Chemistry 9701/52 — October/November 2025

Background Concept

Iron(II) ions, Fe$^{2+}$(aq), are easily oxidised by atmospheric oxygen to form iron(III) ions, Fe$^{3+}$(aq). This is a common side reaction that can interfere with experiments relying on the concentration of Fe$^{2+}$.

Understanding the Question

Flask A contains the equilibrium mixture of Ag$^+$, Fe$^{2+}$, and Fe$^{3+}$. Step 3 involves sealing the flask. We need to suggest why this is done.

Approach

Consider what could react with the contents of the flask if it were open to the air. Fe$^{2+}$ is a reducing agent and can be oxidised by O$_2$ in the air.

Step-by-Step Reasoning

• The equilibrium mixture contains Fe$^{2+}$(aq) ions.
• If the flask is open, oxygen from the air can dissolve in the solution and oxidise Fe$^{2+}$ to Fe$^{3+}$.
• This would alter the equilibrium concentrations and invalidate the $K_c$ calculation.
• Sealing the flask with a bung prevents air (and thus oxygen) from entering.

Key Takeaways

Always consider the stability of reactants and products in equilibrium experiments. Fe$^{2+}$ is particularly susceptible to oxidation by air.

Common Mistakes

• Saying "to prevent evaporation" (not the primary reason here).
• Not specifying what is being oxidised (must mention Fe$^{2+}$ or iron(II)).

Things to Be Careful About

• The mark scheme specifically looks for "avoid oxidation". Be precise.

Background Concept

Many reactions, especially in aqueous solution, are slow. An equilibrium is only established when the forward and reverse reaction rates are equal, and concentrations are constant. Leaving the mixture for an extended period (e.g., 12 hours) ensures this state is achieved.

Understanding the Question

Step 4 involves leaving flask A for twelve hours. We need to suggest the purpose of this waiting period.

Approach

The question states in the preamble that the reaction is "slow". Leaving it for 12 hours allows the slow reaction to proceed until equilibrium is established.

Step-by-Step Reasoning

• The reaction between Ag$^+$ and Fe$^{2+}$ is slow.
• The student needs the system to be at equilibrium to measure the equilibrium concentrations and calculate $K_c$.
• Leaving it for 12 hours ensures that the equilibrium has been reached and concentrations are no longer changing.

Key Takeaways

In equilibrium experiments, sufficient time must be allowed for the system to reach equilibrium, especially if the reaction kinetics are slow.

Common Mistakes

• Saying "to let the precipitate settle" (not the primary reason for the 12-hour wait).
• Not using the word "equilibrium".

Things to Be Careful About

• Use the exact phrasing "equilibrium has been reached" or "established".

Background Concept

In the equilibrium reaction:

Silver ions are reduced to solid silver, which precipitates out of solution. Iron(II) is oxidised to iron(III).

Understanding the Question

Step 5 mentions transferring the mixture without disturbing the precipitate. We need to identify what this precipitate is.

Approach

Look at the products of the reaction in flask A. Ag(s) is a solid and is insoluble in water, so it forms the precipitate.

Step-by-Step Reasoning

• The reaction produces Ag(s) and Fe$^{3+}$(aq).
• Ag(s) is a solid metal that does not dissolve in the aqueous mixture.
• Therefore, the precipitate is silver (Ag).

Key Takeaways

Always check state symbols in the given equations to identify precipitates.

Common Mistakes

• Identifying the precipitate as AgSCN (this is formed later during titration, not in flask A before titration).
• Saying "silver nitrate" or "iron sulfate" (these are the soluble reactants).

Things to Be Careful About

• The question asks for the precipitate in flask A in step 5, before titration. The only solid present is Ag(s).

Background Concept

Concordant titres are results that are within 0.10 cm$^3$ of each other. This indicates that the titration has been performed accurately and reliably. Typically, at least two concordant titres are required to calculate a mean.

Understanding the Question

We are given a table of burette readings and titres. We need to state if concordant titres have been achieved and explain why.

Approach

Calculate the differences between the titres (excluding the rough titration). If at least two titres are within 0.10 cm$^3$ of each other, concordant titres have been achieved.

