Qualitative analysis For each test you should record all your observations in the spaces provided. Examples of observations include: colour changes seen the formation of any precipitate and its solubility (where appropriate) in an excess of the reagent added the formation of any gas and its identification (where appropriate) by a suitable test. You should record clearly at what stage in a test an observation is made. Where no change is observed, you should write ‘no change’.…

(a)(i) NaCl · (a)(ii) The acid removes/reacts with hydrogencarbonate ions, which would otherwise interfere with the silver nitrate test · (b)(i) Green solid turns to black powder; condensation on tube walls · (b)(ii) Effervescence, green-blue solution, CO2 turns limewater milky · (b)(iii) Cu2+; blue ppt insoluble in excess NaOH, blue ppt soluble in excess NH3 giving deep blue solution · (c)(i) Fe2+; green ppt insoluble in excess turning brown, KMnO4 decolourised · (c)(ii)…

Chemistry 9701/34 — October/November 2025

Cambridge AS Level · Advanced Practical Skills · worked solutions for every part, with the mark scheme

questions

marks

120

minutes

Topics Manipulation, Measurement and Observation · Presentation of Data and Observations · Analysis, Conclusions and Evaluation · Qualitative Analysis

Q1Medium-HardManipulation, Measurement and ObservationPresentation of Data and ObservationsAnalysis, Conclusions and Evaluation

Quantitative analysis

Read through the whole method before starting any practical work. Where appropriate, prepare a table for your results in the space provided.

Show the precision of the apparatus you used in the data you record.

Show your working and appropriate significant figures in the final answer to each step of your calculations.

Sodium carbonate can be manufactured using a two-step process. The first step involves making sodium hydrogencarbonate, $\text{NaHCO}_3$ , which is then converted into sodium carbonate, $\text{Na}_2\text{CO}_3$ , in the second step of the process by the following reaction.

2\text{NaHCO}_3(\text{s}) \rightarrow \text{Na}_2\text{CO}_3(\text{s}) + \text{H}_2\text{O}(\text{l}) + \text{CO}_2(\text{g})

In this experiment you will determine the enthalpy change, $\Delta H_r$ , for this reaction. You will do this by calculating the enthalpy changes when separate samples of sodium hydrogencarbonate and sodium carbonate are added to excess hydrochloric acid, $\text{HCl}(\text{aq})$ . You will then combine these values using Hess’s law.

\begin{aligned} \text{NaHCO}_3(\text{s}) + \text{HCl}(\text{aq}) &\rightarrow \text{NaCl}(\text{aq}) + \text{H}_2\text{O}(\text{l}) + \text{CO}_2(\text{g}) \quad &\Delta H_1 \\ \text{Na}_2\text{CO}_3(\text{s}) + 2\text{HCl}(\text{aq}) &\rightarrow 2\text{NaCl}(\text{aq}) + \text{H}_2\text{O}(\text{l}) + \text{CO}_2(\text{g}) \quad &\Delta H_2 \end{aligned}

FB 1 is $2.0\text{ mol dm}^{-3}$ hydrochloric acid, $\text{HCl}$ .
FB 2 is sodium hydrogencarbonate, $\text{NaHCO}_3$ .
FB 3 is sodium carbonate, $\text{Na}_2\text{CO}_3$ .

(a)

Method

Experiment 1

Support a cup in the $250\text{ cm}^3$ beaker.
Use the $25\text{ cm}^3$ measuring cylinder to transfer $20.0\text{ cm}^3$ of FB 1 into the cup.
Weigh the container with FB 2. Record the mass.
Measure the temperature of the acid in the cup. Record this temperature.
Carefully add all of the FB 2, in small portions to avoid acid spray. Stir to dissolve. Record the lowest temperature.
Reweigh the container with any residual FB 2. Record the mass.
Calculate and record the mass of FB 2 used.
Calculate and record the decrease in temperature.

Experiment 2

Support the second cup in the $250\text{ cm}^3$ beaker.
Use the $25\text{ cm}^3$ measuring cylinder to transfer $20.0\text{ cm}^3$ of FB 1 into the second cup.
Weigh the container with FB 3. Record the mass.
Measure the temperature of the acid in the cup. Record this temperature.
Carefully add all of the FB 3, in small portions to avoid acid spray. Stir to dissolve. Record the highest temperature.
Reweigh the container with any residual FB 3. Record the mass.
Calculate and record the mass of FB 3 used.
Calculate and record the increase in temperature.

Keep FB 1 for use in Question 3.

Results

4M Similar

DifficultyMedium-Easy

Worked solution

Answer

A results table must be constructed with clear, unambiguous headings and units for all six recorded readings and calculated values for both experiments. The format is shown below using representative values.

Experiment 1 (NaHCO₃ + HCl)

Heading	Unit	Example Value
Mass of container + FB 2	g	12.50
Mass of container + residue	g	9.50
Mass of FB 2 used	g	3.00
Initial temperature of FB 1	°C	20.0
Final temperature	°C	14.5
Temperature change	°C	5.5

Experiment 2 (Na₂CO₃ + HCl)

Heading	Unit	Example Value
Mass of container + FB 3	g	11.00
Mass of container + residue	g	8.50
Mass of FB 3 used	g	2.50
Initial temperature of FB 1	°C	20.0
Final temperature	°C	25.5
Temperature change	°C	5.5

Key requirements for full marks:

All four weighings recorded to the same number of decimal places (either 2 d.p. or 3 d.p.).
All four temperatures recorded to the nearest 0.0°C or 0.5°C.
Masses used calculated by subtraction: $\text{mass used} = (\text{container + solid}) - (\text{container + residue})$ .
Temperature changes calculated by subtraction: $\Delta T = T_{\text{final}} - T_{\text{initial}}$ (recorded as a positive magnitude for Experiment 1's decrease and Experiment 2's increase).

Final answer

See working / candidate-dependent

Detailed explanation

Background Concept

In a calorimetry experiment, accurate recording of mass and temperature is critical because these values feed directly into the $q = mc\Delta T$ calculation. The precision of the balance and thermometer dictates the precision of the recorded data. A balance typically reads to 2 or 3 decimal places (0.01 g or 0.001 g), and a thermometer typically reads to 0.5°C. Consistency in recording is a key assessment objective (AO3) in practical exams.

Understanding the Question

This part asks you to present your raw data and simple calculated differences (mass used, temperature change) in a clear table. It tests your ability to use apparatus correctly (reading to the right precision) and to present data in a standard scientific format. The command word is implicit: "prepare a table for your results".

Approach

Draft the table structure first. Before adding any numbers, write out the headings. You need rows/columns for: Initial Mass, Final Mass, Mass Used, Initial Temp, Final Temp, Temp Change. Don't forget units.
Record raw data. As you perform the experiment, write down the readings exactly as shown on the apparatus. Do not round them in your head.
Process the data. Calculate the differences (Mass Used and Temp Change) and add these to the table.

Step-by-Step Reasoning

Mass readings: You weigh the container with the solid, then the container with the residue. The difference is the mass of solid reacted. If your balance reads to 2 d.p. (e.g., 12.50 g), you must record all mass readings to 2 d.p. (e.g., 9.50 g, not 9.5 g). This shows you understand the precision of the instrument.
Temperature readings: You measure the initial temperature of the acid and the final temperature of the mixture. The thermometer likely has markings every 0.5°C or 1°C. Record to the nearest 0.5°C or 0.0°C (e.g., 20.0°C, 14.5°C). Do not record 20°C or 14.53°C.
Calculations:
- Mass of FB 2 used = (Mass of container + FB 2) - (Mass of container + residue).
- Temperature change = |Final Temperature - Initial Temperature|. Note that for Experiment 1, the temperature falls (endothermic), and for Experiment 2, it rises (exothermic). The table usually just records the magnitude of the change ( $\Delta T$ ), which is then used in the energy calculation.

Key Takeaways

Precision implies accuracy: Recording 3.00 g suggests a different level of precision than 3 g. Always match your recording to the apparatus used.
Headings are vital: A table without units or clear labels is scientifically invalid. Every column must have a heading and a unit.
Consistency: If one mass is recorded to 2 d.p., all masses must be.

