JPJC Prelim P3 Ans

¤ Jurong Pioneer Junior College [Turn Over 1 (a) Deductions B is soluble in water Not MgO, Al2O3, SiO2 Aq. solution of B reacts with Na2CO3 in 2:3 ratio to form CO2(g) Aq. solution of B is acidic (9) Since the reacting ratio is 2:3, the acid formed must be tribasic ∴ B is P4O10/P4O6 [1] C is soluble in water Not MgO and Al2O3, SiO2 Aq. solution of C reacts with NH4+ to give NH3(g) Aq. solution of C is basic (9) ∴ C is Na2O [1] A reacts with both B and C A is amphoteric (9) ∴ A is Al2O3 [1] 2-39: 1m Equations for reaction with water: B: P4O10(s) + 6H2O(l) o 4H3PO4(aq) [1] C: Na2O(s) + H2O(l) o 2NaOH(aq) [1] For ([DPLQHU¶V Use (b) (i) 3Mg(NH2)2(s) ĺ Mg3N2(s) + 4NH3(g) (ii) Mg(NH2)2 is less thermally stable than Ba(NH2)2. (9) Mg2+ has the same charge but a smaller ionic radius hence a higher charge density than Ba2+. Thus Mg2+ polarises the large NH2³ anion more. (9) This weakens the N-H bond in the Mg(NH2)2 more (9) and thus a lower temperature is needed to decompose magnesium amide. 49: 2m; 2-39: 1m (iii) Rxn 1: Mg3N2(s) + 6H2O(l) ĺ 3Mg(OH)2(s) + 2NH3(aq) Rxn 2: NH3(aq) + HCl(aq) ĺ NH4+(aq) + Cl²(aq) 3 3 32 3 32 Amount of acid From rxn 2: 1 H 1 NH Amount of NH produced from reaction with air 0.00600 mol From rxn 1: 1 Mg N 2 NH Amount of Mg N formed Mass o

{ { 12.0×0.50 0.00600 mol1000 0.006000.00300 mol2 [1] 32 32 f Mg N in 1.00 g sample percentage of Mg N in 1.00 g sample ? 0.00300×100.9 0.303 g 0.303×1001.00 30.3 %[1]

2 ¤ Jurong Pioneer Junior College 9729/03/J2 PRELIMINARY EXAM/2021 1 (c) (i) Precipitate formed is AgCl. initial sp Ag at point of mixing Cl 30 Cl at point of mixing 30 5 For precipitation to take place: ionic product (AgCl) (AgCl) ionic product K

ªº ¬¼ ªº u¬¼ªº ¬¼ t -3 -3 - 0.0100×5 30+5 1.428×10 mol dm 1.428×10 [1] initial C l §·ªº¬¼¨¸ t¨¸ ©¹ ªº?t¬¼ - 3 -10 initial -7 -3 Cl ×30 1.8×1030+5 1.47×10 mol dm [1] (ii) Cream ppt is AgBr, halide present is Br³. (iii)

‡‡ ‡ ‡ ‡ ' ' ' ' ?' [1] [1] -10 4 -1 -1 4 ppt -1 -1 8.31 298 ln 1.8×10 5.56×10 J mol -55.6 kJ mol -5.56×10 =¨+ -178000 J mol -178 kJ mol ppt ppt ppt ppt Using / G GH T S H (iv)

‡ ‡' ' ' ' ? ¦ [1] [1] -1 -1 +178 kJ mol +178=-LE+ -473 + -378 -1030 kJ mol 1 ppt soln soln hyd Since 178 kJ mol , hence Since ions HH HL EH LE (v) This difference indicates that AgCl is not purely ionic / there exists covalent character in the ionic bond in AgCl. [1] The electronegative difference between Ag and Cl is so small that complete transfer of an electron from the silver to the chlorine is not possible. [1] OR Cl³ has a large anion radius allowing it to be readily polarised by Ag+ ions. [1] [Total: 20]

3 ¤ Jurong Pioneer Junior College 9729/03/J2 PRELIMINARY EXAM/2021 [Turn Over 2 (a) (i) Indicator for first end point: cresol red Indicator for second end point: methyl yellow [1] – for both indicators pKa of indicator matches the pH change at equivalence point [1] OR Equivalence point pH lies within working range of the indicator [1] For ([DPLQHU¶V Use (ii) Since CO32² the stronger base, it will react with acid first. Rxn at first end point: CO32²(aq) + H+(aq) ĺ HCO3²(aq) 2 3 23 3 23 amount of H Since 1 CO 1 H amount of CO in 20.0 cm solution Na CO

{ ªº ¬¼ -3 8.40×0.200 0.00168 mol1000 0.00168 mol 10000.00168× ×106 8.90 g dm20.0 [1] [1] Rxn at second end point: HCO3²(aq) + H+(aq) ĺ CO2 + H2O 2 33 3 3 Vol. of acid reacting with HCO formed from CO 8.40 cm Vol. of acid reacting with HCO originally in solution amount of acid reacting amou

