RVHS_H2_CHEM_P3_ANS_RV Prelim

1 Answer key to Paper 3 1 (a) (i) A buffer solution is one that can resist a change in pH when a small amount of acid or base is added to it. (ii) 1. pKb = 5.93 Let [OH] be x x2  (0.10  x) = 105.93 Since x is small, x2  0.10 = 105.93 x = 3.43 × 104 mol dm3 pOH = 3.47 pH = 10.5 2. At equivalence point, 12.5 cm3 of HCl solution is required. pKa of conjugate acid = 8.07 Amt of RNH3 + = (25/1000) × 0.1 = 2.50 × 103 mol dm3 [RNH3 +] = 2.50 × 103  (37.5/1000) = 6.67 × 102 mol dm3 Let the concentration of [H+] be y y2  (6.67 × 102 –y) = 108.07 Since y is small, y = 2.38 × 105 mol dm3 pH = 4.62 3. Amt of HCl remaining = (30/1000) × 0.20 - 2.50 × 103 = 3.50 × 103 mol [HCl] = 3.50 × 103  (55/1000) = 6.36 × 102 mol dm3 pH = 1.20

2 (iii) Maximum buffering capacity occurs at 8.07, which is close to physiological pH of 7.4/ The buffering action occurs near physiological pH of 7.4/ Non-toxic [6] (b) (i) Rate = k[HONO]2[C6H5NH2]0 (ii) Units for k  mol1 dm3 s1 or mol1 dm3 min1 [2] (c) Observations/Reactions Deduction G does not undergo (nucleophilic) substitution with PCl5 to give white fumes G does not contain carboxylic nor alcohol functional group. G undergoes alkaline hydrolysis to give H and salt of I. G is an amide I is a carboxylic acid G has a fishy smell and does not react with HONO. G is an alkyl amine I rotates plane-polarised light I is chiral I undergoes positive iodoform test I contains –COCH3 or –C(OH)CH3 groups I undergoes condensation reaction with 2,4-DNPH I does not have -C(OH)CH3 groups / I must contain –COCH3 functional group I undergoes electrophilic substitution with Br2(aq) I undergoes 1 electrophilic substitution on benzene ring containing phenol functional group /other ortho & para positions are blocked from ES G

3 H I J [9] (d) N2 + 3H2 ⇌ 2NH3 Heterogeneous catalyst: Fe S2O8 2 + 2I   2SO4 2 + I2 Homogeneous catalyst: Fe2+ or Fe3+ Or C6H6 + Br2  C6H5Br + HBr (or C6H6 + Cl2  C6H5Cl + HCl) Homogeneous catalyst: FeBr3 (or FeCl3) [3] [Total: 20]

4 2 (a) Nitrogen is inert and nitrogen compounds are hard to form. Ammonium salts and nitrates are soluble in water and thus woul d only accumulate in dry areas. [2] (b) Low temperature and high pressure. Ammonia gas has hydrogen bonding between its molecules and thus significant intermolecular forces are present at low temperatures. At high pressure, the ammonia molecules would be very close to each other and the volume of the gas molecules would be significant when compared to the volume of the container / gas. [3] (c) pV = nRT pV = mRT/Mr density = m/V = pMr/RT = (1.01  105  17.0) / (8.31  303) = 682 g m3 = 682 mg dm3 Concentration is below lethal value. [3] (d) (i) Gradient = [(5.44  (8.72)] / (3.33  103  2.50  103) = 3.95  103 K H = 3.95  103