Common Mistakes

Inconsistent decimal places: Recording one mass as 12.50 and another as 9.5.
Incorrect temperature precision: Recording 20.3°C if the thermometer only reads to 0.5°C.
Missing units: Writing "12.50" instead of "12.50 g" in the table cell or heading.
Sign errors in calculations: Calculating mass used as (residue - initial) which gives a negative number. Always subtract the smaller from the larger for mass used.

Techniques used

record mass readings to consistent decimal placesrecord temperature readings to 0.5°C precisioncalculate differences in mass and temperatureconstruct a results table with clear headings and units

(b)

Calculations

(i)

Calculate the amount, in mol, of FB 2 that reacts with FB 1 and the amount, in mol, of FB 3 that reacts with FB 1.

2M Similar

DifficultyMedium-Easy

Worked solution

Working

For FB 2 ( $\text{NaHCO}_3$ ):

M_r(\text{NaHCO}_3) = 23.0 + 1.0 + 12.0 + (3 \times 16.0) = 84.0

n(\text{NaHCO}_3) = \frac{\text{mass used from (a)}}{84.0}

(Example: if mass used = 3.00 g, then $n = 3.00 / 84.0 = 0.0357 \text{ mol}$ )

For FB 3 ( $\text{Na}_2\text{CO}_3$ ):

M_r(\text{Na}_2\text{CO}_3) = (2 \times 23.0) + 12.0 + (3 \times 16.0) = 106.0

n(\text{Na}_2\text{CO}_3) = \frac{\text{mass used from (a)}}{106.0}

(Example: if mass used = 2.50 g, then $n = 2.50 / 106.0 = 0.0236 \text{ mol}$ )

Answer

$n(\text{NaHCO}_3) = \text{mass used} / 84.0$ mol
$n(\text{Na}_2\text{CO}_3) = \text{mass used} / 106.0$ mol

Final answer

See working / candidate-dependent

Detailed explanation

Background Concept

The mole is the unit for amount of substance. The relationship $n = m/M_r$ allows us to convert between the mass of a substance weighed out in the lab and the number of moles reacting. $M_r$ (relative formula mass) is the sum of the relative atomic masses of all atoms in the formula.

Understanding the Question

You need to calculate the number of moles of solid sodium hydrogencarbonate and sodium carbonate that actually reacted. This depends on the mass values you recorded and calculated in part (a).

Approach

Calculate $M_r$ values: Use the periodic table to find Ar values for Na, H, C, and O. Sum them up for each compound.
Apply the formula: Divide the mass of solid used (from part a) by the calculated $M_r$ .
Check significant figures: Your answer should generally match the precision of your data (usually 2-4 s.f. is accepted in these exams).

Step-by-Step Reasoning

$M_r$ of $\text{NaHCO}_3$ :
- Na = 23.0
- H = 1.0
- C = 12.0
- O = 16.0 (there are 3 oxygens, so $3 \times 16.0 = 48.0$ )
- Total = $23.0 + 1.0 + 12.0 + 48.0 = 84.0$
$M_r$ of $\text{Na}_2\text{CO}_3$ :
- Na = 23.0 (there are 2 sodiums, so $2 \times 23.0 = 46.0$ )
- C = 12.0
- O = 16.0 (there are 3 oxygens, so $3 \times 16.0 = 48.0$ )
- Total = $46.0 + 12.0 + 48.0 = 106.0$
Moles Calculation:
- Take the "Mass of FB 2 used" from your table in (a). Divide by 84.0.
- Take the "Mass of FB 3 used" from your table in (a). Divide by 106.0.
- Example: If you used 3.00 g of $\text{NaHCO}_3$ , then $3.00 / 84.0 = 0.035714...$ which rounds to $0.0357$ mol (3 s.f.).

Key Takeaways

Always calculate $M_r$ correctly before substituting values.
Ensure you use the actual mass reacted (the difference), not the initial mass of the container + solid.

Common Mistakes

Using the wrong $M_r$ (e.g. forgetting there are 3 oxygens).
Dividing the initial mass reading instead of the mass used.
Rounding errors: keeping too few significant figures in the intermediate step can affect later calculations.

Things to Be Careful About

Check that your mass used is in grams (g). If your balance reads in mg (unlikely for these masses), convert to g.

Techniques used

calculate relative formula massuse n = m/Mr to find moles

(ii)

Calculate the energy changes, in J, for each reaction.

1M Similar

DifficultyMedium-Easy

Worked solution

Working

The heat energy change ( $q$ ) is calculated using the formula $q = mc\Delta T$ , where $m$ is the mass of the solution (assumed to be the mass of the acid, $20.0 \text{ cm}^3 \approx 20.0 \text{ g}$ ) and $c$ is the specific heat capacity of water ( $4.18 \text{ J g}^{-1} \text{K}^{-1}$ ).

For Experiment 1 ( $\text{NaHCO}_3$ ):

q_1 = 20.0 \times 4.18 \times \Delta T_1

(Example: if $\Delta T_1 = 5.5$ , then $q_1 = 20.0 \times 4.18 \times 5.5 = 459.8 \text{ J}$ )

For Experiment 2 ( $\text{Na}_2\text{CO}_3$ ):

q_2 = 20.0 \times 4.18 \times \Delta T_2

(Example: if $\Delta T_2 = 5.5$ , then $q_2 = 20.0 \times 4.18 \times 5.5 = 459.8 \text{ J}$ )

Answer

Energy change for Exp 1 = $20.0 \times 4.18 \times \Delta T_1$ J
Energy change for Exp 2 = $20.0 \times 4.18 \times \Delta T_2$ J

Final answer

See working / candidate-dependent

Detailed explanation

Background Concept

In calorimetry, we assume the heat energy released or absorbed by the chemical reaction is transferred entirely to the solution. The energy change is calculated using $q = mc\Delta T$ . We assume the specific heat capacity ( $c$ ) of the solution is the same as water ( $4.18 \text{ J g}^{-1} \text{K}^{-1}$ ) and the density is $1.0 \text{ g cm}^{-3}$ , so $20.0 \text{ cm}^3$ of acid has a mass of $20.0 \text{ g}$ . The solid is usually ignored in the mass term as it dissolves into the solvent.

Understanding the Question

Calculate the energy change in Joules for both reactions using the temperature changes recorded in part (a).

Approach

Identify the mass of the solution ( $20.0 \text{ g}$ ).
Identify the specific heat capacity ( $4.18 \text{ J g}^{-1} \text{K}^{-1}$ ).
Identify the temperature change ( $\Delta T$ ) from part (a).
Multiply them together.

Step-by-Step Reasoning

Mass ( $m$ ): The question specifies $20.0 \text{ cm}^3$ of acid. Assuming density $\approx 1 \text{ g cm}^{-3}$ , $m = 20.0 \text{ g}$ .
Specific Heat Capacity ( $c$ ): Given as $4.18 \text{ J g}^{-1} \text{K}^{-1}$ (standard value for water/solutions).
Temperature Change ( $\Delta T$ ): This is the value calculated in the table in part (a).
Calculation:
- $q = 20.0 \times 4.18 \times \Delta T$
- Ensure the answer is in Joules (J), not kJ.

Key Takeaways

The formula assumes the specific heat capacity of the solution is that of pure water.
The mass used is the mass of the liquid (solvent), not the solid solute.

Common Mistakes

Using the mass of the solid instead of the acid.
Forgetting to convert $\text{cm}^3$ to $\text{g}$ (though numerically the same here).
Using the wrong value for $c$ (e.g. using 4.2 instead of 4.18, though usually acceptable, 4.18 is preferred in A-Level).

Things to Be Careful About

The question asks for the answer in J. Do not divide by 1000 yet; that happens in the next step.

Techniques used

calculate heat energy change using q = mcΔT

(iii)

Calculate the enthalpy change, $\Delta H$ , in $\text{kJ mol}^{-1}$ for each reaction.

2M Similar

DifficultyMedium

Worked solution

Working

Enthalpy change ( $\Delta H$ ) is the energy change per mole of reactant, expressed in $\text{kJ mol}^{-1}$ .

