2020 HCI Prelim P2 Answers

2020 HCI C2 H2 Chemistry Preliminary Exam / Paper 2 HWA CHONG INSTITUTION 2020 C2 H2 CHEMISTRY PRELIMINARY EXAMINATION SUGGESTED SOLUTIONS Paper 2 1 (a) (i) HCl: does not decompose even on strong heating HBr: gives brown fumes of Br2(g) on strong heating HI: gives large amounts of violet fumes of I2(g) when a red-hot rod is plunged into a jar of HI gas Observation of brown & violet fumes for HBr and HI: 1m Description of trend in ease/rate/extent of decomposition: 1m 1 (a) (ii) The thermal stabilities of these halides decrease in the order HC l > HBr > H I as the H-X bond strength/bond energy decreases in the same order , and less energy is needed to break the H-X bond. [1] 1 (b) (i) HCl(aq) + NaOH(aq)  NaCl(aq) + H2O(l) [1] (Minus 0.5m for wrong or no state symbols, minus 0.5m if acid/base not written as a compound) 1 (b) (ii) HCl(aq) + NH3(aq)  NH4+(aq) + Cl(aq) [1] (Minus 0.5m for wrong or no state symbols, minus 0.5m if acid/base not written as a compound) Acceptable alternatives: HCl(g) + NH3(g)  NH4Cl(s) HCl(aq) + H2O(l)  H3O+(aq) + Cl (aq) HCl(aq) + NaOH(aq)  NaCl(aq) + H2O(l) – provided this answer was not given in (b)(i) 1 (c) (i) From the pH, [H+] = 102.10 = 7.94 × 103 mol dm3 [1] Since [H+] < [HF], HF does not dissociate fully to give H + ions and is a weak acid. [1] OR If HF is a strong acid, pH of a 0.100 mol dm3 HF = lg 0.100 = 1.00 [1] Since the actual pH of HF is higher than if it were a strong acid, HF does not dissociate fully to give H+ ions and is a weak acid. [1] OR

1 Degree of dissociation,  = (102.10)  0.100 = 0.0794 [1] Since the degree of dissociation is less than 1, HF does not dissociate fully to give H+ ions and is a weak acid. [1] 1 (c) (ii) Ka = (102.10)2  (0.100 – 102.10) = 6.85 × 104 mol dm3 [1] 1 (c) (iii) Both titration curves would show a decrease in pH as the acid is added. OR Both titration curves would show a sharp drop in pH when 12.50 cm3 of the acid is added. [1] 1 (c) (iv) 1. The pH at equivalence point is more than 7 when HF is used but equal to 7 when HCl is used [], as F is able to hydrolyse in water, but not Cl[]. 2. After equivalence point, the pH remains relatively constant/curve is more gentle or flatter when HF is used, while the pH drops more when HC l is used []. This is due to formation of a buffer solution consisting of F and excess HF after equivalence point but no buffer is formed for Cl and HCl []. 3. When 50 cm3 of the acid is added, pH of solution is lower when HCl is used [] as HCl is a stronger acid which dissociates to give higher concentration of H+ than HF of the same concentration []. 0.5m for each description [] and each corresponding explanation [] 2 (a) 26Fe: 1s22s22p63s23p63d64s2 26Fe2+: 1s22s22p63s23p63d6 26Fe3+: 1s22s22p63s23p63d5 Small jump from 6 th to 7th IE, indicating 7 th electr