'depends on' the independent variable and is what you record as your results.

Understanding the Question

In the investigation 'how the solubility of calcium hydroxide varies as the temperature increases', the question asks for the dependent variable — what is being measured.

Approach

Since temperature is being changed (independent), the solubility is what is being measured in response — this is the dependent variable.

Step-by-Step Reasoning

The investigation measures how solubility changes as temperature is varied. Solubility is therefore the dependent variable (B1). The mark scheme accepts 'solubility (of calcium hydroxide)'.

Key Takeaways

• The dependent variable is what you measure — it depends on the independent variable.
• In a 'how does X affect Y' investigation, Y is the dependent variable.

Common Mistakes

• Writing 'temperature' (that is the independent variable).
• Writing something vague like 'the amount dissolved' without connecting it to solubility.

Things to Be Careful About

• Be specific: 'solubility of calcium hydroxide' is the expected answer.

Background Concept

Acid-base neutralisation involves a metal hydroxide reacting with an acid to produce a salt and water. For calcium hydroxide (a Group 2 hydroxide with two $\text{OH}^-$ ions per formula unit), two moles of a monoprotic acid like $\text{HCl}$ are required per mole of hydroxide.

Understanding the Question

You are asked to write the balanced equation for the reaction between aqueous calcium hydroxide and hydrochloric acid. This is the reaction that occurs during the titration.

Approach

Write the reactants ($\text{Ca(OH)}_2$ and $\text{HCl}$) and products ($\text{CaCl}_2$ and $\text{H}_2\text{O}$), then balance.

Step-by-Step Reasoning

Calcium hydroxide provides two $\text{OH}^-$ ions, so two $\text{H}^+$ ions from $\text{HCl}$ are needed to form two water molecules. The calcium ion pairs with two chloride ions to form $\text{CaCl}_2$. The balanced equation is:

$\text{Ca(OH)}_2\text{(aq)} + 2\text{HCl(aq)} \rightarrow \text{CaCl}_2\text{(aq)} + 2\text{H}_2\text{O(l)}$

The mark scheme awards B1 for this equation (state symbols are not explicitly required by the mark scheme text but are good practice).

Key Takeaways

• Group 2 hydroxides require 2 moles of monoprotic acid per mole of hydroxide.
• Always check that the equation is balanced for all atoms and charge.

Common Mistakes

• Writing $\text{Ca(OH)}_2 + \text{HCl} \rightarrow \text{CaCl}_2 + \text{H}_2\text{O}$ (unbalanced).
• Writing incorrect products such as $\text{CaCl}$ instead of $\text{CaCl}_2$.

Things to Be Careful About

• Ensure the coefficient of 2 appears before both $\text{HCl}$ and $\text{H}_2\text{O}$.
• The stoichiometric ratio (1:2) is critical for the later calculation in (c)(vii).

Background Concept

In a titration, one solution is delivered from a burette (variable volume, read to 0.05 cm³) and the other is measured using a pipette (fixed volume, high precision). The pipette delivers a known volume of the analyte into the conical flask, while the burette delivers the titrant until the end-point is reached.

Understanding the Question

You must list the apparatus needed for the titration, including volumes where appropriate. The mark scheme specifically looks for a pipette (with a volume of 5, 10, 20, 25 or 50 cm³) and a burette (with a volume of 25, 50 or 100 cm³).

Approach

Identify the two key pieces of volumetric apparatus and state suitable volumes.

Step-by-Step Reasoning

B1: The mark scheme requires both a pipette (with volume from the list 5, 10, 20, 25 or 50 cm³) and a burette (with volume from 25, 50 or 100 cm³). A $25\text{ cm}^3$ pipette and $50\text{ cm}^3$ burette are the most common choices for school titrations.

Key Takeaways

• A pipette is used for the fixed-volume aliquot; a burette for the variable-volume titrant.
• Always specify the volume of the apparatus.

Common Mistakes

• Listing only a measuring cylinder (not precise enough for titration).
• Forgetting to give a volume for the pipette or burette.
• Not mentioning both pieces of apparatus.

Things to Be Careful About

• The mark scheme requires BOTH pipette AND burette with volumes for the single B1 mark.

Background Concept

The choice of indicator depends on the type of titration. For a strong acid (HCl) titrating a weak base (Ca(OH)₂ is only slightly soluble, and the solution is weakly alkaline at ~0.0135 mol dm⁻³), the pH at the equivalence point is slightly acidic (below 7). Indicators with a pH range that includes the steep portion of the titration curve near the equivalence point are suitable. Methyl orange (pH range 3.1–4.4) and methyl red (pH range 4.4–6.2) are appropriate for strong acid–weak base titrations. Phenolphthalein (pH range 8.2–10.0) would be less suitable because the colour change occurs before the equivalence point in this type of titration.

Understanding the Question

You must name a suitable indicator AND state its colour in both acidic and alkaline solutions. The mark scheme requires both the name and the two colours.

