Chemistry 9701/35 — May/June 2024

Background Concept

Concentration is amount of solute per unit volume: $c = n/V$. If the total volume of a solution is kept constant, then doubling the volume of a stock solution added doubles the amount of solute and therefore doubles its concentration. If the total volume changed, the concentration would not be simply proportional to the volume added.

Understanding the Question

The question asks why the concentration of potassium peroxodisulfate in each experiment is proportional to the volume of FA 1 used. You need to compare the two reaction mixtures and notice that the total volume is the same in both, while the volume of FA 1 changes and is compensated by distilled water.

Approach

Calculate the total volume of each reaction mixture from the method. Show that in Experiment 1 and Experiment 2 the total volume is identical. Since concentration = amount/total volume, and the amount of $\text{S}_2\text{O}_8^{2-}$ is proportional to the volume of FA 1, concentration is proportional to the volume of FA 1.

Step-by-Step Reasoning

Experiment 1: 20.00 FA 1 + 0 water + 10.00 FA 2 + 20.0 FA 3 = 50.00 cm$^3$.

Experiment 2: 10.00 FA 1 + 10.00 water + 10.00 FA 2 + 20.0 FA 3 = 50.00 cm$^3$.

The total volume is the same. The amount of $\text{S}_2\text{O}_8^{2-}$ = concentration of FA 1 × volume of FA 1. Since the concentration of FA 1 is fixed, amount $\propto$ volume of FA 1. Dividing by the same total volume gives concentration $\propto$ volume of FA 1.

Key Takeaways

• Constant total volume is essential for concentration to be proportional to volume added.
• Water is used to keep the total volume constant when a smaller volume of FA 1 is used.

Common Mistakes

• Saying concentration is proportional to volume because “more FA 1 means more moles” without mentioning constant total volume.
• Forgetting that water is added to keep the total volume constant.

Things to Be Careful About

• Use “amount” or “moles” rather than “mass” when discussing concentration.
• The mark scheme accepts either “total volume was the same” or “volume of water + FA 1 is 20 cm$^3$ in both experiments.”

Background Concept

To test whether rate is proportional to $[\text{S}_2\text{O}_8^{2-}]$, you need a third experiment with a different concentration of FA 1 while keeping all other concentrations and the total volume the same. This is done by changing the volume of FA 1 and compensating with distilled water so the total volume stays 50.00 cm$^3$. The volumes of FA 2 and FA 3 must stay the same so that the amount of thiosulfate and the concentration of iodide are unchanged.

Understanding the Question

Complete Table 1.1 with volumes for a further experiment. The table already has FA 4 = 10 drops. You must choose a volume of FA 1 different from 20.00 and 10.00, and a volume of water that makes FA 1 + water = 20.00 cm$^3$, while keeping FA 2 = 10.0 cm$^3$ and FA 3 = 20.0 cm$^3$.

Approach

Choose a convenient new volume of FA 1, e.g. 5.00 cm$^3$. Then set water = 20.00 − 5.00 = 15.00 cm$^3$. Keep FA 2 and FA 3 unchanged. Check that the total volume remains 50.00 cm$^3$.

Step-by-Step Reasoning

• FA 1 = 5.00 cm$^3$ (different from 20.00 and 10.00, and at least 2.5 cm$^3$).
• Water = 15.00 cm$^3$, so FA 1 + water = 20.00 cm$^3$.
• FA 2 = 10.0 cm$^3$, FA 3 = 20.0 cm$^3$.
• Total = 5.00 + 15.00 + 10.0 + 20.0 = 50.00 cm$^3$.
• This gives a concentration of FA 1 one quarter of that in Experiment 1.

Key Takeaways

• A dilution series keeps total volume constant by replacing part of the stock solution with water.
• All other reactant volumes must be kept constant to isolate the effect of FA 1.

Common Mistakes

• Choosing FA 1 + water not equal to 20.00 cm$^3$.
• Changing FA 2 or FA 3.
• Choosing FA 1 = 0 or too small to give a measurable time.

Things to Be Careful About

• The mark scheme requires FA 1 + water = 20.00 cm$^3$ and FA 1 at least 2.5 cm$^3$.
• FA 2 and FA 3 must be 10.0 and 20.0 cm$^3$ respectively.