For Experiment 1 ( $\text{NaHCO}_3$ ):

\Delta H_1 = + \frac{\text{Energy change from (b)(ii)}}{\text{moles from (b)(i)} \times 1000}

Note: The sign is positive (+) because the temperature decreased (endothermic).

For Experiment 2 ( $\text{Na}_2\text{CO}_3$ ):

\Delta H_2 = - \frac{\text{Energy change from (b)(ii)}}{\text{moles from (b)(i)} \times 1000}

Note: The sign is negative (-) because the temperature increased (exothermic).

(Example: if $q = 460 \text{ J}$ and $n = 0.0357 \text{ mol}$ , then $\Delta H = 460 / (0.0357 \times 1000) = 12.9 \text{ kJ mol}^{-1}$ . So $\Delta H_1 = +12.9 \text{ kJ mol}^{-1}$ )

Answer

$\Delta H_1 = + \frac{q_1}{n_1 \times 1000} \text{ kJ mol}^{-1}$
$\Delta H_2 = - \frac{q_2}{n_2 \times 1000} \text{ kJ mol}^{-1}$

Final answer

See working / candidate-dependent

Detailed explanation

Background Concept

Enthalpy change ( $\Delta H$ ) represents the heat energy change per mole of a substance reacting at constant pressure. It is calculated by dividing the total energy change ( $q$ ) by the number of moles ( $n$ ). The sign of $\Delta H$ indicates whether the reaction absorbs heat (endothermic, +) or releases heat (exothermic, -).

Understanding the Question

Calculate the molar enthalpy change for both reactions. You need to convert the energy from Joules to Kilojoules and divide by the moles calculated in part (i). Crucially, you must assign the correct sign based on the temperature change.

Approach

Take $q$ from (b)(ii): This is in Joules.
Take $n$ from (b)(i): This is in moles.
Calculate $\Delta H$ : Divide $q$ by $n$ . Since $q$ is in J and we want kJ, divide the result by 1000 (or divide $q$ by 1000 first).
Assign Sign:
- Experiment 1 ( $\text{NaHCO}_3$ ): Temperature fell $\rightarrow$ Endothermic $\rightarrow$ $\Delta H$ is positive (+).
- Experiment 2 ( $\text{Na}_2\text{CO}_3$ ): Temperature rose $\rightarrow$ Exothermic $\rightarrow$ $\Delta H$ is negative (-).

Step-by-Step Reasoning

Formula: $\Delta H = \frac{q}{n \times 1000}$
Experiment 1:
- If $q = 459.8 \text{ J}$ and $n = 0.0357 \text{ mol}$ :
- $\Delta H = 459.8 / (0.0357 \times 1000) = 459.8 / 35.7 = 12.88 \text{ kJ mol}^{-1}$ .
- Since it's endothermic, $\Delta H_1 = +12.9 \text{ kJ mol}^{-1}$ .
Experiment 2:
- If $q = 459.8 \text{ J}$ and $n = 0.0236 \text{ mol}$ :
- $\Delta H = 459.8 / (0.0236 \times 1000) = 459.8 / 23.6 = 19.48 \text{ kJ mol}^{-1}$ .
- Since it's exothermic, $\Delta H_2 = -19.5 \text{ kJ mol}^{-1}$ .

Key Takeaways

Always check the direction of the temperature change to determine the sign of $\Delta H$ .
Ensure units are consistent (J to kJ).

Common Mistakes

Sign Error: Forgetting to make $\Delta H_1$ positive or $\Delta H_2$ negative. This is the most common error in this part.
Unit Error: Forgetting to divide by 1000, giving an answer in $\text{J mol}^{-1}$ instead of $\text{kJ mol}^{-1}$ .
Calculation Error: Dividing moles by energy instead of energy by moles.

Things to Be Careful About

The question asks for the answer in $\text{kJ mol}^{-1}$ .
The sign is part of the answer. A value without a sign is incomplete.

Techniques used

calculate enthalpy change per moleconvert J to kJ

(iv)

Construct an enthalpy cycle and use Hess’s law to determine the enthalpy change, $\Delta H_r$ , for the reaction shown.

Show your working.

2\text{NaHCO}_3(\text{s}) \rightarrow \text{Na}_2\text{CO}_3(\text{s}) + \text{H}_2\text{O}(\text{l}) + \text{CO}_2(\text{g})

(If you were unable to calculate values in (b)(iii) then assume that $\Delta H_1$ is $+27.3\text{ kJ mol}^{-1}$ and that $\Delta H_2$ is $-24.9\text{ kJ mol}^{-1}$ . These may not be the correct values.)

\Delta H_r = \underset{\text{(sign)}}{\text{......}} \underset{\text{(value)}}{\text{......}} \text{ kJ mol}^{-1}

2M Similar

DifficultyMedium-Hard

Worked solution

Working

Hess's Law Cycle:

The target reaction is:

2\text{NaHCO}_3(\text{s}) \rightarrow \text{Na}_2\text{CO}_3(\text{s}) + \text{H}_2\text{O}(\text{l}) + \text{CO}_2(\text{g})

The cycle connects the reactants and products to a common intermediate state (the ions in solution: $2\text{Na}^+ + 2\text{Cl}^- + \text{H}_2\text{O} + \text{CO}_2$ ).

According to Hess's Law:

\Delta H_r + \Delta H_2 = 2\Delta H_1

Rearranging for $\Delta H_r$ :

\Delta H_r = 2\Delta H_1 - \Delta H_2

Calculation:
Substitute the values calculated in (b)(iii).
(Example using provided assumed values: $\Delta H_1 = +27.3$ , $\Delta H_2 = -24.9$ )

\Delta H_r = 2(+27.3) - (-24.9)

\Delta H_r = 54.6 + 24.9

\Delta H_r = +79.5 \text{ kJ mol}^{-1}

Answer

$\Delta H_r = 2\Delta H_1 - \Delta H_2$
(Insert calculated value with correct sign and units here)

Final answer

See working / candidate-dependent

Detailed explanation

Background Concept

Hess's Law states that the total enthalpy change for a reaction is the same regardless of the route taken, provided the initial and final conditions are the same. This allows us to calculate the enthalpy change of a reaction that cannot be measured directly (like the decomposition of sodium hydrogencarbonate) by using indirect reactions that can be measured (reactions with acid).

Understanding the Question

You need to find $\Delta H_r$ for the decomposition of $2\text{NaHCO}_3$ . You have experimental values for $\Delta H_1$ (reaction of $1\text{NaHCO}_3$ with acid) and $\Delta H_2$ (reaction of $1\text{Na}_2\text{CO}_3$ with acid). You must draw the cycle linking these reactions and solve for $\Delta H_r$ .

Approach

Draw the Cycle:
- Top: Reactants ( $2\text{NaHCO}_3$ ) and Products ( $\text{Na}_2\text{CO}_3 + \text{H}_2\text{O} + \text{CO}_2$ ).
- Bottom: The common intermediate (ions in solution after reaction with acid).
- Arrows: Down from reactants to bottom (labelled $2\Delta H_1$ ), Down from products to bottom (labelled $\Delta H_2$ ), and across from reactants to products (labelled $\Delta H_r$ ).
Formulate the Equation:
- The sum of enthalpy changes around a closed loop is zero.
- Going clockwise: $\Delta H_r + \Delta H_2 - 2\Delta H_1 = 0$ .
- Therefore: $\Delta H_r = 2\Delta H_1 - \Delta H_2$ .
Calculate:
- Substitute your experimental values for $\Delta H_1$ and $\Delta H_2$ . Be very careful with the signs (positive and negative).

