Chemistry Answers Chapter 8

  Chapter 8 Exercises 1 2 3 (a) HSO3− (b) CH3NH3+ (c) C2H5COOH (d) HNO3 (e) HF (f) H2SO4 (a) H2PO4− (b) CH3COO− (c) HSO3− (d) SO42− (e) O2− (f) Br− 4 5 HPO42−(aq) + H2O(l) (base behaviour) H2PO4−(aq) + OH−(aq) (a) H2SO4(aq) + CuO(s) → CuSO4(aq) + H2O(l) (c) H3PO4(aq) + 3KOH(aq) → K3PO4(aq) + 3H2O(l) NH3 (base)/NH4+ (acid) (d) 6CH3COOH(aq) + 2Al(s) → 2Al(CH3COO)3(aq) + 3H2(g) (b) CO32− (base)/HCO3− (acid) H3O+ (acid)/H2O (base) NO2− (base)/HNO2 (acid) PO4 3−(aq) + H3O+(aq) (acid (b) HNO3(aq) + NaHCO3(s) → NaNO3(aq) + H2O(l) + CO2(g) (a) CH3COOH (acid)/CH3COO− (base) (c) NH4+ (acid)/NH3 (base) HPO42−(aq) + H2O(l) behaviour) 6 B 7 B 35 10:01 Answers 8 (a) nitric acid + sodium carbonate / sodium hydrogencarbonate / sodium hydroxide 2HNO3(aq) + Na2CO3(aq) → 2NaNO3(aq) + H2O(l) + CO2(g) [OH−] = 5.9 × 10−8 mol dm−3 acidic 24 (a) 0.40 (b) 10.57 (c) 10.00 (b) hydrochloric acid + ammonia solution 25 B 26 (a) Kb = pH increases by 1 unit 27 HNO2 < H3PO4 < H2SO3 10 pH = 4.72 28 11 [H+] = 1.0 × 10−9 mol dm−3, [OH−] = 1.0 × 10−5 mol dm−3 Strong acids and bases are fully dissociated, so it is not useful to think of them in terms of an equilibrium mixture. The pH of their solutions can be derived directly from their concentration. 12 (a) [OH−] = 2.9 × 10−6; basic 29 B 30 Kb = 5.6 × 10−4 31 [H+] = 1.0 × 10−4 mol dm−3; [OH−] = 1.0 × 10−10 mol dm−3 32 pH = 3.22; [H+] = 6.0 × 10−4 mol dm−3 33 A 34 HF is the stronger acid. 35 pKb CN− = 4.79; pKb F− = 10.83 HCl(aq) + NH4OH(aq) → NH4Cl(aq) + H2O(l) (c) copper(II) oxide + sulfuric acid H2SO4(aq) + CuO(s) → CuSO4(aq) + H2O(l) (d) methanoic acid + potassium hydroxide HCOOH(aq) + KOH(aq) → KCOOH(aq) + H2O(l) 9 (b) [H+] = 1.0 × 10−12; basic (c) [H+] = 1.0 × 10−4; acidic (d) [OH−] = 1.2 × 10−10; acidic 13 pH = 2.0 14 (a) pH = 6.9 (b) pH = 2 (c) pH = 4.8 15 pH = 13.17 16 B 18 (a) H2CO3 19 (a) Lewis acid Zn2+; Lewis base NH3 17 A [C2H5NH3+][OH−] [C2H5NH2] [H2SO4][OH−] (b) Kb = [HSO4 −] [HCO3 −][OH−] (c) Kb = [CO32−] CN− is the stronger base. (b) HCOOH 36 (a) pKb CH3COO− = 9.24 (b) Methanoic acid is a stronger acid than ethanoic acid from its lower pKa. Therefore, its conjugate base is weaker. (b) Lewis acid BeCl2; Lewis base Cl− (c) Lewis acid Mg2+; Lewis base H2O 20 D, CH4 because it does not possess a lone pair. 37 B 21 C, there is no exchange of H . 39 22 [H ] = [OH ] = 1.55 × 10 (i) Because it has a higher concentration of the acid and its conjugate base. 40 (a) equal to 7 (b) less than 7 (c) less than 7 (d) greater than 7 + + − −7 pH = pOH = 6.81 pH + pOH = pKw = 13.62 23 38 B neutral 41 B; salt of strong base and weak acid pOH = 7.23 42 (a) less than 7 [H+] = 1.7 × 10−7 mol dm−3 (b) greater than 7 (c) equal to 7 36 10:01 43 D (III only) (c) Particulates act as catalysts in the production of secondary pollutants. 