AQA A-Level Chemistry AS Paper 1, June 2023: Question 2
19 marks · Medium difficulty · State/Explain/Numerical
Answer questions on Group 2 elements including metallic bonding in magnesium, trends in atomic radius, reaction with steam, sulfate solubility, isotopic abundance of strontium, and a back titration of magnesium hydroxide.
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Question text
02 This question is about the elements in Group 2.
02.1 Describe the structure and bonding in magnesium.
[2 marks]
02.2 State the trend in the atomic radius of the elements down Group 2 from Mg to Ba
Give a reason for this trend.
[2 marks]
Trend
Reason
02.3 Give an equation, including state symbols, for the reaction of magnesium with steam.
State two observations for this reaction.
[3 marks]
Equation
Observation 1
Observation 2
02.4 The sulfates of the elements in Group 2 from Mg to Ba have different solubilities.
State the formula of the least soluble of these sulfates.
Give a use for this sulfate.
[2 marks]
Formula
Use
02.5 A sample of strontium is made up of only three isotopes: 86Sr, 87Sr and 88Sr
*03* This sample contains 83.00% by mass of 88Sr
This sample of strontium has Ar = 87.73
Calculate the percentage abundance of each of the other two isotopes in this sample.
[4 marks]
% abundance 87Sr =
% abundance Sr =
02.6 Mg(OH)2 is used as an antacid to treat indigestion.
A student does an experiment to determine the percentage by mass of Mg(OH)2 in an
indigestion tablet.
*04* 3 –3
40.0 cm of 0.200 mol dm HCl (an excess) is added to 0.200 g of a powdered tablet.
The mixture is swirled thoroughly.
All of the Mg(OH)2 reacts with HCl as shown.
Mg(OH)2 + 2HCl → MgCl2 + 2H2O
The amount of HCl remaining after this reaction is determined by titration with
0.100 mol dm–3 NaOH
29.25 cm3 of 0.100 mol dm–3 NaOH are needed.
Calculate the percentage by mass of Mg(OH)2 in the indigestion tablet.
[6 marks]
Percentage by mass
Mark scheme
Show the mark scheme
Question Marking guidance Additional Comments/Guidelines Mark
M1 (giant) lattice of (Mg2+) cations / (giant) lattice of (Mg) atoms Incorrect structure type loses M1
02.1
2+ (2 x AO1)
M2 (Electrostatic) attractions between cations / Mg ions / nuclei and
delocalised electrons
M1 Trend: increases
02.2
M2 Reason: the number of electron energy levels increases Allow: the number of electron shells increases (2 x AO1)
Ignore increase in shielding
Mg(s) + H2O(g) → MgO(s) + H2(g) State symbols essential
Bright/white flame/light 3
02.3 (1 x AO1,
White/grey ash/powder (allow smoke) Do not allow ppt 2 x AO2)
Ignore black solid
Ignore fumes.
M1 BaSO4
02.4
(2 x AO1)
M2 X-rays (of internal organs) / barium meal
M1 Abundance of 87 Sr = X Allow M1 for
and Abundance of 86Sr = 1 – 0.83 – X Abundance of 87Sr = X and Abundance of 86Sr = Y
= 0.17 – X if also states that X + Y = 17
M2 87.73 = (88 × 0.83) + (87 × X) + (86 × (0.17 – X)) 87.73 = (88 × 0.83) + (87 × X) + (86 × Y)
02.5
(4 x AO2)
87.73 = 73.04 + 87X + 14.62 – 86X
87.73 = 87.66 + X
M3 87Sr = 0.07 = 7 %
M4 Abundance of 86Sr = 1 – 0.83 - 0.07 = 0.1 = 10 % M4 = 17 – M3
M1 Amount of HCl added = 0.200 × 0.040 = 0.00800 mol
M2 Amount of NaOH = 0.100 × 0.02925 = 0.002925 mol
(Amount of HCl = 0.002925 mol)
M3 Amount of HCl reacted with Mg(OH)2 = 0.00800 – 0.002925 = M3 = M1 – M2
0.005075 mol 6
02.6
(6 x AO2)
M4 Amount of Mg(OH)2 = 0.005075 ÷ 2 = 0.0025375 mol M4 = M3 ÷ 2
M5 Mass of Mg(OH)2 = 58.3 × 0.0025375 = 0.148 g M5 = M4 × 58.3
0.148 M5
M6 % by mass = × 100 = 74.0 % M6 = × 100
0.200 0.200
Do not allow M6 if >100%
How to answer it
Group 2 Trends, Reactions & Quantitative Titrations
This question brings together foundational physical and inorganic chemistry concepts across Group 2:
- Bonding: Describing metallic bonding accurately using standard AQA nomenclature.
