Pre-U STPM Chemistry Term 1 CC039342a

292 Chemistry Term 1 STPM CHAPTER 7 9 Solid silver chloride is said to be in dynamic equilibrium with its aqueous ions. AgCl(s)  Ag+(aq) + Cl– (aq) 10 Applying the equilibrium law to the above system: [Ag+][Cl– ] [AgCl] = Kc where [ ] = equilibrium concentration Rearranging: [Ag+][Cl– ] = Kc [AgCl] However, concentration of a solid is a constant: Hence, [Ag+][Cl– ] = constant The constant is known as the solubility product, Ksp, of silver chloride. 11 Similarly for the case of lead chloride: PbCl2 (s)  Pb2+(aq) + 2Cl– (aq) [Pb2+][Cl– ]2 = Ksp of PbCl2 12 The solubility product of a sparingly soluble salt is defined as the product of the molar concentration of the constituent ions, in a saturated solution, each raised to the power of their stoichiometric coefficient. 13 The expression for the solubility products, and their units, for some sparingly soluble salts are given below: BaSO4 (s)  Ba2+(aq) + SO4 2–(aq)Ksp = [Ba2+][SO4 2–] mol2 dm–6 Ca(OH)2 (s)  Ca2+(aq) + 2OH– (aq) Ksp = [Ca2+][OH– ]2 mol3 dm–9 Al(OH)3 (s)  Al3+(aq) + 3OH– (aq) Ksp = [Al3+][OH– ]3 mol4 dm–12 Ca3 (PO4 ) 2 (s)  3Ca2+(aq) + 2PO4 3–(aq) Ksp = [Ca2+] 3 [PO4 3–] 2 mol5 dm–15 14 The solubility product, like equilibrium constants, is a function of temperature. 7.9 Calculating K sp from Solubility and Vice Versa Example 7.17 Calculate the solubility products from the solubilities given below: (a) Zn(OH)2 : 1.35  10–4 g dm–3 (b) PbSO4 : 3.90  10–3 g per 100 cm3 Definition of solubility product In aqueous solutions containing silver chloride, the following product is always a constant at constant temperature. [Ag+][Cl– ] = Ksp The magnitude of Ksp depends on temperature. 2016/P1/Q15

293 Chemistry Term 1 STPM CHAPTER 7 Solution (a) Solubility of Zn(OH)2 = 1.35  10–4 65.4 + (17  2) = 1.36  10–6 mol dm–3 Zn(OH)2  Zn2+ + 2OH– a a 2a Ksp = [Zn2+][OH– ]2 = (a)(2a)2 = 4a3 = 4  (1.36  10–6)3 = 1.01  10–17 mol3 dm–9 (b) Solubility of PbSO4 = 3.90  10–3 303  1000 100 mol dm–3 = 1.29  10–4 mol dm–3 PbSO4  Pb2+ + SO4 2– b b b K sp = (b)(b) = (1.29  10–4) 2 = 1.66  10–8 mol2 dm–6 Quick Check 7.11 1 Calculate the solubility (in g dm–3) of the following salts from their respective solubility products. Salt BaSO4 Al(OH)3 Fe2 S3 Ca3 (PO4 ) 2 Ksp/Appropriate unit 1.1  10–10 1.9  10–33 1.0  10–88 1.0  10–25 2 Calculate the solubility products of the following salts from their solubilities. Salt Silver sulphate Barium carbonate Silver iodide Solubility/g dm–3 1.56 0.0178 2.88  10–6 7.10 Common Ion Effect Solubility and Common Ion Effect 1 In a saturated solution of silver chloride, the following holds true: (a) The following equilibrium exists: AgCl(s)  Ag+(aq) + Cl– (aq)

294 Chemistry Term 1 STPM CHAPTER 7 (b) The product of [Ag+][Cl– ] = Ksp = 1.0 × 10–10 mol2 dm–6 (c) [Ag+] = [Cl– ] 2 What would happen if a little sodium chloride is added to the saturated solution of silver chloride? Sodium chloride dissociates completely in water to produce Na+ and Cl– ions: NaCl → Na+(aq) + Cl– (aq) The addition of Cl– (from NaCl) will cause the above equilibrium to shift to the left-hand side and more solid AgCl will precipitate. 3 This means that the solubility of silver chloride is decreased in the presence of sodium chloride because of the Cl– in NaCl. This is known as common ion effect. 4 Common ion effect is the decrease in the solubility of a sparingly soluble salt in a solution that contains an ion common to one of the ions from the dissociation of the salt. 5 The graph shows the variation of the solubility of silver chloride in aqueous sodium chloride of varying concentration. Example 7.18 The solubility product of silver chloride is 1.0  10–10 mol2 dm–6 at 298 K. Calculate the solubility (in g dm–3) of silver chloride in (a) pure water, (b) 0.10 mol dm–3 aqueous sodium chloride. Solution (a) Let the solubility be a mol dm–3 [Ag+][Cl– ] = (a)(a) = 1.0  10–10 ∴ a = 1.0  10–10 = 1.0  10–5 mol dm–3 Solubility = (1.0  10–5)(108 + 35.5) = 1.4  10–3 g dm–3 (b) Let the solubility be b mol dm–3 [Ag+] = b mol dm–3 [Cl – ] = (b + 0.10) mol dm–3 [Ag+][Cl– ] = (b)(b + 0.10) = 1.0  10–10 However, since b is very small, (b + 0.10) ≈ 0.10 Thus, (b)(0.10) = 1.0  10–10 b = 1.0  10–9 mol dm–3 Solubility = (1.0  10–9)(108 + 35.5) = 1.4  10–7 g dm–3 ABBBBBBBB A statement of common ion effect Concentration of NaCl/mol dm–3 Solubility of AgCl/mol dm–3 Info Chem The solubility of AgCI in 0.10 M NaCI decreases. However, the Ksp remains the same.

295 Chemistry Term 1 STPM CHAPTER 7 Example 7.19 The molar solubility of lead carbonate in water is 1.8  10–7 mol dm–3. Calculate the solubility of lead carbonate in 0.050 mol dm–3 sodium carbonate. Solution In water: PbCO3 (s)  Pb2+(aq) + CO3 2–(aq) 1.8  10–7 1.8  10–7 Ksp of PbCO3 = [Pb2+][CO3 2–] = (1.8  10–7)2 = 3.2  10–14 mol2 dm–6 Let the solubility of PbCO3 in Na2 CO3 (aq) = a mol dm–3 PbCO3 (s)  Pb2+(aq) + CO3 2–(aq) a a The total CO3 2– concentration = a + 0.050 ≈ 0.050 ∴ (0.050)(a) = 3.2  10–14 a = 6.5  10–13 mol dm–3 Common Ion Effect and Buffer Solutions 1 Ethanoic acid is a weak acid that dissociates partially in water according to the equation: CH3 COOH(aq) + H2 O(l)  CH3 COO– (aq) + H3 O+(aq) At 25 °C, the degree of dissociation of a 0.10 M ethanoic acid is about 1.6%. 2 On the other hand, sodium ethanoate dissociates 100% in aqueous solution. CH3 COONa(s) + aq → CH3 COO– (aq) + Na+(aq) 3 If some sodium ethanoate is added to an aqueous solution of ethanoic acid, the degree of dissociation of ethanoic acid will be further suppressed due to the presence of the common ion CH3 COO– . 4 As a result, an aqueous mixture of ethanoic acid and sodium ethanoate contains a large amount of undissociated CH3 COOH molecules and a large amount of CH3 COO– ions. This enables the mixture to function as a buffer solution.

296 Chemistry Term 1 STPM CHAPTER 7 7.11 Predicting Precipitation 1 When an aqueous solution of silver nitrate is added to aqueous sodium chloride, will a precipitate of silver chloride be formed? AgNO3 (aq) + NaCl(aq) → AgCl(s) + NaNO3 (aq) or in ionic form: Ag2+(aq) + Cl– (aq) → AgCl(s) 2 To predict whether a precipitate would be formed or not, we have to calculate the solubility quotient (or ionic quotient), Q of the mixture. For the case of silver chloride: Q = [Ag+][Cl– ] 3 Note that Q has the same form as K sp but the concentrations of the ions are not necessarily equilibrium concentrations. 4 If Q  K sp, a precipitate will be formed. If Q  K sp, no precipitate will be formed. If Q = K sp, the solution is just saturated. A precipitate is about to be formed. 5 For the case where Q  K sp, precipitation will occur until the solubility quotient is equal to the solubility product. 6 This is represented by the graph on the left. 7 Any point on the line shows the condition where the solution is just saturated. The area to the right of the curve indicates precipitation will take place. The area to the left of the curve shows that the solution is not saturated, and no precipitate will form. Example 7.20 500 cm3 of 0.0050 mol dm–3 magnesium nitrate is added to 500 cm3 of 0.0060 mol dm–3 sodium carbonate. Predict whether a precipitate of magnesium carbonate would be formed. [Ksp MgCO3 = 1.0  10–5 mol2 dm–6] Solution Total volume of mixture = (500 + 500) = 1000 cm3 Concentration of Mg(NO3 ) 2 in the mixture = 0.0050  500 1000 = 2.5  10–3 mol dm–3 Concentration of Na2 CO3 in the mixture = 0.0060  500 1000 = 3.0  10–3 mol dm–3 Thus, Q = [Mg2+][CO3 2–] = (2.5  10–3)(3.0  10–3) = 7.5  10–6 Since, Q  Ksp, no precipitate would form. [CI– ] Q Ksp (No precipitation) Q Ksp (Precipitation occurs) [Ag+ ] Q = Ksp Exam Tips Exam Tips Since the volume has doubled, the concentration will be halved. 2014/P1/Q15

297 Chemistry Term 1 STPM CHAPTER 7 The volume of the mixture is doubled. To calculate the mass of precipitate formed. Example 7.21 100 cm3 of 1.00  10–5 mol dm–3 magnesium nitrate is added to 100 cm3 of 1.00  10–4 mol dm–3 ammonia. Will a precipitate of magnesium hydroxide be formed? [Ksp of Mg(OH)2 = 2.0  10–13; Kb of NH3 = 1.74  10–5] Solution In the mixture: [Mg2+] = 100 200  (1.00  10–5) = 5.00  10–6 mol dm–3 [NH3 ] = 100 200  (1.00  10–4) = 5.00  10–5 mol dm–3 The [OH– ] concentration is calculated using the equation: [OH– ] = Kb C = (1.74  10–5) (5.00  10–5) = 2.95  10–5 mol dm–3 Mg(OH)2 (s)  Mg2+(aq) + 2OH– (aq) Ionic quotient, Q = [Mg2+][OH– ] 2 = (5.00  10–6)(2.95  10–5) 2 = 4.35  10–15 mol3 dm–9 Since, Q  K sp, no precipitation occurs. ABBB ABBBBBBBBBBBBBBBBBB Example 7.22 10.0 cm3 of aqueous NaCl (0.10 mol dm–3) is added to 10.0 cm3 of aqueous AgNO3 (0.10 mol dm–3). Calculate the mass of silver chloride precipitated. [Ksp AgCl = 1.0  10–10 mol2 dm–6] Solution In the mixture: [Ag+] = 1 2  0.10 = 0.050 mol dm–3 [Cl– ] = 1 2  0.10 = 0.050 mol dm–3 Let the number of moles of AgCl precipitated be a mol dm–3. [Ag+]/mol dm–3 [Cl– ]/mol dm–3 [AgCl]/mol dm–3 Initial 0.050 0.050 0 Final 0.050 – a 0.050 – a a In the saturated solution: [Ag+][Cl– ] = K sp = 1.0  10–10 (0.050 – a) 2 = 1.0  10–10 a = 49.99  10–3 mol dm–3 NH3 is a weak base that dissociates partially: NH3 + H2 O  NH4 + + OH–

