NJC 2023 Solutions to Volumetric Analysis Redox Tutorial

SH1 H2 Chemistry 2023 Solutions to Volumetric Analysis & Redox Tutorial (A) Discussion Questions (Volumetric Analysis) Acid-Base Titration 1. 45 cm3 of concentrated aqueous NH 3 was diluted to 250 cm 3 solution labelled as FA 1. Given that 10.0 cm3 of FA 1 required 23.20 cm3 of 0.18 mol dm-3 of HNO3 for complete neutralisation, calculate the concentration of the concentrated aqueous NH3. [2.32 mol dm−3] Eqn for rxn: NH3 + HNO3 → NH4NO3 3NHη in 10.0 cm3 FA 1 = 3HNOη used = 0.18 × 23.20 1000 = 4.176 × 10–3 mol 3NHη in 250 cm3 of FA 1 = 4.176 × 10–3 × 250 10 = 0.1044 mol [NH3]conc soln = 0.1044 / ( 45 1000 ) = 2.32 mol dm−3 2. A solution of a dibasic acid contains 7.30 g dm −3 of HOOC –(CH2)n–COOH. 20.0 cm 3 of this acid solution was titrated with 25.00 cm3 of NaOH(aq) containing 1.36 g of hydroxide ion per dm3. Calculate (a) the relative molecular mass of the acid; [146.0] Eqn for rxn: HO2C–(CH2)n–CO2H + 2NaOH → NaO2C–(CH2)n–CO2Na + 2H2O [OH−] = 1.36 17.0 = 0.08 mol dm−3 Amount of NaOH reacted = 25 1000 × 0.08 = 2.00 × 10–3 mol Amount of dibasic acid reacted = ½ × 2.00 × 10–3 = 1.00 × 10–3 mol [dibasic acid] = 1.00 ×10-3 20 1000⁄ = 0.0500 mol dm–3 Mr of dibasic acid = 7.30 0.0500 = 146.0 (b) the value of n in the formula. [4] Mr of dibasic acid = 2(12.0 + 16.0 × 2 + 1.0) + n(12.0 + 2.0) = 146.0 n = 4.07 = 4 (to nearest integer)

SH1 H2 Chemistry 2023 3. FA 8 solution contains 20.2 g of the acid HZO4 per dm3 of solution. FA 9 is a 0.100 mol dm -3 NaOH solution. In a titration, 20.0 cm 3 of FA 8 solution reacted with 21.05 cm3 of FA 9 solution. Calculate the relative atomic mass of element Z and identify Z. [126.9; I] NaOH  HZO4 (since HZO4 is monobasic with 1 H+ per molecule) Amount of NaOH = 0.100 × 21.05 1000 = 2.105 × 10–3 mol Amount of HZO4 in 20.0 cm3 FA 8 = 2.105 × 10–3 mol [HZO4] = 2.105×10-3 20 1000⁄ = 0.1053 mol dm–3 Mr of HZO4 = 20.2 0.1053 = 191.9 Ar of Z = 191.9 – 1.0 – 4 × 16.0 = 126.9 Z is Iodine. 4. Washing soda has the formula Na 2CO3∙nH2O. A mass of 1.43 g of washing soda was made up to 250 cm3 with water. 25.0 cm 3 of this solution was neutralised by 20.00 cm 3 of 0.050 mol dm −3 dilute hydrochloric acid. The equation for the reaction is: Na2CO3(aq) + 2HCl (aq) → 2NaCl(aq) + CO2(g) + H2O(l) Find the value of n, and hence the chemical formula of the washing soda. [The equation for the dissolution of washing soda in water is: Na2CO3∙nH2O(s) → Na2CO3(aq) + nH2O] [10] Amount of HCl = 0.050 × 20 1000 = 1.00 × 10–3 mol Amount of Na2CO3 in 25.0 cm3 solution = ½ × 1.00 × 10–3 = 5.00 × 10–4 mol Amount of Na2CO3 in 250 cm3 solution = 5.00 × 10–4 × 250 25 = 5.00 × 10–3 mol Mass of Na2CO3 in 250 cm3 solution = 5.00 × 10–3 × (2×23.0 + 12.0 + 3×16.0) = 0.530 g  mass of H2O in 1.43 g washing soda = 1.43 – 0.530 = 0.900 g  amount of H2O in 1.43 g washing soda = 0.900 18.0 = 0.0500 mol 32 2 CONa OH η η = 0.0500 5.00 x 𝟏𝟎−𝟑 = 10 10 mol of water combine with 1 mol of Na2CO3 n = 10; formula of washing soda is Na2CO3∙10H2O.