- Periodic Trends: Explaining atomic radius changes down a group.
- Redox & Reactivity: Writing balanced equations with state symbols for magnesium reacting with steam, alongside visual observations.
- Solubility & Medical Uses: Recalling sulfate solubility trends and applications (barium meal).
- Isotopic Abundances: Formulating and solving simultaneous equations using relative atomic mass (Aᵣ).
- Back Titration Calculations: Multi-step quantitative analysis determining reactant moles, stoichiometry, mass, and percentage purity.
Structure and Bonding in Magnesium
Describing metallic structure and forces
✅ Correct Answer
- M1: (Giant) lattice of Mg²⁺ cations / (giant) lattice of magnesium atoms.
- M2: Electrostatic attraction between cations / Mg²⁺ ions / nuclei and delocalised electrons.
🧠 Exam Technique
Always state both the type of particle and the arrangement when describing structure:
- Structure = giant metallic lattice.
- Bonding = electrostatic force between positive metal ions and delocalised electrons.
❌ Common Errors & Misconceptions
- Stating "covalent bonds" or "intermolecular forces" instantly forfeits M1.
- Forgetting the word delocalised when referencing electrons loses M2.
- Saying "attraction between magnesium atoms and electrons" (atoms are neutral; the attraction is to cations/nuclei).
💡 Key Knowledge
Magnesium has metallic bonding: Mg atoms each donate 2 valence electrons into a delocalised "sea" of electrons, forming an array of closely packed Mg²⁺ ions held tightly by electrostatic forces.
Trend in Atomic Radius Down Group 2
Trend from Mg to Ba and underlying explanation
✅ Correct Answer
- Trend (M1): Increases.
- Reason (M2): The number of electron shells (or electron energy levels) increases.
❌ Common Errors
- Writing "shielding increases" as the primary reason. While shielding does increase, AQA explicitly requires mentioning the increase in shells / main energy levels for the mark.
- Confusing group trends with period trends (e.g. arguing that nuclear charge increases so radius decreases).
Reaction of Magnesium with Steam
Balanced equation with state symbols and two observations
✅ Correct Answer
Equation:
Mg(s) + H₂O(g) → MgO(s) + H₂(g)
Observations (any two):
- Bright white flame / white light
- White / grey solid / ash / powder (smoke allowed)
💡 Key Knowledge: Steam vs Cold Water
- With steam: Mg burns vigorously to form magnesium oxide (MgO) and hydrogen (H₂). State symbol for steam is (g) !
- With cold water: Reaction is very slow; forms magnesium hydroxide: Mg(s) + 2H₂O(l) → Mg(OH)₂(aq/s) + H₂(g) .
❌ Common Errors
- Writing H₂O(l) instead of H₂O(g) — state symbols are explicitly stated as essential.
- Producing Mg(OH)₂ instead of MgO .
- Calling the solid product a "precipitate" — precipitates only form in aqueous solution, not when a gas reacts with a solid!
- Vague observations like "fumes", "effervescence", or "black solid".
🧠 Exam Technique
Always inspect the question prompt: when it specifies "including state symbols", full equation marks depend strictly on correct states: (s) , (g) .
Sulfate Solubility & Medical Application
Identification of the least soluble Group 2 sulfate
✅ Correct Answer
- Formula: BaSO₄
- Use: Barium meal / X-ray imaging (of digestive system / gut / internal organs).
