NJC Prelims Paper 3_Answers Final

1 NJC/H2 Chem Preliminary Examination/03/2021 [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 27 August 2021 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 /20 2 /16 3 /24 Section B 4 /20 5 /20 Paper 3 Total /80 This document consists of 24 printed pages.

2 NJC/H2 Chem Preliminary Examination/03/2021 Section A Answer all the questions in this section. 1 (a) Nicotine (C10H14N2) is a drug present in tobacco. In aqueous solution, nicotine ionises as shown. C10H14N2 + H2O C10H15N2+ + OH− pKb (C10H14N2) = 6.0 (i) Calculate the pH of a 0.100 mol dm−3 C10H14N2 solution. [2] C10H14N2 is a weak base, [OH−] = √10ି଺×0.100 = 3.162 × 10−4 mol dm−3 pOH = −lg[OH−] = 3.50 pH = 14 – pOH = 10.5 (ii) Suggest a suitable indicator for the titration of aqueous nicotine with HNO3(aq). Explain your answer. [2] Methyl orange [1]. The equivalence point pH would be less than 7 [1/2], the working range of methyl orange (pH 3-5) would overlap with the region of rapid pH change at equivalence point. [1/2] Note: The salt, C10H15N2+, (with Ka value) is the conjugate acid of weak base C10H14N2 (Kb). Hence the salt is acidic and has a pH of < 7 at equivalence point. (iii) Calculate the pH of a 5.00 dm3 solution consisting of 0.100 mol dm‒3 C10H14N2 and 0.200 mol dm‒3 C10H15N2+. [1] The mixture contains a weak base and its conjugate acid, it is a buffer solution. pOH = pKb + lg[௖௢௡௝௨௚௔௧௘ ௔௖௜ௗ] [௪௘௔௞ ௕௔௦௘] = 6.0 + lg଴.ଶ ଴.ଵ = 6.30 [1/2] pH = 14−6.30 = 7.70 [1/2]

3 NJC/H2 Chem Preliminary Examination/03/2021 [Turn over (iv) Calculate the number of moles of HNO3 that needs to be added to the solution in (a)(iii) to obtain a buffer solution of pH 7.40. [2] Weak base C10H14N2 of the buffer will react with the HNO3 added, use ICF table (in mol) to determine limiting reagent. Initial amount of C10H14N2 = 5 × 0.100 = 0.5 mol Initial amount of C10H15N2+ = 5 × 0.200 = 1.0 mol Let amount of HNO3 added = x mol