Chemistry 5070/41 — May/June 2025

Walkthrough

The table has readings at 0 °C and 40 °C. The time 20 °C lies exactly halfway between them, so an appropriate estimate is the average of the two solubilities. The difference between 480 and 300 is 180, and half of this is 90; adding 90 to 300 gives 390 g / dm³.

Key Takeaway

• When a value is wanted between two known points, interpolate by choosing a value between them.
• For equal intervals such as 0 °C and 40 °C, the midpoint is often a simple average.

Common Mistakes

• Using 40 °C and 80 °C instead of the two surrounding points, which would give a different and less suitable estimate.
• Rounding the result unnecessarily; the mark scheme accepts exactly 390.

Things to Be Careful About

The question says "estimate", so a clear number with the unit is enough. Include the unit g / dm³ in the answer line.

Walkthrough

The method for ammonium chloride works by weighing a solid in a bottle before and after adding it to water. Ammonia at room temperature is a gas, so it cannot be measured in the same way. A enough amount of gas cannot simply be put in a weighing bottle and reweighed, because it escapes and cannot be weighed directly on a balance.

The mark scheme's single principle is that "you cannot weigh a gas on a balance".

Key Takeaways

• The solubility method used in the question is designed for solid solutes.
• A gas needs a different technique, such as measuring the volume of gas absorbed or collected.

Common Mistakes

• Answering only "ammonia is different" without mentioning the gas state.
• Saying that ammonia reacts with water; that is not the key reason in this mark scheme.

Things to Be Careful About

Give the reason clearly: ammonia is a gas at ordinary conditions, so it cannot be weighed on a balance.

Walkthrough

To find the number of moles of hydrochloric acid added, use the formula:

Convert the volume from $\text{cm}^3$ to $\text{dm}^3$ by dividing by $1000$:

Multiply by the concentration:

Key Takeaways

• Volume in $\text{cm}^3$ must always be converted to $\text{dm}^3$ before calculating moles using molarity.

Common Mistakes

• Forgetting to convert $\text{cm}^3$ into $\text{dm}^3$, yielding $25$ instead of $0.025$.

Things to Be Careful About

• Ensure correct arithmetic with decimal places.

Walkthrough

Using the average volume of $\text{NaOH}$ from (b)(ii) ($15.3\,\text{cm}^3$) and its concentration ($0.100\,\text{mol / dm}^3$):

Key Takeaways

• Amount of substance in moles is found by multiplying volume in $\text{dm}^3$ by molar concentration.

Common Mistakes

• Using $25.0\,\text{cm}^3$ (volume of B) instead of the titre volume of $\text{NaOH}$.

Things to Be Careful About

• Ensure correct power-of-ten conversion when dividing by $1000$.

Walkthrough

In part (c), the moles of $\text{NaOH}$ reacting with a $25.0\,\text{cm}^3$ portion of mixture B was calculated.
Mixture B has a total volume of $100\,\text{cm}^3$.
To find the moles of $\text{NaOH}$ that would react with the entire $100\,\text{cm}^3$ of B, multiply the moles from (c) by the scaling factor:

Key Takeaways

• When scaling from an aliquot to the original total solution, multiply by the ratio $\frac{\text{total volume}}{\text{aliquot volume}}$.

Common Mistakes

• Dividing by 4 instead of multiplying by 4.

Things to Be Careful About

• Express the answer to at least two significant figures as required by the mark scheme.

Walkthrough

This is a back titration calculation:

• Initial moles of $\text{HCl}$ added to sample A $= 0.025\,\text{mol}$ (from part (a)).
• Moles of unreacted $\text{HCl}$ remaining in $100\,\text{cm}^3$ of B $= 0.00612\,\text{mol}$ (equal to moles of $\text{NaOH}$ in (d) because $\text{HCl}$ and $\text{NaOH}$ react in a $1:1$ ratio).
• Subtract the remaining moles from the initial moles to find the moles of $\text{HCl}$ that reacted with the calcium carbonate:

Rounded to 3 significant figures, this gives $0.0189\,\text{mol}$.

Key Takeaways

• $\text{Moles reacted} = \text{initial moles added} - \text{excess moles remaining}$.

Common Mistakes

• Adding the two values instead of subtracting them.

Things to Be Careful About

• Give the final value to at least two significant figures ($0.0189$ or $0.019$).

Walkthrough

The balanced equation is:

According to the stoichiometry, $2\text{ mol}$ of $\text{HCl}$ reacts with $1\text{ mol}$ of $\text{CaCO}_3$.
Therefore, divide the moles of $\text{HCl}$ reacted (from part (e)) by 2:

Key Takeaways

• Use stoichiometric coefficients from the balanced equation to relate moles of different substances.

Common Mistakes

• Multiplying by 2 instead of dividing by 2.

Things to Be Careful About

• Maintain at least two significant figures in the final value.

Walkthrough

• Explanation: Measuring cylinders have large scale divisions and a wide diameter, which gives them lower precision/resolution compared to specialised volumetric glassware.
• Improvement: To measure an accurate volume of liquid for volumetric analysis, use a volumetric pipette, a burette, or a volumetric flask.

