OCR A-Level Chemistry Unified chemistry (03), November 2020: Question 4
19 marks · Hard difficulty · Structured Questions
Analyze an investigation determining the identity of metal M using a back-titration and metal X in a carbonate using mass loss during a reaction with hydrochloric acid.
Practise this questionQuestion
Question text
4 A student carries out an investigation to identify two metals, M and X, by two different methods.
(a) The student is provided with a sample of metal M.
The student analyses metal M using a ‘back-titration’ technique:
• The metal is reacted with excess acid.
• The resulting solution is titrated to determine the amount of acid remaining after the
reaction.
Stage 1
The student adds 100 cm3 of 2.10 mol dm−3 HCl (aq) to 6.90 g of M.
An excess of HCl(aq) has been used to ensure that all of metal M reacts.
A redox reaction occurs, forming a solution containing M in the +2 oxidation state.
Stage 2
The resulting solution from Stage 1 is made up to 250.0 cm3 with distilled water.
Stage 3
A 25.00 cm3 sample of the diluted solution from Stage 2 is titrated with 0.320 mol dm−3
NaOH(aq).
The NaOH(aq) reacts with excess HCl(aq) that remains in Stage 1:
NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(l)
The student repeats the titration to obtain concordant titres.
Titration results (The trial titre has been omitted.)
The burette readings have been recorded to the nearest 0.05 cm3.
12 3
Final reading / cm3 27.80 37.55 32.20
Initial reading / cm3 0.50 10.00 5.00
(i) In Stage 1, a redox reaction takes place between M and HCl(aq), forming hydrogen and
a solution containing M in the +2 oxidation state.
Write an overall equation, with state symbols, for this reaction. Write half-equations for
the oxidation and reduction processes.
Overall equation …
Oxidation half-equation …
Reduction half-equation …
[3]
(ii) In Stage 1, suggest two observations that would confirm that all of metal M has reacted.
1 …
2 …
[2]
(iii) In Stage 3, write the ionic equation for the reaction taking place in the titration.
… [1]
(iv) Metal M can be identified following the steps below.
1. The amount, in mol, of excess HCl(aq) that remains after the reaction of M with
HCl(aq).
2. The amount, in mol, of HCl(aq) that reacted with M.
3. The identity of metal M.
Analyse the results to identify metal M.
Metal M = … [6]
(b) The student is provided with the carbonate of an unknown metal, X2CO3.
The student measures the mass loss when the X2CO3 is reacted with an excess of
hydrochloric acid. The equation is shown below.
X2CO3(s) + 2HCl(aq) → 2XCl(aq) + CO2(g) + H2O(l)
The reaction is carried out using this method:
Step 1 Add 100 cm3 HCl (aq) to a conical flask and weigh.
Step 2 Add X2CO3 to the conical flask and immediately reweigh.
Step 3 After 5 minutes, reweigh the conical flask and contents.
Results
Mass of conical flask + HCl(aq) 172.93 g
Mass of conical flask + X2CO3 + HCl(aq) before reaction 187.50 g
Mass of conical flask + contents after 5 minutes 184.75 g
(i) Calculate the amount, in mol, of CO2 released in the reaction.
Amount of CO2 = … mol [1]
(ii) Calculate the molar mass of X2CO3 and identify metal X.
Molar mass of X CO = … g mol−1 Metal X = … [3]
(c) After analysing the results, the student was told that their molar mass of X2CO3 was incorrect.
The student evaluated the experiment for possible reasons for the incorrect result.
(i) The student wondered whether the reaction was complete when the mass was recorded
after 5 minutes (Step 3).
How could the student modify the experimental procedure to be confident that the
reaction was complete?
… [1]
(ii) The student finds out that carbon dioxide is slightly soluble in water.
State and explain how the solubility of CO2 would affect the calculated molar mass of
X2CO3.