Step-by-Step Reasoning

Why $2\Delta H_1$ ? The target reaction involves 2 moles of $\text{NaHCO}_3$ . Your experiment measured the enthalpy for 1 mole. So you must multiply $\Delta H_1$ by 2.
Why minus $\Delta H_2$ ? The arrow for $\Delta H_2$ goes down from the products to the intermediate. In the cycle equation, we are going up from the intermediate to the products to close the loop, or simply rearranging the vector addition. Mathematically, $\Delta H_r$ is the difference between the energy of the reactants and products. Both reactants and products go down to the same level. So $\Delta H_r$ is the difference between the two downward arrows: $2\Delta H_1 - \Delta H_2$ .
Calculation Check:
- If $\Delta H_1$ is positive (e.g., +27.3) and $\Delta H_2$ is negative (e.g., -24.9):
- $2(+27.3) = +54.6$
- $-(-24.9) = +24.9$
- Total = $54.6 + 24.9 = +79.5$ .
- The result is positive, indicating the decomposition is endothermic.

Key Takeaways

Hess's Law cycles are powerful tools for finding unmeasurable enthalpy changes.
Always check stoichiometric coefficients (the factor of 2 for $\text{NaHCO}_3$ ).
Sign errors in the final calculation are common; double-check the subtraction of a negative number.

Common Mistakes

Missing the factor of 2: Using $\Delta H_1 - \Delta H_2$ instead of $2\Delta H_1 - \Delta H_2$ .
Sign Error: Calculating $2\Delta H_1 - \Delta H_2$ but forgetting that $\Delta H_2$ is negative, leading to subtraction instead of addition.
Incorrect Cycle Drawing: Arrows pointing in the wrong direction or missing labels.

Things to Be Careful About

Ensure the units are consistent ( $\text{kJ mol}^{-1}$ ).
The final answer must include the sign (+ or -).

Techniques used

construct an enthalpy cycleapply Hess's law

Q2MediumManipulation, Measurement and ObservationPresentation of Data and ObservationsAnalysis, Conclusions and Evaluation

In this experiment you will determine the percentage by mass of a sodium halide impurity present in a sample of sodium hydrogencarbonate by titration.

\text{NaHCO}_3(\text{s}) + \text{HCl}(\text{aq}) \rightarrow \text{NaCl}(\text{aq}) + \text{H}_2\text{O}(\text{l}) + \text{CO}_2(\text{g})

FB 4 is an aqueous solution made by dissolving $17.20\text{ g}$ of the impure sodium hydrogencarbonate in each $\text{dm}^3$ of solution.
FB 5 is $0.200\text{ mol dm}^{-3}$ hydrochloric acid, $\text{HCl}$ .
FB 6 is methyl orange indicator.

(a)

Method

Fill the burette with FB 5.
Pipette $25.0\text{ cm}^3$ of FB 4 into a conical flask.
Add several drops of FB 6 to the conical flask.
Perform a rough titration and record your burette readings in the space below.

The rough titre is .............................. $\text{cm}^3$ .

Carry out as many accurate titrations as you think necessary to obtain consistent results.
Make sure any recorded results show the precision of your practical work.
Record all your burette readings and the volume of FB 5 added in each accurate titration.

Keep FB 4 for use in Question 3.

Results

7M Similar

DifficultyMedium-Easy

Worked solution

Answer

Rough titration

Initial burette reading = 0.00 cm³, final burette reading = 24.90 cm³, rough titre = 24.90 cm³.

Accurate titrations

Titration	Initial burette reading / cm³	Final burette reading / cm³	Titre / cm³
1	0.00	24.60	24.60
2	24.60	49.20	24.60
3	0.00	24.55	24.55

All burette readings are recorded to the nearest 0.05 cm³. The accurate titres are concordant (within 0.10 cm³ of each other).

(These are representative values; your own recorded readings will differ.)

Final answer

Representative example: rough titre 24.90 cm³; accurate titres 24.60, 24.60, 24.55 cm³ (candidate-dependent)

Detailed explanation

Background Concept

A titration is a quantitative technique used to determine the concentration of a solution by reacting it with a solution of known concentration. Here, FB 4 (impure NaHCO₃ solution) is titrated against FB 5 (0.200 mol dm⁻³ HCl) using methyl orange as the indicator. Methyl orange is yellow in alkaline/neutral solution and turns red/pink in acid, so the end-point is the first permanent colour change from yellow to pink.

The burette is the key measuring instrument. It is read to the nearest 0.05 cm³ (half a small division). A rough titration is always performed first to find the approximate end-point quickly, so that subsequent accurate titrations can be done carefully, adding the titrant dropwise near the end-point. Accurate titrations are repeated until two or more concordant results (agreeing within 0.10 cm³) are obtained.

Understanding the Question

This part asks you to perform the titration and record your results in a way that earns all 7 marks. The marks are awarded for:

Recording both burette readings (initial and final) and the titre for the rough titration.
Recording initial and final burette readings for at least two accurate titrations.
Using correct table headings and units (cm³).
Recording all readings to the nearest 0.05 cm³.
Obtaining concordant accurate titres (within 0.10 cm³ of each other).
Three accuracy marks based on how close your mean titre is to the supervisor's value.

Approach

Fill the burette with FB 5, ensuring no air bubbles.
Pipette 25.0 cm³ of FB 4 into a conical flask and add several drops of methyl orange.
Perform a rough titration: add acid quickly, swirling, until the colour changes from yellow to pink. Record initial and final readings and the titre.
Perform accurate titrations: add acid quickly at first, then dropwise near the end-point. Record initial and final readings for each.
Check that your accurate titres agree within 0.10 cm³. If not, perform more titrations.
Record everything in a table with correct headings and units.

Step-by-Step Reasoning

The table below shows a representative set of results (your own values will differ):

Titration	Initial burette reading / cm³	Final burette reading / cm³	Titre / cm³
Rough	0.00	24.90	24.90
1	0.00	24.60	24.60
2	24.60	49.20	24.60
3	0.00	24.55	24.55

The rough titre (24.90 cm³) tells you roughly where the end-point is. The accurate titres (24.60, 24.60, 24.55 cm³) are concordant — they agree within 0.10 cm³. The readings are all recorded to the nearest 0.05 cm³, which is the precision of a burette. The titre is always the difference between the final and initial readings, so it must match the arithmetic of the recorded readings.

The three accuracy marks (Q marks) are awarded by comparing your mean titre with the supervisor's mean titre: within 0.60 cm³, within 0.40 cm³, and within 0.20 cm³ respectively. This rewards both technique (finding the true end-point) and precision (concordant readings).

Key Takeaways

Always perform a rough titration before accurate ones.
Record both initial and final burette readings, not just the titre.
Read the burette to the nearest 0.05 cm³.
Repeat accurate titrations until concordant (within 0.10 cm³).
Use a clear table with headings and units.

Common Mistakes

Recording only the titre, not the initial and final readings — this loses the mark for "two burette readings AND titre".
Reading the burette to only 0.1 cm³ instead of 0.05 cm³.
Not repeating titrations until concordant.
Forgetting units in the table headings.
Including the rough titre in the concordant set.

Things to Be Careful About

Read the burette at eye level to avoid parallax error.
Read the bottom of the meniscus.
Ensure the burette tip is filled (no air bubble) before starting.
Swirl the flask continuously while adding acid.
The end-point is the first permanent colour change from yellow to pink.

Techniques used

perform a rough titrationrecord burette readings to the nearest 0.05 cm^3obtain concordant titres within 0.10 cm^3construct a results table with headings and units

(b)

From your accurate titration results, calculate a suitable mean value to be used in your calculations.

Show clearly how you obtained this value.

$25.0\text{ cm}^3$ of FB 4 required .............................. $\text{cm}^3$ of FB 5.

1M Similar

DifficultyEasy

Worked solution

Working

\text{Mean titre} = \frac{24.60 + 24.60 + 24.55}{3} = 24.58\ \text{cm}^3

Answer

24.58 cm³ (to 2 d.p.)

Final answer

24.58 cm³ (representative mean, to 2 d.p.)

Detailed explanation

Background Concept

The mean titre is the average of the concordant accurate titrations. It is used because it is more reliable than any single reading — random errors in individual readings tend to cancel when averaged. The mark scheme requires the mean to be calculated from two or more titres within a total spread of not more than 0.20 cm³, and quoted to 2 decimal places (the precision of the individual burette readings).

Understanding the Question

From your accurate titration results, select the concordant titres and calculate their mean. You must show your working clearly (or tick the readings you selected) so the examiner can see exactly which values were averaged.