44 (d) SO2(g): CaO(s) + SO2(g) → CaSO3(g) (e) NO: formed from the combination of nitrogen and oxygen at the high temperature of internal combustion engines. pH buffer region 50 equivalence point <7 Wet acid deposition is dispersed over a much larger area and distance from the emission source. 25cm3 volume of HCl added 45 (b) The acid is formed in the air from sulfur dioxide (SO2) and nitrogen oxide (NO) which are emitted by thermal power stations, industry and motor vehicles. A major source is the burning of fossil fuels, particularly in coal-fired power stations. Pollutants are carried by prevailing winds and converted (oxidized) into sulfuric acid (H2SO4) and nitric acid (HNO3). These are then dissolved in cloud droplets (rain, snow, mist, hail) and this precipitation may fall to the ground as dilute forms of sulfuric acid and nitric acid. The dissolved acids consist of sulfate ions, nitrate ions and hydrogen ions. Initial pH of acid ⇒ Ka of acid ⇒ pKa of acid. At half equivalence, pH = pKa, so this can be read directly off the curve. 46 D 47 (a) Strong acid–strong base and strong acid– weak base (b) pKa = 4.6 (midway in endpoint range) (c) yellow at all pHs below 3.8 48 (a) Contains dissolved carbon dioxide which reacts with water to form carbonic acid: CO2(g) + H2O(l) (a) Dry acid deposition typically occurs close to the source of emission. H2CO3(aq) (b) Sulfuric acid: It is formed by the reaction between water and either ozone or atomic oxygen: S(s) + O2(g) → SO2(g)/2SO2(g) + O2(g) → 2SO3(g) H2O(l) + SO3(g) → H2SO4(aq) (c) Nitric acid: reduced by use of catalytic converters, recirculation of exhaust gases Challenge yourself (d) CaCO3(s) + H2SO4(aq) → CaSO4(aq) + H2O(l) + CO2(g) (e) Effects on materials, plant life and human health (see text for details). (f) Use alternative energy source to fossil fuels or use coal with a low sulfur content. 49 (a) SO2 and NO (b) SO2 and particulates 1 Increasing the length of the carbon chain increases the donation/push of electrons towards the carbonyl C atom, known as a positive inductive effect. This causes less electron withdrawal from the O –H bond, so weakening acid strength. The basic strength of amines depends largely on the availability of the 37 10:01 Answers lone pair electrons on nitrogen. C2H5 pushes electrons towards N more than CH3 does, so it is a stronger base. acids). Decomposition of plant material to peat and then coal conserves this sulfur. Additional sources are the depositional environment such as sea water, where sulfates are reduced by bacteria to form H2S, which can react further to form organic sulfur structures. (This will make more sense after studying inductive effects in Chapter 10.) 2 (a) pH increases on dilution of a strong acid as [H+] decreases. 7 Combustion of nitrogen involves the highly endothermic step (+942 kJ mol−1) of breaking the triple N≡N bond, as well as the O=O bond (+498 kJ mol−1). The exothermic step of forming the triple N≡O bond releases less energy – approximately 630 kJ mol−1. Therefore enthalpy change (bonds broken minus bonds formed) is endothermic. So the stability and strength of the nitrogen triple bond creates an unusual situation where the products of combustion are less stable than the reactants. 