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