298 Chemistry Term 1 STPM CHAPTER 7 Mass of AgCl per dm3 of mixture = (49.99  10–3)(108 + 35.5) = 7.17 g ∴ Mass of AgCl precipitated per 20 cm3 of mixture = 7.17  20 1000 = 0.143 g Quick Check 7.12 1 0.40 g of potassium ethanedioate, K2 C2 O4 , is added to 500 cm3 of a saturated solution of barium ethanedioate, BaC2 O4 , at 298 K. What is the mass of barium ethanedioate that would be precipitated? [Ksp of BaC2 O4 = 1.65  10–7 mol2 dm–6 at 298 K] 2 The solubility of silver chloride in 0.20 M silver nitrate is 5.0  10–10 mol dm–3. Calculate the solubility product of silver chloride. 3 Determine whether a precipitate will form when 125 cm3 of 1.0  10–2 M lead nitrate is added to 250 cm3 of 2.0  10–2 M sodium chloride. [Ksp of PbCl2 = 1.8  10–4] 7.12 Solubility Equilibria and Water Softening 1 Water containing dissolved minerals (especially multivalent cations such as Ca2+ and Mg2+) are termed hard water. 2 Hard water forms a precipitate of calcium carbonate and/or magnesium carbonate (called scale) on the inside of boilers, waterheaters, cooking utensils and pipes that carry hot water, causing blockage and damaging the apparatus. 3 The picture below shows the scale build up in a hot water pipe. Scale 4 Hard water also does not lather easily with soap. This is because the dissolved cations form insoluble 'scums' of calcium or magnesium carboxylates. For example, with sodium sterate (a common ingredient for soap): 2CH3 (CH2 )16COO– Na+(aq) + Ca2+(aq)  (CH3 (CH2 )16COO– )2 Ca2+(s) + 2Na+(aq) 2CH3 (CH2 )16COO– Na+(aq) + Mg2+(aq)  (CH3 (CH2 )16COO– )2 Mg2+(s) + 2Na+(aq) Soap Insoluble scums

299 Chemistry Term 1 STPM CHAPTER 7 5 Thus, it is desirable to remove these cations from the water sample. This is called softening of hard water. 6 Hard water can be classified as temporary hardness or permanent hardness. 7 Temporary hardness is caused by the presence of dissolved calcium hydrogencarbonate, Ca(HCO3 )2 or magnesium hydrogencarbonate, Mg(HCO3 )2 . 8 Temporary hardness can be removed by boiling, where the soluble hydrogencarbonates are converted into insoluble carbonates: Ca(HCO3 )2 (aq) heat CaCO3 (s) + CO2 (g) + H2 O(l) Mg(HCO3 )2 (aq) heat MgCO3 (s) + CO2 (g) + H2 O(l) The insoluble carbonates are then filtered off. 9 Permanent hardness is caused by the presence of dissolved calcium sulphate or magnesium sulphate. It cannot be removed via boiling because the solubilities of these sulphates increase with temperature. 10 One way of softening hard water is to treat the sample with aqueous sodium carbonate to precipitate the insoluble calcium carbonate and magnesium carbonate which are then be filtered off. Mg2+(aq) + CO3 2–(aq) → MgCO3 (s) Ca2+(aq) + CO3 2–(aq) → CaCO3 (s) 11 In industries, softening of hard water is usually done by using an ion-exchanger, where the ion-exchange resins are used to replace the magnesium and calcium ions found in hard water with sodium ions. An example of an ion exchange material is sodium aluminium silicate (zeolite), NaAlSi2 O6 . 12 The hard water is made to flow through the ion exchanger. In the exchanger, Na+ ions from zeolite are exchanged for Ca2+ and Mg2+ ions in the hard water: 2NaAlSi2 O6 (s) + Ca2+(aq) → Ca(AlSi2 O6 )2 (s) + 2Na+(aq) 2NaAlSi2 O6 (s) + Mg2+(aq) → Mg(AlSi2 O6 )2 (s) + 2Na+(aq) 13 The water emerging from the ion-exchanger is now free from soluble Ca2+ and Mg2+ ions. Hard water / Ca2+ and Mg2+ Soft water / Na+ Zeolite 14 The exchanger can be ‘recharged’ by flushing it with concentrated sodium chloride solution where the above reactions are reversed, and the zeolite is regenerated. For example, Ca(AlSi2 O6 )(s) + 2Na+(aq) → 2NaAlSi2 O6 (aq) + Ca2+(aq) Info Chem The Ksp of MgCO3 and CaCO3 is much lower than those of MgSO4 and CaSO4 . Info Chem The Ca2+ and Mg2+ are adsorbed on the surface of the zeolite and thus are removed from the water.

300 Chemistry Term 1 STPM CHAPTER 7 Quick Check 7.13 1 Explain why calcium oxalate, CaC2 O4 (Ksp = 2.0  10–9), is soluble in excess of dilute nitric acid, while calcium sulphate does not. 2 Barium(II) aqueous ions are poisonous. The solubility products of barium sulphate and barium carbonate are 1.0  10–10 and 8  10–9 mol2 dm–6 respectively. Ingestion of barium carbonate is dangerous while ingestion of barium sulphate is not. Explain why it is so. SUMMARY SUMMARY 1 Strong electrolytes dissociate completely in aqueous solution. 2 Weak electrolytes dissociate partially in aqueous solution. 3 The degree of dissociation of a weak electrolyte is inversely proportional to the square root of its concentration. 4 Arrhenius acid is a compound that dissolves in water to produce H+ ions. Arrhenius base is a compound that dissolves in water to produce OH– ions. 5 Brønsted-Lowry acid is a proton donor. Brønsted-Lowry base is a proton acceptor. 6 Lewis acid is a lone-pair electron acceptor. Lewis base is a lone-pair donor. 7 Ka and Kb are used to compare the relative strength of acids and bases. 8 pKa = – log Ka pKb = – log Kb 9 pH is defined as the negative logarithm to the base 10 of the H+ molar concentration in the solution. 10 pH = – log [H+] 11 Ionic product of water, Kw = [H+][OH– ] = 1.0  10–14 mol2 dm–6 at 298 K 12 For a conjugate acid-base pair: Ka  Kb = 1.0  10–14 or: pKa + pKa = 14 13 A buffer solution is a solution which has pH that does not change significantly on the addition of a little acid or base. 14 A buffer solution contains a mixture of a weak acid and its conjugate base, or a weak base and its conjugate acid. 15 For an acidic buffer, the pH is given by: pH = pKa – log [Acid] [Salt] pOH = pKb – log [Base] [Salt] 16 The buffer capacity of a buffer solution is the amount of acid or base that has to be added to it to alter its pH value by 1 unit. 17 Maximum buffer capacity occurs when the ratio of [Acid] [Salt] = 1 or [Base] [Salt] = 1. 18 The equivalent point of an acid-base titration is when the amount of acid and base present exactly neutralises one another. 19 End point of an acid-base titration is that point in time when the indicator changes colour. 20 The solubility product of a sparingly salt is defined as the product of the molar concentration of the constituent ions, in a saturated solution, each raised to the power of the stoichiometric coefficient. 21 Common ion effect is the decrease in the solubility of a sparingly soluble salt in a solution that contains an ion common to one of the ions from the dissociation of the salt.

301 Chemistry Term 1 STPM CHAPTER 7 1 The titration curve below is that of the reaction between aqueous ammonia and sulphuric acid. Volume of H2 SO4 P Q R S pH At which point on the curve is the solution a buffer solution with maximum buffer capacity? A P C R B Q D S 2 Barium hydroxide is a strong base. Which of the following represents the concentration of the species in a 0.50 mol dm–3 aqueous barium hydroxide? [Ba(OH)2 ] [Ba2+] [OH– ] [H+] A 0 0.50 0.50 0 B 0.25 0.25 0.10 1  10–13 C 0 0.50 0.10 1  10–13 D 0.50 0.50 0.50 2  10–14 3 The conjugate base of H2 PO4 – is A H3 PO4 C HPO4 2– B PO4 3– D H2 PO3 – 4 Which of the following statement is not true? A An acid is any substance that dissociates to produce H+ ions in aqueous solution. B The pH of water changes with temperature. C Water exhibits amphoteric property. D An aqueous solution of sodium carbonate is basic. 5 Which of the following aqueous mixtures cannot function as a buffer? A Ammonia and ammonium chloride B Sodium carbonate and sodium hydrogen carbonate C Potassium chloride and hydrochloric acid D Benzoic acid and sodium benzoate Objective Questions 6 Consider the dissociation of an acid HA: HA(aq) + H2 O(l)  H3 O+ + A– What would be the pH of the solution if the [HA] [A– ] ratio is 5.56? [The pKa of HA = 4.75] A 3 C 5 B 4 D 7 7 HSO4 – is a weaker acid than H2 SO4 because A HSO4 – is a covalent ion while H2 SO4 is an ionic compound. B it is more difficult to lose an H+ from a negative ion. C H2 SO4 undergoes complete ionisation in water. D HSO4 – is the conjugate base of H2 SO4 . 8 In which of the following reactions is the underlined compound acting as an acid? A NH3 + HCl → NH4 Cl B H2 O2 + 2I– + 2H+ → I2 + 2H2 O C 6CN– + Fe3+ → [Fe(CN)6 ]3– D H2 O + HCO3 – → H2 CO3 + OH– 9 Which of the following is not a Lewis acid? A Cu2+ C NH3 B AlCl3 D SO3 10 Aniline, C6 H5 NH2 , is an organic base. The pKb of aniline is 9.38. C6 H5 NH2 + H2 O  C6 H5 NH3 + + OH– Which statement is not true about aniline? A It is a Bronsted-Lowry base B Aniline is a weaker base than OH– C C6 H5 NH3 + is a weaker acid than H2 O D A mixture of aniline and C6 H5 NH3 Cl is a buffer solution 11 One of the buffer systems in human blood is: CO2 (aq) + H2 O(l)  H2 CO3 (aq)  H+(aq) + HCO3 – (aq) STPM PRACTICE 7

302 Chemistry Term 1 STPM CHAPTER 7 If CO2 is not efficiently removed from the blood, what would be the effect on the blood? A The blood pH would fall. B The blood pH would rise. C There is no effect. D The concentration of blood would decrease. 12 Consider the following ionic equilibrium. HIO3 (aq) + H2 O(l)  IO3 – (aq) + H3 O+(aq) pKa = 0.77 HNO2 (aq) + H2 O(l)  NO2 – (aq) + H3 O+(aq) pKa = 3.32 Which statement is correct about the species in the equations above? A IO3 – will oxidise HNO2 to NO2 – B NO2 – is a stronger acid than H3 O+ C IO3 – is a weaker base compared to NO2 – D HNO2 is a stronger acid than HIO3 13 The solubility product of Ca(OH)2 is 8.0 × 10–6 mol3 dm–9 at 25 °C. What is the pH of a saturated solution of Ca(OH)2 at the same temperature? A 1.59 B 1.89 C 12.1 D 12.4 14 Which of the following is true about buffer solutions? A It can resist changes in pH when an acid is added to it B It contains a mixture of a conjugate base and a conjugate acid C It has pH 7.0 D It has a maximum capacity when [acid] = [conjugate base] 15 The table below gives the ionic product of water at two different temperatures. Temperature/°C 25.0 80.0 Kw 1.0 × 10–14 5.6 × 10–14 Which statement is not true? A The dissociation of water is an endothermic process B The pH of water at 80.0°C is 6.6 C Water is neutral at 80.0°C D The H+ concentration at 80.0°C is 2.8 × 10–14 mol dm–3 16 The degree of dissociation of a 0.40 mol dm–3 HF in a 0.10 mol dm–3 solution of HCl is 0.0068. What is the Ka of HF? A 6.8  10–4 C 2.7  10–8 B 5.6  10–5 D 3.9  10–12 17 Which of the following solutions has the highest degree of dissociation? Solution Dissociation constant/mol dm–3 Molarity/ mol dm–3 A 4.6  10–4 0.0010 B 2.7  10–8 0.0014 C 1.8  10–5 0.12 D 6.8  10–4 0.011 18 In which of the following aqueous solutions would you expect BaSO4 to have the highest solubility? A 0.10 M BaCl2 B 0.25 M Na2 SO4 C 0.050 M H2 SO4 D 0.025 M sodium chloride 19 What happens when a little HCl is added to an aqueous solution of HF? A The F– concentration decreases. B The H+ concentration decreases. C Hydrogen fluoride will be precipitated. D There is no change. 20 The solubility product of copper(II) hydroxide is 4.8 × 10–20 mol3 dm–9 at 25°C. Cu(OH)2 (s)  Cu2+(aq) + 2OH– (aq) Which of the following is true? A The solubility of copper(II) hydroxide is 1.48 × 10–5 mol dm–3 at 25°C B Addition of ammonia increases the solubility of copper(II) hydroxide C Addition of concentrated HCl increases the solubility product of copper(II) hydroxide D The solubility of copper(II) hydroxide increases with the amount of solid present 21 What is the maximum concentration of CO3 2– in an aqueous solution containing Ag+ with concentration of 1.8  10–5 mol dm–3? [Ksp of Ag2 CO3 = 8.1  10–12] A 3.6  10–5 mol dm–3 B 4.5  10–7 mol dm–3 C 2.5  10–2 mol dm–3 D 0.15 mol dm–3