8 HNO2: N = +3, nitric(III) acid (b) pH increases on dilution of a weak acid as [H+] decreases, but the change is less than for a strong acid, as acid dissociation increases with dilution. (c) pH of a buffer stays the same with dilution, as Ka or Kb and [acid]/[salt] ratio stay constant. 3 [acid] = initial concentration of acid; [HA] = equilibrium concentration of acid. These will be equal only for weak acids, in which the extent of dissociation is so low that the acid is considered to be undissociated at equilibrium. HNO3: N = +5, nitric(V) acid H2SO3: S = + 4, sulfuric(IV) acid [salt] = initial concentration of salt solution; [A−] = concentration of anion/conjugate base at equilibrium. These will be equal for fully soluble salts in which the formula unit contains a single anion. 4 5 6 Titration needs to add acid from burette to base (see exam hint on page 400). The pH at half equivalence when half the base has been neutralized and half remains unreacted is recorded. At this point, pOH = pKb. (pOH is usually calculated from the measured pH by assuming the temperature is 298 K, so pH + pOH = 14.) Indicators usually contain structures with delocalized electrons, either benzene rings and/ or multiple bonds. Conjugation of the electrons influences the absorption of light in the visible region, giving rise to specific colours. The gain or loss of H+ changes the delocalization within the structure, which causes changes in absorption maxima and so different colours. Colour is discussed in Topic 13, Chapter 3. Sulfur is present in proteins in living cells (a component of two out of the twenty amino H2SO4: S = +6, sulfuric(VI) acid Practice questions 1 D 2 B 5 9 B 6 C 7 C 8 B C 10 D 11 A 12 A 13 C 14 (a) (Kw =)[H+(aq)][OH−(aq)] 3 C 4 C [1] Do not penalize if (aq) not stated. H3O+ may be given instead of H+. Do not mark award if square brackets are omitted or are incorrect. (b) (i) (pKb =(14.00 − 7.52 =) 6.48 and) Kb = [1] (10−6.48) = 3.3 × 10−7 Do not award mark if answer just left as 10−6.48. x2 [HOCl][OH−] = 3.3 × 10−7 = (ii) Kb = [OCl−] 0.705 − −4 [OH ] = 4.8 × 10 (mol dm−3) Award [2] for correct value of [OH−]. 38 10:01 OCl− only partially hydrolysed / xnegligible (compared to [OCl−]) / OWTEE pKa = 4.7 [3] Accept [HOCl] = [OH−]. 1.00 × 10−14 Kw (iii) [H3O+]/[H+] = = = − 4.8 × 10−4 ] [OH 2.1 × 10−11 pH = (−log10[H3O+]/−log10[H+] = −log10(2.1 × 10−11) =)10.68 (v) [2] Award [2] for correct final answer. (a) (i) Acid: proton/H+ donor and Base: proton/H+ acceptor pH 15 Do not accept OH− for base. [2] Accept in range 4.2 to 5.2. M2 can only be scored if M1 correct (i.e. no marks for just Data Booklet value of 4.76). 