Approach

Choose an indicator whose transition range is appropriate for a strong acid–weak base titration, and state the colour in each medium.

Step-by-Step Reasoning

B1: Methyl orange is the most commonly cited indicator for this type of titration. It is yellow in alkaline solution (the initial colour of the $\text{Ca(OH)}_2$ solution in the flask) and turns red/pink at the end-point when the solution becomes acidic. The mark scheme requires the name AND both colours.

Key Takeaways

• Strong acid vs weak base: use methyl orange or methyl red.
• Always state the colour in BOTH acid and alkali.
• The initial colour in the flask is the alkaline colour (since $\text{Ca(OH)}_2$ is in the flask).

Common Mistakes

• Choosing phenolphthalein (acceptable in some contexts but not ideal here; the mark scheme may accept it but methyl orange is the standard answer).
• Forgetting to state one of the two colours.
• Stating the colours the wrong way round.

Things to Be Careful About

• The question says 'relevant colours' — you must give the colour in acid AND in alkali, not just the end-point colour.

Background Concept

The relationship between mass concentration ($\text{g dm}^{-3}$) and molar concentration ($\text{mol dm}^{-3}$) is:

$c = \frac{\rho}{M_r}$

where $\rho$ is the mass concentration and $M_r$ is the relative formula mass.

Understanding the Question

You are given that the solubility of $\text{Ca(OH)}_2$ is approximately $1\text{ g dm}^{-3}$ at $25\text{ }^\circ\text{C}$, and you must convert this to $\text{mol dm}^{-3}$ using the given $A_r$ values.

Approach

Calculate $M_r$ of $\text{Ca(OH)}_2$, then divide the mass concentration by $M_r$.

Step-by-Step Reasoning

Step 1: Calculate $M_r(\text{Ca(OH)}_2)$:
$M_r = 40.1 + 2(16.0 + 1.0) = 40.1 + 34.0 = 74.1$

Step 2: Convert to molar concentration:
$c = \frac{1}{74.1} = 0.0135\text{ mol dm}^{-3}$

The mark scheme awards B1 for the answer $0.0135\text{ mol dm}^{-3}$.

Key Takeaways

• Mass concentration to molar concentration: divide by $M_r$.
• Always show the $M_r$ calculation to demonstrate working.

Common Mistakes

• Calculating $M_r$ incorrectly (forgetting the 2 in $2(16.0 + 1.0)$).
• Multiplying instead of dividing.
• Using the wrong $A_r$ values.

Things to Be Careful About

• The answer should be to 3 significant figures (0.0135) as given in the mark scheme.
• Include units in the final answer.

Background Concept

When the concentration of the titrant is too high relative to the analyte, the titre volume would be very small, leading to large percentage errors. Diluting the titrant ensures that the titre volume is in a convenient range (typically 20–30 cm³), and for equal reacting volumes, the concentrations should be chosen so that the stoichiometric ratio is satisfied at approximately equal volumes.

A volumetric dilution involves transferring a measured volume of concentrated solution into a volumetric flask and making up to the calibration mark with solvent. This is far more precise than adding a calculated volume of water separately.

Understanding the Question

You must describe how to dilute the $2.00\text{ mol dm}^{-3}$ HCl so that when titrating against a pipetted volume of saturated $\text{Ca(OH)}_2$, the two reacting volumes are approximately equal at the end-point. You must show the relevant calculations and reasoning in full.

The mark scheme requires:

• B1: Describes making a solution in a volumetric flask, using a burette or pipette to measure the HCl volume, and making up to the mark with water.
• B1: The dilution is between 50 and 100 fold (or a mixture giving that dilution).

Approach

1. Assume a pipette volume of $\text{Ca(OH)}_2$ (e.g., $25\text{ cm}^3$).
2. Calculate moles of $\text{Ca(OH)}_2$ in that volume.
3. Use the 1:2 stoichiometry to find moles of HCl needed.
4. Set the volume of HCl equal to the volume of $\text{Ca(OH)}_2$ to find the required concentration.
5. Calculate the dilution factor from 2.00 to that concentration.
6. Describe a proper volumetric flask dilution method achieving that factor.

Step-by-Step Reasoning

Step 1: Assume $25\text{ cm}^3$ of saturated $\text{Ca(OH)}_2$ is pipetted.

Step 2: $n(\text{Ca(OH)}_2) = 0.0135 \times 0.025 = 3.375 \times 10^{-4}$ mol

Step 3: From the equation, $n(\text{HCl}) = 2 \times n(\text{Ca(OH)}_2) = 6.75 \times 10^{-4}$ mol

Step 4: For equal volumes, we want $25\text{ cm}^3$ of HCl to deliver this amount:
$[\text{HCl}] = \frac{6.75 \times 10^{-4}}{0.025} = 0.027\text{ mol dm}^{-3}$

Step 5: Dilution factor $= \frac{2.00}{0.027} \approx 74$. This is between 50 and 100, satisfying the mark scheme.