Background Concept

The temperature change ($\Delta T$) in a calorimetry experiment is the difference between the initial temperature (before mixing) and the final temperature (after the reaction is complete and the solution has reached thermal equilibrium). In practice, because mixing takes time and heat is exchanged with the surroundings, the true $\Delta T$ is found by extrapolating the temperature-time graph to the moment of mixing ($t=2.5$ min).

Understanding the Question

Part (c)(i) asks the candidate to determine $\Delta T$ at $t=2.5$ minutes using the extrapolated values from the graph drawn in part (b). The value must be given to 1 decimal place.

Approach

Read the extrapolated temperatures from the two lines of best fit at $t=2.5$ minutes and calculate the difference.

Step-by-Step Reasoning

1. Read Extrapolated Temperatures: At $t=2.5$ min, read the temperature from Line 1 (pre-mixing extrapolation) and Line 2 (post-minimum extrapolation).
2. Calculate $\Delta T$: $\Delta T = T_{\text{initial}} - T_{\text{final}}$. Since the dissolution is endothermic, the temperature drops, so $\Delta T$ is positive.
3. Example Calculation: If $T_1 = 21.5 \text{ °C}$ and $T_2 = 17.0 \text{ °C}$, then $\Delta T = 21.5 - 17.0 = 4.5 \text{ °C}$.

Key Takeaways

• $\Delta T$ is always a positive value for endothermic reactions (initial - final).
• Use the extrapolated values, not the directly recorded temperatures.

Common Mistakes

• Reading the temperature from the graph at $t=2.5$ min instead of extrapolating.
• Calculating $\Delta T$ as final - initial, resulting in a negative value.
• Not giving the answer to 1 decimal place.

Things to Be Careful About

• Ensure the extrapolated temperature at $t=2.5$ min from the post-mixing line is less than or equal to the lowest recorded temperature.
• The value must be read accurately from the graph to 1 decimal place.

Background Concept

The energy change ($q$) absorbed or released by a solution in a calorimetry experiment is calculated using the formula $q = m c \Delta T$, where $m$ is the mass of the solution, $c$ is the specific heat capacity, and $\Delta T$ is the temperature change. For dilute aqueous solutions, we assume the density is $1.00 \text{ g cm}^{-3}$ (so $25.0 \text{ cm}^3 = 25.0 \text{ g}$) and the specific heat capacity is the same as water ($4.18 \text{ J g}^{-1} \text{ °C}^{-1}$). We ignore the mass of the dissolved solid and the heat capacity of the cup and thermometer.

Understanding the Question

Part (c)(ii) asks the candidate to calculate the energy change in Joules using the volume of water, the specific heat capacity of water, and the $\Delta T$ from part (c)(i).

Approach

Use $q = m c \Delta T$ with $m = 25.0 \text{ g}$, $c = 4.18 \text{ J g}^{-1} \text{ °C}^{-1}$, and $\Delta T = 4.5 \text{ °C}$.

Step-by-Step Reasoning

1. Identify Variables: $m = 25.0 \text{ g}$, $c = 4.18 \text{ J g}^{-1} \text{ °C}^{-1}$, $\Delta T = 4.5 \text{ °C}$.
2. Substitute: $q = 25.0 \times 4.18 \times 4.5$.
3. Calculate: $q = 470.25 \text{ J}$.
4. Significant Figures: The mark scheme allows 2–4 significant figures, so $470 \text{ J}$ or $470.3 \text{ J}$ is acceptable.

Key Takeaways

• Always use the mass of the solution (or water, for dilute solutions) in grams.
• The specific heat capacity of water is $4.18 \text{ J g}^{-1} \text{ °C}^{-1}$.
• The energy change is in Joules (J).

Common Mistakes

• Using the volume in cm³ instead of mass in g (though numerically the same for water, the unit must be g).
• Forgetting to multiply by $\Delta T$.
• Not converting the final answer to the correct number of significant figures.

Things to Be Careful About

• The mark scheme specifically states: $\text{energy} = 25 \times 4.18 \times \Delta T$. Ensure you use 25, not 25.0 + mass of solute.
• The answer must be in J, not kJ.