Approach

Identify the accurate titres that agree within 0.20 cm³ total spread.
Add them together and divide by the number of titres.
Round the result to 2 decimal places.

Step-by-Step Reasoning

Using the representative titres 24.60, 24.60 and 24.55 cm³:

\text{Mean titre} = \frac{24.60 + 24.60 + 24.55}{3} = \frac{73.75}{3} = 24.583... = 24.58\ \text{cm}^3

The spread is 24.60 − 24.55 = 0.05 cm³, which is well within the required 0.20 cm³. The mean is quoted to 2 decimal places, matching the precision of the individual readings. If you had only two concordant titres, you would average just those two; three concordant titres give a more reliable mean.

Key Takeaways

Select only concordant titres for the mean — never include the rough titre.
Quote the mean to 2 d.p.
Show your working or tick the selected readings.

Common Mistakes

Including the rough titre in the mean.
Averaging titres that are not concordant (spread > 0.20 cm³).
Quoting the mean to 1 or 3 decimal places.

Things to Be Careful About

The spread of the titres used must be ≤ 0.20 cm³.
Round to the nearest 0.01 cm³.
The mean is used in all subsequent calculations, so an error here propagates through (c)(ii) and (c)(iii).

Techniques used

select concordant titres within a 0.20 cm^3 spreadcalculate the mean titre to 2 decimal places

(c)

Calculations

(i)

Give your answers to (c)(ii) and (c)(iii) to an appropriate number of significant figures.

1M Similar

(ii)

Use your answer to (b) to calculate the amount, in mol, of hydrochloric acid in your mean titre.

Hence determine the amount, in mol, of sodium hydrogencarbonate present in $25.0\text{ cm}^3$ of FB 4.

1M Similar

DifficultyMedium-Easy

Worked solution

Working

\text{Amount of HCl} = 0.200 \times \frac{24.58}{1000} = 4.92 \times 10^{-3}\ \text{mol}

From the equation, 1 mol of NaHCO₃ reacts with 1 mol of HCl, so:

\text{Amount of NaHCO}_3 = 4.92 \times 10^{-3}\ \text{mol}

Answer

$4.92 \times 10^{-3}\ \text{mol}$ (to 3 s.f.)

Final answer

4.92 × 10^-3 mol (representative, to 3 s.f.)

Detailed explanation

Background Concept

The amount of a substance in moles is given by n = cV, where c is concentration in mol dm⁻³ and V is volume in dm³. Since the volume is given in cm³, it must be divided by 1000 to convert to dm³. The balanced equation for the reaction is:

\text{NaHCO}_3(\text{s}) + \text{HCl}(\text{aq}) \rightarrow \text{NaCl}(\text{aq}) + \text{H}_2\text{O}(\text{l}) + \text{CO}_2(\text{g})

This shows a 1:1 mole ratio between NaHCO₃ and HCl — one mole of each reacts. Therefore, the amount of NaHCO₃ in the 25.0 cm³ sample equals the amount of HCl in the mean titre.

Understanding the Question

Use your mean titre from (b) to calculate the amount of HCl in that volume of FB 5, then determine the amount of NaHCO₃ present in 25.0 cm³ of FB 4. The second part relies on the stoichiometry of the balanced equation.

Approach

Convert the mean titre from cm³ to dm³ by dividing by 1000.
Multiply by the concentration (0.200 mol dm⁻³) to get moles of HCl.
Use the 1:1 stoichiometric ratio from the balanced equation to state the moles of NaHCO₃.

Step-by-Step Reasoning

Using the representative mean titre of 24.58 cm³:

n(\text{HCl}) = c \times V = 0.200 \times \frac{24.58}{1000} = 0.200 \times 0.02458 = 4.916 \times 10^{-3}\ \text{mol}

Rounded to 3 significant figures: 4.92 × 10⁻³ mol.

From the balanced equation, the mole ratio NaHCO₃ : HCl is 1:1, so:

n(\text{NaHCO}_3) = 4.916 \times 10^{-3}\ \text{mol} \approx 4.92 \times 10^{-3}\ \text{mol}

This is the amount of NaHCO₃ in the 25.0 cm³ portion of FB 4 pipetted into the flask.

Key Takeaways

n = cV with volume in dm³ (divide cm³ by 1000).
Use the balanced equation to find the stoichiometric ratio.
The 1:1 ratio here means the moles of NaHCO₃ equal the moles of HCl.

Common Mistakes

Forgetting to divide the volume by 1000 (using 24.58 instead of 0.02458).
Using the wrong stoichiometric ratio (e.g., 2:1 or 1:2).
Not quoting to 3 or 4 significant figures.

Things to Be Careful About

The volume must be in dm³ before multiplying by concentration.
The 1:1 ratio comes directly from the balanced equation — check the coefficients.
Keep the unrounded value (4.916 × 10⁻³ mol) for use in (c)(iii) to avoid rounding errors.

Techniques used

calculate moles from concentration and volumeapply the 1:1 stoichiometric ratio from the balanced equation

(iii)

Calculate the mass of sodium hydrogencarbonate present in each $\text{dm}^3$ of solution.

Hence calculate the percentage by mass of the sodium halide impurity in FB 4.

Show your working.

2M Similar

DifficultyMedium

Worked solution

Working

\text{Mass of NaHCO}_3\ \text{in}\ 25.0\ \text{cm}^3 = 4.916 \times 10^{-3} \times 84.0 = 0.413\ \text{g}

\text{Mass of NaHCO}_3\ \text{in}\ 1\ \text{dm}^3 = 0.413 \times 40 = 16.5\ \text{g}

\text{Mass of impurity in}\ 1\ \text{dm}^3 = 17.20 - 16.5 = 0.7\ \text{g}

\text{Percentage by mass of impurity} = \frac{0.7}{17.20} \times 100 = 4.1\%

Answer

4.1% (to 3 s.f.)

Final answer

4.1% (representative, to 3 s.f.)

Detailed explanation

Background Concept

Mass = moles × molar mass (m = n × Mr). Mr(NaHCO₃) = 23.0 + 1.0 + 12.0 + 3(16.0) = 84.0. To scale from 25.0 cm³ to 1 dm³ (1000 cm³), multiply by 40 (since 1000 ÷ 25 = 40). The percentage by mass of the impurity is (mass of impurity ÷ mass of sample) × 100, where the sample mass is 17.20 g per dm³ of FB 4.

Understanding the Question

Calculate the mass of NaHCO₃ in each dm³ of solution, then the percentage by mass of the sodium halide impurity in FB 4. The total mass per dm³ (17.20 g) is the sum of the NaHCO₃ and the impurity, so the impurity mass is found by subtraction.

Approach

Mass of NaHCO₃ in 25.0 cm³ = n(NaHCO₃) × Mr(NaHCO₃).
Mass in 1 dm³ = mass in 25 cm³ × 40.
Mass of impurity = 17.20 − mass of NaHCO₃.
Percentage = (mass of impurity ÷ 17.20) × 100.

Step-by-Step Reasoning

Using the unrounded amount from (c)(ii), n(NaHCO₃) = 4.916 × 10⁻³ mol:

\text{Mass of NaHCO}_3\ \text{in}\ 25.0\ \text{cm}^3 = 4.916 \times 10^{-3} \times 84.0 = 0.413\ \text{g}

\text{Mass of NaHCO}_3\ \text{in}\ 1\ \text{dm}^3 = 0.413 \times 40 = 16.5\ \text{g}

\text{Mass of impurity} = 17.20 - 16.5 = 0.7\ \text{g}

\text{Percentage by mass of impurity} = \frac{0.7}{17.20} \times 100 = 4.1\%

Using the unrounded intermediate values throughout gives 3.95%, which rounds to 4.0% to 2 s.f. or 3.95% to 3 s.f. The mark scheme accepts 3 or 4 significant figures, so either 3.95% or 4.0% (3 s.f.) is acceptable. The key is that the final answer is the percentage of the IMPURITY, not of the NaHCO₃.

Key Takeaways

m = n × Mr.
Scaling factor 40 converts 25 cm³ to 1 dm³.
Percentage by mass = (mass of component ÷ total mass) × 100.
The impurity mass is found by subtracting the NaHCO₃ mass from the total sample mass.

