NJC 2025 H2 Chemistry Prelim P3 Ans

[Turn over NATIONAL JUNIOR COLLEGE SH2 PRELIMINARY EXAMINATION Higher 2 CANDIDATE NAME SUBJECT CLASS REGISTRATION NUMBER CHEMISTRY Paper 3 Free Response Candidates answer on Question Paper. Additional Materials: Data Booklet 9729/03 23 September 2025 2 hours READ THE INSTRUCTIONS FIRST Write your subject class, registration number and name on all the work you hand in. Write in dark blue or black pen. You may use a soft pencil for any diagrams or graphs. Do not use staples, paper clips, glue or correction fluid. Section A Answer all questions. Section B Answer one question. A Data Booklet is provided. The use of an approved scientific calculator is expected, where appropriate. The number of marks is given in brackets [ ] at the end of each question or part question. For Examiner’s Use Section A 1 /19 2 /20 3 /21 Section B (*circle the question you attempted) 4 /20 5 /20 Paper 3 Total /80 This document consists of 28 printed pages.

2 NJC/H2 Chem Prelim/03/2025 Section A Answer all the questions in this section. 1 (a) A student investigated the thermal decomposition of Group 2 compounds. He heated a mixture of 2.50 g of magnesium nitrate and magnesium carbonate using the setup shown in Fig. 1.1, till no further change was observed. A colourless gas of a volume of 80.8 c m3 was collected at 30 C and atmospheric pressure. Fig 1.1 (i) Write balanced equations for the thermal decomposition of MgCO3 and Mg(NO3)2. [1] MgCO3  MgO + CO2 Mg(NO3)2  MgO + 2 NO2 + ½ O2 (ii) Explain the purpose of sodium hydroxide in the above setup and hence identify the gas collected in the gas syringe. [2] NaOH(aq) reacts with the acidic gases (CO2 and NO2). The gas collected in the gas syringe is O2. (iii) Calculate the percentage by mass of magnesium nitrate present in the mixture. [3] Mg(NO3)2  MgO + 2 NO2 + ½ O2 pV = nRT Amount of O2 gas collected = 101325 × 80.8 × 10−6 8.31×(30+273) = 0.003252 mol Amount of Mg(NO3)2 = 2  0.003252 = 0.006504 mol Mass of Mg(NO3)2 = 0.006504  (24.3+142+166) = 0.9644 g % by mass of Mg(NO3)2 in mixture = 0.9644 2.50  100% = 38.6 %

3 NJC/H2 Chem Prelim/03/2025 [Turn over (b) The student carried out another experiment by heating equal amounts of carbonates of magnesium, calcium and barium for two minutes using a Bunsen burner. Table 1.1 shows the volume of gas collected. Table 1.1 Group 2 carbonate MgCO3 CaCO3 BaCO3 Volume of gas collected / cm3 80 25 5 Using relevant data from the Data Booklet, explain the results obtained by the student. [3] The ionic radius of Mg2+, Ca2+ and Ba2+ are 0.065 nm, 0.099 nm and 0.135 nm respectively. Down Group 2, the ionic charge remains the same while the ionic radius increases, hence charge/size ratio decreases down the group. Mg2+ has the greatest polarising power, thus it can distort the electron cloud of CO 32 to a larger extent. The C O covalent bond in MgCO3 is weakened more significantly and less amount of energy is required to decompose MgCO3. As a result, MgCO3 decomposes more readily than CaCO3 and BaCO3 as it produces the largest amount of CO2 in the same time period. (c) Table 1.2 shows the bond length of various nitrogen-oxygen bonds. Table 1.2 Bond N O N=O nitrogen-oxygen bond in NO3 Bond Length (nm) 0.136 0.115 0.128 Suggest an explanation for the observed NO bond length in nitrate ion. [2] In NO3 , the presence of continuous overlap of p orbitals across the 3 oxygen atoms and the nitrogen atom allows the  electrons and the lone pair of electrons on oxygen to be delocalised. As a result, all the three NO bonds in NO 3 are an intermediate bond between an N O bond and N=O bond (partial double bond) or bond order is between 1 and 2