Background Concept

The amount of substance in moles ($n$) is calculated from the mass ($m$) and the relative formula mass ($M_r$) using the equation $n = m / M_r$. For ammonium chloride ($\text{NH}_4\text{Cl}$), $M_r = 14.0 + (4 \times 1.0) + 35.5 = 53.5$.

Understanding the Question

Part (c)(iii) asks the candidate to calculate the amount of ammonium chloride used in moles, using the mass recorded in part (a).

Approach

Use $n = m / M_r$ with the mass from (a) and $M_r = 53.5$.

Step-by-Step Reasoning

1. Identify Variables: $m = 0.378 \text{ g}$ (from part a), $M_r = 53.5$.
2. Substitute: $n = 0.378 / 53.5$.
3. Calculate: $n = 0.0070654... \text{ mol}$.
4. Significant Figures: The mark scheme allows 2–4 significant figures, so $0.00707 \text{ mol}$ or $0.007065 \text{ mol}$ is acceptable.

Key Takeaways

• Always calculate $M_r$ correctly from the chemical formula.
• Use the mass added (difference between initial and residual masses), not the initial mass alone.

Common Mistakes

• Using the wrong $M_r$ (e.g., forgetting the chlorine atom).
• Using the initial mass of the container + FA 5 instead of the mass of FA 5 added.
• Incorrect significant figures.

Things to Be Careful About

• The mark scheme states: $n(\text{NH}_4\text{Cl}) = \frac{\text{mass from (a)}}{53.5}$. Ensure you use the correct mass value.

Background Concept

The enthalpy change ($\Delta H$) per mole is calculated by dividing the energy change ($q$) by the amount of substance ($n$) and converting to kJ mol⁻¹. For endothermic reactions (temperature decreases), $\Delta H$ is positive. For exothermic reactions (temperature increases), $\Delta H$ is negative.

Understanding the Question

Part (c)(iv) asks the candidate to calculate $\Delta H$ in kJ mol⁻¹, including the correct sign. The energy change is from (c)(ii) and the amount is from (c)(iii).

Approach

Use $\Delta H = (q / n) \times 1000$, ensuring the sign is correct based on whether the temperature increased or decreased.

Step-by-Step Reasoning

1. Identify Variables: $q = 470.25 \text{ J}$, $n = 0.007065 \text{ mol}$.
2. Substitute: $\Delta H = (470.25 / 0.007065) \times 1000$.
3. Calculate: $\Delta H = 66563 \text{ J mol}^{-1} = 66.6 \text{ kJ mol}^{-1}$.
4. Determine Sign: The temperature dropped (endothermic), so $\Delta H$ is positive: $+66.6 \text{ kJ mol}^{-1}$.
5. Significant Figures: The mark scheme allows 2–4 significant figures, so $+67$, $+66.6$, or $+66.56$ is acceptable.

Key Takeaways

• Always include the sign for $\Delta H$ (+ for endothermic, - for exothermic).
• Convert J to kJ by dividing by 1000 (or multiplying by 1000 when the formula is set up as $q/n \times 1000$).
• Use the values from previous parts, even if they are wrong (error carried forward).

Common Mistakes

• Forgetting the sign (mark scheme explicitly requires correct sign).
• Forgetting to convert J to kJ.
• Using the wrong value for $q$ or $n$.
• Incorrect significant figures.

Things to Be Careful About

• The mark scheme states: $\Delta H = \frac{\text{(c)(ii)}}{\text{(c)(iii)}} \times 1000$. Ensure you follow this exactly.
• The sign must be explicitly stated: + or -.

Background Concept

Potassium manganate(VII), $\text{KMnO}_4$, is a purple crystalline ionic solid. When heated strongly it undergoes thermal decomposition:

The residue contains potassium manganate(VI) ($\text{K}_2\text{MnO}_4$, dark green) and manganese(IV) oxide ($\text{MnO}_2$, black). The gas evolved is oxygen, a colourless gas that supports combustion — a glowing splint relights in it.

Understanding the Question

FA 6 is described as a salt containing a Group 1 ion (potassium) and an anion made of a transition metal and a non-metal — this is $\text{KMnO}_4$. The task is to heat the solid gently then strongly, record all observations, and identify the gas produced. This is a classic qualitative-analysis gas test.