Common Mistakes

Using Mr of NaCl (58.5) instead of NaHCO₃ (84.0).
Forgetting the ×40 scaling factor (giving the mass in 25 cm³, not 1 dm³).
Calculating the percentage of NaHCO₃ instead of the impurity.
Not quoting to 3 or 4 significant figures.

Things to Be Careful About

Mr(NaHCO₃) = 84.0.
The final answer must be the percentage of the impurity, so subtract the NaHCO₃ mass from 17.20 g first.
Carry unrounded intermediate values through your working to avoid rounding drift.

Techniques used

convert moles to mass using molar massscale from 25.0 cm^3 to 1 dm^3calculate percentage by mass of impurity

(d)

A student carries out the experiment in Question 1 using the impure sodium hydrogencarbonate dissolved to make FB 4.

State how this affects the value of $\Delta H_1$ determined in 1(b)(iii) compared to the value the student would get if pure sodium hydrogencarbonate were used.

State what assumption you have made about the sodium halide impurity.

2M Similar

DifficultyMedium

Worked solution

Answer

$\Delta H_1$ will be smaller / less positive for the impure sample than for pure sodium hydrogencarbonate.

Assumption: the sodium halide impurity does not react with the acid (HCl / FB 5).

Final answer

ΔH1 smaller / less positive; assumption: impurity does not react with the acid

Detailed explanation

Background Concept

ΔH₁ is the enthalpy change for the reaction of sodium hydrogencarbonate with hydrochloric acid:

\text{NaHCO}_3(\text{s}) + \text{HCl}(\text{aq}) \rightarrow \text{NaCl}(\text{aq}) + \text{H}_2\text{O}(\text{l}) + \text{CO}_2(\text{g})

This reaction is endothermic (ΔH positive), because energy is absorbed to decompose the carbonic acid formed into CO₂ and water. In Question 1, ΔH₁ was determined by calorimetry — measuring the temperature change when a known mass of NaHCO₃ reacts with excess HCl.

Understanding the Question

This question links to Question 1, where ΔH₁ was determined for the reaction of NaHCO₃ with HCl. Now, using the impure sample (which contains a sodium halide impurity) instead of pure NaHCO₃, how does the determined ΔH₁ change? You must also state the assumption made about the impurity's behaviour.

Approach

Consider what fraction of the impure sample actually reacts with the acid. The sodium halide impurity (NaX) does not react with HCl — halide ions are the conjugate bases of strong acids and do not accept protons. Therefore, per gram of impure sample, less NaHCO₃ reacts, producing a smaller temperature change and hence a smaller (less positive) ΔH₁.

Step-by-Step Reasoning

The impure sample contains NaHCO₃ plus a sodium halide. When the impure sample reacts with HCl, only the NaHCO₃ reacts; the halide is inert. Therefore, per gram of impure sample, there is less NaHCO₃ than per gram of pure NaHCO₃. Since this reaction is endothermic, less NaHCO₃ means less heat is absorbed per gram of sample, so the measured ΔH₁ is smaller / less positive.

The assumption required is that the sodium halide impurity does not react with the acid (HCl / FB 5). If it did react, it would contribute to the measured heat change, and the calculated ΔH₁ would not reflect the NaHCO₃ reaction alone.

Key Takeaways

An inert impurity dilutes the reactive component, reducing the measured enthalpy change per gram of sample.
Always state the assumption about the impurity's behaviour explicitly — it is a separate mark.
For an endothermic reaction, less reactant means a smaller (less positive) ΔH.

Common Mistakes

Saying ΔH₁ is larger or more positive (confusing the direction of the effect).
Not stating the assumption about the impurity (this is a separate mark).
Confusing the sign of ΔH — this reaction is endothermic, so ΔH₁ is positive; the impurity makes it less positive.

Things to Be Careful About

The sign convention: this reaction is endothermic, so ΔH₁ is positive; the impurity makes it smaller / less positive.
The assumption must be explicitly stated to earn the second mark — "impurity does not react with the acid" is the required point.
The mark scheme accepts "smaller" or "less positive" — either wording scores, but "larger" or "more positive" does not.

Techniques used

reason about the effect of an inert impurity on the measured enthalpy changestate the assumption about the impurity's reactivity

Q3MediumQualitative Analysis

Qualitative analysis

For each test you should record all your observations in the spaces provided.

Examples of observations include:

colour changes seen
the formation of any precipitate and its solubility (where appropriate) in an excess of the reagent added
the formation of any gas and its identification (where appropriate) by a suitable test.

You should record clearly at what stage in a test an observation is made.

Where no change is observed, you should write ‘no change’.

Where reagents are selected for use in a test, the name or correct formula of the element or compound must be given.

If any solution is warmed, a boiling tube must be used. If any solid is heated a hard-glass test-tube must be used.

Rinse and reuse test-tubes and boiling tubes where possible.

No additional tests should be attempted.

(a)

(i)

To a $1\text{ cm}$ depth of FB 4 in a boiling tube, slowly add a $1\text{ cm}$ depth of dilute nitric acid. Stir gently until the reaction is complete. Use this solution for your tests.

Select reagents to identify which sodium halide is present.
Record details of the reagents used and your observations.
Identify the sodium halide that is present as an impurity in FB 4.

The formula of the impurity is ...............

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

Reagents used: aqueous silver nitrate, followed by aqueous ammonia.

Observations: a white precipitate formed with $\text{AgNO}_3$ (aq); the precipitate dissolved in excess aqueous ammonia giving a colourless solution.

The impurity is $\text{NaCl}$ .

Final answer

NaCl

Detailed explanation

Background Concept

Halide ions ( $\text{Cl}^-$ , $\text{Br}^-$ , $\text{I}^-$ ) are identified by adding aqueous silver nitrate, which forms silver halide precipitates: AgCl is white, AgBr is cream, AgI is pale yellow. Because the colours of the last two are close, the precipitate's solubility in aqueous ammonia confirms the identity: AgCl dissolves in dilute NH3(aq), AgBr dissolves only in concentrated NH3(aq), and AgI is insoluble in ammonia of any concentration.

Understanding the Question

FB 4 is a solution containing a sodium halide impurity. After acidifying with dilute nitric acid, you must choose your own reagents (the command 'select reagents' means you must name them), record observations at each stage, and deduce the halide. The mark scheme awards marks in pairs: reagent selection, precipitate observation, solubility observation, and the identity.

Approach

Use the standard halide test: AgNO3(aq) to form the precipitate, then NH3(aq) to test solubility. A white precipitate soluble in NH3(aq) points uniquely to chloride, so the salt is NaCl.

Step-by-Step Reasoning

Selecting AgNO3(aq) is the only sensible reagent for halide identification; selecting NH3(aq) as the second reagent is needed for the solubility stage. Reagent selection plus observations are bundled: any two of the starred points earns each mark.
The white precipitate is AgCl: $\text{Ag}^+ + \text{Cl}^- \rightarrow \text{AgCl}$ .
AgCl dissolves in aqueous ammonia because $\text{NH}_3$ forms a complex with $\text{Ag}^+$ , $[\text{Ag}(\text{NH}_3)_2]^+$ , pulling the equilibrium back into solution. Solubility in NH3(aq) confirms chloride rather than bromide or iodide.
Therefore the sodium halide is NaCl.

Key Takeaways

The halide test is a two-stage test: precipitate with AgNO3, then confirm with NH3 solubility.
White + soluble in dilute NH3 = Cl⁻; cream + soluble only in concentrated NH3 = Br⁻; pale yellow + insoluble = I⁻.

Common Mistakes

Adding dilute hydrochloric acid instead of nitric acid — this introduces chloride ions and invalidates the test.
Writing only 'a precipitate formed' without the colour; the colour is the mark.
Omitting the solubility-in-excess observation, which is what distinguishes the halides.

Things to Be Careful About

Record observations stage by stage: what is seen on adding AgNO3, then what is seen on adding NH3(aq).
Name reagents by name or correct formula — 'silver nitrate solution' or $\text{AgNO}_3$ (aq), not just 'the reagent'.

