ACJC H1 CHEM P2 ans

2 © ACJC 2017 JC2 H1 Preliminary Exam 2017 Section A Answer all questions in this section in the spaces provided. 1 Aluminium is the most abundant Group 13 element and constitutes about 8% of the Earth’s crust. The extraction of aluminium is done by processing aluminium ore, bauxite to produce aluminium oxide also known as alumina. A variety of aluminium compounds, for example aluminium chloride and aluminium hydroxide, are used for different purposes such as food additives, colouring and pharmaceuticals. Aluminium hydroxide and magnesium hydroxide are antacids. They are used to treat symptoms of increased stomach acid, such as heartburn, upset stomach, sour stomach, or acid indigestion. Once ingested, they react with the hydrochloric acid in the stomach. One label of a commercial product, Mintox™ is shown below. (a) (i) Write down the electronic configuration of Al. [1] 1s22s22p63s23p1 (ii) Why is the ionic radius of aluminium far smaller than its atomic radius? Al 3+ ion has a higher proton to electron ratio, hence the remaining electrons experience stronger electrostatic forces of attraction to the nucleus. In addition, Al 3+ ion has one less principal quantum shell as compared to the neutral atom, hence the ionic radius is smaller than its atomic radius. (iii) Explain why aluminium forms compounds with an oxidation state of +3 but not sodium. [1] As compared to sodium, aluminium would require smaller amount of energy to remove the (2 nd and 3rd) electrons which are from the outermost principle quantum shell.

3 © ACJC 2017 JC2 H1 Preliminary Exam 2017 (b) (i) Which antacid in the tablet is more effective in reacting with the hydrochloric acid in the stomach? Show relevant working to support your answer. [2] Relevant working based on masses in the tablet No of moles of Mg(OH) 2 = 200 x 10-3 / 58.3 = 3.43 x 10-3 mol No of moles of Al(OH)3 = 200 x 10-3 / 78 = 2.56 x 10-3 mol Al(OH)3 will produces more number of moles of hydroxide ions (7.68 x10- 3) than Mg(OH)2(6.86 x10-3) . Thus aluminium hydroxide is more effective. (ii) Calculate the maximum number of chewable tablets that a person can take in a week. [1] 4x 4 x 7 = 112 (iii) Assuming that a typical adult has a body mass of 70 kg, determine the maximum weekly intake of aluminium hydroxide in grams per kg of body mass. [2] Maximum intake of aluminium hydroxide is 200 mg x 112 Maximum intake per body mass = 0.320 g per kg (c) (i) Aluminium chloride is an active ingredient used in skin medication to control excessive sweating. Aluminium chloride is often describe as electron deficient. Explain what is meant by electron deficient. [1] Electron deficient implies that the central atom, Al has empty orbital in AlCl3 (ii) In the vapour phase, aluminium chloride forms a gaseous product with a molar mass of 267 g mol -1. With an aid of a clearly labelled diagram, explain how this product is formed from aluminium chloride. [2] Dative Bond OR [Total: 11]