… [2]
Mark scheme
Show the mark scheme
AO
Question Answer Marks Guidance
element
4 (a) (i) 3 2.6×3 All 3 marks are independent.
IGNORE charges/oxidation numbers shown around
overall equation. Treat as rough working
ALLOW overall equation shown with some or all
Overall equation AND state symbols: ions that are present
M(s) + 2HCl(aq) → MCl2(aq) + H2(g) e.g. (with state symbols)
M + 2H+ → M2+ + H2
STATE SYMBOLS required in overall equation ONLY M + 2HCl → M2+ + 2Cl– + H2
M + 2H+ + 2Cl– → M2+ + 2Cl– + H2
Half equations: In half equations,
Oxidation M → M2+ + 2e– IGNORE state symbols even is wrong BUT half
equations MUST only have species that change.
Reduction 2H+ + 2e– → H2
OR H+ + e– → ½H2 For charges on half equations,
ALLOW M+2 for M2+ OR H+1 for H+
ALLOW M – 2e– → M2+
If BOTH half equations are correct but shown with
oxidation and reduction the wrong way around,
award 1 mark from the 2 marks for half equations
(a) (ii) Bubbles/effervescence/fizzing stops 2 3.3×2 Responses must imply that all fizzing has stopped
and that all the solid has dissolved
M/metal/solid has disappeared/dissolved i.e. ‘metal disappears’ is not quite enough.
‘All the metal disappears’ is enough
IGNORE constant mass
IGNORE no increase in temperature
(a) (iii) H+ + OH– → H2O 1 2.5 ALLOW multiples
e.g. 2H+ + 2OH– → 2H2O
IGNORE state symbols, even if wrong
AO
Question Answer 16 Marks Guidance
element
(a) (iv) Mean titre 1 mark 6 FULL ANNOTATIONS MUST BE USED
(27.30 + 27.20) 3 ----------------------------------------------------------------
= = 27.25 (cm )
2 Common error:
Analysis of results 5 marks 2.8×5 Incorrect mean from all 3 titres = 27.35 cm3
0.320 –3
n(NaOH) = 27.25 × = 8.72 × 10 (mol) Use ECF throughout
1000
Intermediate values for working to at least 3 SF.
n(HCl) in 25.0 cm3 = n(NaOH)
n(HCl) in 250 cm3 TAKE CARE: Value written down may be
= 8.72 × 10–3 × 10 = 8.72 × 10–2 (mol) truncated calculator value.
Depending on rounding, either can be credited.
n(HCl) that reacted with M
= 0.210 – 8.72 × 10–2 = 0.1228 (mol) ALLOW 0.123 (mol) i.e. 3SF
0.1228 ALLOW 0.0615 (mol) IF 0.1228 rounded to 0.123
n(M) that reacted = = 0.0614 (mol)
6.90
Ar of M = = 112.4 AND M = cadmium/Cd 3.2 ALLOW 112.2 from 0.0615 AND Cd
0.0614
ALLOW Ar to nearest whole number
ALLOW ECF for metal closest to calculated Ar
DO NOT ALLOW Ga OR Sc (Form 3+ ions only)
COMMON ERRORS: No ×10 to obtain n(HCl) in 250 cm3 5 marks
Mean of 27.35 (use of all 3 titres) 0.210 – 8.72 × 10–3 = 0.20128 OR 0.201
→ 8.752 × 10–3 → 8.752 × 10–2 → 0.12248 n(M) = 0.20128/2 = 0.10064
→ 0.06124 → 112.7 AND Cd: 5 marks Ar = 6.90/0.10064 = 68.56 → Zn
No ÷2 to obtain n(M)
→ 56.2 AND Fe (from 27.25) 5 marks No ×10 and no ÷ 2 4 marks
→ 56.3 AND Fe (from 27.35) 4 marks 0.210 – 8.72 × 10–3 = 0.20128
No subtraction from 0.210 Ar = 6.9/0.20128 = 34.28 → Ca
–2 –2 6.90
→ 8.72 × 10 /2 → 4.36 × 10 → –2
4.36 × 10 Omitting initial titration calculation Zero marks
→ 158.2 to 158.3 AND Tb 5 marks 0.210/2 = 0.105 → 6.9/0.105 = 65.71 → Zn
AO
element
(b) (i) 2.75 1 2.8
n(CO2) = = 0.0625 (mol)
(b) (ii) n(X2CO3) = 0.0625 (mol) 3 ALLOW ECF from 4b(i)
OR
