RI 2022 Prelim P3 (Ans)

1 © Raffles Institution 2022 9729/03/S/22 2022 Y6 H2 Chemistry Preliminary Exam Paper 3 – Suggested Solutions Section A 1(a)(i) [CO32−] = 0.010  20 1000 20 + 20 1000 = 5.00  10–3 mol dm–3 As maximum amount of AgCl have been precipitated without precipitating Ag2CO3, the solution is saturated with Ag2CO3. Hence, ionic product = Ksp(Ag2CO3) [Ag+]2(5.00  10–3) = 8.1  10–12 [Ag+] = 4.025  10–5 mol dm–3 As the solution is also saturated with AgCl, [Ag+][Cl–]remaining = (4.025  10–5)[Cl–]remaining = 2.0  10–10 [Cl–]remaining = 4.97  10–6 mol dm–3 Comments: • Instead of using the concentration of CO32−, s everal students incorrectly substituted the amount/number of moles of CO32− into the Ksp expression. • A handful of students did not consider the dilution of CO32− upon the mixing of the two solution, KCl(aq) and Na2CO3(aq). • Some students attempted to calculate the solubility of Ag2CO3 in water, without realising that there was already CO32− present before the addition of AgNO3. 1(a)(ii) Initial [Cl–] in mixture = 0.010  20 1000 20 + 20 1000 = 5.00  10–3 mol dm–3 Percentage of Cl– precipitated = (5.00  10–3) – (4.97  10–6) 5.00  10–3  100% = 99.9% This is an effective method. Comments: • Several students erroneously subtracted the concentration of Cl– remaining from the amount of Cl– present initially.

2 © Raffles Institution 2022 9729/03/S/22 1(b)(i) Co-ordination number is the number of nearest neighbouring ions that surrounds an ion of opposite charge. Comments: • Students are advised to read the question carefully . The question requires students to define the term co-ordination number when used to refer to a crystal lattice. For the crystal lattices given in this question, the bonds formed are ionic in nature (rather than dative covalent, which are present in trans ition metal complexes). • Weaker responses referred to the number of ionic bonds that an ion forms with other oppositely charged ions. Students should note that ionic bonds are non - directional (i.e. ionic bonds can also be formed between oppositely charged ions that are not in close vicinity). 1(b)(ii) As the Ba2+ ion is larger than the Ca 2+ ion, the Ba2+ ion is able to accommodate a greater number of chloride ions surrounding it. Comments: • Several students have the misconception that the larger Ba2+ ion is less sterically hindered. • A handful of students wrongly concluded that the lower charge density of Ba 2+ allowed for a larger coordination number. 1(b)(iii) The ions in CaCl2 are less tightly packed in the lattice structure. Hence, less energy is required to overcome the weaker electrostatic forces of attraction between Ca 2+ and Cl– ions. OR Due to its smaller size/radius, Ca2+ has a higher charge density and polarising power than Ba2+. As such, CaCl2 has a greater covalent character than BaCl2, resulting in an unexpectedly lower m