303 Chemistry Term 1 STPM CHAPTER 7 22 The solubility product of silver carbonate is 6.4  10–12 mol3 dm–9. What is the concentration of the carbonate ions in a saturated solution of silver carbonate? A 1.17  10–4 mol dm–3 B 2.34  10–4 mol dm–3 C 1.37  10–8 mol dm–3 D 2.70  10–8 mol dm–3 23 The pH of 3.5  10–4 mol dm–3 of a monoprotic acid is 3.11. What is the numerical value of the acid dissociation constant of the acid? A 7.76  10–4 B 1.74  10–3 C 3.50  10–2 D 1.29  10–1 24 In aqueous solution, ethanoate ions, CH3 COO– , are hydrolysed according to the equation: CH3 COO– (aq) + H2 O(l)   CH3 COOH(aq) + OH– (aq) What is the pH of a 0.50 mol dm–3 solution of sodium ethanoate, CH3 COONa, at 298 K? [Ka for CH3 COOH = 1.85  10–5 mol dm–3, Kw = 1.00  10–14 mol2 dm–6] A 4.78 B 5.41 C 9.22 D 10.19 25 Which of the underlined species reacts as a Lewis base? A CH3 COO– + H2 O → CH3 COOH + OH– B CaO + SO3 → CaSO4 C OH– + CO2 → HCO3 – D BeCl2 + 2NH3 → BeCl2 (NH3 )2 26 When solid sodium ethanoate is added to an aqueous solution of ethanoic acid, I the pH of the solution increases II the degree of dissociation of ethanoic acid decreases III the acid dissociation constant of ethanoic acid decreases A I only B III only C I and II D II and III 27 The base dissociation constants for two bases are given below: CH3 NH2 + H2 O  CH3 NH3 + + OH– Kb = 4.4 × 10–4 mol dm–3 N2 H4 + H2 O  N2 H5 + + OH– Kb = 1.0 × 10–6 mol dm–3 Which of the following statements about the species in the above equations are true? I N2 H5 + is a stronger acid than CH3 NH3 +. II CH3 NH2 and OH– is an acid-base conjugate pair. III CH3 NH2 is a stronger base than OH– . IV The acid dissociation constant for CH3 NH3 + is 2.3 × 10–11 mol dm–3. A I and IV C II, III and IV B II and III D I, II, III and IV 28 Which pair of solutions will yield a buffer solution with a pH of about 5 when solutions of equal volume and concentration are mixed? A NaOH and H2 SO4 B NH3 and (NH4 )2 SO4 C NaH2 PO4 and H3 PO4 D CH3 NH2 and CH3 NH3 Cl 29 The solubility product of lead(II) sulphate, PbSO4 is 1.6 × 10–8 mol2 dm–6 at 298 K. What will happen if a few drops of lead(II) ethanoate are added to a saturated solution of lead(II) sulphate at the same temperature? A The solubility product of lead(II) sulphate increases. B The concentration of Pb2+ ion is more than 1.6 × 10–8 mol dm–3. C The concentration of SO4 2– ion is less than 1.26 × 10–4 mol dm–3. D The white solid dissolves. 30 A mixture of hydrochloric(I) acid, HOCl, and sodium chlorate(I), NaOCl, is a buffer solution. Which statement(s) is/are true of the mixture? I The pH of the mixture is less than 7. II On the addition of a little base, the concentration of chlorate(I) increases. III On the addition of a little acid, the concentration of HOCl decreases. A I only C I and II B II only D I, II and III

304 Chemistry Term 1 STPM CHAPTER 7 31 The solubility of cadmium phosphate, Cd3 (PO4 )2 is x mol dm–3 at 25 °C. What is the expression of the solubility product of cadmium phosphate? A 6x2 C 36x5 B 12x3 D 108x5 32 A saturated solution of lead(II) hydroxide has a pH of 8.8. What is the numerical value of the solubility product of lead(II) hydroxide? A 1.0 × 10–18 C 1.8 × 10–16 B 1.3 × 10–16 D 1.3 × 10–8 33 Which of the following are the buffer that regulate the pH of the human blood? I CH3 COOH  CH3 COO– + H+ II H2 CO3  HCO3 – + H+ III NH3 + H2 O  NH4 + + OH– IV H2 PO4 –  HPO4 2– + H+ A I and III C I, II and IV B II and IV D I, II, III and IV 34 Which mixture of ammonia and ammonium chloride will produce a buffer solution of maximum buffer capacity? [NH3 ] [NH4 Cl] A 0.6 0.8 B 0.9 0.2 C 0.8 0.8 D 0.2 1.0 35 The solubility product of silver sulphate, Ag2 SO4 is x mol3 dm–9 at 298 K. What is the concentration of the silver ion, in mol dm–3, for a saturated solution of silver sulphate at the same temperature? A 2x2 B 4x3 C 3 x 4 AB D 3 AB2x Structured and Essay Questions 1 (a) Define solubility product. (b) 20.0 cm3 of a saturated solution of calcium hydroxide requires 20.2 cm3 of 0.045 M hydrochloric acid for neutralisation. (i) Calculate the pH of saturated calcium hydroxide. (ii) Write an expression for the solubility product of calcium hydroxide. (iii) Calculate the solubility product of calcium hydroxide. State its unit. (iv) Explain why calcium hydroxide is not precipitated when aqueous ammonia is added to aqueous calcium nitrate. (v) In what volume of water will 2.33  10–3 g of calcium hydroxide just dissolve at the temperature of the experiment? 2 Citric acid (represented by RCOOH) together with its sodium salt (represented by RCOONa) is used as an acid regulator. (a) Assuming that citric acid is a monoprotic acid and dissociates in water according to the equation: RCOOH + H2 O  RCOO– + H3 O+ (i) Write an expression for the acid dissociation constant for the acid. (ii) The acid dissociation constant for citric acid is 7.5  10–4 mol dm–3 at 300 K. Calculate the pH of lime juice which contains 0.18 mol dm–3 of citric acid. (b) A mixture of citric acid and its sodium salt acts as a buffer. (i) What is the function of a buffer solution? (ii) Using balanced equations to show how the mixture acts as a buffer. (iii) Calculate the pH of a buffer solution containing 0.18 mol dm–3 of citric acid and 0.35 mol dm–3 of sodium citrate. (c) Explain why the pH you obtained in (b)(iii) differs from the value in (a)(ii).

305 Chemistry Term 1 STPM CHAPTER 7 3 Propanoic acid, CH3 CH2 COOH, is a Bronsted-Lowry acid. (a) Using propanoic acid as example, explain what is meant by Bronsted-Lowry acid. (b) An aqueous mixture of propanoic acid and sodium propanoate, CH3 CH2 COONa, acts as a buffer solution. Explain the action of the buffer solution on the addition of a little acid or base. (c) Calculate the mass of sodium propanoate that needs to be added to 500 cm3 of 0.050 mol dm–3 propanoic acid to produce a buffer with pH = 5.9. [The acid dissociation constant of propanoic acid = 1.34 × 10–5 mol dm–3] 4 Phenylboronic acid, C6 H5 B(OH)2 and benzoic acid, C6 H5 COOH are both weak monoprotic acids. The pKa for the two acids are 8.8 and 4.2 respectively. (a) What is the relationship between pKa of an acid and its acid dissociation constant, Ka ? (b) Give the equations for the dissociation of phenylboronic acid and benzoic acid. (c) Calculate the pH of a 0.12 mol dm–3 phenylboronic acid. (d) Would the pH of a 0.12 mol dm–3 benzoic acid higher or lower than the value you obtained in (c)? Explain your answer. (e) A 25.0 cm3 sample of a mixture of the two acids was titrated with aqueous sodium hydroxide of concentration 0.10 mol dm–3. It was found that 12.0 cm3 of sodium hydroxide is required to change the colour of the first indicator, and a further 9.4 cm3 was needed to change the colour of the second indicator. (i) Sketch the pH curve of the titration. Include in the diagram the appropriate volumes and pH values. (ii) Calculate the molar concentration of phenylboronic acid and benzoic acid in the mixture. 5 Explain the following observations. (a) Aqueous ammonium chloride has a pH of less than 7. (b) Benzoic acid is a weak acid. (c) Lead(II) chloride is insoluble in dilute hydrochloric acid but dissolves in concentrated hydrochloric acid. (d) When aqueous ammonia is added to aqueous magnesium chloride, a white precipitate is obtained. However, the white precipitate dissolves on the addition of solid ammonium chloride. 6 A weak monoprotic acid, HA, has a pH of 3.10. (a) Define pH. (b) Calculate the concentration of hydrogen ion in the sample of the acid. (c) 25.00 cm3 of HA requires 22.50 cm3 of 0.100 mol dm–3 sodium hydroxide for complete neutralisation. (i) Suggest an indicator that is suitable for the titration. (ii) Calculate the concentration of HA. (d) Explain why the value in (b) and (c)(ii) are different. (e) Calculate the acid dissociation constant, Ka of HA. (f) Sketch a titration graph for the titration in (c).

CHAPTER PHASE EQUILIBRIA 8 Concept Map Non-ideal Solution • Negative deviation from Raoult’s Law • Solutions that show positive deviation • Fractional distillation under reduced pressures Mixture of Two Miscible Liquids • Vapour pressure and Raoult’s Law Miscible Liquids Ideal Solution • Composition of vapour in equilibrium with an ideal solution • Vapour pressure/composition graph • Boiling point and vapour pressure • Boiling point and solutions • Fractional distillation Phase Equilibria Students should be able to: Phase equilibria • state and apply Raoult’s law for two miscible liquids; • interpret the boiling point-composition curves for mixtures of two miscible liquids in terms of ‘ideal’ behaviour or positive or negative deviations from Raoult’s law; • explain the principles involved in fractional distillation of ideal and non-ideal liquid mixtures; • explain the term azeotropic mixture; • explain the limitations on the separation of two components forming an azeotropic mixture; • explain qualitatively the advantages and disadvantages of fractional distillation under reduced pressure. Learning earning Outcomes

307 Chemistry Term 1 STPM CHAPTER 8 8.1 Miscible Liquids 1 Examples of miscible liquids are water and ethanol, benzene and methylbenzene. 2 One convenient way to express the relative amount of each component in the solution is by using the mole fraction. 3 The mole fraction of a liquid A in a mixture of liquids A and B is given by No. of moles of A (No. of moles of A) + (No. of moles of B) Or, XA = nA nA + nB where X = mole fraction n = number of moles Example 8.1 A solution contains 2.6 moles of benzene and 0.98 moles of cyclohexane. Calculate the mole fraction of benzene and cyclohexane in the mixture. Solution Mole fraction of benzene = 2.6 (2.6 + 0.98) = 0.726 ∴ Mole fraction of cyclohexane = 1 – 0.726 = 0.274 Example 8.2 A mixture contains 10 g of water and 10 g of ethanol. Calculate the mole fraction of ethanol in the mixture. Solution No. of moles of H2 O = 10 18 = 0.556 mol No. of moles of C2 H5 OH = 10 46 = 0.217 mol Mole fraction of C2 H5 OH = 0.217 (0.217 + 0.556) = 0.556 Definition of mole fraction Sum of all the mole fractions of all the components = 1 2010/P1/Q16