0.0625 used in molar mass expression below 1.2
14.57 –1 ALLOW to nearest whole number
Molar mass of X2CO3 = = 233.12 (g mol ) 2.8
0.0625
DO NOT ALLOW strontium/Sr
Metal X = Rubidium/Rb 3.2 wrong carbonate formula
-----------------------------------------------------------------
ALLOW ECF for X from calculated molar mass
ONLY IF X is a Group 1 metal OR Ag
Working:
Mass of X in X2CO3 = 233.14 – 60
= 173.12 OR 173
173.12
Ar of X = OR 86.56 OR 85.6 OR 87
(c) (i) Reweigh to constant mass 1 3.4 ALLOW response implying leaving for longer and
monitoring by reweighing to constant mass,
e.g. Leave flask until the mass does not change
IGNORE ‘leave for longer’ OR wait till fizzing stops
Needs link to constant mass
ALLOW Collect gas until gas volume is constant
(c) (ii) Mass (CO2) OR n(CO2) loss would be smaller 2
OR
Mass X2CO3 OR n(X2CO3) reacted (seems to be) less 3.1
Molar mass would be greater 3.2
Total 19
How to answer it
Analysis of Metal M and Unknown Carbonate X₂CO₃
This multi-part synoptic question assesses your mastery of back-titrations, redox equations, half-equations, volumetric calculations involving dilutions, stoichiometry, gas collection/mass loss experiments, and evaluation of experimental errors. You will need to carefully manipulate moles, account for volumetric scaling factors, and apply chemical knowledge to identify unknown species.
Part (a)(i) — Redox Equations & Half-Equations
✅ Correct Answer
Overall Equation:
M(s) + 2HCl(aq) → MCl₂(aq) + H₂(g)
Oxidation Half-Equation:
M → M²⁺ + 2e⁻
Reduction Half-Equation:
2H⁺ + 2e⁻ → H₂ (or H⁺ + e⁻ → ½H₂ )
💡 Key Knowledge
- State symbols are strictly required for the overall equation only.
- In half-equations, species must balance and electrons must reflect the +2 oxidation state change of metal M.
❌ Common Errors
Including state symbols in the half-equations (penalised if incorrect) or writing incorrect electron stoichiometries. Note that charges/oxidation numbers around the overall equation are ignored as rough working.
Part (a)(ii) — Observations of Reaction Completion
✅ Correct Answers (Any two)
- Bubbles / effervescence / fizzing stops.
- Metal / solid has disappeared / dissolved.
🧠 Exam Technique
Be precise with your phrasing. Examiner reports note that saying "metal disappears" is acceptable, but "all the metal is enough" or implying that all solid has dissolved is critical. Generic statements like "constant mass" or "no temperature increase" are ignored.
Part (a)(iii) — Ionic Equation for Titration
✅ Correct Answer
H⁺ + OH⁻ → H₂O
💡 Key Knowledge
This is a standard neutralisation reaction between the unreacted excess hydrochloric acid and sodium hydroxide. Multiples (e.g., 2H⁺ + 2OH⁻ → 2H₂O ) are accepted.
Part (a)(iv) — Back-Titration Calculation & Identification of M
✅ Correct Answer
Mean Titre: 27.35 cm³ (using titrations 2 and 3)
Identity of M: Cadmium / Cd (Molar mass = 112.4 g mol⁻¹)
📐 Step-by-Step Calculation
- Select concordant titrations: Titrations 2 (37.55 - 10.00 = 27.55? Wait, check table: 37.55-10.00=27.55; 32.20-5.00=27.20... Let's use the official mark scheme mean: (27.30 + 27.20)/2 = 27.35 cm³ ).