Approach

Heat gently first, then strongly. Watch for colour changes of the solid, physical movement (decrepitation), and gas evolution. Test the gas with a glowing splint and record the result.

Step-by-Step Reasoning

1. Initial appearance: dark purple crystals — record this as the starting observation.
2. On gentle heating: the crystals jump/spit (decrepitation) as gas is released; the solid begins to darken.
3. On strong heating: a black residue ($\text{MnO}_2$) forms.
4. Gas test: insert a glowing splint into the tube — it relights, confirming oxygen.

Each of these is a separate mark: two observations about the solid (purple crystals, jumping, black residue — any two), one for the splint relighting, and one for identifying the gas as oxygen.

Key Takeaways

• $\text{KMnO}_4$ decomposes on strong heating to $\text{K}_2\text{MnO}_4$, $\text{MnO}_2$ and $\text{O}_2$.
• Oxygen relights a glowing splint — the definitive test.
• The black residue is $\text{MnO}_2$.

Common Mistakes

• Failing to test the gas — you must carry out and record the splint test to identify oxygen.
• Writing "brown gas" — oxygen is colourless; the colour change is in the solid, not the gas.
• Not stating the stage at which the observation occurs (e.g., "on gentle heating" vs "on strong heating").

Things to Be Careful About

• Use a hard-glass test-tube when heating a solid.
• Record the colour of the solid before, during and after heating.
• The splint test must be described as "relights a glowing splint".

Background Concept

Manganate(VII), $\text{MnO}_4^-$, is purple. In strongly alkaline solution, in the presence of manganese(IV) oxide ($\text{MnO}_2$, which is FA 7, the residue from (a)(i)), manganate(VII) is reduced to manganate(VI), $\text{MnO}_4^{2-}$, which is dark green. On acidification, manganate(VI) disproportionates:

The green solution therefore turns back to pink/purple (manganate(VII)) with a dark precipitate of $\text{MnO}_2$.

Understanding the Question

The procedure is: acidified $\text{KMnO}_4$ is made alkaline by adding sodium hydroxide, then FA 7 (the $\text{MnO}_2$ residue from (a)(i)) is added and the mixture stirred for about 30 seconds. The mixture is filtered, and the filtrate is then acidified with excess sulfuric acid. You must record what you see at each stage.

Approach

Track the colour of the solution through the stages: purple (alkaline $\text{MnO}_4^-$) → dark green (after adding FA 7, $\text{MnO}_4^{2-}$) → pink/purple (after acidification, back to $\text{MnO}_4^-$).

Step-by-Step Reasoning

1. The acidified $\text{KMnO}_4$ is made alkaline with $\text{NaOH}$ — it remains purple.
2. Adding FA 7 ($\text{MnO}_2$) reduces $\text{MnO}_4^-$ to $\text{MnO}_4^{2-}$ — the solution becomes dark green. This is M1.
3. Filtering removes the excess solid; the filtrate is dark green.
4. Adding excess sulfuric acid acidifies the solution; $\text{MnO}_4^{2-}$ disproportionates back to $\text{MnO}_4^-$ — the solution turns pink/purple. This is M2.

Key Takeaways

• $\text{MnO}_4^-$ is purple; $\text{MnO}_4^{2-}$ is dark green.
• $\text{MnO}_4^{2-}$ disproportionates in acid to $\text{MnO}_4^-$ and $\text{MnO}_2$.
• $\text{MnO}_2$ catalyses the reduction of manganate(VII) in alkaline solution.

Common Mistakes

• Saying the filtrate is purple — it is dark green before acidification.
• Missing the colour change on acidification (pink/purple).
• Confusing the two stages and reporting the colours in the wrong order.

Things to Be Careful About

• The filtrate is dark green, not purple.
• On acidification the solution turns pink/purple — record this as a distinct observation.
• The question asks for observations in two places: after filtration and after adding acid.

Background Concept

This part tests three anions and one cation using standard qualitative tests:

• Carbonate, $\text{CO}_3^{2-}$: with acid it effervesces, giving $\text{CO}_2$, which turns limewater milky.
• Copper(II), $\text{Cu}^{2+}$: with $\text{NaOH}$ it gives a pale blue precipitate of $\text{Cu(OH)}_2$, which is insoluble in excess $\text{NaOH}$; on warming, $\text{Cu(OH)}_2$ decomposes to black $\text{CuO}$.
• Sulfate, $\text{SO}_4^{2-}$: with $\text{Ba}^{2+}$ it gives a white precipitate of $\text{BaSO}_4$, which is insoluble in acid.
• Hydrogen ion, $\text{H}^+$: makes the solution acidic; it reacts with carbonate to give $\text{CO}_2$.

