NJC 2023 Volumetric Analysis Redox Student

National Junior College SH1 Chemistry 1 Content • Reacting volumes (of solutions) • Redox processes; electron transfer and/or of changes in oxidation number (oxidation state) Learning Outcomes required for H2 (9729) and H1 (8873) Chemistry: [The term relative formula mass or Mr will be used for ionic compounds] Candidates should be able to (a) Write and /or construct balanced equations; (b) Perform calculations, including use of the mole concept, involving volumes and concentrations of solutions; [when performing calculations, candidates’ answers should reflect the no. of significant figures given or asked for in the question] (c) Deduce stoichiometric relationships from calculations such as those in (b). (d) Describe and explain redox processes in terms of electron transfer and/or of changes in oxidation number (oxidation state). (e) Construct redox equations using the relevant half-equations. Volumetric Analysis & Redox All Rights Reserved. No part of this publication may be reproduced or transmitted in any form or by any means, electronic or mechanical, including photocopy, recording or any other information storage and retrieval system, without prior permission in writing from the copyright owner. Copyright © 2023 National Junior College

National Junior College SH1 Chemistry 2 1. VOLUMETRIC ANALYSIS (TITRATION) Note: Standard solution is a solution of known concentration. The standard solution may also be placed in the conical flask, and the reactant of unknown concentration in the burette. Success Criteria: • Understand what the following terms mean: analyte; standard solution; titrant; end-point; equivalence point Volumetric analysis (or titrimetric analysis) is carried out to obtain quantitative information about chemical reactions, such as the concentration of a solution and/or establish the stoichiometry of a reaction. General overview of the titration method: 1. Pipette a portion of the solution of unknown concentration (called the analyte) into a conical flask. 2. Add a few drops of a suitable indicator into the conical flask. (Some titrations do not require any indicator) 3. Titrate the solu tion by adding a standard solution from a burette slowly, with constant swirling until a first permanent colour change is noted (this is the end-point). End-point (experimental): The point during titration at which a sudden sharp colour change (of indicator) is observed. Equivalence point (theoretical): The po int during titration at which stoichiometric amounts of the two reactants have reacted completely with one another, i.e., none of the reactants is in excess. For a titration result to be accurate, the end-point must be close to the equivalence point. E.g. Titration between dilute H2SO4 and NaOH solution: 2NaOH + H2SO4 ⎯→ Na2SO4 + 2H2O Note: Notation for amount of H2SO4 is nH2SO4

National Junior College SH1 Chemistry 3 Common types of titration include: - Acid-base titration - Back titration - Redox titration 2. ACID-BASE TITRATION Success Criteria: • Able to state the colour changes (and end -point colours) of common indicators; namely methyl orange & phenolphthalein • Able to write/construct balanced equations • Able to perform calculations, using mole concept involving volumes and concentrations of solutions; including scaling up when dilution is performed. Acid-Base titration usually involve s the use of an indicator to determine the end-point. An indicator consists of either a weak acid or base and is added in a small amount to the solution in the conical flask at the start of an acid -base titration. The indicator causes the colour of the solution in the conical flask to change at the end-point depending on the pH of solution. This table shows the colour changes of some common indicators used in acid- base titrations. Different indicators are chosen for specific titrations, so that their end-point corresponds to the equivalence point of the titration. Colour Indicator In base (or alkali) At end-point In acid thymol blue blue yellow red bromothymol blue blue green yellow methyl orange yellow orange red screened methyl orange green grey purple phenolphthalein (no longer used in practical) pink pale pink or colourless* colourless thymolphthalein blue pale blue or colourless* colourless *depending on the original colour of indicator in conical flask Worked Example 1 (Acid-Base Titration) In a titration experiment, 25.0 cm 3 of dilute H 2SO4 reacts with 18.00 cm 3 of 0.100 mol dm-3 NaOH(aq). Calculate the concentration of H2SO4 in (a) mol dm−3, and (b) g dm−3. (a) Amount of NaOH used in titration = 0.100 × 18.00 1000 = 1.800 × 10–3 mol (4 s.f.) H2SO4(aq) + 2NaOH(aq) ⎯→ Na2SO4(aq) + 2H2O(l) Amount of H2SO4 in 25.0 cm3 = 1 2 × 1.800 × 10–3 = 9.000 × 10–4 mol [H2SO4] = 9.000 × 10−4 (25.0 1000⁄ ) = 3.60 × 10−2 mol dm−3 (3 s.f.) (b) conc in g dm−3 = 3.60 × 10−2 × (2 × 1.0 + 32.1 + 4 × 16.0) = 3.53 g dm−3

National Junior College SH1 Chemistry 4 Note: Remember to scale up to find the no. of moles in the original solution Worked Example 2 (Acid-Base Titration) 0.982 g of an impure sample of solid sodium hydroxide was dissolved in deionised water and made up to 250 cm 3 in a graduated flask. 25.0 cm 3 of this solution was neutralised by 23.50 cm3 of 0.100 mol dm−3 dilute hydrochloric acid. Calculate the percentage purity of sodium hydroxide in the sample. NaOH(aq) + HCl(aq) ⎯→ NaCl(aq) + H2O(l) nHCl in 23.50 cm3 = 0.100 × 23.50 1000 = 2.350 × 10−3 mol (4.s.f.) nNaOH in 25.0 cm3 = 2.350 × 10−3 mol nNaOH in 250 cm3 = 2.350 × 10−3 × 250 25 = 2.350 × 10−2 mol Molar mass of NaOH = 40.0 g mol–1 Mass of pure NaOH = 2.350 × 10−2 × 40.0 = 0.940 g percentage purity of NaOH in sample = 0.940 0.982 × 100% = 95.7% Pipette 25.0 cm3 out for titration 25.0 cm3 of impure NaOH(aq) 23.50 cm3 of 0.100 mol dm−3 HCl 0.982 g of impure NaOH(s) Made up to 250 cm3

National Junior College SH1 Chemistry 5 Checkpoint 1 (Acid-Base Titration) 1) Calculate the volume of 0.12 mol dm −3 KOH required to react with 25.0 cm 3 of H 3PO4, containing 4.90g of H3PO4 per dm3 solution. Equation for the reaction: 2KOH(aq) + H3PO4(aq) ⎯→ K2HPO4(aq) + 2H2O(l) [20.8 cm3] [H3PO4] = 4.90 3.0 + 31.0 + (4)(16.0) = 0.0500 mol dm–3 Amount of H3PO4 = 25.0 1000 × 0.0500 = 1.25 × 10−3 mol Amount of KOH = 2 × 1.25 × 10−3 = 2.50 × 10−3 mol Vol of aqueous KOH required = 2.50 × 10-3 0.12 = 0.0208 dm3 = 20.8 cm3 2) 3.60 g of an impure sample of solid potassium hydroxide was dissolved in deionised water and made up to 250 cm3 in a graduated flask. 25.0 cm3 of this solution was neutralised by 21.70 cm3 of 0.100 mol dm−3 dilute sulfuric acid. Calculate the percentage purity of potassium hydroxide in the sample. [67.6%] 2KOH(aq) + H2SO4(aq) ⎯→ K2SO4(aq) + 2