Techniques used

select appropriate reagents for halide testingrecord precipitate colour and solubility in excess reagentdeduce the identity of the halide from observations

(ii)

Suggest why it is necessary to add the nitric acid to FB 4 before carrying out your tests in (a)(i).

1M Similar

DifficultyMedium-Easy

Worked solution

Answer

The nitric acid removes / reacts with any hydrogencarbonate (carbonate) ions present, which would otherwise react with silver ions and interfere with the halide test.

Final answer

The acid removes/reacts with hydrogencarbonate ions, which would otherwise interfere with the silver nitrate test

Detailed explanation

Background Concept

Silver ions form insoluble precipitates not only with halides but also with carbonate and hydrogencarbonate ions (Ag2CO3 is a white precipitate). If carbonate/hydrogencarbonate is present, a white precipitate with AgNO3 could be misidentified as AgCl.

Understanding the Question

FB 4 evidently contains hydrogencarbonate (the context of the paper — FB 4 is a mixture, here sodium hydrogencarbonate contaminated with NaCl). The question asks why acid must be added before the halide test; the command word is 'suggest', so a one-line chemical reason is expected.

Approach

Link the acid to the interfering ion: acid destroys hydrogencarbonate (effervescence of CO2), so it cannot form a precipitate with Ag+.

Step-by-Step Reasoning

Dilute nitric acid reacts with hydrogencarbonate ions: $\text{HCO}_3^- + \text{H}^+ \rightarrow \text{CO}_2 + \text{H}_2\text{O}$ , removing them from solution.
Without this step, $\text{Ag}^+$ would react with hydrogencarbonate/carbonate to give a white precipitate, giving a false positive for chloride.
Nitric acid is used (not HCl or H2SO4) because it introduces no halide or sulfate ions that would themselves precipitate with silver ions.

Key Takeaways

Acidify with nitric acid before testing for halides to remove carbonate/hydrogencarbonate interference.
The acid must be nitric acid specifically to avoid introducing other interfering anions.

Common Mistakes

Writing vaguely that 'the acid removes impurities' without naming hydrogencarbonate/carbonate ions — the named ion is the mark.
Saying the acid 'catalyses' the test or 'makes the solution acidic for the AgNO3' — no chemical reason given.

Things to Be Careful About

Either wording scores: acid removes/reacts with hydrogencarbonate ions, OR silver ions would react with hydrogencarbonate ions. Both are on the mark scheme.

Techniques used

explain the purpose of acidification before halide testingidentify interfering ions

(b)

FB 7 contains a metal cation which is listed in the Qualitative analysis notes.

(i)

Place half the sample of FB 7 in a hard-glass test-tube. Heat gently at first and then more strongly.

Record your observations.

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

The green solid turns into a black powder. The powder is very fluid / moves around inside the test-tube as it is heated. Condensation forms on the cooler walls of the hard-glass test-tube.

Final answer

Green solid turns to black powder; condensation on tube walls

Detailed explanation

Background Concept

Copper(II) carbonate is a green solid that decomposes on heating:
$\text{CuCO}_3(\text{s}) \rightarrow \text{CuO}(\text{s}) + \text{CO}_2(\text{g})$
Copper(II) oxide is black. The gaseous CO2 released can condense water vapour on the cool parts of the tube if moisture is present, and the freshly formed oxide powder is fine and free-flowing.

Understanding the Question

You heat half the sample of FB 7 gently then strongly in a hard-glass test-tube and record observations. Two marks, each from a pair of starred points, so at least four distinct observations are available and you should record several.

Approach

Record: initial colour, final colour/physical form, behaviour of the solid, and any condensation. Do not just write 'it changed colour'.

Step-by-Step Reasoning

FB 7 is CuCO3, which is green — note the starting colour ('green at start').
On heating it decomposes to CuO, a black solid — 'turns to a black powder'.
The fine oxide powder is very mobile/fluid, swirling around the tube — a distinctive observation examiners look for.
CO2 (with moisture) causes condensation on the cooler upper walls of the tube.

Key Takeaways

Thermal decomposition of a carbonate: metal oxide + CO2; colour changes identify the metal (CuCO3 green → CuO black).
Observations should include physical state/behaviour, not only colour.

Common Mistakes

Writing 'turned black' without noting the starting green colour or the powder form.
Missing the condensation observation.
Heating in a soft test-tube instead of a hard-glass tube (instruction in the stem).

Things to Be Careful About

'Black powder' is required, not just 'black'; the fluidity of the powder is a separate creditable point.
Any two of the four starred observations earn each mark, so include all of them to be safe.

Techniques used

heat a solid and record colour changesobserve physical behaviour of the decomposition productnote condensation

(ii)

Use the $25\text{ cm}^3$ measuring cylinder to measure $20\text{ cm}^3$ of FB 1.

Place the remaining sample of FB 7 in a boiling tube and add portions of FB 1 from the measuring cylinder until all the FB 7 has reacted. Stir until the reaction is complete.

Record your observations.

Keep this solution for use in (b)(iii).

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

Effervescence occurs; the reaction is vigorous. The solid reacts to form a green-blue solution. The gas was tested with limewater, which turned milky / gave a white precipitate, showing the gas is carbon dioxide.

Final answer

Effervescence, green-blue solution, CO2 turns limewater milky

Detailed explanation

Background Concept

Carbonates react with acids: metal carbonate + acid → salt + water + carbon dioxide. The CO2 is identified by bubbling it through limewater, $\text{Ca(OH)}_2$ (aq), which turns milky due to precipitated CaCO3. The salt formed from copper(II) carbonate and sulfuric acid (FB 1) is copper(II) sulfate, which is blue in solution.

Understanding the Question

FB 1 is dilute sulfuric acid. Adding it to the remaining FB 7 (CuCO3) gives CuSO4 solution + CO2. Record observations; two marks from pairs of starred points.

Approach

Record: vigour of reaction, effervescence, colour of resulting solution, and the limewater test result.

Step-by-Step Reasoning

$\text{CuCO}_3 + \text{H}_2\text{SO}_4 \rightarrow \text{CuSO}_4 + \text{CO}_2 + \text{H}_2\text{O}$ .
Effervescence (fizzing) as CO2 escapes; the reaction is vigorous.
The solution formed is green-blue — characteristic of dilute copper(II) sulfate.
Attempting/testing with limewater: the gas turns limewater milky (white precipitate of CaCO3), confirming CO2. Even an attempt at the limewater test earns credit.

Key Takeaways

Acid + carbonate → effervescence of CO2, confirmed by limewater.
CuSO4(aq) is blue/green-blue in solution.

Common Mistakes

Recording only 'fizzing' without the solution colour or the gas test.
Saying 'hydrogen gas produced' — carbonates release CO2 with acids, not H2.
Forgetting to keep the solution for part (b)(iii) as instructed.

Things to Be Careful About

Any two starred pairs earn the two marks; include effervescence, colour, vigour, and limewater result to cover all bases.

Techniques used

observe effervescence and solution colourtest a gas with limewater

(iii)

Carry out the following tests. For each test use a $1\text{ cm}$ depth of the solution from (b)(ii) in a test-tube.

Record your observations in Table 3.1.
Identify the metal ion present in FB 7.

Table 3.1

test	observations
Test 1 Add aqueous sodium hydroxide.
Test 2 Add aqueous ammonia.
Test 3 Add a piece of aluminium foil.

The formula of the metal ion in FB 7 is ...............

4M Similar

DifficultyMedium

Worked solution

Answer

Table 3.1

test	observations
Test 1 — add NaOH(aq)	(pale) blue precipitate, insoluble in excess NaOH(aq)
Test 2 — add NH3(aq)	(pale) blue precipitate; precipitate dissolves in excess NH3(aq) forming a dark / deep blue solution
Test 3 — add aluminium foil	effervescence / fizzing; gas pops with a lighted splint; pink-brown solid formed; solution turns colourless / pale blue; mixture gets hotter

The metal ion in FB 7 is $\text{Cu}^{2+}$ .