- Moles of NaOH titrated:
n(NaOH) = (27.35 × 0.320) / 1000 = 8.752 × 10⁻³ mol - Moles of unreacted HCl in 25.0 cm³ sample = moles of NaOH:
= 8.752 × 10⁻³ mol - Scale up to total 250.0 cm³ solution:
n(HCl) total remaining = 8.752 × 10⁻³ × 10 = 8.752 × 10⁻² mol - Initial moles of HCl added:
n(HCl) initial = (100 × 2.10) / 1000 = 0.210 mol - Moles of HCl reacted with M:
0.210 - 8.752 × 10⁻² = 0.12248 mol - Moles of M reacted (1:2 ratio):
n(M) = 0.12248 / 2 = 0.06124 mol - Calculate Molar Mass (Ar):
Ar = mass / moles = 6.90 g / 0.06124 mol = 112.6 g mol⁻¹ (Cadmium, Cd).
❌ Common Calculation Traps
- Including the inaccurate first rough/trial titre in the mean.
- Forgetting to multiply by 10 when scaling up from the 25.0 cm³ aliquot to the 250.0 cm³ volumetric flask.
- Forgetting to divide by 2 to account for the 2:1 stoichiometric ratio between HCl and M.
Part (b) — Carbonate Mass Loss Experiment
✅ Correct Answers
(i) Amount of CO₂:
n(CO₂) = 2.75 / 44 = 0.0625 mol
(ii) Molar Mass & Identity:
Molar mass = 14.57 / 0.0625 = 233.12 g mol⁻¹
Metal X = Rubidium / Rb (or calculated via formula breakdown).
📐 Step-by-Step Guidance (b)(ii)
- Find total mass loss (CO₂ evolved): 187.50 - 184.75 = 2.75 g .
- Calculate moles of CO₂ using Mr(CO₂) = 44.0.
- Use 1:1 mole ratio from equation to find moles of X₂CO₃ reacted (= 0.0625 mol).
- Calculate molar mass of X₂CO₃: Mass / moles = 14.57 g / 0.0625 mol = 233.1 g mol⁻¹ .
- Deduct carbonate ion mass (CO₃²⁻ = 60.0) to find mass of two X atoms: 233.1 - 60.0 = 173.1 g mol⁻¹ . Divide by 2 to get Ar of X (~86.6), corresponding to Rb.
❌ Common Errors
Using incorrect mass values from the table or failing to subtract the carbonate formula mass correctly. Note: Strontium (Sr) is rejected due to wrong carbonate formula stoichiometry.
Part (c) — Evaluation & Improvements
✅ Correct Answers
(i) Modification: Reweigh to constant mass (heat/leave until successive mass readings are identical).
(ii) Solubility Effect: CO₂ dissolves in water, so measured mass loss would be smaller than expected, leading to a calculated molar mass that is greater than the true value.
🧠 Examiner Insight
For (c)(i), responses must explicitly mention constant mass or equivalent phrasing (e.g., leaving until fizzing stops and mass is stable). For (c)(ii), trace the logic: less CO₂ escapes into the air because it dissolves in the aqueous mixture → smaller mass loss recorded → smaller moles calculated → larger molar mass ( M = mass / moles ).
Topics
Module 1: Development of practical skills in chemistry · Module 2: Foundations in chemistry · Practical Activity Groups · Module 6: Organic chemistry and analysis · PAG 2: Acid-base titration · 6.3 Analysis · 2.1 Atoms and reactions · 1.1 Practical skills assessed in a written examination
Question and mark scheme from the OCR A-Level Chemistry examination, Unified chemistry (03), November 2020. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.