# Suggested answers to in-text activities and unit-end exercises. Topic 16 Unit 55

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1 Suggested answers to in-text activities and unit-end exercises In-text activities Discussion (page 117) Some possible ways for minimizing possible sources of error in the experiment: Add a slight excess of precipitating agent to precipitate out all the phosphorus. (Avoid adding a large excess of precipitating agent as this increases chances of adsorption on the surface of the precipitate.) Use the minimum amount of solvent to wash the precipitate. Avoid splashes. Keep the precipitate in a desiccator when left for drying overnight. Weigh the precipitate and the filter paper rapidly. Discussion (page 121) Step 1 The distilled water remaining in the beaker would not affect the amount of Ca 2+ (aq) ions in the sample solution. Step 3 The mass of the precipitate obtained would be lower than the true value. Hence the calculated result would be lower than the true value. Step 4 Some of the precipitate might remain in the beaker. So, the mass of the precipitate obtained would be lower than the true value. Step 4 Some impurities might adhere to the precipitate. So, the mass of the precipitate obtained would be higher than the true value. Step 5 The precipitate might take up moisture from the atmosphere. So, the mass of the precipitate obtained would be higher than the true value. Checkpoint (page 122) 1 Zn in foot powder ZnNH 4 PO 4 ZnP 2 O g of sample g Number of moles of ZnP 2 O 7 = = 1.54 x 10 3 mol = number of moles of Zn in foot powder Mass of Zn in foot powder = 1.54 x 10 3 mol x 65.4 g mol 1 = g g g mol g Percentage by mass of Zn in foot powder = x 100% g = 9.35% Suggested answers to in-text activities and unit-end exercises 1 Jing Kung. All rights reserved.

2 2 a) Ba 2+ (aq) + SO 4 2 (aq) BaSO 4 (s) b) Add a little BaCl 2 (aq) to the mother liquor after filtering off the precipitate. If some sulphate is still present, a precipitate will form. Then add more precipitating agent to precipitate the remaining sulphate and re-filter. c) To remove the impurities. d) Mass of BaSO 4 = ( ) g = g Number of moles of BaSO 4 = g g mol 1 = 1.44 x 10 3 mol = number of moles of sulphate Mass of sulphate in fertilizer = 1.44 x 10 3 mol x 96.1 g mol 1 = g g Percentage by mass of sulphate in fertilizer = x 100% g = 62.7% e) Mass of (NH 4 ) 2 SO 4 in fertilizer = 1.44 x 10 3 mol x g mol 1 = g g Percentage by mass of (NH 4 ) 2 SO 4 in fertilizer = x 100% g = 86.4% f) The following sources of error make the experimental result for the percentage by mass of sulphate in the fertilizer slightly lower than the true value: Not all sulphate is precipitated out as we may not have added enough barium chloride solution. As no ionic substance is completely insoluble in water, a little barium sulphate will remain dissolved in the solution. A little of the precipitate will be lost as we wash the precipitate. Little splashes, inefficient rinsing out of the beaker, and inefficient filtering can cause some loss of the precipitate. The following sources of error make the experimental result for the percentage by mass of sulphate in the fertilizer slightly higher than the true value: Impurities may be present. The precipitate may not be dried completely and so residual moisture can add to its mass. Suggested answers to in-text activities and unit-end exercises 2 Jing Kung. All rights reserved.

4 There is no air oxidation of iron(ii) ions to iron(iii) ions. 2 a) 5H 2 O 2 (aq) + 2MnO 4 (aq) + 6H + (aq) 5O 2 (g) + 2Mn 2+ (aq) + 8H 2 O(l) 10.0 cm mol dm cm cm 3 (used) 25.0 cm 3 Number of moles of MnO 4 ions in 35.0 cm 3 solution = mol dm = 3.50 x 10 4 mol According to the equation, 2 moles of MnO 4 ions react with 5 moles of H 2 O 2. i.e. number of moles of H 2 O 2 in 25.0 cm 3 diluted solution = x 3.50 x 10 4 mol = 8.75 x 10 4 mol Number of moles of H 2 O 2 in cm 3 diluted solution = 8.75 x 10 4 x = 1.75 x 10 2 mol Mass of H 2 O 2 in cm 3 diluted solution = 1.75 x 10 2 mol x 34.0 g mol 1 = g 0.595g Concentration of original H 2 O 2 solution = dm = 59.5 g dm 3 the concentration of the aqueous solution of hydrogen peroxide is 59.5 g dm mol b) The concentration of hydrogen peroxide in a rainwater sample is too low. Checkpoint (page 144) 1 a) To minimize air oxidation of the iodide ions. b) Use starch solution as an indicator. The iodine added will react with the sulphur dioxide. Once all the sulphur dioxide has reacted, the excess iodine added will react with the starch to form a dark blue complex. This colour change indicates the end point. c) SO 2 (g) + I 2 (aq) + 2H 2 O(l) SO 4 2 (aq) + 2I (aq) + 4H + (aq) 25.0 cm mol dm 3 Suggested answers to in-text activities and unit-end exercises 4 Jing Kung. All rights reserved.