308 Chemistry Term 1 STPM CHAPTER 8 Exam Tips Quick Check 8.1 1 A mixture contains 56 g of benzene, C6 H6 , and 72 g of methylbenzene, C6 H5 CH3 . Calculate the mole fraction of benzene in the mixture. 2 An aqueous solution contains 100 g of water and 34.5 g of methanol, CH3 OH. (a) Calculate the mole fraction of methanol in the mixture. (b) Calculate the mass of methanol that needs to be added to the above solution so as to increase the mole fraction of methanol to 0.85. 3 The mole fraction of methylbenzene, C6 H5 CH3 , in a mixture containing 100 g of methylbenzene and an unknown mass of ethylbenzene, C6 H5 CH2 CH3 is 0.438. Calculate the mass of ethylbenzene in the mixture. Exam Tips The vapour pressure of a pure liquid depends only on temperature. The vapour pressure exerted by vapour A is called the partial pressure of A. A statement of Raoultʼs law xA + xB = 1 A(g) + B(g) PA = ? PB = ? A(I) + B(I) 8.2 Mixture of Two Miscible Liquids 1 A solution formed from mixing two miscible liquids can be an ideal solution or a non-ideal solution. 2 An ideal solution is one that obeys Raoult’s Law. Vapour Pressure and Raoult’s Law 1 Consider two pure liquids A and B. In their pure forms, each liquid exerts a vapour pressure which is dependent on the temperature and not on the amount of liquid present, as long as there is sufficient liquid to establish equilibrium with its vapour. Vapour of A Liquid A Vapour of B Liquid B 2 Let us say that the vapour pressure of pure A is PA o and that of pure B is PB o at room temperature. 3 Now, if we were to mix the two liquids together to form a solution, what will be the vapour pressure exerted by A, and what will be the pressure exerted by B in the mixture? 4 If the solution is an ideal solution, then the vapour pressure of A or B is given by Raoult’s law. 5 Raoult’s law states that the partial vapour pressure of liquid A in a solution is equal to the product of the vapour pressure of pure A and the mole fraction of A in the mixture, at the same temperature. PA = XAPA o 6 Similarly, PB = XBPB o The total pressure of the solution = PA + PB

309 Chemistry Term 1 STPM CHAPTER 8 7 However, for a non-ideal solution, the vapour pressures exerted by the component liquids are either too high or too low from that calculated from Raoult’s Law. Ideal Solutions 1 An ideal solution is formed if the bonds holding the different molecules together in the mixture are of the same nature and strength as those holding like molecules together in the pure liquids. A ------ B ≈ A -----A and B -----B 2 Since energy is neither absorbed nor released during mixing, there is no enthalpy change. Also there is no change in the total volume on mixing as the molecules are neither being pulled closer (as would be the case if the bonds in the mixture are stronger than those in the pure liquids) or drifted further apart (as would be the case if the bonds in the mixture are weaker than those in the pure liquids). 3 The tendency for the molecules to escape as vapour in the mixture is the same as that in their pure liquids. 4 Consider the following situation of a mixture of two liquids, A and B. In the mixture, a part of the liquid surface is occupied by molecules of A as well as B. The fraction of the total surface for each liquid to evaporate is reduced. As a result, the vapour pressure exerted by the liquids in the mixture is less than the vapour pressures of the pure liquids at the same temperature. 5 For example, a mixture containing 0.5 mole of A and 0.8 mole of B at 300 K. [P°A = 15.0 kPa and P°B = 18.0 kPa] The vapour pressure of A in the mixture = 0.5 (0.5 + 0.8)  15 kPa = 5.77 kPa The vapour pressure of B in the mixture = 0.8 (0.5 + 0.8)  18 = 11.1 kPa The total pressure of the mixture = (5.77 + 11.1) kPa = 16.87 kPa Non-ideal solutions Exam Tips Exam Tips On mixing: Enthalpy change = 0 Volume change = 0 The surface is occupied by both A molecules and B molecules. A(g) + B(g) A(l) + B(l) Molecules of A Molecules of B 2011/P1/Q11 Quick Check 8.2 1 Calculate the total pressure of a benzene-octane mixture if the mole fraction of octane is 0.36. [P o (benzene) = 45.6 kPa and P o (octane) = 23.6 kPa at the temperature of the experiment] 2 X and Y are two miscible liquids which form an ideal solution. The vapour pressures, at 298 K, of X and Y are 23.6 kPa and 46.2 kPa respectively. (a) What do you understand by the term ideal solution? (b) Calculate the total pressure of a mixture containing: (i) 2.3 moles of X and 4.5 moles of Y, (ii) equimolar quantities of X and Y, (iii) 0.82 mole fraction of X.

310 Chemistry Term 1 STPM CHAPTER 8 Composition of Vapour in Equilibrium with an Ideal Solution 1 Consider a liquid mixture containing equimolar quantities of A and B that obeys Raoult’s law. 2 Let us assumed that the vapour pressures of pure A and B at 300 K are 40 kPa and 60 kPa. 3 Since the vapour pressure of B is higher than that of A at the same temperature, B is more volatile than A. 4 The vapour pressures of A and B in the mixture are calculated using Raoult’s law: PA = 0.5  40 = 20 kPa PB = 0.5  60 = 30 kPa Thus, the total pressure = (20 + 30) kPa = 50 kPa 5 Now, what is the composition of the vapour? Is it the same as that of the liquid mixture? 6 The composition of the vapour can be calculated using Dalton’s Law of partial pressure. XA(g) = PA PT [where PT = total pressure] ∴ XA(g) = 20 50 = 0.40 And, XB(g) = 30 50 = 0.60 [Or XB(g) = 1 – 0.40 = 0.60] 7 This can be summarised in the diagram: 8 It can be seen that the vapour contains a higher percentage of B (the more volatile component) than its liquid mixture. This is because B is more volatile than A, and under the same conditions, more B will evaporate compared to A. XA (g) = 0.40; XB (g) = 0.60 XA (I) = 0.50; XB (I) = 0.50 Liquid Vapour Quick Check 8.3 1 Benzene and methylbenzene form an ideal solution. Calculate the composition of the vapour that is in equilibrium with the following liquid mixtures at 60 °C: [P o benzene = 80 kPa and P o methylbenzene = 64 kPa at 60 °C] (a) 0.25 mol benzene and 0.70 mol methylbenzene (b) 0.86 mol benzene and 0.35 mol methylbenzene 2 Hexane and heptane form an ideal solution. For equimolar quantities of hexane and heptane, calculate (a) the total pressure of the mixture, (b) the mole fractions of hexane and heptane in the vapour that is in equilibrium with the liquid mixture, at 300 K. [Vapour pressures of heptane = 16.5 kPa; hexane = 35.7 kPa, at 300 K]

311 Chemistry Term 1 STPM CHAPTER 8 Vapour Pressure/Composition Graph 1 From the examples above, we see that the total pressure of a mixture of two immiscible liquids, at constant temperature, is dependent on the composition of the solution. 2 Using Raoult’s law and Dalton’s law of partial pressure, we can work out how the total pressure and the composition of the vapour vary with the composition of the solution. 3 The table below lists the data for a benzene-methylbenzene mixture at T K. Xbenzene(l) 0.0 0.2 0.4 0.6 0.8 1.0 Pbenzene/kPa 0.0 12.0 24.0 36.0 48.0 60.0 Xmethylbenzene(l) 1.0 0.8 0.6 0.4 0.2 0.0 P methylbenzene /kPa 40.0 32.0 24.0 16.0 8.0 0.0 PTotal 40.0 44.0 48.0 52.0 56.0 60.0 Xbenzene(g) 0.0 0.27 0.50 0.69 0.86 1.0 4 A graph of vapour pressure against composition of the solution, at constant temperature, is shown below. 7 6 5 4 3 2 1 0.2 0.4 0.6 0.8 1.0 0 8 9 Xbenzene Pressure (fi 104 )/Pa Ptotal Pbenzene Pmethylbenzene 5 For an ideal solution, the total pressure curve is a straight line joining the vapour pressures of the two liquids. Exam Tips Exam Tips Xbenzene(g) = Pbenzene PTotal Info Chem In practice, PTotal is a slight curve as there is no such solution as a perfect ideal solution.

312 Chemistry Term 1 STPM CHAPTER 8 6 A vapour pressure/composition graph for both the liquid and vapour, at constant temperature is shown below: Example 8.3 The graph below shows the vapour pressure/composition curve (at 300 K) for a mixture of two liquids X and Y which forms an ideal solution. 30 20 10 0.2 0.4 0.6 0.8 1.0 0 40 50 Mole fraction of X Pressure (× 104)/Pa b(ii) b(i) (a) Which liquid is more volatile, X or Y? Explain your answer. (b) What is the composition of the vapour in equilibrium with a liquid mixture containing (i) 50% mole fraction, and (ii) 30% mole fraction, of X? (c) What conclusion can you infer from the answers in (b)(i) and (ii)? (d) Determine the partial pressure of Y in a mixture containing 60% mole fraction of X. 40 30 0.20 0.4 0.6 0.8 1.0 Pressure/kPa Liquid line Vapour line Mole fraction of benzene 50 60 Mole fraction of benzene in solution Mole fraction of benzene in vapour 48 Pressure/kPa Liquid line Vapour line 44 0.2 0.27 0 0.4 0.50

313 Chemistry Term 1 STPM CHAPTER 8 Solution (a) X. It has a higher vapour pressure than Y at the same temperature. (b) (i) 62% X and 38% Y (ii) 42% X and 58% Y (c) The vapour contains a higher percentage of X compared to the liquid mixture. (d) Vapour pressure of pure Y = 45 kPa ∴ Partial pressure of Y = 0.4  45 Pa = 18 Pa Definition of boiling point Ethanol is more volatile than water. Quick Check 8.4 Two liquids P and Q form an ideal solution. The vapour pressures of P and Q are 65 kPa and 40 kPa at 298 K. (a) Plot on a graph paper; (i) The vapour pressure/composition curve for mixtures of P and Q. (ii) Sketch on the same graph, the line that corresponds to the composition of the vapour in the mixtures. (b) (i) What is the total vapour pressure of a mixture containing 0.4 mole fraction of P? (ii) What is the composition of the vapour that is in equilibrium with the mixture at the same temperature? (c) The vapour of a mixture contains 0.7 mole fraction of P. What is the composition of the solution? Boiling Point and Vapour Pressure 1 When a liquid is heated, the vapour pressure of the liquid increases as more liquid particles have sufficient energy to escape as vapour. 2 When the vapour pressure of the liquid reaches 1 atm (or 101 kPa), the liquid boils. 3 The boiling point of a liquid is defined as the temperature when the vapour pressure of the liquid is equal to the external pressure (usually 1 atm or 101 kPa). 4 The more volatile a liquid (that is one with a higher vapour pressure at room temperature), the lower is the boiling point and vice versa. The boiling points of water and ethanol are shown in the diagram. Temperature/°C Pressure/atm 1 78.5 Ethanol Water 100

314 Chemistry Term 1 STPM CHAPTER 8 Boiling Points and Solutions 1 Consider a mixture of benzene and methylbenzene. Benzene is more volatile than methylbenzene and hence its boiling point is lower than that of methylbenzene. Liquid Boiling point/°C Benzene 80.0 Methylbenzene 110 2 The total vapour pressure of mixtures of benzene and methylbenzene of any composition is less than the vapour pressure of pure benzene, but higher than that of pure methylbenzene. Hence, their boiling points are between 80.0 °C and 110 °C. 3 A graph of boiling point against composition of a benzenemethylbenzene mixture is shown in the diagram: Mole fraction of benzene Temperature(°C) 0 1.0 80 110 4 We can add to the graph above another curve which corresponds to the composition of the vapour in equilibrium with a solution of a particular composition at the boiling point of the solution. Keeping in mind that, the vapour will contain a higher percentage of benzene (the more volatile component) compared to the liquid mixture at the same temperature. 5 Note that the vapour pressure/composition and boiling point/ composition graphs are reflection of one another. This is because, the higher the vapour pressure, the lower is the boiling point. Composition Vapour pressure Vapour Liquid 0 100 % Benzene Composition % Benzene Boiling point/°C Liquid 110 80 0 100 Vapour Mole fraction of benzene Vapour pressure 0 1.0 Mole fraction of benzene Temperature(°C) 0 Vapour t 1 a b Liquid 1.0 80 110 At temperature t1 , a solution with composition a will have a vapour with composition b.

315 Chemistry Term 1 STPM CHAPTER 8 Example 8.4 The boiling point/composition curve of a hexane-octane mixture is as shown below: 70 0 Boiling point/ o C 0.2 Mole fraction of hexane 0.4 0.6 0.8 1.0 (b)(i) (b)(ii) (a) (c)(i) c(ii) 80 90 100 110 120 (a) Use your graph to determine the following. (a) THe boiling point of hexane and octane. (b) THe boiling point of a liquid mixture of the following composition: (i) 20% mole fraction of octane (ii) 10% mole fraction of hexane (c) (i) The boiling point of a mixture containing equimolar quantity of hexane and octane. (ii) The composition of the vapour in equilibrium with the liquid mixture at its boiling point. Solution (a) Hexane: 70.0 °C Octane: 125.0 °C (b) (i) 78.0 °C (ii) 118.0 °C (c) (i) 92.0 °C (ii) 0.68 mole fraction of hexane and 0.32 mole fraction of octane.

