2019 VJC H2 Chem Prelim P3 Ans

1 Victoria Junior College 2019 H2 Chemistry Prelim Exam 9729/3 Suggested Answers Section A Answer all the questions in this section. 1 (a) Account for the reactions that occur when MgC l2 and PCl5 are separately dissolved in water. Predict the pH of the resulting solutions formed and write equations for the reactions that occur. [4] Mg2+ has high charge density due to small ionic radius (and high charge). Mg2+ has high polarising power and hence MgC l2 undergoes hydration and partial hydrolysis, producing a slightly acidic solution. MgCl2 + 6H2O  [Mg(H2O)6]2+ + 2Cl– hydration [Mg(H2O)6]2+ + H2O ⇌ [Mg(H2O)5OH]+ + H3O+ hydrolysis pH of resulting solution = 6.5 PCl5 undergoes complete hydrolysis due to the presence of energetically accessible vacant 3d orbitals on phosphorus which can accommodate lone pair from water molecules. The solution produced is strongly acidic. PCl5 + 4H2O  H3PO4 + 5HCl pH of resulting solution = 2 (b) A sample consists of a solid mixture of MgO and Al2O3. Describe briefly an experimental procedure that will enable you to separate the mixture and recover each of the oxides in its pure form. [3] 1. Add excess NaOH(aq) to the sample. 2. Filter the resulting mixture. The residue collected is mainly MgO. 3. Wash the residue with distilled water and press between pieces of filter paper to obtain dry solid MgO. 4. To the filtrate, add HCl(aq) dropwise till the maximum mass of precipitate is formed. 5. Filter the resulting mixture. The residue collected is mainly Al(OH)3. 6. Wash the residue with distilled water and press between pieces of filter paper to obtain dry solid Al(OH)3. 7. Heat the residue till constant mass to obtain dry solid Al2O3. [Note: Al(OH)3 is expected to decompose on heating to form Al2O3, similar to Group 2 hydroxides.] (c) The highest fluoride of xenon, XeF6, can be obtained by heating the octafluoroxenates of the Group 1 metals, M2XeF8, where M represents the Group 1 metal. M2XeF8 → 2MF + XeF6 Suggest reasons why the sodium salt (M = Na) decomposes below 100 °C, whereas the caesium salt (M = Cs) requires a temperature of 400 °C. Hence explain why MgXeF 8 is not known to exist. [3]

2 Down group 1, as ionic radius increases, charge density of cations decreases, polarising power of cations decreases, which polarises the XeF82– anion to a lesser extent. The anion becomes more thermally stable and hence higher temperature is required to decompose caesium salt compared to sodium salt. Mg2+ has higher charge density than Na+, making MgXeF8 very unstable and hence does not exist. (d) Suggest identities for the following substances A to D, writing equations where appropriate. When magnesium is heated with nitrogen under inert conditions, an ionic compound, A is produced. When water is added to A, a colourless gas B which turns damp red litmus paper blue is produced. B reacts with chlorate(I) ion, ClO– in a 2 : 1 mole ratio to form a colourless liquid C with empirical formula NH2. The reaction of C with sulfuric acid in a 1 : 1 mole ratio produces a salt D, N2H6SO4, which contains one cation and one anion per formula unit. [4] A: Mg3N2 B: NH3 C: N2H4 D: [N2H5+][HSO4–] (accept [N2H62+][SO42–]) 3Mg + N2 → Mg3N2 Mg3N2 + 3H2O → 3MgO + 2NH3 (Accept Mg3N2 + 6H2O → 3Mg(OH)2 + 2NH3) 2NH3 + ClO– → N2H4 + Cl– + H2O N2H4 + H2SO4 → [N2H5+][HSO4–] (accept [N2H62+][SO42–]) (e) Real gases do not obey the ideal gas equation exactly. Many chemists have tried to come up with gas equations that describe the behaviour of real gases. In 1873 J D van der Waals introduced an approximate gas equation that is applicable for all real gases. The van der Waals equation is P = nRT V−nb − a n2 V2 where a and b are constants which are characteristic of each gas. The other symbols carry their usual meaning and units as in the ideal gas equation. (i) Using what you have learnt about the differences between ideal and real gases, suggest what the constants a and b represent. [2] The constant a takes into account real gas particles have intermolecular forces of attraction , hence the pressure would be lower than expected if assuming ideal gas behaviour (the equation involves subtracting an2/V2) The constant b takes into account real gas particles are of finite size and occupy a significant volume compared to the volume of the container, hence the volume in which the gas particles can freely move about would be lower than expected if assuming ideal gas behaviour (the equation involves subtracting nb from V)

3 (ii) The values of the constants a and b for CO 2 are a = 0.3658 Pa m 6 mol–2 and b = 4.29 x 10–5 m3 mol–1. Use your answer in (e)(i) to suggest how the value of the constant a for xenon (Xe) will compare with CO2. Explain your answer briefly. [1] Xe will have larger value of a, since it will have stronger instantaneous dipole- induced dipole interactions and have larger particle size than CO2. (iii) Use the  ideal gas equation and  van der Waals equation to calculate the pressure exerted by 1 mol of CO 2 at a temperature of 30 °C and volume of 1 dm3. [3] Using ideal gas equation, pressure = 1 x 8.31 x (30 + 273) / (1 x 10–3) = 2.52 x 106 Pa Using van der Waals equation, pressure = [1 x 8.31 x (30 + 273)] / (1 x 10–3 – 1 x 4.29 x 10–5) – 0.3658 [12 / (1 x 10–3)2] = 2.26 x 106 Pa [Total: 20]

4 2 (a) Malonic acid, CH 2(CO2H)2 is an organic weak dibasic acid . It is a building block chemical to produce numerous valuable compounds, including the flavo ur and fragrance compound, cinnamic acid, and the pharmaceutical compound , valproate. The two pKa values of CH2(CO2H)2 are 2.83 and 5.69. (i) Define the term weak acid. [1] Weak acid partially dissociates in water to give H+ ions. (ii) Calculate the pH of 25.0 cm3 solution of 0.100 mol dm–3 CH2(CO2H)2. [1] Ka = [H+][CH2(COOH)(COO–)] [CH2(COOH)2] pH = –lg√Ka×[CH2(COOH)2] = –lg√10–2.83×0.100 = 1.92 (iii) Calculate pH of the resulting solution when 50 cm3 of 0.100 mol dm –3 NaOH was added to the solution in (a)(ii). [2] When 50 cm3 of NaOH added, complete neutralisation has taken place, product is CH2(COO–)2. [CH2(COO–)2] = 25 ÷ 75  0.100 = 0.0333 mol dm–3 pKb of (CH2COO–)2 = 14 – 5.69 = 8.31 pOH = –lg√Kb×[CH2(COO-)2] = –lg√10–8.31×0.0333 = 4.89 pH = 14 – pOH = 9.11 (iv) Using your answer s in (a)(ii) and (a)(iii), as well as the pKa values provided, sketch a graph to show how the pH of the solution changes as 50 cm 3 of 0.100 mol dm –3 NaOH is gradually added to 25.0 cm 3 of 0.100 mol dm –3 CH2(CO2H)2. Clearly indicate the corresponding volumes of NaOH in your graph. [2]

5 (b) Malonic acid can be converted to its corresponding β–diester. β–diesters are commonly us