Step-by-Step Reasoning

• Titres are: rough = 22.50, 1 = 21.75, 2 = 21.90, 3 = 22.00.
• Ignore the rough titration.
• Difference between tit 2 and tit 3: $|22.00 - 21.90| = 0.10$ cm$^3$.
• Since 0.10 cm$^3 \leq 0.10$ cm$^3$, titres 2 and 3 are concordant.
• (Note: tit 1 is 21.75, which is 0.15 cm$^3$ from tit 2, so it is not concordant with them, but we only need two concordant values).

Key Takeaways

Concordant titres are within 0.10 cm$^3$. Always check the differences between the accurate titres (not the rough).

Common Mistakes

• Including the rough titration in the concordance check.
• Saying "yes, all titres are the same" (they are not).

Things to Be Careful About

• The mark scheme accepts "yes" and the explanation that two titres are within 0.10 cm$^3$. Be specific about which titres.

Background Concept

Burette readings are typically read to the nearest 0.05 cm$^3$. Therefore, the absolute error for a single reading is $\pm 0.05$ cm$^3$. A titre is the difference between two readings (final - initial), so the maximum absolute error in a titre is $2 \times 0.05 = 0.10$ cm$^3$.

Percentage error is calculated as:

Understanding the Question

Calculate the percentage error in the titre volume for titration 2, which is 21.90 cm$^3$.

Approach

1. Identify the absolute error for a titre (0.10 cm$^3$).
2. Apply the percentage error formula using the titre value from titration 2.

Step-by-Step Reasoning

• Absolute error = $2 \times 0.05 = 0.10$ cm$^3$.
• Titre value = 21.90 cm$^3$.
• Percentage error = $(0.10 / 21.90) \times 100 = 0.4566...\% \approx 0.457\%$.

Key Takeaways

Always remember that a titre involves two burette readings, so the absolute error is doubled. Show your working clearly.

Common Mistakes

• Using an absolute error of 0.05 cm$^3$ (forgetting to double it for two readings).
• Not showing working (the mark scheme requires it).
• Rounding too early or using wrong significant figures.

Things to Be Careful About

• The mark scheme specifically shows the working as $\frac{2 \times 0.05}{21.90} \times 100$. Follow this format.

Background Concept

The absolute error in a burette reading is fixed (e.g., $\pm 0.10$ cm$^3$ for a titre). The percentage error is inversely proportional to the measured value:

Therefore, a larger measured value (larger titre) results in a smaller percentage error, making the result more precise.

Understanding the Question

The student uses a higher concentration of KSCN, resulting in smaller titres. We need to suggest why a larger titre is better.

Approach

Relate the titre volume to the percentage error. A larger titre means the fixed absolute error is a smaller fraction of the total volume.

Step-by-Step Reasoning

• The absolute error from the burette is constant (e.g., 0.10 cm$^3$).
• If the titre is small (e.g., 5.00 cm$^3$), the percentage error is $(0.10 / 5.00) \times 100 = 2.0\%$.
• If the titre is large (e.g., 25.00 cm$^3$), the percentage error is $(0.10 / 25.00) \times 100 = 0.4\%$.
• Therefore, a larger titre gives a lower percentage error and is more accurate/precise.

Key Takeaways

Larger titre volumes reduce the relative (percentage) error in titrations. This is why we often use more concentrated solutions or larger sample volumes to get titres around 20-30 cm$^3$.

Common Mistakes

• Saying "it is more accurate" without explaining why (need to mention percentage error).
• Saying "it saves time" (not relevant to accuracy).

Things to Be Careful About

• The mark scheme specifically looks for "percentage error is lower". Use this terminology.

Background Concept

The titration determines the concentration of Ag$^+$ ions remaining in the equilibrium mixture. The reaction is:

This is a 1:1 stoichiometric reaction. The moles of SCN$^-$ used equal the moles of Ag$^+$ in the aliquot (sample) taken.

Understanding the Question

Using a mean titre of 21.85 cm$^3$ of 0.0200 mol dm$^{-3}$ KSCN, calculate the concentration of Ag$^+$ in the equilibrium mixture in flask A. The sample titrated was 10.0 cm$^3$ taken from flask A.

Approach

1. Calculate moles of SCN$^-$ used in the titration.
2. Use the 1:1 ratio to find moles of Ag$^+$ in the 10.0 cm$^3$ sample.
3. Calculate the concentration of Ag$^+$ using the sample volume.