SH1 H2 Chemistry 2023 Back Titration 5. 3.92 g of an oxide of formula MO was completely dissolved in 30.0 cm3 of 2.00 mol dm−3 sulfuric acid. The resulting solution was made up to 100 cm 3. 25.0 cm 3 of this solution was neutralised by 27.50 cm3 of 0.10 mol dm−3 NaOH(aq). What is the relative atomic mass of M? Identify the metal. [55.9; Fe] Reaction 1: MO + H2SO4 → MSO4 + H2O Reaction 2: 2NaOH + H2SO4 → Na2SO4 + 2H2O Total starting amount of H2SO4 (S) = 2.00 × 30 1000 = 0.0600 mol For Reaction 2, Amount of NaOH reacted = 0.10 × 27.50 1000 = 2.75 × 10–3 mol Amount of H2SO4 in the 25 cm3 solution = 2.75 × 10–3 × 1 2 = 1.375 × 10–3 mol Amount of H2SO4 in the 100 cm3 solution (E) = 1.375 × 10–3 × 100 25 = 5.50 × 10–3 mol For back titration, R = S − E Amount of H2SO4 that reacted with MO (R) = 0.0600 – 5.50 × 10–3 = 0.0545 mol Amount of MO = 0.0545 mol Mr of MO = 0.0545 3.920 = 71.9 Ar of M = 71.9 – 16.0 = 55.9 M is Fe (Ar of 55.8). + 3.92 g MO(s) H2SO4(aq) 30.0 cm3 2.00 mol dm−3 Some unreacted H2SO4(aq) remained → → H2SO4(aq) diluted to 100 cm3 27.50 cm3 of 0.10 mol dm-3 NaOH(aq) → 25.0 cm3 diluted H2SO4(aq) titrate reaction → dilute pipette →

SH1 H2 Chemistry 2023 6. 5.00 g of ammonium chloride contaminated with sodium chloride was boiled with 100.0 cm 3 of 2 mol dm −3 NaOH solution until no ammonia was evolved. The residual solution was made up to 250 cm 3 with water. 25.0 cm 3 of this solution required 22.40 cm 3 of 0.5 0 mol dm −3 HCl for neutralisation. (a) What was the mass of sodium chloride in the ammonium chloride sample? [0.292 g] Reaction 1: NH4Cl + NaOH → NH3 + NaCl + H2O Reaction 2: NaOH + HCl → NaCl + H2O For reaction 2, amount of HCl = 22.40 1000 × 0.5 = 0.0112 mol Amount of unreacted NaOH in 25 cm3 = 0.0112 mol Amount of unreacted NaOH in 250 cm3 (E) = 0.0112 × 250 25 = 0.112 mol Initial amount of NaOH added (S) = 100 1000 × 2.0 = 0.2 mol For back titration, R = S − E Amount of NaOH reacted with NH4Cl (R) = 0.2 – 0.112 = 0.088 mol Mass of NH4Cl = 0.088 × 53.5 = 4.708 g Mass of NaCl = 5.00 – 4.708 = 0.292 g (b) Hence, calculate the percentage by mass of sodium chloride in the ammonium chloride sample. [5.84%] Percentage by mass of NaCl in the sample = 0.292 5.00 × 100% = 5.84% + 5.00 g mixture of NH4Cl(s) + NaCl (s) NaOH(aq) 100.0 cm3 2.00 mol dm−3 Some unreacted NaOH(aq) remained → reaction → NaOH(aq) diluted to 250 cm3 22.40 cm3 of 0.50 mol dm-3 HCl(aq) 25.0 cm3 diluted NaOH(aq) titrate → dilute pipette →

SH1 H2 Chemistry 2023 Redox Titration 7. FA 4 contains 10.0 g of Fe2+ and Fe3+, dissolved in 500 cm3 of solution. FA 5 contains 0.015 mol dm−3 KMnO4. In an experiment, 10.0 cm3 of solution of FA 4 was pipetted into a titration flask. Excess dilute sulfuric acid was added and the mixture titrated with FA 5. 26.65 cm3 of FA 5 was required to reach end-point. (a) Calculate the mass of Fe2+ in 500 cm3 of FA 4. [5.58 g] MnO4– + 8H+ + 5Fe2+ → 5Fe3+ + Mn2+ + 4H2O Amount of KMnO4 reacted = 0.015 × 26.65 1000 = 3.998 × 10–4 mol Amount of Fe2+ in 10.0 cm3 FA 4 = 5 × 3.998 × 10–4 = 1.999 × 10–3 mol Amount of Fe2+ in 500 cm3 FA 4 = 1.999 × 10–3 × 500 10 = 0.09995 mol Mass of Fe2+ in 500 cm3 FA 4 = 0.09995 × 55.8 = 5.58 g (b) Calculate the percentage by mass of Fe3+ ions in FA 4. [44.2 %] % by mass of Fe3+ in FA 4 = 10.00 - 5.58 10 × 100 = 44.2 % 8. Chlorate(V) ions, ClO3–, act as an oxidising agent according to the following half-equation: ClO3– + 6H+ + 6e– → Cl– + 3H2O (a) Chlorate(V) ions are reduced by Fe2+ ions. Write a balanced equation for the reaction between Fe2+ and ClO3−. [Hint: Write the half equation for Fe2+ oxidized to Fe3+ before combining the 2 half-equations] ClO3– + 6H+ + 6Fe2+ → Cl– + 3H2O + 6Fe3+ (b) In an experiment, 25.0 cm 3 of a solution of KClO3 was titrated with a solution of iron( II) sulfate containing 6.72 g dm −3 of Fe 2+. In the titration, 20.0 cm 3 of iron( II) sulfate was used for the reaction. (i) What other chemical is required for the titration? Suggest a reactant that can be added to the reaction mixture for the titration. acid to provide H+; dilute H2SO4 Note: HNO3 and HCl are not suitable acids as they will take part in redox reaction. HNO3 is an oxidising agent (NO3− can undergo reduction) and Cl− can undergo oxidation.

SH1 H2 Chemistry 2023 (ii) Calculate the conc