WJEC A-Level Chemistry 1410QS
Formulae and Equations
Basic Ideas About Atoms
Chemical Calculations
Bonding
Solid Structures
The Periodic Table
Simple Equilibria and Acid-Base Reactions
Thermochemistry
Rates of Reaction
The Wider Impact of Chemistry
Organic Compounds
Hydrocarbons
Halogenoalkanes
Alcohols and Carboxylic Acids
Instrumental Analysis
Redox and Standard Electrode Potential
Redox Reactions
Chemistry of the P-Block
Chemistry of the D-Block Transition Metals
Chemical Kinetics
Enthalpy Changes for Solids and Solutions
Entropy and Feasibility of Reactions
Equilibrium Constants
Acid-Base Equilibria
Stereoisomerism
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Alcohols and Phenols
Aldehydes and Ketones
Carboxylic Acids and Their Derivatives
Amines
Amino Acids, Peptides and Proteins
Organic Synthesis and Analysis
Practical Skills and Techniques
Planning, Analysis and Evaluation
Core Experiments and Their Chemistry
Mathematical Skills in Chemistry
Redox Reactions
Carrying out and calculating a redox titration: redox titration calculations
- 3 things to know
- A2 Unit 3
Overview
Sixteen recall questions on Carrying out and calculating a redox titration: redox titration calculations.
What you need to know
Answer each one before you reveal it.
Calculate the volume of 0.0200 mol dm⁻³ potassium manganate(VII) needed to react with 25.0 cm³ of 0.0500 mol dm⁻³ sodium ethanedioate in acid solution.
Calculate the volume of 0.100 mol dm⁻³ iron(II) sulfate solution that would react with 25.0 cm³ of 0.0250 mol dm⁻³ potassium dichromate(VI) in acid solution.
Questions from this subtopic
Each one with the model answer underneath.
- 1
Explain how each piece of apparatus should be rinsed before a titration.
Model answer
The burette is rinsed with the solution it will hold and the pipette with the solution it will deliver, so that no residual water dilutes them. The conical flask is rinsed with distilled water only, because extra water there changes the volume but not the number of moles present.
- 2
25.0 cm³ of an iron(II) solution requires 22.40 cm³ of 0.0200 mol dm⁻³ potassium manganate(VII). Calculate the concentration of the iron(II) solution.
Model answer
n(MnO₄⁻) = 0.02240 dm³ × 0.0200 mol dm⁻³ = 4.480 × 10⁻⁴ mol. The equation MnO₄⁻ + 5Fe²⁺ + 8H⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O gives a 1 : 5 ratio, so n(Fe²⁺) = 5 × 4.480 × 10⁻⁴ = 2.240 × 10⁻³ mol. Concentration = 2.240 × 10⁻³ / 0.0250 dm³ = 0.0896 mol dm⁻³.
- 3
Five iron tablets are dissolved in dilute sulfuric acid and the solution made up to 250 cm³. A 25.0 cm³ portion requires 12.50 cm³ of 0.0100 mol dm⁻³ potassium manganate(VII). Calculate the mass of iron in one tablet. Use A(Fe) = 55.8.
Model answer
n(MnO₄⁻) = 0.01250 dm³ × 0.0100 mol dm⁻³ = 1.250 × 10⁻⁴ mol. n(Fe²⁺) in the portion = 5 × 1.250 × 10⁻⁴ = 6.250 × 10⁻⁴ mol. In the whole 250 cm³, n(Fe²⁺) = 10 × 6.250 × 10⁻⁴ = 6.250 × 10⁻³ mol. Mass of iron = 6.250 × 10⁻³ × 55.8 = 0.34875 g for five tablets, so each tablet contains 0.34875 / 5 = 0.0698 g, that is 69.8 mg.
- 4
1.00 g of iron wire is dissolved in excess dilute sulfuric acid and the solution made up to 250 cm³. A 25.0 cm³ portion needs 17.20 cm³ of 0.0200 mol dm⁻³ potassium manganate(VII). Calculate the percentage of iron in the wire. Use A(Fe) = 55.8.
Model answer
n(MnO₄⁻) = 0.01720 dm³ × 0.0200 mol dm⁻³ = 3.440 × 10⁻⁴ mol. n(Fe²⁺) in the portion = 5 × 3.440 × 10⁻⁴ = 1.720 × 10⁻³ mol. In 250 cm³, n(Fe²⁺) = 1.720 × 10⁻² mol. Mass of iron = 1.720 × 10⁻² × 55.8 = 0.960 g. Percentage = (0.960 / 1.00) × 100 = 96.0%.
- 5
25.0 cm³ of a diluted hydrogen peroxide solution requires 21.00 cm³ of 0.0200 mol dm⁻³ potassium manganate(VII) in acid. Calculate the concentration of the hydrogen peroxide.
Model answer
n(MnO₄⁻) = 0.02100 dm³ × 0.0200 mol dm⁻³ = 4.200 × 10⁻⁴ mol. The equation 2MnO₄⁻ + 5H₂O₂ + 6H⁺ → 2Mn²⁺ + 5O₂ + 8H₂O gives a 2 : 5 ratio, so n(H₂O₂) = (5/2) × 4.200 × 10⁻⁴ = 1.050 × 10⁻³ mol. Concentration = 1.050 × 10⁻³ / 0.0250 dm³ = 0.0420 mol dm⁻³.
- 6
Calculate the volume of 0.0200 mol dm⁻³ potassium manganate(VII) needed to react with 25.0 cm³ of 0.0500 mol dm⁻³ sodium ethanedioate in acid solution.
Model answer
n(C₂O₄²⁻) = 0.0250 dm³ × 0.0500 mol dm⁻³ = 1.25 × 10⁻³ mol. The ratio of manganate(VII) to ethanedioate is 2 : 5, so n(MnO₄⁻) = (2/5) × 1.25 × 10⁻³ = 5.00 × 10⁻⁴ mol. Volume = 5.00 × 10⁻⁴ / 0.0200 = 0.0250 dm³, which is 25.0 cm³.
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