Step-by-Step Reasoning

• Moles of SCN$^-$: $n = 0.0200 \times (21.85 / 1000) = 4.37 \times 10^{-4}$ mol.
• Moles of Ag$^+$ in sample: Since ratio is 1:1, $n(\text{Ag}^+) = 4.37 \times 10^{-4}$ mol.
• Concentration of Ag$^+$: $c = n / V = (4.37 \times 10^{-4}) / (10.0 / 1000) = 0.0437$ mol dm$^{-3}$.
• Note: The concentration in the 10.0 cm$^3$ sample is the same as the concentration in the entire mixture in flask A (the solution is homogeneous). We do not need to scale up to the full 50 cm$^3$ because concentration is an intensive property.

Key Takeaways

Concentration is the same throughout a homogeneous solution. Calculate the concentration directly from the sample volume and moles; no dilution factor is needed for concentration.

Common Mistakes

• Multiplying the moles by 5 (to scale up to 50 cm$^3$) and then dividing by 50 cm$^3$ (which is correct but unnecessary; you can just use the sample values directly).
• Forgetting to convert cm$^3$ to dm$^3$.

Things to Be Careful About

• The concentration calculated is for the equilibrium mixture, which is what is needed for the $K_c$ expression.

Background Concept

In the equilibrium reaction:

For every 1 mole of Ag$^+$ that reacts, 1 mole of Fe$^{3+}$ is produced. The concentration of Fe$^{3+}$ at equilibrium is equal to the amount of Ag$^+$ that has been consumed from its initial concentration.

Understanding the Question

Calculate [Fe$^{3+}$]$_{\text{eqm}}$ in flask A. We know the initial concentrations and the equilibrium concentration of Ag$^+$.

Approach

1. Calculate the initial concentrations of Ag$^+$ and Fe$^{2+}$ after mixing (accounting for the doubling of volume).
2. Use the difference between initial and equilibrium [Ag$^+$] to find how much reacted.
3. This difference equals [Fe$^{3+}$]$_{\text{eqm}}$ due to the 1:1 stoichiometry.

Step-by-Step Reasoning

• Initial [Ag$^+$]: 25.0 cm$^3$ of 0.100 mol dm$^{-3}$ diluted to 50.0 cm$^3$. $c = 0.100 \times (25.0 / 50.0) = 0.0500$ mol dm$^{-3}$.
• Initial [Fe$^{2+}$]: Similarly, $0.0500$ mol dm$^{-3}$.
• Equilibrium [Ag$^+$]: Calculated in part (e)(i) as $0.0437$ mol dm$^{-3}$.
• Amount of Ag$^+$ reacted: $0.0500 - 0.0437 = 0.0063$ mol dm$^{-3}$.
• Equilibrium [Fe$^{3+}$]: Since 1 mol Ag$^+$ produces 1 mol Fe$^{3+}$, $[\text{Fe}^{3+}]_{\text{eqm}} = 0.00630$ mol dm$^{-3}$.

Key Takeaways

Use an ICE (Initial, Change, Equilibrium) table approach mentally. The change in reactant concentration equals the change in product concentration (adjusted for stoichiometry).

Common Mistakes

• Forgetting to calculate the initial concentrations after mixing (using 0.100 instead of 0.0500).
• Not using the correct number of significant figures (0.00630, not 0.0063).

Things to Be Careful About

• The volume doubles when mixing equal volumes, so initial concentrations are halved.
• Use the value from part (e)(i) directly.

Background Concept

The equilibrium constant $K_c$ is calculated using the equilibrium concentrations of the aqueous species. Solids (like Ag(s)) are not included in the $K_c$ expression.

Units are derived by substituting the units of concentration (mol dm$^{-3}$) into the expression.

Understanding the Question

Calculate $K_c$ and its units using the equilibrium concentrations. Note that [Fe$^{2+}$]$_{\text{eqm}}$ must be calculated first.

Approach

1. Calculate [Fe$^{2+}$]$_{\text{eqm}}$ using the initial concentration and the amount reacted (which equals [Fe$^{3+}$]$_{\text{eqm}}$).
2. Substitute all equilibrium concentrations into the $K_c$ expression.
3. Derive the units algebraically.

