OCR A-Level Chemistry AS Breadth in chemistry (01), November 2020: Question 23
11 marks · Medium difficulty · Structured Questions
Calculate concentrations of barium hydroxide and hydroxide ions from mass and titration data, and outline preparation routes for barium hydroxide from barium metal.
Practise this questionQuestion
Question text
23 This question is about barium hydroxide.
(a) Barium hydroxide is an alkali which releases hydroxide ions, OH−, in aqueous solution.
A barium hydroxide solution contains 3.89 g of Ba(OH) in 100 cm3 at 20 °C.
Calculate the concentration of hydroxide ions, OH−, in mol dm−3, of this solution at 20 °C.
Give your answer to 3 significant figures.
concentration of OH− ions = … mol dm−3 [3]
(b) A student carries out a titration to determine the concentration of an aqueous solution of
Ba(OH)2.
The student adds 25.0 cm3 of the Ba(OH) (aq) solution to a conical flask.
The student titrates this solution by adding 0.160 mol dm−3 HNO (aq) from the burette.
The equation is shown below.
Ba(OH)2(aq) + 2HNO3(aq) → Ba(NO3)2(aq) + 2H2O(l)
The student repeats the titration until concordant titres are obtained.
The mean titre of 0.160 mol dm−3 HNO (aq) is 26.75 cm3.
(i) What is meant by concordant titres?
… [1]
(ii) Calculate the concentration, in mol dm−3, of the Ba(OH) (aq) solution.
concentration of Ba(OH) (aq) = … mol dm−3 [3]
(c) A student plans to prepare a solution of Ba(OH)2 from barium by two different reaction routes.
Outline 2 reaction routes for preparing a solution of Ba(OH)2 from barium in the laboratory.
Include relevant equations.
… [4]
Mark scheme
Show the mark scheme
AO
Question Answer Marks Guidance
element
23 (a) FIRST CHECK ANSWER ON THE ANSWER LINE 3 ALLOW ECF throughout
If answer = 0.454 (mol dm–3) award 3 marks
If answer = 0.227 (mol dm–3) award first 2 marks
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ALLOW use of 171 within working
(Use of Ar: Ba 137 rather than 137.3)
n(Ba(OH) ) in 100 cm3 1 mark
3.89 Calculator: 0.02270869819
= = 0.0227…. (mol)
171.3 3.1×2 IGNORE figures after 7 in 0.0227
3 SF or more
ALLOW working with ×10 before ×2
Concentration of OH– 2 marks
n(Ba(OH)2) ×2 = 2 × 0.0227 Use of ×10 = 10 × 0.0227
= 0.0454…. (mol) = 0.227…. (mol)
Use of ×10 = 10 × 0.0454 Use of ×2 = 2 × 0.227
Concentration of OH– = 0.454 (mol dm–3) 3.2 Concentration of OH– = 0.454 (mol dm–3)
3 SF required 3 SF required
Common error
0.227 no × 2 2 marks
(b) (i) (Titres that agree) within 0.1 cm3 1 2.3 ALLOW within 0.05 cm3
ALLOW ml for cm3
If cm3 units are absent, ASSUME cm3
BUT
DO NOT ALLOW incorrect units,
e.g. dm3; mol dm–3
AO
Question Answer 9 Marks Guidance
element
(b) (ii) FIRST CHECK ANSWER ON THE ANSWER LINE 3 Use ECF throughout
If answer = 0.0856 (mol dm–3) award 3 marks
----------------------------------------------------------------------
26.75 DO NOT ALLOW 4.3 × 10–3
n(HNO3) = 0.160 × = 4.28 × 10–3 (mol)
1000 2.8×2 BUT remaining marks available by ECF
e.g.
4.28 × 10–3 4.3 × 10–3 ÷ 2 = 2.15 × 10–3 ECF
n(Ba(OH)2) in 25.0 cm3 =
= 2.14 × 10–3 (mol) 1000
2.15 × 10–3 × = 0.086 ECF
–3 1000
Concentration = 2.14 × 10 ×
= 0.0856 (mol dm–3) 2.4
(c) Route 1 4
Reactant:
Add water (to Ba) OR H2O in equation 3.3 ALLOW multiples in equations
Balanced equation: Balanced equation automatically collects 2
Ba + 2H2O → Ba(OH)2 + H2 2.6 marks for Route 1
Route 2
Balanced equation with O2 ALLOW 1 mark for BOTH reactants in route 2:
2Ba + O2 → 2BaO 3.3 i.e. React with O2 AND then with H2O
Balanced equation with H2O NOTE
BaO + H2O → Ba(OH)2 3.3 3 correct balanced equations → 4 marks
How to answer it
Barium Hydroxide Calculations and Preparation Study Guide
This question assesses core quantitative chemistry and inorganic reactions: calculating moles and solution concentrations, understanding stoichiometry in acid-base titrations, defining analytical terminology (concordant titres), and writing balanced synthetic pathways involving Group 2 metals (barium).