Key Takeaways

• Measuring cylinders are suitable for approximate volumes, but volumetric analysis requires higher-precision apparatus such as pipettes or burettes.

Common Mistakes

• Suggesting a beaker as an improvement (a beaker is even less accurate).
• Saying the measuring cylinder is 'unreliable' without stating that it lacks resolution or precision.

Things to Be Careful About

• Ensure both the explanation and the improvement are clearly stated.

Walkthrough

When a burette is rinsed with water, droplets of water cling to the inside walls. If sodium hydroxide solution is added immediately without rinsing with the solution first, these water droplets will dilute the sodium hydroxide solution, decreasing its concentration and causing an inaccurate titration result.

Key Takeaways

• Burettes and pipettes must be rinsed with the solution they are to contain to prevent dilution from residual water.
• Conical flasks are only rinsed with distilled water, as leftover water does not change the number of moles of reactant placed inside them.

Common Mistakes

• Stating that it is 'to clean the burette' rather than specifically mentioning preventing dilution or washing away residual water.

Things to Be Careful About

• Clearly mention either removing the water or preventing the dilution of the $\text{NaOH}$.

Walkthrough

A flame test is done by dipping a clean wire or splint into the solution and holding it in a hot, non-luminous Bunsen flame. The blue flame is used because a yellow luminous flame would mask the colour produced by the metal ion. The initial colour of the flame is observed; for potassium it is lilac.

Key Takeaways

• Flame test procedure: clean wire, dip in sample, place in blue flame, observe initial colour.
• Potassium gives a lilac flame.

Common Mistakes

• Using a yellow/luminous flame, which obscures the test colour.
• Not cleaning the wire, so contamination from previous tests appears.
• Observing the flame after the colour has faded instead of the initial flash.

Things to Be Careful About

• The mark scheme accepts dipping a rod/splint/wire or spraying the solution into the flame.
• The flame must be blue/roaring/non-luminous (air-hole open).
• The observation should be the initial colour.

Walkthrough

The lilac flame is characteristic of potassium ions. From part (a), the solution contains iodide ions. Combining these, W must be potassium iodide, formula KI.

Key Takeaways

• Flame colour identifies the cation.
• The anion is identified by a separate test.
• A salt is named by its cation and anion.

Common Mistakes

• Writing 'potassium iodine' instead of 'potassium iodide'.
• Forgetting the formula KI.

Things to Be Careful About

• Iodide is the ion $\text{I}^-$; iodine is the element $\text{I}_2$.
• The formula of potassium iodide is KI, not $\text{KI}_2$.

Walkthrough

When sodium hydroxide is added, a white precipitate soluble in excess is a known result for both $\text{Al}^{3+}$ and $\text{Zn}^{2+}$. Their hydroxides are amphoteric and dissolve in excess NaOH. So the cation may be $\text{Al}^{3+}$ or $\text{Zn}^{2+}$. The damp red litmus paper not changing colour means no ammonia gas was produced. If ammonium ions were present, warming with NaOH would release ammonia, which turns damp red litmus blue. Therefore $\text{NH}_4^+$ is absent.

Key Takeaways

• $\text{Al}^{3+}$ and $\text{Zn}^{2+}$ both give white precipitates with NaOH that dissolve in excess.
• $\text{NH}_4^+$ ions give ammonia gas when warmed with NaOH.
• Damp red litmus turning blue is the test for ammonia.

Common Mistakes

• Stating the cation is definitely $\text{Al}^{3+}$ or definitely $\text{Zn}^{2+}$; the test only shows it may be either.
• Forgetting to conclude that ammonium ions are absent.
• Saying the litmus turned blue when it did not.

Things to Be Careful About

• The question says 'one anion and one cation', so it cannot contain both $\text{Al}^{3+}$ and $\text{Zn}^{2+}$.
• 'Soluble in excess' is the key phrase that narrows the possibilities.
• The litmus result is a negative test: no ammonia, so no $\text{NH}_4^+$.

Walkthrough

To distinguish $\text{Al}^{3+}$ from $\text{Zn}^{2+}$, add aqueous ammonia until it is in excess. Both form white precipitates with ammonia, but aluminium hydroxide is insoluble in excess ammonia, while zinc hydroxide dissolves in excess ammonia. So the solubility in excess ammonia identifies the cation.

Key Takeaways

• Ammonia is used to distinguish $\text{Al}^{3+}$ from $\text{Zn}^{2+}$.
• $\text{Al(OH)}_3$ is insoluble in excess ammonia; $\text{Zn(OH)}_2$ is soluble.

Common Mistakes

• Using sodium hydroxide instead of ammonia; both give the same result for $\text{Al}^{3+}$ and $\text{Zn}^{2+}$, so it would not distinguish them.
• Not adding excess ammonia.
• Saying both precipitates dissolve or both are insoluble.

Things to Be Careful About

• The test must be 'aqueous ammonia until in excess'.
• The observation must state what happens in excess ammonia.
• Mark scheme: Al gives white ppt insoluble in excess; Zn gives white ppt soluble in excess.