Step-by-Step Reasoning

• [Fe$^{2+}$]$_{\text{eqm}}$: Initial was 0.0500 mol dm$^{-3}$. Amount reacted = [Fe$^{3+}$]$_{\text{eqm}}$ = 0.00630 mol dm$^{-3}$. So, $[\text{Fe}^{2+}]_{\text{eqm}} = 0.0500 - 0.00630 = 0.0437$ mol dm$^{-3}$.
• $K_c$ calculation:

• Units:

Key Takeaways

Always calculate all equilibrium concentrations before substituting into $K_c$. Units must be derived systematically.

Common Mistakes

• Forgetting to calculate [Fe$^{2+}$]$_{\text{eqm}}$ and using the initial concentration (0.0500) instead.
• Including Ag(s) in the $K_c$ expression.
• Getting the units wrong (e.g., writing mol dm$^{-3}$ instead of dm$^3$ mol$^{-1}$).

Things to Be Careful About

• The mark scheme shows the calculation as $0.00630 / (0.04370)^2 = 3.30$. Note that [Fe$^{2+}$]$_{\text{eqm}}$ happens to be numerically equal to [Ag$^+$]$_{\text{eqm}}$ in this case (0.0437), but you must show the correct reasoning.
• Give $K_c$ to 3 significant figures.

Background Concept

Direct proportionality between two variables $y$ and $x$ means $y = kx$, where $k$ is a constant. A graph of $y$ against $x$ would be a straight line passing through the origin (0,0) with a positive gradient.

Understanding the Question

A student suggests $K_c$ is directly proportional to temperature. We need to state if the graph supports this and explain.

Approach

Look at the graph in Fig 1.1. Does $K_c$ increase linearly with temperature? No, it decreases. Therefore, they are not directly proportional.

Step-by-Step Reasoning

• The graph shows $K_c$ on the y-axis and temperature on the x-axis.
• As temperature increases (moves right), $K_c$ decreases (moves down).
• Direct proportionality requires that as one increases, the other increases (positive correlation).
• Therefore, the results do not support the suggestion.

Key Takeaways

Direct proportionality means a linear relationship with a positive gradient through the origin. A decreasing relationship is not directly proportional.

Common Mistakes

• Saying "yes, it's a curve" (not addressing the direction of change).
• Not explaining why it's not directly proportional (must mention that $K_c$ decreases as T increases).

Things to Be Careful About

• The mark scheme looks for "No" and the explanation that $K_c$ decreases as temperature increases.

Background Concept

Reliable data in an experiment is characterised by consistency and the absence of anomalies. When plotting a graph, if all data points lie close to a line or curve of best fit without any points that deviate significantly (anomalies), the data is considered reliable.

Understanding the Question

Another student suggests the data in Fig 1.1 is reliable. Explain how the graph supports this.

Approach

Look at the scatter of the data points around the curve of best fit. Are there any outliers?

Step-by-Step Reasoning

• The graph shows multiple data points (crosses) plotted.
• A smooth curve has been drawn through them.
• All points lie very close to the curve; there are no points that are far away from the trend (no anomalies).
• This consistency suggests the measurements are reliable and precise.

Key Takeaways

Reliable data shows a clear trend with no outliers. Anomalous points reduce confidence in the results.

Common Mistakes

• Saying "the line is smooth" (the line is drawn by the student; the data points following the line is what matters).
• Not mentioning the absence of anomalies.

Things to Be Careful About

• Focus on the data points and their relationship to the trend, not just the appearance of the curve.

Background Concept

Le Chatelier's principle states that if a change is made to a system at equilibrium, the position of equilibrium shifts to counteract the change.

• For an exothermic forward reaction ($\Delta H < 0$): increasing temperature favours the reverse (endothermic) reaction, so $K_c$ decreases.
• For an endothermic forward reaction ($\Delta H > 0$): increasing temperature favours the forward reaction, so $K_c$ increases.

Understanding the Question

Use the graph to determine if the forward reaction is exothermic or endothermic and explain.

Approach

1. Observe the trend in the graph: as T increases, $K_c$ decreases.
2. Relate this to the position of equilibrium: a decreasing $K_c$ means the equilibrium shifts left (towards reactants).
3. Apply Le Chatelier's principle: increasing temperature shifts equilibrium in the endothermic direction. Since it shifts left, the reverse reaction is endothermic, so the forward reaction is exothermic.