4 © ACJC 2017 JC2 H1 Preliminary Exam 2017 2 In a university laboratory, the percentage purity of a sample of complex iron salt, K3Fe(C2O4)3.3H2O can be determined by analyzing the C 2O42- content through titrating with acidified KMnO4. 1.20 g of impure K 3Fe(C2O4)3.3H2O sample was dissolved and made up to 100cm 3. 10.0 cm3 of this solution was pipetted into a conical flask and 10.0cm 3 of 1 mol dm -3 sulfuric acid was added. The mixture was heated and titrated with 0.0200 mol dm -3 KMnO4. CO2 is produced during the reaction. It was determined that 12.30 cm3 of KMnO4 was required to reach the end-point. (a) (i) Suggest why hydrochloric acid is not used to acidify the mixture. [2] Cl- ions can possibly be oxidized to Cl2 and hence will cause an increase in the titration readings. (ii) In the acidic medium, C2O42- ions exist as H2C2O4. Write a half equation to show the conversion of H2C2O4 to CO2. [1] H2C2O4  CO2 + 2H+ + 2e (b) (i) Calculate the amount of KMnO4 used to react with 10.0cm3 of the iron complex salt solution. [1] Amt of MnO4- = 12.3/1000 x 0.02 = 2.46 x 10-4 mol (ii) Using the half-equation below and that in (b)(ii), calculate the amount of C2O42- present in 10.0 cm3 of the iron complex salt solution. MnO4- + 8H+ + 5e → Mn2+ + 4H2O [1] Amount of e involved = 5 x 2.46 x 10-4= 1.23 x 10-3 Amount of H2C2O4 = amount of C2O42- = 5/2 x 2.46 x 10-4 = 6.15 x 10-4 (iii) Hence, determine the mass of K3Fe(C2O4)3.3H2O in 100 cm3 of iron complex salt solution. (molar mass of K3Fe(C2O4)3.3H2O = 491.1 g mol-1) [2] K3Fe(C2O4)3.3H2O ≡ 3C2O42- amount of K3Fe(C2O4)3.3H2O in 100cm3 = 5/2/3 x 2.46 x 10-4 x 10 Mass of K3Fe(C2O4)3.3H2O = 5/2/3 x 2.46 x 10-4 x 10 x 491.1 = 1.01g (iv) Calculate the percentage purity of the iron complex salt. [1] 1.01/1.20 x 100% = 84.2% [Total: 8]

5 © ACJC 2017 JC2 H1 Preliminary Exam 2017 3 The emergence of multidrug-resistant bacteria has encouraged vigorous efforts to develop antibacterial agents. N-methylhydroxylamine has been found to show vast potential as an antibacterial agent. N-methylhydroxylamine (pKb = 8.04) N-methylhydroxylamine has properties similar to ammonia and it dissolves in water as shown below: CH 3NHOH(aq) + H2O(l) CH3NH2OH+(aq) + OH-(aq) (a) Write the expression for the base dissociation constant of N- methylhydroxylamine in water. [1] K b = NHOH][CH ]][OHOHNH[CH 3 23  (b) Calculate the base dissociation constant of the N-methylhydroxylamine solution. [1] K b = 10-8.04 = 9.12 x 10-9 mol dm-3 An aqueous solution of 0.05 mol dm -3 hydrochloric acid was gradually added to 50.0 cm3 of 0.02 mol dm-3 aqueous N-methylhydroxylamine. (c) Determine the initial pH of N-methylhydroxylamine solution. [2] CH3NHOH(aq) + H2O(l) CH3NH2OH+(aq) + OH-(aq) K b = NHOH][CH ]][OHOHNH[CH 3 23  Since [OH -] = [CH3NH2OH+], K b = NHOH][CH ][OH 3 2 9.12 x 10 -9 = 0.02 ][OH 2 [OH-] = 1.35  10-5 mol dm-3 pH = 14 – [-log (1.48  10-5)] = 9.13 (d) Calculate the volume of hydrochloric acid needed at the equivalence point. [1] CH3NHOH + HCl  CH3NH2OH +Cl- Volume of HC l needed = (50.0  0.02)  0.05 = 20.0 cm 3

6 © ACJC 2017 JC2 H1 Preliminary Exam 2017 (e) State the volume of hydrochloric acid required to be added to another identical solution of N-methylhydroxylamine to obtain a solution which best resists pH change. [1] 10.0 cm 3 (f) Calculate the pH of that solution. [1] pOH = pKb = 8.04 pH = 14-8.04 = 5.96 (g) Write two equations to show how the solution in (e) resists change in pH when small amounts of acid and alkali are added. [2] CH3NHOH + H+  CH3NH2OH+ CH 3NH2OH+ + OH-  CH3NHOH + H2O [Total: 9] 4 (a) Draw the structures of the organic product(s) formed when compound A below reacts with each of the following reagents. CH H C CH3 C CH3 H compound A Reagents and Conditions Organic Product(s) formed (