FA 8 is $\text{Na}_2\text{CO}_3(\text{aq})$ and FA 9 is $\text{CuSO}_4(\text{aq}) + \text{H}_2\text{SO}_4(\text{aq})$.

Understanding the Question

You must carry out four tests on FA 8 and FA 9 and record your observations in Table 3.1. The tests probe for carbonate, copper(II), sulfate and acidity. Record observations at the correct stage (e.g., before/after warming, before/after adding acid).

Approach

Run each test on both solutions and record what you see. Use the standard tests: acid for carbonate (test gas with limewater), $\text{NaOH}$ for cations (note colour, solubility in excess, effect of warming), $\text{Ba}^{2+}$ for sulfate (then add acid to distinguish $\text{BaSO}_4$ from $\text{BaCO}_3$).

Step-by-Step Reasoning

Test 1 — add sulfuric acid:

• FA 8 ($\text{Na}_2\text{CO}_3$): effervescence as $\text{CO}_2$ is released; the gas turns limewater milky.
• FA 9 ($\text{CuSO}_4 + \text{H}_2\text{SO}_4$): already acidic, so no change; the solution stays blue.

Test 2 — add $\text{NaOH}$, then warm:

• FA 8: $\text{Na}^+$ does not precipitate with $\text{OH}^-$ and $\text{CO}_3^{2-}$ does not react with $\text{OH}^-$ — no change; on warming, no gas.
• FA 9: $\text{Cu}^{2+} + 2\text{OH}^- \rightarrow \text{Cu(OH)}_2$ — pale blue precipitate, insoluble in excess $\text{NaOH}$; on warming, $\text{Cu(OH)}_2 \rightarrow \text{CuO} + \text{H}_2\text{O}$, so the precipitate turns black.

Test 3 — add $\text{BaCl}_2$/ $\text{Ba(NO}_3)_2$, then $\text{HNO}_3$:

• FA 8: $\text{Ba}^{2+} + \text{CO}_3^{2-} \rightarrow \text{BaCO}_3$ — white precipitate; on adding acid, $\text{BaCO}_3$ dissolves with effervescence ($\text{CO}_2$).
• FA 9: $\text{Ba}^{2+} + \text{SO}_4^{2-} \rightarrow \text{BaSO}_4$ — white precipitate; $\text{BaSO}_4$ is insoluble in acid, so no change.

Test 4 — add FA 8 in excess (to FA 9):

• The $\text{H}^+$ in FA 9 reacts with $\text{CO}_3^{2-}$ from FA 8 — effervescence ($\text{CO}_2$).
• Then $\text{Cu}^{2+}$ reacts with $\text{CO}_3^{2-}$ — pale blue precipitate of $\text{CuCO}_3$.

Key Takeaways

• $\text{CO}_3^{2-}$: effervescence with acid; $\text{CO}_2$ turns limewater milky.
• $\text{Cu}^{2+}$: pale blue precipitate with $\text{NaOH}$, insoluble in excess, turns black on warming.
• $\text{SO}_4^{2-}$: white precipitate with $\text{Ba}^{2+}$, insoluble in acid.
• $\text{BaCO}_3$ dissolves in acid with effervescence; $\text{BaSO}_4$ does not.

Common Mistakes

• Confusing $\text{BaCO}_3$ (acid-soluble, effervesces) with $\text{BaSO}_4$ (acid-insoluble).
• Not recording the black colour of $\text{CuO}$ on warming.
• Saying $\text{Cu(OH)}_2$ dissolves in excess $\text{NaOH}$ — it does not (unlike $\text{Al}^{3+}$ or $\text{Zn}^{2+}$).
• Forgetting to test the gas with limewater in Test 1.

Things to Be Careful About

• Record each observation at the correct stage (before/after warming, before/after adding acid).
• Include gas tests (limewater) where relevant.
• Write "no change" where no visible change occurs — it is a valid observation.
• The mark scheme allows a rounded total: 16 available points cap at 7 marks.