316 Chemistry Term 1 STPM CHAPTER 8 8.3 Fractional Distillation 1 Fractional distillation is a technique used to separate liquid mixtures based on their boiling points. 2 Let us look at a mixture of A and B which obeys Raoult’s law. Assuming that A is more volatile than B. The boiling point/ composition graph is shown in the diagram. Since A is more volatile than B, the boiling point of A is lower than that of B. 3 Starting with a mixture with composition C1 . This mixture will boil at temperature t 1 , and the vapour given off when condensed will produce a liquid of composition C2 which is richer in A compared to the original solution. Vapour Condensation Composition C Composition C1 2 What is fractional distillation? % A Temperature TA TB C t 1 t 2 1 C2 C3 C4 0 100 Quick Check 8.5 1 Two liquids X and Y (which form an ideal solution) have vapour pressures of 35 kPa and 46 kPa respectively at 28 °C. (a) Sketch (i) the vapour pressure/composition graph, (ii) the boiling point/composition graph, showing both the liquid and vapour line. (b) By drawing appropriate lines to your sketch in (a)(ii), show how you would determine the (i) boiling point, and (ii) the composition of the vapour, of a solution containing equimolar quantities of X and Y. 2 The table below gives information for mixtures of two miscible liquids X and Y. Mole fraction of X in solution 0.0 0.2 0.3 0.8 1.0 Boiling point/°C 90.0 82.0 78.0 64.0 60.0 Mole fraction of X in vapour 0.0 0.37 0.50 0.88 1.0 (a) Plot a boiling point/composition graph for mixtures of X and Y. (b) For a liquid mixture containing 0.5 mole fraction of X, determine (i) the boiling point of the mixture, and (ii) the mole fraction of X in the vapour above the liquid mixture at the boiling point of the solution. INFO Fractional Distillation

317 Chemistry Term 1 STPM CHAPTER 8 4 The liquid of composition C2 , when heated, will boil at temperature t 2 , and the vapour given off when condensed forms a solution of composition C3 . Vapour Condensation Composition C Composition C2 3 5 This solution, when boiled, will give a vapour composition C4 that condenses to a liquid of composition C4 . 6 With repeated boiling and condensing, the percentage of A in the vapour increases after every step. 7 Eventually, the vapour given off will contain only A, which when condensed produces pure liquid A. Vapour Condensation Composition C Pure A n 8 As more A is removed from the solution, the percentage of B remaining in the flask increases. When all A is removed, pure B will be left in the flask. A and B are thus separated. 9 However, this process of boiling and condensing is tedious and time consuming. Instead a fractional distillation column is used. 10 A simplified diagram of a fractional distillation column is shown below. Fractionating column Round-bottomed flask Thermometer Condenser Heat Out In

318 Chemistry Term 1 STPM CHAPTER 8 11 The fractionating column is usually made of glass and packed with small glass beads or other suitable substances to increase the surface area where the hot vapour can condense. 12 When the mixture of composition C1 is boiled, the vapour rises up the column and condenses on the glass beads to produce a liquid of composition C2 . 13 This liquid would then be heated by the ascending hot vapour and will boil to produce a vapour of composition C3 . 14 This process is repeated in the fractionating column. 15 As the vapour ascends the column, it gets richer in A. When the vapour reaches the top of the column, it contains only pure A which distilled over at temperature TA. 16 As more and more A distilled over, the liquid remaining in the flask gets richer in B, until only pure B is left and will distill over at temperature TB. 17 The variation of the thermometer reading with time during the distillation process is shown below. Temperature/ °C Time A distills over T B distills over B TA 18 To get an efficient separation, the following features should be observed. (a) The column must be long enough. This is especially important for liquids having very close boiling points. (b) The rate of distillation should be carried out slowly (by heating the mixture slowly) to enable the liquid and vapour to come into equilibrium. (c) The column should be lagged on the outside to prevent heat lost. Example 8.5 Water and methanol form an ideal solution. The vapour pressures of water and methanol are 2.4 kPa and 12.5 kPa respectively at 298 K. (a) Which liquid is expected to have a higher boiling point? (b) Sketch the vapour pressure/composition diagram for the water/methanol mixture. Label both the liquid and vapour lines. (c) Sketch the boiling point/composition diagram for the water/ methanol mixture. (d) A mixture of water and methanol containing 0.8 mole fraction of water is subjected to fractional distillation. Name the liquid that distilled over first. Large surface area available for the hot vapour to condense. The vapour ascending the column gets richer in component A (the more volatile component). Features of a good distillation column Exam Tips Exam Tips The component with lower boiling point will distil over first.

319 Chemistry Term 1 STPM CHAPTER 8 Solution (a) Water (because it has a lower vapour pressure compared to methanol). [The boiling points of water and methanol are 100 °C and 65 °C respectively.] (b) (c) Mole fraction of water Vapour pressure Liquid Vapour 0 1 Mole fraction of water Boiling point/°C 100 Vapour Liquid 65 0 1 (d) Methanol (with a lower boiling point) will be distilled over first. 8.4 Non-ideal Solutions 1 As mentioned earlier, there are some liquid mixtures that do not obey Raoult’s Law. These are called non-ideal solutions or solutions that show deviation from Raoult’s law. 2 Non-ideal solutions can be equated to non-ideal gas, which are gases that do not obey the ideal gas law, pV = nRT. 3 There are two types of deviation: positive deviation and negative deviation. 4 If the measured total vapour pressure of the solution is lower than that calculated from Raoult’s law, it is said to show negative deviation from Raoult’s law. PT (actual)  PT(theoretical) 5 If the measured total vapour pressure of the solution is higher than that calculated from Raoult’s law, it is said to show positive deviation from Raoult’s law. PT(actual)  PT(theoretical) Negative Deviation from Raoult’s Law 1 A pair of miscible liquids, A and B, will show negative deviation if the forces holding A and B together in the mixture is stronger than the forces holding A and A or B and B in their respective pure liquids. A B  A A or B B 2 As a result, the molecules in the mixture have a lower tendency to escape as vapour compared to that in their pure liquids. Their partial pressures are thus lower than that predicted by Raoult’s law. Negative deviation Non-ideal solution: Negative deviation and positive deviation Positive deviation 2015/P1/Q15 2016/P1/Q17 2008/P1/Q41 2009/P2/Q2(a) 2013/P1/Q13

320 Chemistry Term 1 STPM CHAPTER 8CHAPTER 8 3 Due to the stronger intermolecular forces in the mixture, the molecules are pulled closer to one another compared to that in their respective pure liquids. The mixing process is accompanied by a decrease in volume. 4 Since less energy is absorbed to break the bonds holding the molecules of the pure liquids, and more energy is released when new bonds (which are stronger) are formed in the mixture, the mixing process is exothermic. There is an increase in temperature. 5 An example of a mixture that shows negative deviation from Raoult’s law is a mixture of trichloromethane (chloroform) and propanone (acetone). 6 Both are polar molecules. The intermolecular forces between their respective molecules are the weak van der Waals forces. van der Waals force 3 CCH CH3 O 3 CCH CH3 O van der Waals force CHCl3 CHCl3 7 However, in the mixture, the intermolecular force is as strong as hydrogen bonds. CH3 C CH3 O Cl H Cl C Hydrogen bond Cl 8 The vapour pressure-composition graph of a solution that shows negative deviation has a minimum while the boiling pointcomposition graph shows a maximum. [Assuming that B is more volatile than A]. Exam Tips Negative deviation: (∆V = negative), heat is given off (∆H = negative). Mole fraction of A Pressure 0 1 P° A P° B Minimum vapour pressure Ideal solution z Mole fraction of A Temperature 0 1 Maximum boiling point z Vapour Liquid Liquid Vapour Info Chem The electron withdrawing effect of 3 chlorine atoms is equivalent to that of one fluorine atom.

321 Chemistry Term 1 STPM CHAPTER 8CHAPTER 8 9 The mixture with composition z which has the highest boiling point compared to other mixtures is called the azeotropic mixture or azeotrope. 10 Other examples of maximum boiling point mixture are: Components Boiling point/°C Azeotrope % by mass Boiling point/°C Hydrogen chloride Water –84.0 100.0 20.2 79.8 108.6 Nitric acid Water 86.0 100.0 68.0 32.0 120.5 Hydrogen fluoride Water 19.4 100.0 35.6 64.4 111.4 Trichloromethane Propanone 61.2 56.2 80.0 20.0 64.7 Example 8.6 (a) Explain why a mixture of nitric acid and water forms a non-ideal solution that shows negative deviation from Raoult’s law. (b) Sketch the boiling point/composition graph for the nitric acid-water system. Solution The intermolecular forces between pure nitric acid molecules, and between pure water molecules are the hydrogen bonds. N O O O O N O O H O H H H H H O H O H However, in the mixture, nitric acid dissociates and combined with water to form hydrated ions: HNO3 (l) + H2 O(l)  H3 O+(aq) + NO3 – (aq) The attractions between the ions are stronger than the hydrogen bonds in water and nitric acid. The tendency for the H2 O molecules and HNO3 molecules to escape as vapour is reduced. This results in the vapour pressure of the mixture being lower than expected. % HNO3 Temperature/°C 0 100 120.5 86 68 100 Exam Tips The boiling point of B is lower because it is more volatile.

322 Chemistry Term 1 STPM CHAPTER 8 Fractional Distillation of Solutions That Show Negative Deviation 1 The boiling point/composition graph of the water/nitric acid system is shown below. % HNO3 Temperature 0 68 120.5 A 100 100 % H2 100 O 32 0 B 86 2 When a mixture of nitric acid and water is subjected to fractional distillation, the products of distillation depend on whether the composition of mixture is more or less than 68% HNO3 by mass. Distilling a Mixture That Contains Less Than 68% Nitric Acid by Mass (Region A) % HNO3 Temperature 0 68 120.5 A 100 100 % H2 100 O 32 0 B 1 When a mixture containing less than 68% by mass of nitric acid is subjected to fractional distillation, the vapour given off contains more and more water as it ascends the column (follow the arrows in diagram above). 2 Eventually, the vapour that reaches the top of the distillation column contains only pure water which distill over at 100 °C. 3 As more and more water distilled over, the concentration (or the percentage) of the nitric acid remaining in the distillation flask increases. 4 When the concentration of the acid remaining in the flask increases to 68%, the whole mixture will distill over at 120.5 °C. 5 The products of the distillation are pure water, followed by the azeotrope. H2 O is distilled over first because it has a lower boiling point than the azeotrope.

323 Chemistry Term 1 STPM CHAPTER 8 Distilling a Mixture That Contains More Than 68% Nitric Acid by Mass (Region B) % HNO3 Temperature 0 68 120.5 A 100 % H2 100 O 32 0 B 86 1 When a mixture containing more than 68% by mass of nitric acid is subjected to fractional distillation, the vapour given off contains more and more nitric acid as it ascends the column (follow the arrows in diagram above). 2 Eventually, the vapour that reaches the top of the distillation column contains only pure nitric acid which distills over at 86 °C. 3 As more and more nitric acid distilled over, the concentration (or the percentage) of the nitric acid remaining in the distillation flask decreases. 4 When the concentration of the acid remaining in the flask decreases to 68%, the whole mixture will distill over at 120.5 °C. 5 The products of the distillation are pure nitric acid, followed by the azeotrope. Distilling a Mixture Containing 68% Nitric Acid by Mass (Point Z) % HNO3 Temperature 0 68 120.5 A 100 % H2 100 O 32 0 B Z 1 When a mixture containing 68% by mass of nitric acid is subjected to fractional distillation, the vapour given off has the same composition as the solution. 2 There is no change in the composition of the vapour as distillation continues. What distills over is the azeotrope at 120.5 °C. HNO3 is distilled over first because it has a lower boiling point than the azeotrope. The azeotropic mixture is distilled unchange.

