2019 DHS Prelim H2 Chem P3 ANS

© DHS 2019 Name: Answers Index Number: Class: DUNMAN HIGH SCHOOL Preliminary Examination 2019 Year 6 H2 CHEMISTRY 9729/03 Paper 3 Free Response Additional Materials: Data Booklet Answer Booklet INSTRUCTIONS TO CANDIDATES 1 Write your name, index number and class on this cover page. 2 Write your answers on the separate answer booklet provided. 3 Write in dark blue or black pen. 4 You may use an HB pencil for any diagrams or graphs. 5 Start each question on a fresh sheet of paper. *[Marks will be deducted if you fail to do so.] 6 Do not use staples, paper clips, glue or correction fluid. Section A 7 Answer all questions Section B 8 Answer one question. The number of marks is given in brackets [ ] at the end of each question or part question. You are advised to show all workings in calculations. You are reminded of the need for good English and clear presentation in your answers.

2 © DHS 2019 9729/03 Section A 1 To facilitate better plant growth, there is widespread use of fertilisers and soil additives in the agricultural sector. (a) Agricultural lime is a soil additive used to increase soil pH so as to facilitate uptake of plant nutrients such as nitrogen and phosphorous. It is usually made up of a combination of calcium carbonate and magnesium carbonate. A manufacturer claims that a 2.00 g sample of agricultural lime contains 92 % calcium carbonate (Mr = 100.1) and 8 % magnesium carbonate (Mr = 84.3) by mass. To verify the manufacturer’s claim, the following steps were carried out: Step 1 The solid carbonate mixture was dissolved in 250 cm 3 of 0.25 mol dm–3 hydrochloric acid. Step 2 A 25 cm 3 aliquot of this resultant solution was titrated with 0.14 mol dm–3 potassium hydroxide. It was found that 12.50 cm 3 of potassium hydroxide was required for complete neutralisation. (i) Based on the manufacturer’s claim of 92% calcium carbonate and 8% magnesium carbonate by mass, calculate the amount, in moles, of each carbonate present in the sample of agricultural lime. [1] Mass of CaCO3 present = 92 % × 2.00 g = 1.84 g Mass of MgCO3 present = 8 % × 2.00 g = 0.16 g Moles of CaCO3 present = 1.84 g ÷ 100.1 g mol–1 = 0.018382 = 0.0184 mol Moles of MgCO3 present = 0.16 g ÷ 84.3 g mol–1 = 0.0018980 = 0.00190 mol (ii) Use the information above to calculate the amount of HC l that reacted with the carbonate mixture in Step 1. [2] Moles of KOH reacted with excess HCl = 12.50 1000 × 0.14 = 0.00175 mol Moles of HCl present in 25 cm3 aliquot = 0.00175 mol Moles of excess HCl present in 250 cm3 solution = 0.00175 × 250 25 = 0.0175 mol Total moles of HCl used = 250 1000 × 0.25 = 0.0625 mol Moles of HCl reacted with carbonates = 0.0625 – 0.0175 = 0.045 mol (iii) The manufacturer’s claim is considered valid if the difference between the actual and theoretical total amount of carbonate present in the sample is less than 0.0010 mol. Using your answers in (a)(i) and (a)(ii), determine if the manufacturer’s claim is valid.