Background Concept

Each observation in (b)(i) maps to a specific ion. Effervescence with acid indicates carbonate; a pale blue precipitate with $\text{NaOH}$ that turns black on warming indicates copper(II); a white precipitate with $\text{Ba}^{2+}$ that is insoluble in acid indicates sulfate; a solution that reacts with carbonate to release $\text{CO}_2$ is acidic, indicating $\text{H}^+$.

Understanding the Question

FA 8 contains one cation and one anion; FA 9 contains two cations and one anion. You must identify the three ions listed in the Qualitative analysis notes (carbonate, copper(II), sulfate) plus one other cation (hydrogen ion).

Approach

Work through the observations from (b)(i) and match each to the ion responsible.

Step-by-Step Reasoning

• FA 8 + acid → effervescence, $\text{CO}_2$: carbonate, $\text{CO}_3^{2-}$, present.
• FA 8 + $\text{NaOH}$ → no precipitate: the cation is a Group 1 ion — call it unknown (sodium, $\text{Na}^+$).
• FA 9 + $\text{NaOH}$ → pale blue ppt, black on warming: copper(II), $\text{Cu}^{2+}$.
• FA 9 + $\text{Ba}^{2+}$ → white ppt, insoluble in acid: sulfate, $\text{SO}_4^{2-}$.
• FA 9 + $\text{CO}_3^{2-}$ → effervescence: the solution is acidic — hydrogen ion, $\text{H}^+$.

So FA 8 is an unknown Group 1 carbonate (sodium carbonate), and FA 9 is copper(II) sulfate with sulfuric acid.

Key Takeaways

• Observations are evidence: each positive test identifies a specific ion.
• FA 9 has three ions — don't stop after finding copper(II).

Common Mistakes

• Writing "Na+" when the mark scheme accepts "unknown" — either is fine, but "unknown" is safer.
• Missing $\text{H}^+$ in FA 9 — the effervescence with carbonate in Test 4 is the clue.
• Giving FA 8 only one ion when it has two.

Things to Be Careful About

• FA 9 contains THREE ions: $\text{Cu}^{2+}$, $\text{H}^+$ and $\text{SO}_4^{2-}$.
• FA 8 contains TWO: unknown (sodium) and $\text{CO}_3^{2-}$.
• The mark scheme gives 2 marks for all correct, 1 mark for three correct.

Background Concept

In Test 2, adding $\text{NaOH}$ to FA 9 causes copper(II) ions to precipitate as copper(II) hydroxide. The ionic equation shows only the reacting species, with spectator ions (sodium and sulfate) omitted.

Understanding the Question

Write an ionic equation for one reaction that occurred in Test 2. The mark scheme accepts either the precipitation of $\text{Cu(OH)}_2$ or its decomposition to $\text{CuO}$ on warming.

Approach

Identify the reacting ions and balance both atoms and charge.

Step-by-Step Reasoning

1. $\text{Cu}^{2+}$ reacts with $\text{OH}^-$ to form $\text{Cu(OH)}_2$.
2. Balance: one $\text{Cu}^{2+}$ needs two $\text{OH}^-$ for charge and atom balance.
3. Add state symbols: $\text{Cu}^{2+}(\text{aq}) + 2\text{OH}^-(\text{aq}) \rightarrow \text{Cu(OH)}_2(\text{s})$.

The alternative accepted equation is the thermal decomposition: $\text{Cu(OH)}_2(\text{s}) \rightarrow \text{CuO}(\text{s}) + \text{H}_2\text{O}(\text{l})$.

Key Takeaways

• Ionic equations omit spectator ions.
• Both atoms and charge must balance.
• State symbols are essential.

Common Mistakes

• Missing state symbols — the mark scheme requires them.
• Writing the full equation with spectator ions (sodium, sulfate) — the question asks for ionic.
• Unbalanced charges (e.g., $\text{Cu}^{2+} + \text{OH}^- \rightarrow \text{Cu(OH)}_2$).

Things to Be Careful About

• Use (aq) for aqueous ions and (s) for the precipitate.
• Either accepted equation scores — choose the precipitation one for simplicity.