Step-by-Step Reasoning

• Graph trend: As temperature increases from 280 K to 370 K, $K_c$ decreases from ~3.42 to ~2.56.
• Meaning of decreasing $K_c$: $K_c = [\text{products}] / [\text{reactants}]$. If $K_c$ decreases, the numerator decreases and the denominator increases, meaning equilibrium shifts to the left (towards Ag$^+$ and Fe$^{2+}$).
• Le Chatelier's principle: Increasing temperature adds heat. The system counteracts this by absorbing heat, which means favouring the endothermic direction. Since the equilibrium shifts left, the leftward (reverse) reaction is endothermic.
• Conclusion: If the reverse reaction is endothermic, the forward reaction must be exothermic.

Key Takeaways

• $K_c$ decreases with increasing T $\rightarrow$ exothermic forward reaction.
• $K_c$ increases with increasing T $\rightarrow$ endothermic forward reaction.

Common Mistakes

• Getting the direction of the shift wrong (confusing which way $K_c$ change indicates).
• Not mentioning Le Chatelier's principle or the endothermic/exothermic relationship.
• Saying "$K_c$ decreases so it's exothermic" without explaining why.

Things to Be Careful About

• The explanation must link the graph trend to the shift in equilibrium and then to the enthalpy change. All three steps are important for full marks.

Background Concept

In quantitative analysis, transferring a precise, fixed volume of liquid requires apparatus calibrated to deliver or contain that exact volume at a specific temperature. A volumetric pipette is designed to deliver a specific volume (e.g., 25.00 cm³) with high accuracy (typically ±0.05 cm³ or better). Burettes and measuring cylinders lack the necessary precision for this level of analytical work.

Understanding the Question

The question asks for a suitable piece of apparatus to transfer exactly 25.00 cm³ of lake water in step 1. The value '25.00' implies a need for four significant figures and high precision.

Approach

Identify the standard piece of equipment used in titrations and conductivity experiments to accurately measure and transfer a fixed volume of liquid.

Step-by-Step Reasoning

The volume required is 25.00 cm³, which is a standard volumetric pipette size. A measuring cylinder or beaker would not provide the required precision (usually ±0.5 cm³ or worse). A burette is for variable volumes, not a fixed 25.00 cm³ transfer. Therefore, a 25.00 cm³ volumetric pipette is the correct choice.

Key Takeaways

Always match the apparatus to the required precision: volumetric pipettes for fixed volumes, burettes for variable precise volumes, measuring cylinders for rough volumes.

Common Mistakes

Writing 'pipette' without specifying 'volumetric' (a graduated pipette is less accurate). Writing 'syringe' or 'beaker'.

Things to Be Careful About

Ensure the answer matches the mark scheme exactly: 'volumetric pipette' is the standard term.

Background Concept

Barium hydroxide, Ba(OH)₂(aq), is a strong base and is classified as an irritant to skin and eyes. Standard laboratory risk assessments require personal protective equipment (PPE) to prevent contact with irritants. While safety goggles protect the eyes, the skin is also vulnerable.

Understanding the Question

The question states Ba(OH)₂ is an irritant to skin and eyes and asks for one safety precaution other than wearing safety goggles.

Approach

Identify PPE that protects the skin from chemical irritants.

Step-by-Step Reasoning

Since the hazard is irritation to skin and eyes, and goggles cover the eyes, the remaining primary exposure route is skin contact. Wearing chemically resistant gloves (e.g., nitrile or latex gloves) will prevent the irritant from touching the skin. Wearing a lab coat is also acceptable, but gloves are the most direct answer for skin irritation.

Key Takeaways

Always consider all routes of exposure (eyes, skin, inhalation) when selecting PPE. If eyes are covered, protect skin.

Common Mistakes

Saying 'wear a lab coat' (often accepted, but gloves are more specific to skin irritation). Saying 'avoid contact' (not a specific precaution). Saying 'wear a mask' (not relevant for skin/eye irritants in solution form).

Things to Be Careful About

The mark scheme specifically accepts 'chemically resistant gloves'. 'Gloves' alone is usually acceptable, but 'chemically resistant' is more precise.

Background Concept

In conductivity experiments where a reagent is added to a fixed volume of sample, the total volume increases. This dilution effect reduces the concentration of all ions, lowering the measured conductivity even if no reaction is occurring. To compare conductivity values fairly across different volumes, a correction factor is applied to normalize the values back to the original concentration, as if no dilution had occurred.

Understanding the Question

The student has a table with missing 'total volume in beaker' and 'corrected conductivity' values. The formulas for both are given in the question stem.