5 15.8 cm 3 Number of moles of I 2 in 15.8 cm 3 solution = mol dm = 7.11 x 10 5 mol According to the equation, 1 mole of SO 2 reacts with 1 mole of I 2. i.e. number of moles of SO 2 in 25.0 cm 3 solution = 7.11 x 10 5 mol Mass of SO 2 in 25.0 cm 3 solution = 7.11 x 10 5 mol x 64.1 g mol 1 = 4.56 x 10 3 g = 4.56 mg Concentration of SO 2 in the sample = = 182 mg dm 3 the concentration of sulphur dioxide in the sample of white wine is 182 mg dm 3. 2 a) i) I 2 (aq) + 2S 2 O 3 2 (aq) 2I (aq) + S 4 O 6 2 (aq) ii) Number of moles of S 2 O 2 3 ions reacted with the iodine liberated = mol dm 3 x 24.6 dm 3 = 2.46 x 10 3 mol According to the equation, 1 mole of I 2 reacts with 2 moles of S 2 O 2 3 ions. i.e. number of moles of I 2 formed in Stage 1 = 2.46 x mol = 1.23 x 10 3 mol b) i) Cl 2 (aq) + 2I (aq) I 2 (aq) + 2Cl (aq) 1.23 x 10 3 mol According to the equation, 1 mole of Cl 2 reacts with I ions to give 1 mole of I 2. i.e. number of moles of Cl 2 in bleaching powder = 1.23 x 10 3 mol Mass of Cl 2 in bleaching powder = 1.23 x 10 3 mol x 71.0 g mol 1 = 8.73 x 10 2 g the mass of available chlorine in the bleaching powder is 8.73 x 10 2 g x 10 2 g ii) Percentage by mass of Cl 2 in bleaching powder = x 100% g = 30.6% the percentage by mass of available chlorine in the bleaching powder is 30.6%. Suggested answers to in-text activities and unit-end exercises 5 Jing Kung. All rights reserved.

6 Unit-end exercises (pages ) Answers for the HKCEE and HKALE questions are not provided. (Extension) 1 a) Gravimetric analysis involves determining the mass of the species being analyzed or the mass of a compound chemically related to that species. b) Any two of the following: Analyze the concentration of chloride ions in a sample solution by precipitating the chloride ions as silver chloride. Determine the phosphorus content in a sample of fertilizer by precipitating the phosphorus as magnesium ammonium phosphate (MgNH 4 PO 4 6H 2 O). Determine the calcium content in a sample solution by precipitating the calcium ions as calcium oxalate. 2 Any five of the following: the sample type the sample size available concentration range of the species being analyzed the accuracy required separation that may be required to eliminate interferences the instruments available the cost the speed (Extension) 3 a) A pipette / burette may hold / deliver a volume slightly different from the volume indicated by the manufacturer. b) During a titration, a student who is insensitive to colour changes may use more reagent. c) During a titration, a small excess of the reagent is added to cause an indicator to change colour. 4 a) An aqueous solution of potassium permanganate is unstable as it tends to decompose slowly at room temperature. b) Standardize an aqueous solution of potassium permanganate by titrating it against an aqueous solution of sodium oxalate. In an acidic solution, the oxalate ions are converted to oxalic acid. Potassium permanganate reacts with the oxalic acid according to the following equation: 5H 2 C 2 O 4 (aq) + 2MnO 4 (aq) + 6H + (aq) 10CO 2 (g) + 2Mn 2+ (aq) + 8H 2 O(l) Suggested answers to in-text activities and unit-end exercises 6 Jing Kung. All rights reserved.