Final answer

Cu2+; blue ppt insoluble in excess NaOH, blue ppt soluble in excess NH3 giving deep blue solution

Detailed explanation

Background Concept

Cation identification with NaOH(aq) and NH3(aq) relies on the precipitation of metal hydroxides and their differing solubility in excess reagent. Cu2+ gives a pale blue precipitate of Cu(OH)2 with both reagents; it is insoluble in excess NaOH but dissolves in excess NH3 because ammonia forms the deep-blue complex $[\text{Cu}(\text{NH}_3)_4]^{2+}$ . Aluminium foil in an acidic solution of a less reactive metal ion displaces the metal (Al is more reactive) and, with the acid present, also generates hydrogen gas: this is the classic 'aluminium foil' test used to confirm Cu2+ (and nitrates).

Understanding the Question

The solution from (b)(ii) contains CuSO4 (acidified). You carry out three tests and use the combined evidence to name the cation. Four marks from pairs of starred observations plus one for the ion.

Approach

Record full observations for each test, including behaviour in excess reagent and all changes in Test 3, then match to the Qualitative analysis notes.

Step-by-Step Reasoning

Test 1: $\text{Cu}^{2+} + 2\text{OH}^- \rightarrow \text{Cu(OH)}_2$ — a pale blue precipitate that does not dissolve in excess NaOH.
Test 2: same pale blue precipitate initially; in excess NH3 it dissolves giving a deep/dark blue solution due to the ammine complex — this is diagnostic for Cu2+.
Test 3: aluminium is more reactive than copper, so it displaces Cu2+: redox reaction. The pink-brown solid is copper metal. The acid present reacts with Al releasing H2, which effervesces and pops with a lighted splint. The blue colour fades as Cu2+ is removed (colourless/pale blue solution). The reaction is exothermic — the tube gets hotter.
All evidence confirms $\text{Cu}^{2+}$ .

Key Takeaways

Cu2+ identification: blue ppt with NaOH (insoluble in excess) and with NH3 (soluble in excess, deep blue solution).
Al foil test: displacement of the metal plus H2 evolution confirms the cation; colour change of solution shows reduction of the metal ion.

Common Mistakes

Omitting 'insoluble in excess' for NaOH — this is a required point.
Saying the NH3 precipitate is 'insoluble in excess' — for Cu2+ it dissolves, giving the deep blue solution.
Missing the pop test or the pink-brown copper deposit in Test 3.

Things to Be Careful About

Each mark is a pair of starred points, so record several observations per test.
Use precise colour terms: 'pale blue precipitate', 'deep/dark blue solution', 'pink-brown solid'.

Techniques used

perform cation tests with NaOH and NH3test gas with a lighted splintobserve displacement by aluminium foildeduce the cation from observations

(c)

FB 8 is an acidified aqueous solution containing a metal cation which is listed in the Qualitative analysis notes.

(i)

Carry out the following tests. For each test use a $1\text{ cm}$ depth of FB 8 in a test-tube.

Record your observations in Table 3.2.
Identify the metal ion present in FB 8.

Table 3.2

test	observations
Test 1 Add aqueous sodium hydroxide.
Test 2 Add a $1\text{ cm}$ depth of acidified aqueous potassium manganate(VII).

The formula of the metal ion in FB 8 is ...............

3M Similar

DifficultyMedium

Worked solution

Answer

Table 3.2

test	observations
Test 1 — add NaOH(aq)	green precipitate, insoluble in excess; precipitate turns brown on its surface
Test 2 — add acidified KMnO4	purple colour disappears; solution becomes colourless / (pale) yellow — KMnO4 is decolourised

The metal ion in FB 8 is $\text{Fe}^{2+}$ .

Final answer

Fe2+; green ppt insoluble in excess turning brown, KMnO4 decolourised

Detailed explanation

Background Concept

Fe2+ and Fe3+ are distinguished by their hydroxide colours (Fe(OH)2 is green, Fe(OH)3 is brown) and by redox behaviour: Fe2+ is a reducing agent and decolourises acidified potassium manganate(VII), being oxidised to Fe3+. Fe2+(aq) solutions are pale green; Fe3+(aq) is yellow-brown.

Understanding the Question

FB 8 is an acidified solution of a listed cation. Two tests — NaOH and acidified KMnO4 — plus the identity of the ion. Two marks from pairs of observations, one mark for the ion.

Approach

Use the NaOH test for the hydroxide colour, and the KMnO4 test to distinguish Fe2+ (decolourises purple MnO4⁻) from Fe3+ (no reaction).

Step-by-Step Reasoning

Test 1: $\text{Fe}^{2+} + 2\text{OH}^- \rightarrow \text{Fe(OH)}_2$ (s), a green precipitate, insoluble in excess. In air, Fe(OH)2 is slowly oxidised by oxygen to brown Fe(OH)3, so the precipitate turns brown on its surface — a key confirming observation.
Test 2: MnO4⁻ (purple) is reduced to Mn2+ (very pale pink/colourless) while Fe2+ is oxidised to Fe3+ (yellow): $\text{MnO}_4^- + 8\text{H}^+ + 5\text{Fe}^{2+} \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O} + 5\text{Fe}^{3+}$ . The purple colour disappears, giving a colourless to pale yellow solution.
Together: green ppt turning brown + decolourised KMnO4 = Fe2+.

Key Takeaways

Green ppt with NaOH = Fe2+; brown ppt = Fe3+; surface browning of a green ppt shows air oxidation of Fe2+.
Acidified KMnO4 decolourised = presence of a reducing ion, here Fe2+.

Common Mistakes

Writing 'brown precipitate' for Test 1 immediately — the initial precipitate is green; the brown colour develops on the surface.
Omitting 'insoluble in excess'.
Confusing the KMnO4 observation: the purple manganate(VII) is decolourised; the Fe3+ formed gives the pale yellow tint.

Things to Be Careful About

Record the colour change direction: 'purple to colourless/pale yellow', not just 'colour change'.
The ion identity mark depends on consistent observations, so make them precise.

Techniques used

perform cation test with NaOHtest reducing ability with acidified KMnO4deduce the cation from observations

(ii)

State the type of reaction between the metal ion in FB 8 and acidified aqueous potassium manganate(VII).

Explain how your observations support your answer.

1M Similar

DifficultyMedium-Easy

Worked solution

Answer

The reaction is a redox reaction. The purple manganate(VII) ions are reduced to colourless $\text{Mn}^{2+}$ (the purple → colourless change), while $\text{Fe}^{2+}$ is oxidised to $\text{Fe}^{3+}$ , seen as the yellow colour of the solution.

Final answer

Redox; decolourisation shows reduction of MnO4−, yellow colour shows oxidation of Fe2+ to Fe3+

Detailed explanation

Background Concept

A redox reaction involves simultaneous oxidation (loss of electrons) and reduction (gain of electrons). In acidified KMnO4 titrations/tests, MnO4⁻ (+7) is reduced to Mn2+ (+2), losing its intense purple colour, while the reducing agent is oxidised.

Understanding the Question

'State the type of reaction' requires the single word 'redox'; 'explain how your observations support your answer' requires linking each colour change to a specific half-reaction.

Approach

Name the reaction type, then map: purple → colourless = reduction of MnO4⁻; colourless/pale yellow = oxidation of Fe2+ to yellow Fe3+.

Step-by-Step Reasoning

The purple colour of MnO4⁻ disappears because it gains electrons (reduction) forming near-colourless Mn2+.
The solution becomes pale yellow because Fe2+ loses electrons (oxidation) forming Fe3+, which is yellow in solution.
Both processes occur together, so the reaction is redox.

Key Takeaways

Always link the observation to the species oxidised or reduced — naming 'redox' alone without the explanatory link may not score the full mark.

Common Mistakes

Writing only 'redox' with no explanation — the question explicitly asks for the explanatory link.
Saying 'displacement' or 'neutralisation' — neither fits; the electron transfer is the defining feature.

Things to Be Careful About

Either link scores: colourless change ↔ reduction of manganate(VII), OR yellow colour ↔ oxidation of Fe2+/formation of Fe3+. Include both for safety.

Techniques used

classify the reaction type as redoxlink observations to oxidation and reduction