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0 Weak base: (base/electrolyte) partially dissociated/ionized (in solution/ water) and Strong base: (base/ electrolyte assumed to be almost) completely/100% dissociated/ionized (in solution/water) / OWTTE Starting pH: 1 NH3 / CH3CH2NH2 Equivalence point: pH = 7 and 25 cm3 [3] Allow either name or formula or other suitable example. (ii) sulfurous acid/H2SO3 corrodes marble/limestone buildings/ statues / leaching in soils / harms/kills plants OR nitrous acid/HNO2 corrodes marble/limestone buildings/ statues / leaching in soils / harms/kills plants OR carbonic acid/H2CO3 corrodes marble/limestone buildings/ statues / acidification of lakes [2] Do not allow oxides (e.g. CO2 etc.). Do not accept just corrodes or damages. (iii) Volume of KOH: 20 (cm3) Allow any value between 20 and 21 (cm3). pH at the equivalence point: 8.0–10.0 [2] (iv) At half-equivalence point, [CH3COOH] = [CH3COO−] so pH = pKa; 0 10 20 30 Volume / cm3 40 50 Final pH reached: 12–13 correct curve shape [4] Do not award M4 if turn in curve is seen at low volumes (suggesting weak acid– strong base titration). Award [4] if values corresponding to M1, M2 and M3 are labelled on graph (e.g. using X) and correct shape of curve shown. (b) (i) HIn is a weak acid / weak base HIn colour 1 H+ + In− colour 2 required. Award [2] for M2 alone. in base equilibrium moves to right / in acid equilibrium moves to left [3] (ii) phenolphthalein indicator colour change occurs in range of pH at the equivalence point / OWTTE [2] M2 can be scored independently even if indicator is incorrect. 39 10:01 Answers 16 (a) strong acid completely dissociated/ionized and weak acid partially dissociated/ionized HNO3(aq) → H+(aq) + NO3 −(aq) HCN(aq) H+(aq) + CN−(aq) [3] Insist on both arrows as shown. (Conjugate) base: HPO42− Accept H2O and H3O+. [H+][CN−] (b) Ka = [HCN] Allow H3O+ instead of H+. No mark for NaH2PO4 or Na2HPO4. H2PO4 −(aq) H+(aq) + HPO42−(aq) [3] Accept reverse equation or reaction with water. Ka = 10−9.21 = 6.17 × 10−10 (c) [H+] = √Ka[HCN] / √(6.17 × 10−10 × 0.108 Ignore state symbols, but equilibrium sign is required. = 8.16 × 10−6 Allow in the range 8.13 × 10−6 to 8.16 × 10−6. pH = 5.09 pH = 12(pKa − log[HCN])/12(9.21 − log0.108) = 5.09 [H+] = 10−5.09 = 8.16 × 10−6 −6 −6 Allow in the range 8.13 × 10 to 8.16 × 10 . + If expression for [H ] missing but both answer correct, award [3], if one answer correct, award [2]. assume [H+] << 0.108/negligible dissociation (a) (i) (Ka) = [H+][OH−]/(Kw) = [H3O+][OH−] [4] [1] Do not award mark if [ ] omitted or other brackets are used. (ii) [H ] increases, [OH ] decreases but still some present (Kw constant) / [OH−] cannot go to zero as equilibrium K present/ [OH−] = +w , thus [OH−] cannot [H ] be zero / OWTTE [1] − (iii) (changing T disturbs equilibrium) endothermic reaction / forward reaction favoured / equilibrium shifts to the right to use up (some of the) heat supplied Kw increases (as both [H+] and [OH−] increases) (iv) (as [H+] increases) pH decreases / pH<7 Accept OH− (ions) react with H+ (ions) to form H2O. (ii) strong base/OH− replaced by weak base (HPO42−, and effect minimized) / strong base reacts with acid of buffer / equilibrium in (i) shifts in forward direction OR + inverse relationship between pH and 1 [2] [H+]/pH = −log[H+]/pH = log10 + [H ] Accept [H3O+] in place of [H+]. (b) (i) Acid: H2PO4 − State symbols not needed. 