💡 Key Knowledge: Group 2 Solubility Trends
- Hydroxides [Mg(OH)₂ → Ba(OH)₂]: Solubility increases down the group. (Mg(OH)₂ is sparingly soluble; Ba(OH)₂ is soluble).
- Sulfates [MgSO₄ → BaSO₄]: Solubility decreases down the group. (MgSO₄ is soluble; BaSO₄ is insoluble).
- Even though Ba²⁺ ions are toxic, BaSO₄ can be ingested because its extreme insolubility prevents it from entering the bloodstream.
Isotope Abundance Calculation
Determining unknown percentages from relative atomic mass
📐 Step-by-Step Calculation
Total abundance = 100%
Abundance of ⁸⁸Sr = 83.00%
Remaining abundance for (⁸⁶Sr + ⁸⁷Sr) = 100 − 83.00 = 17.00%
Let % abundance of ⁸⁷Sr = x, then % abundance of ⁸⁶Sr = 17.00 − x
Aᵣ = [ (88 × 83.00) + (87 × x) + (86 × (17.00 − x)) ] / 100 = 87.73
8773 = 7304 + 87x + 1462 − 86x
8773 = 8766 + x
x = 8773 − 8766 = 7.00%
% ⁸⁶Sr = 17.00 − 7.00 = 10.00%
✅ Final Answers
% abundance ⁸⁷Sr = 7% (or 7.00%)
% abundance ⁸⁶Sr = 10% (or 10.00%)
❌ Common Errors
- Algebraic slip when expanding 86(17 − x) .
- Forgetting to divide by 100 or mixing fractional abundances with percentages.
- Swapping the final two answers around at the end.
Back Titration: Percentage Mass of Mg(OH)₂
Determining purity using excess acid and standard base titration
📐 Step-by-Step Back Titration Method
n(HCl) initial = conc × vol = 0.200 mol dm⁻³ × (40.0 / 1000) dm³ = 0.00800 mol
Reaction: HCl + NaOH → NaCl + H₂O (1 : 1 ratio)
n(NaOH) = 0.100 mol dm⁻³ × (29.25 / 1000) dm³ = 0.002925 mol
Therefore, n(HCl) unreacted = 0.002925 mol
n(HCl) reacted = 0.00800 − 0.002925 = 0.005075 mol
From equation: Mg(OH)₂ + 2HCl → MgCl₂ + 2H₂O (1 : 2 ratio)
n(Mg(OH)₂) = 0.005075 / 2 = 0.0025375 mol
Mᵣ(Mg(OH)₂) = 24.3 + (2 × 16.0) + (2 × 1.0) = 58.3 g mol⁻¹
mass = 0.0025375 mol × 58.3 g mol⁻¹ = 0.147936 g (0.148 g)
% by mass = (0.147936 g / 0.200 g) × 100 = 73.97% ≈ 74.0%
❌ Common Traps in Back Titrations
- Missing the 1 : 2 ratio: Forgetting to divide moles of HCl by 2 when converting to Mg(OH)₂ (M4).
- Wrong subtraction: Adding the titration moles instead of subtracting them from initial moles to find the reacted quantity.
- Mᵣ Errors: Calculating Mᵣ of Mg(OH)₂ without doubling both oxygen and hydrogen atoms (must be 58.3).
- Sanity Check: Any final percentage over 100% automatically scores 0 for M6.
🧠 Top Tip: The Back Titration Map
Always draw a quick mental timeline:
[Total Acid Added] − [Excess Acid Reacted with NaOH] = [Acid Reacted with Tablet]
Once you have moles of acid consumed by the tablet, apply the chemical equation's stoichiometric ratio to get tablet active ingredient moles.
Topics
Inorganic Chemistry · Physical Chemistry · 3.2.2 Group 2, The Alkaline Earth Metals · 3.1.1 Atomic Structure · 3.1.2 Amount of Substance · 3.1.3 Bonding
Question and mark scheme from the AQA A-Level Chemistry examination, AS Paper 1, June 2023. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.