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  mas