NJC Prelim P3 Solutions

1 NJC H2 Chemistry Prelim Paper 3 Suggested Answers 1 (a) (i) BaCO3(s)  BaO(s) + CO2(g) (ii)  Down the group, cations of Group 2 metals have the same charge but their size increases. The 𝑐ℎ𝑎𝑟𝑔𝑒 𝑠𝑖𝑧𝑒 ratio of the cation decreases. This leads to a decrease in polarising power.  The electron cl ouds in the carbonate anion is being distorted to a smaller extent down the group and the weakening of C –O bond is less significant down the group.  More energy is required to break the C –O bonds in the carbonate anion down the group.  Therefore, the ease of thermal decomposition of the carbonates decreases down the group. (b) (i) Shape: trigonal planar Bond angle: 120 o (ii) Reaction 1: (NH2)2CO + H2O  2NH3 + CO2 Reaction 2: Ba(OH)2 + CO2  BaCO3 + H2O (c) (i) Ka = [HCO3 - ][H3O+] [H2CO3] Ka = (x)2 0.100−x = 4.3 × 10–7 mol dm–3 [H+] = √4.3 × 10−7 × 0.1 = 2.074 × 10–4 mol dm–3 pH = 3.68 (ii) H2CO3 + 2NaOH  Na2CO3 + 2H2O Amount of H2CO3 = 25 1000×0.1 = 0.0025 mol Amount of NaOH required = 0.005 mol Volume of NaOH required = 0.005 0.125 = 0.04dm3 = 40.0 cm3 (iii) At maximum buffering capacity, pH = pKa1 = 6.37

2 (iv) From part (ii), 40 cm3 of NaOH is required to completely neutralise the 2 H+ from H2CO3. When 20 cm3 of NaOH is added, only 1 H+ from H2CO3 will react with NaOH OH + H2CO3  HCO3 + H2O Initial / mol 0.0025 0.0025 0 - Change / mol 0.0025 0.0025 +0.0025 - Final / mol 0 0 0.0025 - [HCO3–] = 2.50 × 10–3/ (0.02 + 0.025) = 0.05556 mol dm–3 HCO3− acts as a weak base that dissociates partially. HCO3− + H2O H2CO3 + OH− Kb = [H2CO3][OH-] [HCO3-] Kb = (x)2 0.05556−x = 1×10−14 4.3×10−7 = 2.325 × 10–8 mol dm–3 x = [OH–] = 3.595 × 10–5 pOH = −lg[OH–] = 4.44 pH = 9.56 (v) (d) (i) BaCO3(s) Ba2+(aq) + CO32(aq) Let solubility of BaCO3(s) be s Ksp =[Ba2+][CO32–] = s2 = 5.5 × 1010 s = 2.35 × 10−5 mol dm−3

3 (ii) BaCO3(s) Ba2+(aq) + CO32(aq) --- eqm (1) When ingested by mouth, the H+ in stomach will react with CO32– to form H2CO3 or CO 2. [CO 32–] will decrease and equilibrium (1) will shift right. BaCO 3 will be more soluble in presence of H + and produces more Ba 2+, resulting in concentration of Ba2+ to exceed 0.100 mol dm–3. BaSO4(s) Ba2+(aq) + SO42(aq) --- eqm (2) [SO42–] remains unchanged as SO42− does not react with H+. Hence, there is no shift in the position of equilibrium in (2). Concentration of Ba 2+ is not affected and will not exceed the lethal level of 0.100 mol dm–3 (e) Hsolo = Hhydo − LE Hsolo of CaSO4 = −1577 −1099 + 2374 = −302 kJ mol−1 [1] Hsolo of BaSO4 = −1305 −1099 + 2480 = +76 kJ mol−1 [1] The Hsolo of CaSO4 is more negative compared to BaSO 4. CaSO4 is more soluble than BaSO4. A negative Hsolo shows that the aq ions of CaSO 4 are more stable than the ionic solid. Linking to G = H  TS, a more negative H value will give a more negative G value. Hence the reaction is more likely to be energetically feasible. 2 (a) (i) Item to be anodised placed at anode Correct inert metal / graphite used as cathode Clear and labelled diagram Use of battery (ii) 2H2O(l)  O2(g) + 4H+(aq) + 4e  2Al(s) + 3/2O2(g)  Al2O3(s) Pt or graphite (cathode) dilute H2SO4 iPhone 7®