Approach

Use the given formulas to calculate the missing values for each reading. Total volume is simply the sum of the initial sample volume and the added reagent volume. Corrected conductivity multiplies the measured value by the ratio of total volume to initial volume.

Step-by-Step Reasoning

Total volume column:
Start with 25.00 cm³. Add the volume of Ba(OH)₂ added for each reading.

• Reading 2: $25.00 + 5.00 = 30.00$
• Reading 3: $25.00 + 10.00 = 35.00$
• Reading 4: $25.00 + 15.00 = 40.00$
• Reading 5: $25.00 + 20.00 = 45.00$
• Reading 6: $25.00 + 25.00 = 50.00$
• Reading 7: $25.00 + 30.00 = 55.00$
• Reading 8: $25.00 + 35.00 = 60.00$
• Reading 9: $25.00 + 40.00 = 65.00$

Corrected conductivity column:
Apply $\text{corrected} = \text{measured} \times (\text{total volume} / 25.00)$.

• Reading 5: $5000 \times (45.00 / 25.00) = 5000 \times 1.8 = 9000$
• Reading 6: $12000 \times (50.00 / 25.00) = 12000 \times 2 = 24000$
• Reading 7: $18000 \times (55.00 / 25.00) = 18000 \times 2.2 = 39600$
• Reading 8: $21000 \times (60.00 / 25.00) = 21000 \times 2.4 = 50400$
• Reading 9: $24500 \times (65.00 / 25.00) = 24500 \times 2.6 = 63700$

Key Takeaways

Always check if a correction factor is needed when volumes change during an experiment. The correction normalizes the data to a common baseline.

Common Mistakes

Forgetting to add the initial 25.00 cm³ when calculating total volume (e.g., writing 5.00 instead of 30.00 for reading 2). Using the wrong ratio in the correction formula (e.g., $25.00 / \text{total volume}$ instead of $\text{total volume} / 25.00$).

Things to Be Careful About

Ensure decimal places are consistent (two decimal places for volume, e.g., 30.00, not 30). Check significant figures in calculations; the mark scheme accepts the exact calculated values.

Background Concept

In an experiment, the independent variable is the quantity that the experimenter deliberately changes or controls to observe its effect on the dependent variable. The dependent variable is the one being measured.

Understanding the Question

The student adds increasing volumes of Ba(OH)₂(aq) to the beaker and measures the conductivity. We need to identify what is being changed.

Approach

Look at the table headers and the method. The volume of Ba(OH)₂ added is set by the student (0, 5, 10, 15... cm³). The conductivity is measured as a result.

Step-by-Step Reasoning

The independent variable is the volume of 0.100 mol dm⁻³ Ba(OH)₂(aq) added to the beaker. This is the variable plotted on the x-axis of the graph in part (c). The dependent variable is the corrected conductivity.

Key Takeaways

The independent variable is what you change; the dependent variable is what you measure.

Common Mistakes

Saying 'conductivity' (this is the dependent variable). Saying 'concentration of sulfate' (this is what we are trying to find, not what we change directly in the table).

Things to Be Careful About

Be precise with the name. 'Volume of Ba(OH)₂' is acceptable, but 'volume of 0.100 mol dm⁻³ Ba(OH)₂(aq) added to beaker' is the most complete and accurate answer.

Background Concept

The intersection of the two lines of best fit in a conductivity titration graph represents the equivalence point. At this point, the amount of titrant added is exactly stoichiometric with the analyte. Before this point, conductivity is dominated by the removal of ions (negative gradient). After this point, conductivity is dominated by the addition of excess ions (positive gradient). The minimum conductivity occurs near this point, but the intersection of the extrapolated linear regions gives the most accurate equivalence volume.

Understanding the Question

The question asks to use the graph from part (c) to determine the volume of Ba(OH)₂ required to react exactly with the SO₄²⁻ in the lake water sample. This is the x-coordinate of the intersection point.

Approach

Read the x-value (volume of Ba(OH)₂ added) where the two lines of best fit intersect on the graph.

Step-by-Step Reasoning

Looking at the completed graph, the line through readings 1-4 (decreasing conductivity) and the line through readings 5-9 (increasing conductivity) intersect at an x-value. Based on the data, the minimum corrected conductivity is at 15.00 cm³ (3680) and the next point is 20.00 cm³ (9000). The intersection will be between 15 and 20 cm³. A typical reading from such a graph would be approximately 17.0 cm³. Any value in the range 15.0-19.0 cm³ is acceptable as it is read from the grid.