4 NJC/H2 Chem Prelim/03/2025 (d) Sodium carbonate, a Group 1 carbonate, can be used to maintain the pH of swimming pool water to the ideal range of 7.0−7.6. If the concentration of sodium carbonate is too high, it will cause skin irritation to swimmers. The management committee of a public swimming pool hired a chemist to advise them on the need to adjust the pH of pool water. The chemist titrated a 10.0 cm 3 sample of pool water (assume it contains only Na2CO3) against 0.05 mol dm−3 HCl. 20.00 cm3 of the HCl solution was required to turn the phenolphthalein indicator from pink to colourless. When the titration is repeated using methyl orange indicator, 40.00 cm 3 of HCl was required to reach the end point. [The Kb1 and Kb2 values of Na2CO3 at 25 C are 2.13 × 10−4 and 2.25 × 10−8 respectively.] (i) The pH at the first equivalence point is found to be greater than 7. Write an equation to explain this observed pH. [1] HCO3 + H2O ⇌ H2CO3 + OH Note: must be reversible arrow for partial dissociation of weak base. (ii) Calculate the concentration of Na2CO3 in the pool water sample. [1] Working range of phenolphthalein is pH 8 10. When phenolphthalein changes colour, it indicates the first reaction between CO32 and H+ is completed. CO32 + H+  HCO3 Amount of HCl used = 20 1000  0.05 = 0.001 mol Amount of CO32 reacted = 0.001 mol Concentration of Na2CO3 in 10.0 cm3 sample = 0.001 ÷ 10 1000= 0.1 mol dm 3 OR Working range of methyl orange is pH 3 5. When methyl orange changes colour, it indicates the complete reaction between CO32 and two H+. CO32 + 2H+  CO2 + H2O Amount of HCl used = 40 1000  0.05 = 0.002 mol Amount of CO32 reacted = 0.001 mol Concentration of Na2CO3 in 10.0 cm3 sample = 0.001 ÷ 10 1000= 0.1 mol dm 3 (iii) Using your answer to (ii), calculate the initial pH of the pool water sample. [2] CO32 + H2O HCO3 + OH [OH ] = √𝐾𝑏 × [𝐶𝑂3 2−] [OH ] = √2.13 × 10−4 × 0.1 [OH ] = 0.004615 mol dm 3 pOH = 2.336 pH = 14 pOH = 11.66 (2 d.p.) allow ecf from (ii)

5 NJC/H2 Chem Prelim/03/2025 [Turn over (iv) Sketch the pH volume added curve you would expect to obtain when 50.00 cm 3 of the HCl solution is added to 10.0 cm 3 of the pool water sample. Label the various key points on the curve. [2] First maximum buffering capacity occurs when there is 1:1 of CO32 and HCO3 pOH = pKb1 pH = 14 – pOH = 14 –[–lg(2.13 × 10−4)] = 10.33 (2 d.p.) Second maximum buffering capacity occurs when there is 1:1 of HCO3 and H2CO3 pOH = pKb2 pH = 14 – pOH = 14 –[–lg(2.25 × 10−8)] = 6.35 (2 d.p.) Shape of pH curve ● Correct shape with 2 equivalence points ● sharp change in pH at the equivalence points ● pH is approximately constant near the point of maximum buffering capacity ● Final pH would approach low pH Label ● Axis with units ● initial pH ● pH and volume for maximum buffer regions ● pH >7 at the first equivalence point (e) Lithium carbonate is a sparingly soluble salt with a Ksp value of 8.15 × 10 4. Calculate the solubility of lithium carbonate. [2] Let the solubility of Li2CO3 be s Li2CO3(s) 2Li+(aq) + CO32–(aq) I /mol dm−3 0 0 C /mol dm−3 −s +2s +s E /mol dm–3 2s s

6 NJC/H2 Chem Prelim/03/2025 Ksp of Li2CO3 = [Li+]2[CO32–] = (2s)2(s) 8.15 × 10–4 = 4s3 s = 5.88 × 10−2 mol dm 3 [Total : 19] 2 Cobalt is a transition element that can form coloured ions of various oxidation states in aqueous solutions. (a) (i) Explain what is meant by the term transition element. [1] Transition elements are d−block elements which form one or more stable ions with a partially filled d subshell. (ii) Explain why aqueous solution of Co3+ ions is coloured. [3] In the presen