4 (iii) Volume of Al2O3 required = 93.0 × 0.02 = 1.86 cm3 Mass of Al2O3 = 3.95 × 1.86 = 7.347 g No. of moles of O2 = 3/2 × No of moles of Al2O3 = 3/2 × 7.347 102 = 0.1080 mol No of moles of electrons passed = 0.1080 × 4 = 0.4322 mol Q = 0.4322 × 96500 = 41705 C Time needed = 41705  2.0 = 20853 = 2.09 × 104 s (3 s.f.) (b) (i) Fe3+ has high charge/size and is able to distort the electron cloud of water molecules, breaking the O-H bonds and release H+ OR [Fe(H2O)6]3+ [Fe(H2O)5(OH)]2+ + H+ H+ reacts with carbonate to form CO2. Fe(OH)3 is the red-brown ppt. (ii) Fe2+(aq) + 2OH–(aq) ⇌ Fe(OH)2(s) green ppt [Fe(H2O)6]2+(aq) + 6CN–(aq) ⇌ [Fe(CN)6]4– (aq) + 6H2O(l) formation of coloured solution OR Fe(OH)2(s) + 6CN–(aq) ⇌ [Fe(CN)6]4– (aq) + 2 OH−(aq) (c) Ru N Cl H Mass in 100 g 36.9 30.7 25.8 6.6 Amount 36.9 / 101.1 = 0.365 30.7 / 14 = 2.193 25.8 / 35.5 = 0.7268 6.6 / 1.0 = 6.6 Simplest ratio 0.365 / 0.365 = 1 2.193 / 0.365 = 6 0.7268 / 0.365 = 2 6.6 / 0.365 = 18 There are 2 moles of free Cl– ions. The remaining N and H must be from 6 NH3 ligands. Hence, molecular formula of orange compound is [Ru(NH3)6]Cl2. accept RuN6H18Cl2 (d) (i) Ethene

5 (ii) Two possible answers: Step 4: Ru + Ru Step 4: Ru + Ru (iii) and (e) 3 (a) (i) Ka= ]H[ ]H][[   Cy Cy (ii) [H+]=10–5.3 Ka= ]H[ ]H][[   Cy Cy 5 × 10–5 = ]H[ ]10][[ 3.5   Cy Cy ][ ]H[ Cy Cy  = ]105[ ]10[ 5 3.5    =0.100 OR pH = pKa + lg ][ ][ CyH Cy

6 5.3 = –lg(5 × 10−5) + lg ][ ][ CyH Cy lg ][ ][ CyH Cy = 0.9990 ][ ][ CyH Cy = 100.9990 = 9.976 ][ ]H[ Cy Cy  =(19.976) = 0.100 (3sf) Since concentration of Cy is much greater than CyH+, the dominant species is Cy. Hence, the colour of the solution will be purple colour. (b) (i) Kc= ]][H[ ]H][[ 2 23 SOCy HCySO   [H+] = 10–3.0 mol dm–3 Let initial concentration of CyH+ be x mol dm–3 [CyH+]eqm = x10 1 mol dm–3 [CySO3H2] eqm = x10 9 mol dm–3 Kc= ]][H[ ]H][[ 2 23 SOCy HCySO   = ]101][)101[( ]10][)109[( 2 0.3   x x = 0.9 (ii) The decolourisation of the preserved fruit juice will be more severe at pH = 4.0. At higher pH, [H+] is lower. According to Le Chatelier’s principle, position of equilibrium (I) will shift to the right to increase [H+], thereby reducing the amount of cyanidin to a greater extent. (c) Information Deduction Both J and K react with sodium carbonate J and K undergoes acid base reaction with sodium carbonate.  Both compounds contain – carboxyl (COOH) group. Both J and K react with acidified K2Cr2O7, but not with 2,4– dinitrophenylhydrazine. J and K undergoes oxidation reaction with K2Cr2O7.  Both compounds are not carbonyl compound and contain 1°/2°alcohol.

7 Both J and K react with excess hot concentrated H 2SO4, but only J gives a mixture with a pair of cis-trans isomers. J and K undergoes elimination of H2O to form alkene. Alkenes obtained from J exhibited cis-trans isomerism, but not that of K (terminal alkene). A 0.234 g sample of J reacts completely with 35 cm3 of 0.10 mol dm–3 NaOH. J undergoes neutralisation/acid base reaction with NaOH. Amt of J= 134 234.0 =1.746 x10–3 mol Amt of NaOH = 3.5 x 10–3 mol Mole ratio of J: NaOH is 1:2 J is dibasic acid which contains two COOH groups. Given Mr of J = 134, 2xMr(COOH)+Mr(OH)+Mr(CxHy) = 134 2(45) + (17) + x(12) + y(1) = 134 107 + x(12) + y(1) = 134 x(12) + y(1) = 27 x = 2 and y = 3 K give a yellow precipitate with alkaline aqueous iodine. K undergoes mild oxidation with alkaline aqueous iodine. K contains the following structure CH3 C OH H A 7.5 x 10 –4 mol of K produces 18 cm 3 H2 gas at r.t.p. when excess Na is added. K undergoes redox/ acid metal reaction with Na. Amt of H2 liberated = 18/24000 =7.5 x 10–4 mol Mole ratio of K:H2 is 1:1 K contains two OH groups (make up of 1 COOH and 1 OH group). Given Mr of K = 90, 1xMr(COOH)+Mr(CH3CH(OH))=90

8 J: C C H H C OH H C O O H O OH K: H C H H C OH H C O O H (d) In conjugate base of pyruvic acid, the negative charge on O of COO – can delocalise over 3 oxygen atoms w