Key Takeaways

The intersection of lines in a conductivity graph gives the equivalence point volume, which is more accurate than reading the minimum data point directly.

Common Mistakes

Reading the y-value instead of the x-value. Reading the volume of total solution instead of the volume of Ba(OH)₂ added.

Things to Be Careful About

Ensure the value is read from the x-axis (volume of Ba(OH)₂ added), not the total volume in the beaker. The mark scheme allows a range because it's a graphical reading.

Background Concept

To find the concentration of an analyte in a titration or precipitation reaction, we use the stoichiometry of the balanced equation. The moles of titrant added at the equivalence point are calculated using $n = c \times V$. The moles of analyte are then found using the molar ratio from the equation. Finally, concentration is calculated using $c = n / V$ of the original sample.

Understanding the Question

Using the volume of Ba(OH)₂ from part (d)(i) (e.g., 17.0 cm³), calculate the concentration of SO₄²⁻ in the original 25.00 cm³ lake water sample. The reaction is $\text{Ba}^{2+}(\text{aq}) + \text{SO}_4^{2-}(\text{aq}) \rightarrow \text{BaSO}_4(\text{s})$.

Approach

1. Calculate moles of Ba(OH)₂ (or Ba²⁺) added at the equivalence point.
2. Use the 1:1 molar ratio to find moles of SO₄²⁻.
3. Divide moles of SO₄²⁻ by the volume of the lake water sample (in dm³) to get concentration.

Step-by-Step Reasoning

Step 1: Moles of Ba²⁺
Volume of Ba(OH)₂ = 17.0 cm³ = 0.0170 dm³ (using 17.0 as an example from the graph).
$n(\text{Ba}^{2+}) = 0.100 \text{ mol dm}^{-3} \times 0.0170 \text{ dm}^3 = 1.70 \times 10^{-3} \text{ mol}$

Step 2: Moles of SO₄²⁻
The equation shows 1 mol Ba²⁺ reacts with 1 mol SO₄²⁻.
$n(\text{SO}_4^{2-}) = 1.70 \times 10^{-3} \text{ mol}$

Step 3: Concentration of SO₄²⁻
Volume of lake water = 25.00 cm³ = 0.02500 dm³.
$c(\text{SO}_4^{2-}) = \frac{1.70 \times 10^{-3} \text{ mol}}{0.02500 \text{ dm}^3} = 0.0680 \text{ mol dm}^{-3}$

If the candidate read 16.0 cm³ from the graph: $n = 0.100 \times 0.016 = 0.0016$ mol. $c = 0.0016 / 0.025 = 0.064$ mol dm⁻³. The mark scheme allows error carried forward (ecf) from (d)(i).

Key Takeaways

Always convert volumes to dm³ when using concentration in mol dm⁻³. Remember the 1:1 stoichiometry for Ba²⁺ and SO₄²⁻.

Common Mistakes

• Forgetting to divide volume by 1000 to convert cm³ to dm³.
• Using the total volume in the beaker (e.g., 25 + 17 = 42 cm³) instead of the original sample volume (25.00 cm³) for the concentration calculation.
• Incorrect stoichiometry (e.g., assuming 1:2 ratio because of Ba(OH)₂, but the net ionic equation shows 1:1 for Ba²⁺:SO₄²⁻).

Things to Be Careful About

• Significant figures: The concentration should be given to 3 significant figures (matching the 0.100 mol dm⁻³ and 25.00 cm³ data).
• The mark scheme shows the formula: $[\text{SO}_4^{2-}] = [0.100 \times \text{ans(d)(i)} / 1000] / 0.025$. Ensure this structure is followed.

Background Concept

Filtration is used to separate an insoluble solid (precipitate) from a liquid. The apparatus consists of a funnel containing filter paper, supported above a receiving vessel (conical flask or beaker). The precipitate (residue) is retained on the filter paper, and the liquid (filtrate) passes through.

Understanding the Question

The question asks for a labelled diagram of the apparatus needed to complete step 3 (filter the contents of the flask). This is the precipitation and filtration method for determining sulfate concentration.

Approach

Draw a simple line diagram showing a funnel with filter paper inside it, positioned above a beaker or conical flask to collect the filtrate. Label the key components.