324 Chemistry Term 1 STPM CHAPTER 8 3 An azeotrope is defined as a liquid mixture that has a constant boiling point and is distilled unchanged. 4 The products of fractional distillation is summarised in the table below. Composition First distillate Second distillate  68% nitric acid Water Azeotrope  68% nitric acid Nitric acid Azeotrope = 68% nitric acid Azeotrope Azeotrope % Nitric acid Temperature 0 68 120.5 A 100 100 100 % Water 32 0 B Pure water followed by 86 the azeotrope Pure nitric acid followed by the azeotrope Example 8.7 The graph below shows the boiling point/composition curve for two miscible liquids A and B. 50 40 200 40 60 80 100 Temperature/o C % A by mass 60 70 80 90 50 40 60 70 80 90 v (e) (e) (d)(ii) (d)(i) v l l (a) Which is more volatile, A or B? Explain your answer. (b) Label the liquid and vapour line on the graph. (c) What are the composition and the boiling point of the azeotrope? Definition of ‘azeotrope’

325 Chemistry Term 1 STPM CHAPTER 8 (d) (i) What is the boiling point of a mixture containing 60% A? (ii) What is the percentage of A in the vapour on boiling this mixture? (e) Give the composition of two mixtures that have a boiling point of 80 °C. (f) What will be the composition of the first and last distillate when mixtures with the following composition are subjected to fractional distillation? (i) 30% A (ii) 30% B (iii) 40% A (g) Describe the change (if any) in temperature and volume on mixing A and B. Solution (a) A. Because it has a lower boiling point (b) On the graph (c) Boiling point = 90.0 °C Composition: 40% A and 60% B (d) (i) 73.0 °C (ii) 71.0% A (e) 22% A and 52% A (f) First distillate Last distillate (i) Pure B The azeotrope (ii) Pure A The azeotrope (iii) The azeotrope The azeotrope (g) Volume decreases; Temperature increases Example 8.8 The boiling point/composition graph for a pair of miscible liquids X and Y is given below. 100 50 40 200 40 60 80 Temperature/o C Percentage of Y by mass 60 70 80

326 Chemistry Term 1 STPM CHAPTER 8 A mixture containing 50 g of X and 50 g of Y is distilled by fractional distillation. Calculate the mass of pure X that will be collected. Solution Distillation of the above mixture will produce pure X and the azeotrope (80% Y and 20% X). Let the mass of pure X distilled over = a g ∴ Mass of X left in the azeotrope = 50 – a g 50 50 – a = 80 20 ∴ a = 37.5 g Solutions that Show Positive Deviation 1 A pair of liquids, A and B, will show positive deviation if the forces holding A and B together in the mixture is weaker than the forces holding A and A or B and B together in their respective pure liquids. A B  A A or B B 2 As a result, the tendency for the molecules to escape as vapour in the mixture is higher compared to their pure liquids. The vapour pressure produced is higher than that predicted by Raoult’s law. 3 On mixing, temperature decreases. This is because more energy is absorbed to break the stronger bonds holding the molecules of the pure liquids, but less energy is released when weaker bonds are formed in the mixture. The mixing process is endothermic. 4 Weaker bonds allow the molecules to drift further apart. Thus mixing is accompanied by an increase in volume. 5 The vapour pressure-composition graph of a solution that shows positive deviation shows a maximum while the boiling pointcomposition graph shows a minimum. [Assuming that A is more volatile than B]. Mole fraction of A Pressure 0 1 P° A P° B Maximum vapour pressure Ideal solution z pH 7 22.5 V Temperature Minimum boiling point Mole fraction of A 0 z 1 Exam Tips Exam Tips Positive deviation: ∆V = positive ∆H = positive In the azeotrope: Mass of Y ————–– Mass of X = 50 ——— 50 – a 2014/P1/Q12 2017/P1/Q15 2018/P1/Q15

327 Chemistry Term 1 STPM CHAPTER 8 6 The mixture with composition z which has the lowest boiling point compared to other mixtures is called an azeotropic mixture or azeotrope. 7 Other examples of minimum boiling point mixture are: Components Boiling point/°C Azeotrope % by mass Boiling point/°C Benzene Ethanol 80.0 78.5 67.6 32.4 67.8 Benzene Water 80.0 100.0 91.0 9.0 69.4 Ethanoic acid Water 118.1 100.0 3.0 97.0 76.6 Ethanoic acid Benzene 118.1 80.0 2.0 98.0 80.1 Ethanol Water 78.5 100.0 95.6 4.4 78.2 Trichloromethane Water 61.2 100.0 97.0 3.0 56.3 Tetrachloromethane Ethanol 76.8 78.5 84.2 15.8 65.0 Example 8.9 Explain why a mixture of benzene and ethanol shows positive deviation from Raoult’s law. Solution The intermolecular force between molecules of ethanol is the hydrogen bonds, while the intermolecular force between benzene molecules is the van der Waals force. C2 H5 OH Hydrogen bond C2 H5 OH van der Waals force However, in the mixture, the intermolecular forces that hold the ethanol molecules and benzene molecules together are the weaker van der Waals forces. C2 H5 OH van der Waals force Hence the tendency for evaporation in the mixture is higher than that in their respective pure liquids. Due to larger number of electrons in the C6 H6 molecule, the van der Waals force in C6 H6 is stronger than the hydrogen bond in ethanol.

328 Chemistry Term 1 STPM CHAPTER 8 Fractional Distillation of Solutions That Show Positive Deviation The boiling point-composition diagram of ethanol/benzene mixture is shown below: % by mass of ethanol Temperature/°C 0 78.5 32.4 A 80 100 B 100 % by mass of benzene 0 67.8 The product of fractional distillation depends on the composition of the starting mixture, whether its composition is more or less than 32.4% ethanol. Distilling a Mixture That Contains Less Than 32.4% Ethanol by Mass (Region A) % by mass of ethanol Temperature/°C 0 78.5 32.4 A 80 100 B 100 % by mass of benzene 0 67.8 1 When a mixture of composition in region A (less than 32.4% ethanol) is subjected to fractional distillation, the vapour given off gets richer and richer in ethanol as it ascends the column (see the arrows above). 2 Finally, the vapour that reaches the top of the column is the azeotrope (which has the lowest boiling point that contains 32.4% ethanol and 67.6% benzene by mass), which distills over at 67.8 °C. 3 As the distillation continues, the liquid left in the flask gets richer in benzene. Finally, only pure benzene is left and will distill over at 80 °C as the second distillate. 4 The products of fractional distillation are the azeotrope mixture and pure benzene. 5 The azeotrope is distilled over first because it has a lower boiling point. The azeotrope is distilled over first because its boiling point is lower than benzene.

329 Chemistry Term 1 STPM CHAPTER 8 Distilling a Mixture That Contains More Than 32.4% Ethanol by Mass (Region B) % by mass of ethanol Temperature/°C 0 78.5 32.4 A 80 100 B 100 % by mass of benzene 0 67.8 1 When a mixture with composition in region B is subjected to fractional distillation, the vapour given off gets richer and richer in benzene as it ascends the column. 2 Finally, the vapour that reaches the top of the column is the azeotrope (which has the lowest boiling point that contains 32.4% ethanol and 67.6% benzene by mass), which distills over at 67.8 °C. 3 As the distillation continues, the liquid left in the flask gets richer in ethanol. Finally, only pure ethanol is left and distills over at 78 °C as the second distillate. Distilling a Mixture That Contains 32.4% Ethanol (Point z) 1 When a mixture with composition z is subjected to fractional distillation, it boils at 67.8 °C. The vapour given off also has the same composition, i.e. 32.4% ethanol. 2 There is no change in the composition of the vapour as distillation continues. What distills over is the azeotrope at 67.8 °C. 3 The products of fractional distillation is summarised in the table below. Composition First distillate Second distillate  32.4% ethanol Azeotrope Benzene  32.4% ethanol Azeotrope Ethanol = 32.4% ethanol Azeotrope Azeotrope % by mass of ethanol Temperature/°C 0 78.5 32.4 A 80 100 B 100 % by mass of benzene 0 67.8 Azeotrope followed by pure ethanol Azeotrope followed by pure benzene The azeotrope is distilled over first because it has a lower boiling point than ethanol.

330 Chemistry Term 1 STPM CHAPTER 8 8.5 Fractional Distillation under Reduced Pressures 1 The boiling point of a liquid or liquid mixture is the temperature where the vapour pressure of the liquid is equal to the external pressure (usually at 1 atm or 101 kPa). 2 However, if the external pressure is reduced, the boiling point of the liquid will be lowered as shown by the graph below: Pressure/atm Temperature p1 p2 t 2 t 1 3 The boiling point of the liquid is t 1 at a pressure of p1 , but is t 2 at a pressure of p2 . 4 Hence by reducing the pressure above a liquid mixture, it is then possible to carry out the distillation at a lower temperature. 5 This is of utmost important when distilling liquid compounds that decompose before their boiling points are reached. 6 For example, the ‘normal boiling point’ of nitrobenzene is 293 °C. However, nitrobenzene boils at 179 °C when the pressure is reduced to 5.05 kPa (0.05 atm). 7 However, reduced pressure distillation has a disadvantage in that the apparatus must be strong enough to withstand breakage under reduced pressure. Furthermore, impurities with low boiling points may also distil over. The lower the external pressure, the lower the boiling point. Prevents the decomposition of certain organic liquids. Disadvantage of reduced pressure distillation.

331 Chemistry Term 1 STPM CHAPTER 8 Quick Check 8.6 1 The boiling point/composition for two immiscible liquids P and Q is given below. 60 0.2 0 0.4 0.6 0.8 1.0 Temperature/o C Mole fraction of P 70 80 90 50 (a) Which liquid is more volatile, P or Q? Explain your answer. (b) A mixture contains 0.2 mole fraction of P is subjected to fractional distillation. (i) Determine the boiling point of the mixture. (ii) What is the composition of the first distillate? (iii) What is the composition of the final distillate? (c) With reference to the graph explain what is meant by an azeotrope. 2 The data below refers to the ethanol/water mixture: Boiling point of ethanol = 78.5° C Boiling point of water = 100.0 °C Composition of azeotrope = 95% by mass of ethanol Boiling point of azeotrope = 78.0 °C (a) State whether a mixture of ethanol and water shows positive or negative deviation from Raoult’s law. Explain how you arrived at the conclusion. (b) Explain in terms of intermolecular forces why a mixture of ethanol and water shows such deviation. (c) Sketch (i) the vapour pressure/composition graph, and (ii) the boiling point/composition graph for the ethanol/water system. (d) A liquid mixture contains 60 g of ethanol and 40 g of water is subjected to fractional distillation. What is the composition of (i) the first distillate, (ii) the final distillate?

