MJC H2 Chem 2012 Prelim P2 Soln

Meridian JC 2012 H2 Chemistry Paper 2 Suggested Answers 1. (ai) 1. Using a measuring cylinder, measure 100 cm3 of water and place it in a 250 cm3 beaker. 2. Add solid KClO3 to water and stir with a glass rod to dissolve the solids. 3. Add more solid KClO3 to the water until excess solids remains / no more solid dissolves. 4. Allow the mixture to stand for a period of time (eg. 30 min) to establish equilibrium. 5. Filter the solution to mixture to obtain the saturated solution as a filtrate. (ii) 1. Weigh an empty evaporating basin using a mass balance. 2. Using a measuring cylinder, transfer 20 cm 3 of the saturated solution from (b)(i) into the weighed evaporating basin. (Any volume from 20 – 50 cm3) 3. Weigh the evaporating basin and saturated solution. 4. Evaporate the water from the solution by placing the evaporating basin on top of a beaker of boiling water. 5. Cool and reweigh the evaporating basin. 6. Repeat the heating, cooling and weighing until a constant mass is obtained. (iii) Mass of evaporating basin = m1 g Mass of evaporating basin and 20 cm3 saturated solution = m2 g Final mass of solid KClO3 and evaporating basin = m3 g Mass of water in saturated KClO3 solution = m2 – m3 g Mass of solid KClO3 in saturated solution = m3 – m1 g Solubility of KClO3 = m 3 -m 1 ×100 m2 -m3 (iv) KClO3 solid is corrosive or caustic. Wear gloves when handling the reagent. copyright chem dept @ meridian jc 1

copyright chem dept @ meridian jc 2 (b) Temperature / oC (i) 52.0 oC (ii) above the transition temperature, The solubility of NaBr increases gradually with temperature. below the transition temperature. The gradient of the graph is steeper, hence the solubility of NaBr.2H 2O increases more rapidly with temperature. Dissolving sodium bromide under equilibrium conditions is endothermic. 10 20 30 40 50 60 70 80 90 100 70 80 90 0 100 110 120 130 60 x x x x x x x x x x ‘Solubility’ of sodium bromide / g per 100g H2O

copyright chem dept @ meridian jc 3 2 (a) Elimination (b) (i) Electrophilic addition CH2=C=O CH3C=O + :OH- CH3COOH H- OH (ii) (iii) (iv) sp 3 (a) A B C D E d+ d- slow + O O O CH3 O N H C C O O C C C N H H H H H CH3 CH3 C CH3O H C H3 CH3 C H2 O C H3 CH3 Cl CH3 CH3 C Cl CH3 O H OHO CH3 O C C H3 C H3 R S

copyright chem dept @ meridian jc 4 (b) (i) Add concentrated sulfuric acid to sodium chloride. (ii) I- is a stronger reducing agent than Cl- Most of the H I can be oxidised by concentrated H 2SO4 to form violet fumes of I2 gas. NaI (s) + H2SO4 (l) ¾¾® HI (g) + NaHSO4 (s) 8 HI (g) + H2SO4 (l) ¾¾® 4 I2 (g) + H2S (g) + 4 H2O (l) (iii) P (s) + 3/2 I2 (s) ¾¾® PI3 (s) or P4 (s) + 6 I2 (s) ¾¾® 4 PI3 (s) P I3 (s) + 3 H2O (l) ¾¾® H3PO3 (aq) + 3 HI (g) (c) Benzene in G is more resonance stabilized than alkene in B (d) (i) (ii) 4 (a) (i) O O O O H H CH3 CH2CH2 C H3 CH3 C H2 CH3 C H2 CH2 CHH2N C O O O

copyright chem dept @ meridian jc 5 (ii) + - (iii) Aspartic acid and t yrosine formed negative ions and they migrate towards the positive terminal. Aspartic acid is nearer to the positive terminal than tyrosine as the aspartic acid has a negative charge (-2) and a lower molecular mass than tyrosine. (b) (i) Aspartame exist as zwitterions. A large amount of energy needed to break the ionic bonds between the oppositely charged ions (ii) 1 acidic hydrolysis 2 HC C O OH NH3 + H2CHOOC CH3OH Asp Tyr Phe Lys After electric current applied H3N+ CH C CH2 O OH

copyright chem dept @ meridian jc 6 (c) Step Reagent(s) and conditions I Aqueous Cl2 II (Excess) conc. H2SO4 170oC III KMnO4 , dilute H2SO4, heat 5 (a) The electron withdrawing hydroxyl grp decreases the electron density on the lone pair of N atom. Hence, the lone pair on N atom is less available to accept a proton Hence, Tris is less basic than tert-butylamine. (b) (HOCH 2)3CNH2 + H+ (HOCH2)3CNH3+ (HOCH 2)3CNH3+ + OH- (HOCH2)3CNH2 + H2O (c) (i) No. of mole of (unreacted) HCl in 105 cm 3 = 4.99 × 10-3 [H+] = 4.99 × 10-3/0.105 = 0.04752 mol dm-3 pH = 1.32 (ii) pOH = 5.93 + lg (0.025/𝑉𝑉) (0.015/𝑉𝑉) where V = total volume = 6.15 pH = 7.85 (d) Solution Y has a larger buffering capacity and will much better resist changes in pH upon the addition of a strong acid or base as the concentrations of its buffer components are much larger than that of solution X. X Y Cl OH NH2

copyright chem dept @ meridian jc 7 6 (a) (i) (ii) E cell = +1.36 – (+0.77)= +0.59 V (b) E qFe3+/Fe2+ is more positive than E qRu3+/Ru2+, the relative stability of + 2 oxidation state compared to +3 oxidation state is larger for iron than ruthenium. (c) (i) Complex A [Ru(NH3)6]3+ Compound B [Ru[(NH3)6]2+.2Cl- or Ru[(NH3)6].2Cl or Ru(NH3)6Cl2 (ii) E cell = -0.56 – (-0.76) = +0.20V Since Fe(OH)3 can be reduced by Zn, the compound C is Fe(OH)2 (d) Fe3+ + 6F- [FeF6]3-Or [Fe(H2O)6]3+ + 6F- [FeF6]3- + 6H2O (e) (f) (i) K c of [Fe(edta)]- is larger than Kc of [Fe(SCN)(H2O)5]2+. Edta is a stronger ligand than H 2O and SCN- and the deep red [Fe(SCN)(H 2O)5]2+ solution changes to yellow [Fe(edta)]- (ii) Addition of edta results in forming [Fe(edta)]2- and [Fe(edta)]-. Since Kc of [Fe(edta)]- is greater K c of [Fe(edta)]2-, edta ligand stabilises Fe(III) relative to Fe( II) to a larger extent than water ligand. Hence Eo value is less than +0.77V/decreases. (g) [Fe(CN)6]3- has a partially -filled d orbitals; undergoes d- orbital splitting . During the transition, the d electron absorbs the blue wavelength of light from the visible region of the electromagnetic spectrum and emits the remaining wavelengths which appear as the orange colour of [Fe(CN)6]3- observed. Pt(s) [Cl - (aq)] = 1 mol dm-3 [Fe3+ (aq)] = 1 mol dm-3 [Fe2+ (aq)] = 1 mol dm-3 Pt (s) Cl2 (g) at 298K and 1 atm Pt (s) ο [F-]6 [FeF6]3- [[FeF6]3-] [F-]6 [[Fe(H2O)6]3+]