3 -4 18.80-8.40 -8.40 2.00 cm 2.00×0.2001000 4.00×10 mol [1] 3 3 nt of HCO originally in solution HCO originally in solution

ªº? ¬¼ -4 -3 10004.00×10 × ×84.020.0 1.68 g dm [1] OR 3 3 2 33 Vol. of acid used to react with total amount of HCO Total amount of acid used Total amount of HCO present At first end point, 1 CO 1 HCO Amount o

{ 318.80-8.40 10.40 cm 10.40×0.2001000 0.00208 mol 2 33 3 3 f HCO formed from CO 0.00168 mol original amount of HCO HCO originally in solution

ªº ¬¼ -4 -4 -3 0.00208-0.00168 4.00×10 mol 10004.00×10 × ×84.020.0 1.68 g dm [1] [1]

4 ¤ Jurong Pioneer Junior College 9729/03/J2 PRELIMINARY EXAM/2021 2 (b) (i) It is more difficult/energy required to remove a positively charged H+ ion from anion than a neutral molecule due to greater electrostatic attraction. [1] OR Doubly charged anion is more unstable than a singly charged anion. [1] (ii) A higher pKa1 value for succinic acid implies it is the weaker acid than malonic acid, indicating the anion of succinic acid is less stable/anion of tartaric acid is more stable (9)1. Any one of the following reasons: 1. Two electron-withdrawing ²OH groups that helps to disperse the negative charge on O² in the monoanion of tartaric acid (9)2, making it more stable. There is electron-donating alkyl group that intensifies the negative charge on O² in the monoanion of succinic acid, (9)3 making it less stable. 2. Monoanion of tartaric acid can form (more extensive) intramolecular hydrogen bonding forming 5² or 6²membered rings. (9)2 Monoanion of succinic acid cannot form (have less extensive) intramolecular hydrogen bond as it forms an unstable 7-membered ring. (9)3 O- O C OH H H O H OH O Monoanion formed from tartaric acid O- O C C O H H O H H H Monoanion formed from succinic acid structural formulae of mono-anions 39: 2m; 29: 1m (iii) HO2CCH(OH)CH(OH)CO2² + H+ ĺ HO2CCH(OH)CH(OH)CO2H [1] HO2CCH(OH)CH(OH)CO2² + OH² ĺ ²O2CCH(OH)CH(OH)CO2² + H2O [1] Accept also : HO2CCH(OH)CH(OH)CO2– + H2O –O2CCH(OH)CH(OH)CO2– + H3O+ HO2CCH(OH)CH(OH)CO2– + H2O HO2CCH(OH)CH(OH)CO2H + OH–

5 ¤ Jurong Pioneer Junior College 9729/03/J2 PRELIMINARY EXAM/2021 [Turn Over 2 (b) (vi) At point X: initial pH of tartaric acid (weak acid)

aH acid pH Kªº ªº u¬¼¬¼ -2.95 -3 10 0.20 0.0150 mol dm -lg 0.0150 1.82 [1] At point Y: pH of amphiprotic species ²O2C(CH(OH))2(CO2H) a1 a2 1pH 2pK pK 12.95+4.25 3.602 [1] At point Z: solution is at maximum buffering capacity when [²O2C(CH(OH))2(CO2H)] = [²O2C(CH(OH))2(CO2²)] a2pHpK 4.25 [1] (c) (i) D: CH2=CHCO2H [1] E: CH2BrCH(OH)CO2H [1] (ii) Step II: Br2(aq) [1] Step IV: H2SO4(aq)/HCl(aq), heat [1] (iii) Use aqueous bromine to test the reaction mixture. [1] If aqueous bromine remains orange, reaction is complete. / If orange aqueous bromine decolourises, reaction is incomplete. [1] [Total: 20] 3 (a) (i) NaOH is required to generate the nucleophile CN³. [1] HCN is a weak acid/ionises only partially. Thus [CN³] is low and reaction is slow. [1] For ([DPLQHU¶V Use (ii) Comparing experiments 1 and 2: When [CH3CHO] increases by 2x, rate increases by 2x Î rate ∝ [CH3CHO] ∴ order of reaction wrt CH3CHO is 1. [1] Comparing experiments 1 and 3: Let rate = k[CH3CHO][NaOH]a order of reaction wrt NaOHa? a-2 -4-14 -14 a -2 -4 k 1.25×10 1.25×101.15×10= 6.90×10k 3.75×10 2.50×10 1 [1]

6 ¤ Jurong Pioneer Junior College 9729/03/J2 PRELIMINARY EXAM/2021 3 (b) (i) The rate equation defines the slow step and shows that there is only 1 molecule of carbonyl compound and one CN௅ involved. Hence, the rate-determining step is Step 1. (ii) The bigger Kc value indicates position of equilibrium to form the cyanohydrin compound lies more to the right, compound is more susceptible to nucleophilic substitution. [1] Comparing reactions I and II: The electron donating ‒CH3 group make the carbonyl C less electron deficient, thus less susceptible to nucleophilic attack. [1] Comparing reactions I and III: The electron withdr