Calculation of Hydroxide Ion Concentration
Calculate the concentration of OH⁻ in mol dm⁻³ from 3.89 g of Ba(OH)₂ in 100 cm³ at 20 °C (3 sig fig).
✅ Correct Answer
0.454 mol dm⁻³
📐 Step-by-Step Calculation
- Find molar mass of Ba(OH)₂: 137.3 + (2 × 17.0) = 171.3 g mol⁻¹
- Calculate moles in 100 cm³: 3.89 / 171.3 = 0.022708 mol
- Scale to 1 dm³ (1000 cm³): 0.022708 × 10 = 0.22708 mol dm⁻³ (concentration of Ba(OH)₂ solution)
- Account for OH⁻ stoichiometry: Ba(OH)₂ → Ba²⁺ + 2OH⁻. Multiply by 2: 0.22708 × 2 = 0.45416 mol dm⁻³
- Round to 3 significant figures: 0.454 mol dm⁻³
❌ Common Errors
- Forgetting the 2:1 ratio: Students often calculate the concentration of the barium hydroxide solution ( 0.227 ) and forget that each mole of Ba(OH)₂ releases two moles of OH⁻ ions.
- Volume scaling errors: Dividing by 100 instead of multiplying by 10 when converting from 100 cm³ to 1 dm³.
Definition of Concordant Titres
What is meant by concordant titres?
✅ Correct Answer
Titres that agree closely with each other, specifically within 0.1 cm³ (or 0.05 cm³).
💡 Key Knowledge
In titrations, concordant titres ensure that random errors are minimised and that consistent, reliable data is used to calculate a mean titre. Titres outside this range (such as rough titres) are excluded.
Titration Calculation
Calculate the concentration of Ba(OH)₂(aq) using a mean titre of 26.75 cm³ of 0.160 mol dm⁻³ HNO₃(aq) added to 25.0 cm³ of Ba(OH)₂.
✅ Correct Answer
0.0856 mol dm⁻³
📐 Step-by-Step Calculation
- Moles of HNO₃ added: (26.75 / 1000) × 0.160 = 4.28 × 10⁻³ mol
- Moles of Ba(OH)₂ reacting: Using the 1:2 stoichiometric ratio from Ba(OH)₂ + 2HNO₃ → Ba(NO₃)₂ + 2H₂O , divide by 2: (4.28 × 10⁻³) / 2 = 2.14 × 10⁻³ mol
- Concentration of Ba(OH)₂: Scale from 25.0 cm³ to 1 dm³: (2.14 × 10⁻³ / 25.0) × 1000 = 0.0856 mol dm⁻³
🧠 Exam Technique & ECF
Error Carried Forward (ECF) applies! If you misread the stoichiometry ratio in step 2 (e.g., forgot to divide by 2), subsequent calculations following your incorrect mole value can still pick up partial credit.
Synthesis Routes for Barium Hydroxide
Outline 2 reaction routes for preparing a solution of Ba(OH)₂ from barium in the laboratory, including equations. (4 marks)
✅ Correct Answer & Mark Breakdown
Route 1 (Direct hydration): React barium directly with water.
Ba + 2H₂O → Ba(OH)₂ + H₂
(2 marks: 1 for correct reactant/water inclusion, 1 for balanced equation)
Route 2 (Two-step oxidation/hydration): React barium with oxygen to form barium oxide, then add water.
2Ba + O₂ → 2BaO
BaO + H₂O → Ba(OH)₂
(2 marks: 1 for each balanced equation in the sequence)
💡 Key Knowledge
Group 2 metals react increasingly vigorously down the group with water to form metal hydroxides and hydrogen gas. Alternatively, burning the metal in oxygen produces the metal oxide, which acts as a basic oxide and dissolves/reacts with water to form the hydroxide alkaline solution.
Topics
Module 2: Foundations in chemistry · Module 3: Periodic table and energy · Practical Activity Groups · PAG 2: Acid-base titration · 2.1 Atoms and reactions · 3.1 The periodic table
Question and mark scheme from the OCR A-Level Chemistry examination, AS Breadth in chemistry (01), November 2020. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.