332 Chemistry Term 1 STPM CHAPTER 8 3 The vapour pressures of two miscible liquids, A and B, are 45 kPa and 60 kPa respectively at 300 K. (a) Assuming that the mixture obeys Raoult’s law, calculate the total pressure of a mixture containing 0.35 mole fraction of A. (b) However, the actual vapour pressure measured is lower than that calculated from (a). Explain why it is so. 4 Water (b.p. = 100.0 °C) and hydrogen chloride (b.p. = – 84.0 °C) forms an azeotrope with a boiling point of 108.6 °C containing 20% hydrogen chloride by mass. (a) Calculate the mole fraction of hydrogen chloride in the azeotrope. (b) Sketch (i) the vapour pressure/composition (in mole fraction), and (ii) the boiling point/composition (in mole fraction) for the water/hydrogen chloride system. (c) Explain why the water/hydrogen chloride system does not obey Raoult’s law. (d) Use your sketch in (b)(ii) to determine the composition of the first distillate when a mixture containing 0.6 mole fraction of hydrogen chloride is subjected to fractional distillation. SUMMARY SUMMARY 1 A solution is a homogeneous mixture of two or more substances. The components are uniformly distributed in the mixture. 2 An ideal solution is one that obeys Raoult’s law. 3 Raoult’s law states that the partial vapour pressure exerted by a liquid in a solution is given by the product of the mole fraction of the liquid in the mixture and the vapour pressure of the pure liquid at the same temperature. PA = XAPA o 4 In an ideal solution, (a) the intermolecular forces between like molecules (in their pure liquids) and between unlike molecules (in the solution) are the same. (b) there is no change in temperature or volume on mixing. 5 Ideal mixtures can be separated into their pure components by fractional distillation. 6 A solution with vapour pressure that is higher than that calculated from Raoult’s law is said to show positive deviation. (a) The intermolecular forces between unlike molecules in the mixture are weaker than those between like molecules in the pure liquids. (b) Mixing is accompanied by an increase in the total volume. (c) The mixing process is endothermic (a drop in temperature). (d) The vapour pressure/composition diagram shows a maximum (e) The boiling point/composition diagram shows a minimum 7 A solution with vapour pressure that is lower than that calculated from Raoult’s law is said to show negative deviation. (a) The intermolecular forces between unlike molecules in the mixture are stronger than those between like molecules in the pure liquids. (b) Mixing is accompanied by a decrease in the total volume. (c) The mixing process is endothermic (a rise in temperature) (d) The vapour pressure/composition diagram shows a minimum

333 Chemistry Term 1 STPM CHAPTER 8 1 Which of the following is true for an azeotropic mixture of a solution with positive deviation from Raoult’s law? A The boiling point of the azeotrope is higher than the boiling points of the pure components. B It has a constant boiling point. C It has the lowest vapour pressure compared to other mixtures. D It can be separated by reduced pressure distillation. 2 Benzene (b.p. = 80°C) and ethanol (b.p. = 78°C) forms an azeotrope with a boiling point of 68°C and containing 0.6 mole fraction of benzene. What is the composition of the first distillate obtained if a solution containing equimolar quantity of benzene and ethanol is fractionally distilled? A 100% benzene B 100 H2 O C A mixture containing 0.6 mole fraction of benzene D A mixture containing 0.6 mole fraction of ethanol 3 The vapour pressures of liquid X and liquid Y at 300 K are 37.5 kPa and 15.5 kPa respectively. Calculate the total pressure exerted by a mixture consisting of 0.96 mole of X and 0.48 mole of Y at the same temperature. A 15 kPa C 30.2 kPa B 23.5 kPa D 53.0 kPa 4 When solid iodine is heated in an open container at room conditions, it sublimes. What conclusion can be drawn from this observation alone? Objective Questions (e) The boiling point/composition diagram shows a maximum 8 The different components in a non-ideal solution cannot be completely separated by fractional distillation. 9 An azeotrope is a solution with constant boiling point. The composition of the solution and that of the vapour is the same. An azeotrope is distilled unchanged. A Iodine is a Group 17 element. B The triple point temperature of iodine is above room temperature. C The triple point pressure of iodine is above room pressure. D The critical point temperature of iodine is below room temperature. 5 What is the correct definition of an azeotrope? A A mixture with boiling point that is lower than that of the pure components. B A mixture with boiling point that is higher than that of the pure components. C A mixture containing equimolar quantities of both components. D A mixture that distilled without change in composition. 6 The boiling point-composition curve of a mixture of liquids A and B at different mole fraction of B is shown below. T 0 X 1 Which of the following is correct? A The mixture shows negative deviation from Raoult’s law STPM PRACTICE 8

334 Chemistry Term 1 STPM CHAPTER 8 B The enthalpy of solution is positive C The intermolecular forces in the mixture are weaker than that in the pure liquids D Fractional distillation of a mixture of composition X would produce pure B and a mixture of A and B 7 Which of the following substances is expected to have the highest critical temperature? A H2 C CO2 B NH3 D P4 O10 8 What is the vapour pressure of a substance at its normal boiling point? A Equal to the external pressure B Equal to atmospheric pressure C 100 Pa D 1 atm 9 Vapour pressure and volatility of a liquid are A inversely proportional to one another B directly proportional to one another C independent of temperatur D not related 10 Under normal conditions, a solid would sublime rather than melt if A its critical temperature is above room temperature. B its critical pressure is above atmospheric pressure. C its triple point temperature is below room temperature. D its triple point pressure is above atmospheric pressure. 11 Liquids X and Y form an ideal solution. A mixture containing 40% mole fraction of Y has a total pressure of 6.0  104 Pa. If the vapour pressure of pure X is 4.0  104 Pa, what is the vapour pressure of pure Y? All measurements are done at room temperature. A 2.0  104 Pa C 9.0  104 Pa B 5.0  104 Pa D 10.0  104 Pa 12 The critical pressure is A the pressure required to condense a vapour at its critical temperature. B the pressure where the solid melts at room temperature. C the pressure to prevent sublimation at room conditions. D the pressure where the substance is a gas at all temperatures. 13 The vapour pressures of liquids X and Y at 298 K are 35.6 kPa and 58.3 kPa respectively. A mixture containing equimolar quantities of X and Y has a pressure of 43.8 kPa at 298 K. Which statement(s) is/are true of the mixture? I The solution obeys Raoult’s law. II X and Y cannot be separated completely by fractional distillation. III An azeotrope is formed. A I C II and III B II D I, II and III 14 The vapour pressure-composition curve of a solution consisting of two liquid components X and Y is shown below: Vapour pressure 1.0 Mole fraction 1.0 X Y Which statements about the graph are true? I The boiling point of Y is lower than that of X. II The solution shows positive deviation from Raoult’s law. III The boiling point/composition graph would show a maximum. IV The temperature decreases when X and Y are mixed. A I and II C I, II and IV B II and IV D I, II, III and IV 15 The boiling point-composition graph for two miscible liquid X and Y is shown below: Boiling point / °C % Composition 100X P Q 100Y What is the composition of the distillate if a mixture with composition P is subjected to fractional distillation? A X C X and Y B Y D X and Q

335 Chemistry Term 1 STPM CHAPTER 8 Structured and Essay Questions 1 A mixture of heptane (boiling point = 98 °C) and octane (boiling point = 126 °C) forms an ideal solution. (a) Name the type of intermolecular forces present in each of the liquid. (b) What is an ideal solution? (c) Explain, in terms of intermolecular forces, why this mixture shows ideal behaviour. (d) Sketch a graph to show how the boiling points of mixtures of the two liquids vary with the composition of the mixture. (e) Name the process by which a mixture containing 0.35 mole fraction of heptane can be separated into pure heptane and pure octane. 2 Nitric acid (boiling point of 87 °C) and water (boiling point of 100 °C) form an azeotrope with a boiling point of 122 °C and contains 65% by mass of nitric acid. (a) What do you understand by an azeotrope? (b) Sketch the boiling point/composition curve for the nitric acid/water system, labelling both the liquid and vapour lines. (c) A nitric acid/water mixture containing 30% by mass of nitric acid is subjected to fractional distillation. State the composition of (i) the first distillate, and (ii) the final distillate (d) (i) State Raoult’s law. (ii) Explain why a mixture of nitric acid and water deviates from Raoult’s law. 3 (a) An ideal solution is one that obeys Raoult’s law. State Raoult’s law. (b) A mixture of benzene and methylbenzene forms an ideal solution. [Vapour pressures of benzene and methylbenzene at 298 K are 12.7 kPa and 3.9 kPa respectively.] (i) Explain why a mixture of benzene and methylbenzene shows ideal behaviour. (ii) Sketch the vapour pressure/composition graph for mixtures of benzene and methylbenzene. (iii) Sketch the boiling point/composition graph for the mixtures of benzene and methylbenzene. Label the liquid and vapour lines. (iv) A benzene/methylbenzene mixture containing initially 30% by mass of benzene is heated in a fractional distillation apparatus. State what composition will be (I) the first distillate, (II) the final distillate. 4 Hydrogen bromide (b.p. = –67 °C) and water (b.p.= 100 °C) form an azeotropic mixture with a boiling point of 12 °C. The azeotrope contains 47% by mass of hydrogen bromide. (a) What do you understand by the term azeotrope? (b) Sketch the vapour pressure/composition graph for mixtures of hydrogen bromide and water. Explain the shape of your curve. (c) Sketch the boiling point/composition graph for mixtures of hydrogen bromide and water. Label the liquid and vapour lines. (d) A mixture containing 60% by mass of hydrogen bromide is subjected to fractional distillation. Use your graph to determine the composition of (i) the first distillate, (ii) the final distillate. 5 The boiling point of a mixture of two liquids X and Y are given below. Mole fraction of X 0.0 0.2 0.4 0.6 0.8 1.0 Boiling point/°C 81.0 74.3 69.5 67.5 67.5 79.0 (a) Plot on a graph paper the boiling point/composition curve for the mixture of X and Y.

336 Chemistry Term 1 STPM CHAPTER 8 (b) What type of deviation does a mixture of X and Y show? (c) Determine the boiling point and composition of the azeotrope. (d) A mixture containing 0.3 mole fraction of X is subjected to fractional distillation. (i) Determine the boiling point of this mixture. (ii) What is the composition of the vapour at the boiling point in (i)? (iii) What is the composition of the first distillate that comes over? (iv) What is the composition of the final distillate collected? 6 (a) A mixture of ethanol (boiling point of 78 °C) and cyclohexane (boiling point of 82 °C) shows a positive deviation from Raoult’s law. For the mixture, sketch (i) the vapour pressure/composition curve, and (ii) the boiling point/composition curve. (b) Explain why there is a drop in temperature when ethanol and cyclohexane are mixed together. 7 The boiling point-composition graph for the mixture of two liquids P and Q is shown below. 60 0.2 0 0.4 0.6 0.8 1.0 Temperature/o C Mole fraction of P 70 80 90 50 (a) State the type of deviation from Raoult’s law exhibited by the mixture of P and Q. (b) Define azeotrope. (c) From the graph, determine the composition and the boiling point of the azeotrope. (d) A solution containing 0.2 mole fraction of P is subjected to fractional distillation. (i) Determine the boiling point of the mixture. (ii) State the composition of the first distillate and the last distillate.

337 Chemistry Term 1 STPM STPM Model Paper (962/1) STPM Model Paper (962/1) Section A [15 marks] Bahagian A [15 markah] Answer all questions in this section. Jawab semua soalan dalam bahagian ini. 1 If n is the number of atoms in 12 g of 12C, what is the number of atoms in 32 g of oxygen? Jika n adalah bilangan atom dalam 12 g 12C, berapakah bilangan atom dalam 32 g oksigen? A 1 2 n C 2n B n D 2.7n 2 A solution contains 20.8 g dm–3 of XCl2 . When react with excess of silver nitrate solution, 28.7 g of silver chloride is formed. What is the relative atomic mass of X? [Relative atomic mass Ag = 108, Cl = 35.5] Suatu larutan mengandungi 20.8 g dm–3 XCl2 . Apabila bertindak balas dengan larutan argentum nitrat berlebihan, 28.7 g argentum klorida terbentuk. Apakah jisim atom relatif X? [Jisim atom relatif Ag = 108, Cl = 35.5] A 52 C 156 B 104.5 D 208 3 The molecular peaks in the mass spectrum of 1,2-dibromoethane, C2 H4 Br2 is shown below. Which of the graph is correct? [Bromine consists of two isotopes, 79Br and 81Br in the ratio of 1 : 1] Puncak-puncak molekul dalam spektrum jisim 1,2-dibrometana, C2 H4 Br2 ditunjukkan di bawah. Graf yang manakah adalah benar? [Bromin terdiri daripada dua isotop, 79Br dan 81Br dalam nisbah 1 : 1] A 190 B 186 190 C 186 188 190 D 186 190188 4 In the hydrogen atom, how many electronic transitions are possible between the first seven energy levels? Berapa bilangan peralihan elektron yang mungkin berlaku antara tujuh aras tenaga yang pertama dalam atom hidrogen? A 6 C 21 B 7 D 29 5 Which statement is true about the emission spectrum of hydrogen? Pernyataan yang manakah adalah benar mengenai spektrum pancaran atom hidrogen? A There are more lines in the Lyman series than that in the Balmer series Terdapat lebih banyak garisan pada siri Lyman berbanding dengan siri Balmer B The ionisation energy of hydrogen can be calculated from the line with the longest wavelength