Step 6 (B1 — method): Use a burette to measure a suitable volume of $2.00\text{ mol dm}^{-3}$ HCl (e.g., $3.5\text{ cm}^3$) into a $250\text{ cm}^3$ volumetric flask. Add distilled water to the mark. This gives a dilution of $250/3.5 \approx 71$-fold, yielding $[\text{HCl}] \approx 0.028\text{ mol dm}^{-3}$.

Alternative: pipette $5.0\text{ cm}^3$ of $2.00\text{ mol dm}^{-3}$ HCl into a $250\text{ cm}^3$ volumetric flask → dilution factor = 50, $[\text{HCl}] = 0.04\text{ mol dm}^{-3}$. This is at the boundary of the acceptable range.

Key Takeaways

• The dilution factor must be calculated from the stoichiometry and the desire for equal volumes.
• A volumetric flask with 'make up to the mark' is the correct technique — not adding a calculated volume of water.
• The burette or pipette is used to measure the concentrated solution precisely.

Common Mistakes

• Adding a specific volume of water rather than making up to the mark in a volumetric flask.
• Choosing a dilution factor outside the 50–100 range.
• Not showing the calculation linking the dilution factor to the equal-volume requirement.
• Using a measuring cylinder instead of a burette or pipette to measure the concentrated acid.

Things to Be Careful About

• The mark scheme specifically requires mention of a volumetric flask and 'making up to the mark'.
• The dilution must be between 50 and 100 fold — not 10 fold or 200 fold.
• Show the reasoning for WHY this dilution gives equal volumes.

Background Concept

Titration technique requires a rough run to identify the approximate end-point volume, followed by accurate runs where the titrant is added dropwise near the end-point to avoid overshooting. Concordant results (typically within $0.10\text{ cm}^3$) are averaged to minimise random error. The rough titre is not included in the average.

Understanding the Question

You must describe how to carry out sufficient titrations to obtain an accurate end-point. The mark scheme awards B1 for stating that the titration is repeated to achieve concordant results or an average titre.

Approach

Describe the standard titration procedure: rough → accurate repeats → concordance → mean.

Step-by-Step Reasoning

B1: The key point the mark scheme looks for is repetition to achieve concordant titres or an average. The full procedure includes:

• A rough titration to find the approximate end-point.
• Accurate titrations adding the acid slowly (dropwise) near the expected end-point.
• Continuing until at least two results agree within $0.10\text{ cm}^3$.
• Averaging the concordant titres (excluding the rough).

Key Takeaways

• Always do a rough titration first.
• Add dropwise near the end-point.
• 'Concordant' means within $0.10\text{ cm}^3$.
• The mean of concordant titres is used for calculations.

Common Mistakes

• Not mentioning a rough titration.
• Averaging all titres including the rough one.
• Not specifying what 'concordant' means.
• Saying 'repeat until the same answer is obtained' without the $0.10\text{ cm}^3$ tolerance.

Things to Be Careful About

• The mark scheme is looking for the concept of repetition and concordance — this is the single B1 mark.

Background Concept

In a titration calculation, the known concentration and measured volume of the titrant give the moles of titrant used. The balanced equation provides the stoichiometric ratio between titrant and analyte, allowing the moles of analyte to be deduced. Dividing by the known volume of analyte (from the pipette) gives its concentration.

Understanding the Question

You must outline how the titration results are processed to determine the accurate concentration of the saturated $\text{Ca(OH)}_2$ solution. The mark scheme requires three linked steps in a single B1 mark.

Approach

Work through the calculation chain: titre → moles HCl → moles Ca(OH)₂ → concentration.

Step-by-Step Reasoning

Step 1: Moles of HCl = concentration of dilute HCl × (mean titre / 1000)

Step 2: From the equation $\text{Ca(OH)}_2 + 2\text{HCl} \rightarrow \text{CaCl}_2 + 2\text{H}_2\text{O}$, the ratio is 1:2, so:
$n(\text{Ca(OH)}_2) = \frac{1}{2} \times n(\text{HCl})$

Step 3: Concentration of $\text{Ca(OH)}_2$:
$[\text{Ca(OH)}_2] = \frac{n(\text{Ca(OH)}_2)}{V_{\text{pipette}}}$

All three steps must be present for the B1 mark.

Key Takeaways

• The calculation chain in titrations is always: volume × concentration → moles → stoichiometric ratio → moles of unknown → concentration of unknown.
• The 1:2 ratio is critical and comes from the balanced equation.

Common Mistakes

• Forgetting the 1:2 ratio (using 1:1 instead).
• Not dividing by 1000 when converting cm³ to dm³.
• Dividing by the titre volume instead of the pipette volume in the final step.

Things to Be Careful About

• The mark scheme requires all three elements (moles HCl from titre, the 0.5 ratio, and concentration deduced) for the single B1 mark.