3 © DHS 2019 9729/03 [2] 2H+ + CO32– → CO2 + H2O Total moles of carbonates present = 0.045 ÷ 2 = 0.0225 mol Total moles of carbonates based on manufacturer’s claim = 0.0184 + 0.00190 = 0.0203 mol Since the difference between the calculated moles and theoretical moles of carbonates is more than 0.0010 mol, the manufacturer’s claim is false. (iii) When solid magnesium chloride is dissolved in water, it produces a slightly acidic solution of pH 6.5. Using appropriate equation(s), explain the above observation. [2] MgCl2 + 6H2O → [Mg(H2O)6]2+ + 2Cl– [Mg(H2O)6]2+ + H2O [Mg(H2O)5(OH)]+ + H3O+ [1] MgCl2 dissolves in water and Mg2+ undergoes partial hydrolysis in water to produce H3O+, giving a slightly acidic solution. (iv) However, the resultant solution is neutral when solid calcium chloride is dissolved in water. By quoting relevant values from the Data Booklet, explain why a solution of calcium chloride is neutral while a solution of magnesium chloride is slightly acidic. [3] Cationic radius of Mg2+ = 0.065 mn Cationic radius of Ca2+ = 0.099 mn Since the cationic radius of Ca2+ is larger than that of Mg2+, Ca2+ has a lower charge density. As such, it polarises the O –H bond of the water molecule to a lesser extent and no H + ions are produced. OR Ca 2+ does not hydrolyse in water to produce H+ ions. Hence, the solution is neutral. (b) Nitrogen fertilisers are important in enhancing the growth of plants. Almost all nitrogen fertilisers are produced from ammonia, which in turn, is manufactured in the Haber Process. This process combines nitrogen and hydrogen gas to form ammonia according to the following equilibrium. N2(g) + 3H2(g) 2NH3(g) (i) State how the percentage yield of ammonia will change with increasing pressure. [1]

4 © DHS 2019 9729/03 Percentage yield of ammonia will increase with increasing pressure. (ii) N2 and H2 was introduced into an enclosed vessel at 300 C. The initial partial pressures of N 2 and H 2 were 2 atm and 3 atm respectively. After an equilibrium has been established, the total pressure in the vessel was 4.7 atm. Calculate the value of Kp at 300 C, stating its units. [3] N2(g) + 3H2(g) 2NH3(g) Initial partial pressure / atm 2 3 0 Change / atm – x – 3x +2x Eqm partial pressure / atm 2 – x 3 – 3x 2x Total pressure = 4.7 atm Hence, 2 – x + 3 – 3x + 2x = 4.7 x = 0.15 atm N2(g) + 3H2(g) 2NH3(g) Eqm partial pressure / atm 1.85 2.55 0.30 Kp = 𝑃𝑁𝐻3 2 𝑃𝑁2[𝐻2]3 = (0.30)2 (1.85)(2.55)3 = 0.0029339 = 0.00293 atm–2 (iii) 1 atm of N2 was added to the equilibrium mixture in (b)(ii) at 300 C and time was allowed for a new equilibrium to be established. State and explain how the Kp value at this point will compare to that in (b)(ii). [1] The Kp value will remain constant / be the same as Kp is only dependent on temperature. [Total: 15] % yield of NH3 pressure

5 © DHS 2019 9729/03 2 Mandelic acid, C6H5CH(OH)COOH, is a white crystalline solid that is soluble in water. (a) Benzaldehyde is a useful starting material for the preparation of mandelic acid. Suggest a two –step synthesis of mandelic acid from benzaldehyde, stating clearly the reagents and conditions used and the structure of the intermediate. [3] CNOHO HCN, trace NaCN cold COOHOH HCl(aq) heat under reflux (b) Suggest how you would distinguish between aqueous solutions of mandelic acid and nitric acid of the same concentration by means of a physical method. Explain your reasoning. [3] Measure the pH of the aqueous solutions using pH paper / meter. The solution of a lower pH is that of nitric acid. Nitric acid is a strong acid while mandelic acid is a weak acid . Nitric acid dissociates to a greater extent in water to produce a higher concentration of H+ ions which results in a lower pH for its aqueous solution. (c) (i) 20 cm3 of 1.0 mol dm3 mandelic acid solution was titrated with 2.0 mol dm3 aqueous sodium hydroxide. Calculate the concentration of the salt, sodium mandelate, formed at equivalence point. [1] Let mandelic acid be HA. HA  A moles of A = 0.020  1.0 = 0.0200 mol total volume at equivalence point = 20 + 10 = 30 cm3 [NaA] = [A] = 0.0200 / 0.030 = 0.667 mol dm3 (ii) Hence calculate the pH of the solution at equivalence point, given p Ka of mandelic acid is 3.41. [2] At equivalence point, salt hydrolysis occurs: OH O O + OH2 OH OH O + OH Kb (A) = 1014 / 103.41 = 2.5704  1011 mol dm3 [OH] = √(2.5704  1011)(0.66667) = 4.1396  106 mol dm3 pH = 1