338 Chemistry Term 1 STPM STPM Model Paper (962/1) Tenaga pengionan hidrogen boleh dihitung menggunakan garis dengan jarak gelombang yang paling tinggi C The spectrum is similar to that of the He2+ ion Spektrum adalah serupa dengan ion He2+ D It shows that the electron in the hydrogen atom can be located anywhere around the nucleus Ia menunjukkan elektron dalam atom hidrogen boleh wujud mana-mana di sekeliling nukleus 6 The following valence configurations are for atoms that form ions with charge +3 except Konfigurasi-konfigurasi elektron berikut adalah untuk atom-atom yang membentuk ion-ion dengan cas +3 kecuali A 2s2 2p1 B 3d4 4s2 C 3d10 4s1 D 3s2 3p3 7 In magnesium sulfide, MgS, both the ions have the same Dalam magnesium sulfida, MgS, kedua-dua ion mempunyai persamaan A charge cas B electronic configuration konfigurasi elektron C nuclear charge cas nukleus D ionic radius jejari ion 8 Ammonia reacts with aluminium chloride to form an addition compound, AlCl3 (NH3 ). Which of the following is not true of the reaction? Ammonia bertindak balas dengan aluminium klorida untuk menghasilkan sebatian tambahan, AlCl3 (NH3 ). Antara yang berikut,yang manakah tidak benar mengenai tindak balas ini? A Ammonia acts as a Lewis base Ammonia bertindak sebagai bes Lewis B The general shape of the product is tetrahedral Rupa bentuk hasil adalah tetrahedron C The product is covalent Hasil bersifat kovalen D Ammonia uses its empty orbital to form a coordinate (dative) bond with aluminium chloride Ammonia menggunakan salah satu orbital kosong untuk membentuk ikatan kordinat(datif) dengan aluminium klorida 9 The bond angles in ammonia and water are 107° and 104.5° respectively. Which of the following best explain this difference? Sudut ikatan pada ammonia dan air masingmasing adalah 107° dan 104.5°. Antara yang berikut, yang manakah paling sesuai untuk menerangkan perbezaan ini? A Ammonia is a gas whereas water is a liquid at room conditions Ammonia adalah gas pada keadaan bilik sedangkan air adalah cecair dalam keadaan bilik B Oxygen is more electronegative than nitrogen Oksigen lebih elektronegatif daripada nitrogen C Oxygen atom is smaller than nitrogen atom Atom oksigen lebih kecil daripada atom nitrogen D There are more lone-pair electrons in water than in ammonia Terdapat lebih elektron pasangan tersendiri pada air berbanding dengan ammonia 10 The rate constant of a first order reaction is 3.0 × 10–2 s–1 and the enthalpy change of reaction is –45.9 kJ mol–1. Which statement is not necessarily true of the reaction? Pemalar kadar suatu tindak balas tertib pertama ialah 3.0 × 10–2 s–1 dan perubahan entalpi tindak balas ialah –45.9 kJ mol–1. Pernyataan yang manakah tidak semestinya benar mengenai tindak balas ini? A The half-life of the reaction is 23.1 s Setengah hayat tindak balas ini ialah 23.1 s B It is a one-step reaction Ia merupakan tindak balas satu langkah C The rate constant increases with temperature Pemalar kadar bertambah dengan suhu D The activation energy of the forward reaction is lower than that of the reverse reaction Tenaga pengaktifan tindak balas ke hadapan lebih rendah daripada yang bagi tindak balas berbalik

339 Chemistry Term 1 STPM STPM Model Paper (962/1) 11 Which of the following is correct when a system in dynamic equilibrium? Antara yang berikut, yang manakah adalah benar apabila suatu sistem berada dalam keseimbangan dinamik? A The rate constant for the forward reaction is the same as the rate constant for the reverse reaction Pemalar kadar bagi tindak balas ke hadapan adalah sama dengan pemalar kadar bagi tindak balas berbalik B The overall enthalpy change of reaction is zero Perubahan entalpi keseluruhan adalah sifar C The activation energy of the forward reaction is the same as the activation energy for the reverse reaction Tenaga pengaktifan untuk tindak balas ke hadapan adalah sama dengan tenaga pengaktifan tindak balas berbalik D The concentration of the product is always higher than the concentration of the reactant Kepekatan hasil selalunya lebih tinggi daripada kepekatan bahan tindak balas 12 Consider the reaction below: Pertimbangkan tindak balas di bawah: A(g)  2B(g) The equilibrium constant is 5.6 × 102 Pa at 50°C and 8.4 × 102 Pa at 120°C. It can be concluded that Pemalar kesimbangan adalah 5.6 × 102 Pa pada 50°C dan 8.4 × 102 Pa pada 120°C. Ini boleh disimpulkan bahawa A the reverse reaction is endothermic tindak balas berbalik adalah endotermik B increasing pressure will increase the yield of B hasil B bertambah apabila tekanan ditambahkan C the production of B is favoured by low pressure penghasilan B digalakkan oleh tekanan yang rendah D the rate constant for the reverse reaction decreases with increasing temperature pemalar kadar untuk tindak balas berbalik berkurang dengan penambahan suhu 13 The rate constant of a chemical reaction is 4.05 × 10–7 at 556°C and 4.49 × 10–2 at 79°C. What is the activation energy (in kJ mol–1) of the reaction? Pemalar kadar suatu tindak balas adalah 4.05 × 10–7 pada 556°C dan 4.49 × 10–2 pada 791°C. Berapakah tenaga pengaktifan (dalam kJ mol–1) bagi tindak balas tersebut? A 57 C 178 B 143 D 201 14 Which of the following statements is correct? Pernyataan yang manakah adalah benar? A A buffer solution is obtained by mixing a weak acid and a strong base Suatu larutan tampan diperoleh dengan mencampurkan suatu asid lemah dan suatu bes kuat B The pH at the equivalent point of an acid-base titration is always 7 pH pada takat setara pentitratan asid-bes adalah sentiasa 7 C Anions of strong acids are not hydrolysed in water Anion-anion bagi asid-asid kuat tidak dihidrolisiskan oleh air D The solubility product of barium sulphate is lower in dilute sulphuric acid than in water Hasil darab keterlarutan barium sulfat adalah lebih rendah dalam asid sulfurik cair berbanding dengan dalam air 15 For a non-ideal solution Bagi suata larutan tak unggul A the boiling point of the azeotrope is always higher than the boiling points of the pure liquids takat didih campuran azeotrop selalunya lebih tinggi daripada takat didih cecair-cecair tulen B heat is evolved on mixing haba dibebaskan semasa percampuran C the components cannot be separated completely by reduced-pressure fractional distillation komponen-komponen tidak dapat diasingkan secara lengkap melalui penyulingan berperingkat tekanan-rendah D the components are immiscible komponen-komponen tidak tarlarutcampur

340 Chemistry Term 1 STPM STPM Model Paper (962/1) Section B [15 marks] Bahagian B [15 markah] Answer all questions in this section. Jawab semua soalan dalam bahagian ini. 16 Ammonia is amphoteric, which can accept a proton to form the ammonium ion, or lose a proton to form the amide ion, NH2 – . Ammonia bersifat amfoterik di mana ia boleh menerima suatu proton untuk membentuk ion ammonium, atau kehilangan satu proton untuk membentuk ion amida, NH2 – . (a) Write a balance equation for the formation of the ammonium ion and the amide ion. [2 marks] Tulis persamaan seimbang bagi pembentukan ion ammonium dan ion amida. [2 markah] (b) Draw the Lewis diagrams for ammonium ion and amide ion. [2 marks] Lukis struktur Lewis bagi ion ammonium dan ion amida. [2 markah] (c) Which ion is expected to have a larger bond angle? Explain your answer. [2 marks] Ion yang manakah dijangka mempunyai sudut ikatan yang lebih besar? Jelaskan jawapan anda. [2 markah] 17 (a) Define pH. [1 mark] Takrifkan pH. [1 markah] (b) Calculate the concentration of the following solutions that have pH = 3.1 Hitung kepekatan larutan-larutan berikut yang masing-masing mempunyai pH = 3.1 (i) Hydrochloric acid [2 marks] Asid hidroklorik [2 markah] (ii) Ethanoic acid [2 marks] Asid etanoik [2 markah] [Ka CH3 COOH = 1.8 × 10–5 mol dm–3] (c) Explain the difference in the concentration of the two acids. [1 mark] Jelaskan perbezaan kepekatan kedua-dua asid. [1 markah] (d) Under suitable conditions, ethyne reacts with the amide ion to produce ammonia. Di bawah keadaan yang sesuai, etuna bertindak balas dengan ion amida untuk membentuk ammonia. (i) Write a balanced equation for the reaction. [1 mark] Tulis suatu persamaan seimbang bagi tindak balas itu. [1 markah] (ii) Explain the reaction according to Bronsted-Lowry’s theory of acids and bases. [2 marks] Terangkan tindak balas tersebut mengikut teori asid-bes Bronsted-Lowry. [2 markah] Section C [30 marks] Bahagian C [30 markah] Answer any two questions from this section. Jawab mana-mana dua soalan daripada bahagian ini. 18 (a) Describe and explain how the rate of a chemical reaction is affected by Huraikan dan terangkan bagaimana kadar suatu tindak balas kimia dipenggaruhi oleh (i) a change in temperature [4 marks] perubahan suhu [4 markah] (ii) the presence of a catalyst [4 marks] kehadiran suatu mangkin [4 markah]

341 Chemistry Term 1 STPM STPM Model Paper (962/1) (b) Define order of reaction. [2 marks] Takrifkan tertib tindak balas. [2 markah] (c) The thermal decomposition of nitryl chloride, a yellow gas, is second order with respect to nitryl chloride is shown below. Penguraian terma nitril klorida, suatu gas kuning, mengikut tertib kedua terhadap nitryl klorida ditunjukkan di bawah. 2NO2 Cl : 2NO2 + Cl2 (i) Write the rate equation for the reaction [2 marks] Tulis persamaan kadar bagi tindak balas itu. [2 markah] (ii) Suggest a mechanism that is in consistent with the kinetic information. Indicating the rate-determining step in your answer. [2 marks] Cadangkan suatu mekanisma yang konsisten dengan maklumat kadar yang diberikan. Tunjukkan langkah penentuan kadar dalam jawapan anda. [2 markah] (iii) State any observable change during the reaction. [1 mark] Nyata perubahan-perubahan yang boleh diperhatikan dalam tindak balas ini. [1 markah] 19 (a) Define buffer solution [1 mark] Takrifkan larutan tampan. [1 markah] (b) A buffer solution is prepared by mixing a 0.10 mol dm–3 ethanoic acid and 0.10 mol dm–3 sodium ethanoate. Explain, with the aids of balanced equations, how the buffer solution works. [3 marks] Suatu larutan tampan disediakan dengan mencampurkan 0.10 mol dm–3 asid etanoik dan 0.10 mol dm–3 natrium etanoat. Terangkan dengan persamaan-persamaan seimbang bagaimana larutan tampan ini berfungsi. [3 markah] (c) The pH of mixtures of 0.10 M ethanoic acid 0.10 M sodium ethanoate is shown below: pH campuran-campuran 0.10 M asid etanoik dan 0.10 M natrium etanoat ditunjukkan di bawah: Volume of ethanoic acid/ cm3 Isi padu asid etanoik/ cm3 45.0 40.0 30.0 Volume of sodium ethanoate/ cm3 Isi padu natrium etanoat/ cm3 5 10 20 pH 3.92 4.28 4.70 Plot a suitable graph to determine the acid dissociation constant Ka of ethanoic acid. [5 marks] Plot suatu graf yang sesuai untuk menentukan nilai pemalar penceraian, Ka untuk asid etnaoik. [5 markah] (d) (i) Define solubility product, Ksp. [1 mark] Takrifkan hasil darab keterlarutan, Ksp. [1 markah] (ii) Using silver chloride as example, explain what do you understand by common ion effect? [2 marks] Dengan menggunakan argentum klorida sebagai contoh, terangkan apa yang anda faham dengan kesan ion sepunya. [2 markah] (iii) 50.0 cm3 of 0.020 M manganese(II) sulphate is added to 50.0 cm3 of 0.050 M ammonia solution. Determine whether a precipitate of manganese(II) hydroxide will be formed or not. [Ksp for manganese(II) hydroxide = 2.00 × 10–13 mol3 dm–9. Ka for ammonia = 1.80 × 10–5 mol dm–3] [3 marks] 50.0 cm3 0.020 M magnesium(II) sulfat dicampur ke dalam 50.0 cm3 0.050 M larutan ammonia. Tentukan sama ada suatu mendakan magnesium hidoksida akan terbentuk atau tidak. [Ksp magnesium(II) hidroksida = 2.00 × 10–13 mol3 dm–9. Ka ammonia = 1.80 × 10–5 mol dm–3] [3 markah]