Step-by-Step Reasoning

Apparatus needed:

1. Funnel: A glass or plastic funnel with a long stem.
2. Filter paper: Folded into a cone shape and placed inside the funnel.
3. Receiving vessel: A conical flask or beaker placed under the funnel to catch the liquid (filtrate).
4. Support: The funnel may be supported by a funnel stand (optional but good practice), though the question just asks for the arrangement.

Diagram description:

• Draw a V-shaped funnel with a long vertical stem.
• Inside the funnel, draw a cone shape representing the filter paper.
• Below the funnel, draw a beaker or conical flask to catch the liquid.
• Label 'funnel' and 'filter paper'.

Key Takeaways

Filtration setups are common in quantitative analysis. Always label the funnel and filter paper clearly.

Common Mistakes

• Forgetting to label the components.
• Drawing the filter paper outside the funnel.
• Not including a receiving vessel (beaker/flask) to catch the filtrate.
• Drawing a Büchner funnel or vacuum filtration setup (too complex for this simple gravity filtration).

Things to Be Careful About

The mark scheme specifically looks for 'labelled filter paper in a labelled (filter) funnel'. Ensure both are clearly labelled. A simple line diagram is sufficient; it doesn't need to be artistic.

Background Concept

In gravimetric analysis, after filtration, the precipitate (residue) on the filter paper may still contain traces of the original solution or the excess reagent (Ba(OH)₂). These soluble impurities would add to the mass of the residue if not removed, leading to an overestimation of the analyte concentration. Washing with distilled water removes these soluble impurities.

Understanding the Question

The student's method involves filtering, then drying, then weighing. We need to suggest a step between filtering (step 3) and drying (step 4) to improve the method.

Approach

Identify a standard step in gravimetric analysis that occurs after filtration and before drying to ensure purity of the precipitate.

Step-by-Step Reasoning

After filtering, the precipitate (BaSO₄) is wet and may contain dissolved salts (e.g., excess Ba(OH)₂, or other ions from the lake water). To ensure only BaSO₄ is weighed, the residue must be washed with distilled water (or deionized water) to remove soluble impurities. This is a crucial step for accuracy.

Key Takeaways

Always wash the precipitate in gravimetric analysis to remove soluble impurities.

Common Mistakes

Saying 'rinse with tap water' (tap water contains ions that would contaminate the precipitate). Saying 'dry the residue' (this is already step 4). Saying 'add more Ba(OH)₂' (this is already done in step 2).

Things to Be Careful About

Specify 'distilled water' or 'deionized water', not just 'water'. 'Wash the residue' or 'rinse the precipitate' are acceptable verbs.

Background Concept

In gravimetric analysis, it is essential to ensure that the precipitate is completely dry before weighing, as any remaining water will add to the mass and cause a positive error. The standard technique to ensure complete drying is to heat the residue, cool it, weigh it, and repeat until the mass no longer changes. This is called 'heating to constant mass'.

Understanding the Question

The question asks what the student can do to ensure the residue weighed in step 5 is completely dry.

Approach

Identify the standard laboratory technique for confirming complete drying of a solid.

Step-by-Step Reasoning

Instead of just drying in a warm oven once, the student should heat the residue (e.g., in an oven or using a Bunsen burner if thermally stable, though BaSO₄ is stable), allow it to cool in a desiccator (to prevent reabsorption of moisture), and weigh it. This process is repeated until two consecutive readings are the same (or within a small tolerance, e.g., 0.001 g). This confirms all water has been removed.

Key Takeaways

'Heat to constant mass' is the key phrase for ensuring complete drying in gravimetric analysis.

Common Mistakes

Saying 'leave it to dry overnight' (not rigorous enough). Saying 'use a hot oven' (doesn't confirm it's dry). Saying 'weigh it again' (doesn't specify the condition for knowing it's dry).

Things to Be Careful About

The exact phrase 'heat to constant mass' is the standard mark scheme answer. Mentioning 'cool in a desiccator' is good practice but 'heat to constant mass' is the core requirement.

Background Concept

In gravimetric analysis, the concentration of the analyte is calculated from the mass of the precipitate. If the precipitate is not pure (e.g., contains water or other impurities), the measured mass will be incorrect. An overestimation of mass leads to an overestimation of moles, and thus an overestimation of concentration.

Understanding the Question