17 No mark for more acidic. [3] OH−(aq) + H2PO4−(aq) → H2O(l) + HPO42−(aq) [2] Ignore state symbols, accept equilibrium sign. Accept OH– added reacts with H+ to form H2O. (iii) strong acid/H+ replaced by weak acid (H2PO4−, and effect minimized) / strong acid reacts with base of buffer / equilibrium in (i) shifts in reverse direction H+(aq) + HPO4 2− (aq) → H2PO4−(aq) [2] Accept reaction with H3O+. Ignore state symbols. (c) (i) NH3 weak(er) base/partial dissociation [OH−] < 0.1(0)/pOH > 1 (thus pH < 13 / pH + pOH = 14) [2] (ii) around pH = 5 Accept a value between 4 and 6. strong acid–weak base titration, (thus acidic) / at equivalence point, NH4+ present is acidic / NH4+ NH3 + H+ [2] (iii) NH3(aq) + H2O(l) NH4 +(aq) + OH−(aq) Ignore state symbols, but equilibrium sign required. 40 10:01 Kb = [NH4+][OH−] [NH3] Ignore units. [2] (iv) [NH3] = [NH4+] Award [3] for correct final answer. [1] (vi) optimum/most effective/highest buffer capacity/50%–50% buffer/equally effective as an acidic buffer and a basic buffer / OWTTE [1] (v) pOH = 14.00 − 9.25 = 4.75 pKb (= pOH) = 4.75 Kb = 1.78 × 10−5 [3] Chapter 9 Exercises 1 (a) NH4+ = N −3, H +1 3 (b) CuCl2 = Cu +2, Cl −1 (a) Ca(s) + 2H+(aq) → Ca2+(aq) + H2(g) 0 +1 +2 0 oxidation: Ca(s) → Ca2+(aq) + 2e− (c) H2O = H +1, O −2 reduction: 2H+(aq) + 2e− → H2(g) (d) SO2 = S +4, O −2 (b) 2Fe2+(aq) + Cl2(aq) → 2Fe3+(aq) + 2Cl−(aq) +2 0 +3 −1 (e) Fe2O3 = Fe +3, O −2 (f) NO3− = N +5, O −2 oxidation: 2Fe2+(aq) → 2Fe3+(aq) +2e− (g) MnO2 = Mn +4, O −2 reduction: Cl2(g) + 2e− → 2Cl−(aq) (h) PO43− = P +5, O −2 (c) Sn2+(aq) + 2Fe3+(aq) → Sn4+(aq) + 2Fe2+(aq) +2 +3 +4 +2 (i) K2Cr2O7 = K +1, Cr +6, O −2 (j) MnO4− = Mn +7, O −2 2 (a) oxidation: Sn2+(aq) → Sn4+(aq) + 2e− reduction reduction: 2Fe3+(aq) + 2e− → 2Fe2+(aq) (d) Cl2(aq) + 2Br−(aq) → 2Cl−(aq) + Br2(aq) 0 −1 −1 0 Sn (aq) 1 2Fe (aq) → Sn (aq) 1 2Fe (aq) 12 13 14 12 21 31 41 21 oxidation (b) oxidation: 2Br−(aq) → Br2(aq) + 2e− reduction Cl2(aq) 1 2NaBr(aq) → Br2(aq) 1 2NaCl(aq) 0 1121 0 11 21 reduction: Cl2(aq) + 2e− → 2Cl−(aq) 4 oxidation reduction (c) (b) 2I−(aq) + HSO4 −(aq) + 3H+(aq) → I2(aq) + SO2(g) + 2H2O(l) 2FeCl2(aq) 1 Cl2(aq) → 2FeCl3(aq) 1221 0 13 21 (c) NO3−(aq) + 4Zn(s) + 10H+(aq) → NH4+(aq) + 4Zn2+(aq) + 3H2O(l) oxidation reduction (d) (d) I2(aq) + 5OCl−(aq) + H2O(l) → 2IO3 −(aq) + 5Cl−(aq) + 2H+(aq) 2H2O(l) 1 2F2(aq) → 4HF(aq) 1 O2(g) 1122 0 1121 0 (e) 2MnO4−(aq) + 5H2SO3(aq) → 2Mn2+(aq) + 3H2O(l) + 5SO42−(aq) + 4H+(aq) oxidation (e) reduction I2(aq) 1 SO3 22(aq) 1 H2O(l) → 2I2(aq) 1 SO422(aq) 1 2H1(aq) 0 1422 1122 21 1622 11 (a) Zn(s) + SO42−(aq) + 4H+(aq) → Zn2+(aq) + SO2(g) + 2H2O(l) 5 B 6 D oxidation 41 10:01