7 Follow the steps below in the standardization process: Dissolve a known mass of sodium oxalate in dilute sulphuric acid. Heat the oxalate solution to C and titrate with the aqueous solution of potassium permanganate until a pale pink colour persists, showing that all the oxalic acid has reacted. 5 a) Iodine is a strong oxidizing agent. It reacts with other reducing agents in the environment. b) i) IO 3 (aq) + 5I (aq) + 6H + (aq) 3I 2 (aq) + 3H 2 O(l) ii) Number of moles of KIO 3 in cm 3 of solution = Number of moles of I 2 produced = 3 x 5.00 x 10 3 mol mol Molarity of iodine solution = dm = 6.00 x 10 2 mol dm g g mol 1 = 5.00 x 10 3 mol 6 Chloride ions react with silver ions according to the following reaction: Ag + (aq) + Cl (aq) AgCl(s) Number of moles of AgCl obtained = = 2.03 x 10 3 mol Number of moles of Cl ions = 2.03 x 10 3 mol Mass of KCl in the sample = 2.03 x 10 3 mol x 74.6 g mol 1 = 1.51 x 10 1 g g g mol x 10 1 g Percentage by mass of KCl in the sample = x 100% g = 60.4% the percentage by mass of KCl in the sample is 60.4%. 7 a) Zn(s) + Sn 4+ (aq) Zn 2+ (aq) + Sn 2+( aq) 5 b) i) x mol dm = 6.75 x 10 4 mol 2 5 ii) x mol dm = 1.02 x 10 3 mol 2 c) Number of moles of Sn4+ = (1.02 x x 10 4) mol = 3.45 x 10 4 mol Suggested answers to in-text activities and unit-end exercises 7 Jing Kung. All rights reserved.

8 The ratio of Sn 2+ / Sn in oxide A = 2 : the chemical formula of A is 2SnO + SnO 2, i.e. Sn 3 O 4. 8 a) i) Chromate indicator ii) When all the chloride ions are precipitated, the first excess aqueous solution of silver nitrate gives a reddish brown silver chromate precipitate with the chromate indicator. This signals the end point of the titration. b) Ag + (aq) + Cl (aq) AgCl(s) mol dm cm cm cm 3 (used) 25.0 cm 3 Number of moles of Ag + ions in 19.8 cm 3 solution = mol dm = 1.49 x 10 3 mol According to the equation, 1 mole of Ag + ions reacts with 1 mole of Cl ions. i.e. number of moles of Cl ions in 25.0 cm 3 diluted sea water = 1.49 x 10 3 mol Number of moles of Cl ions in cm 3 diluted sea water = 10 x 1.49 x 10 3 mol = 1.49 x 10 2 mol = number of moles of Cl ions in 25.0 cm 3 sea water sample mol Concentration of Cl ions in sea water sample = dm = mol dm 3 = mol dm 3 x 35.5 g mol 1 = 21.2 g dm 3 the concentration of chloride ions in the sea water sample is 21.2 g dm 3. 9 a) H 2 O 2 (aq) + 2H + (aq) + 2e 2H 2 O(l) b) i) Use a pipette to deliver the hydrogen peroxide solution to a conical flask. Add dilute sulphuric acid to the conical flask. Fill a burette with the aqueous solution of potassium permanganate. Suggested answers to in-text activities and unit-end exercises 8 Jing Kung. All rights reserved.

9 Run the aqueous solution of potassium permanganate into the hydrogen peroxide solution until the first appearance of a persistent pale pink colour. Repeat to obtain at least two consistent titres. ii) 2MnO 4 (aq) + 6H + (aq) + 5H 2 O 2 (aq) 2Mn 2+ (aq) + 8H 2 O(l) + 5O 2 (g) mol dm 3 (used) 10.0 cm cm 3 Number of moles of MnO 4 ions in 18.2 cm 3 solution = mol dm = 3.64 x 10 4 mol According to the equation, 2 moles of MnO 4 ions react with 5 moles of H 2 O 2. i.e. number of moles of H 2 O 2 ions in 10.0 cm 3 solution = mol Concentration of the diluted H 2 O 2 solution = dm x 3.64 x 10 4 mol = 9.10 x 10 4 mol = 9.10 x 10 2 mol dm 3 Concentration of the undiluted H 2 O 2 solution = 10 x 9.10 x 10 2 mol dm 3 = mol dm 3 the concentration of the undiluted H 2 O 2 solution is mol dm iii) Concentration of the undiluted H 2 O 2 solution = mol dm 3 x 34.0 g mol 1 = 30.9 g dm 3 Mass of H 2 O 2 in 100 cm 3 of solution = 30.9 g dm 3 x 100 dm 3 = 3.09 g the undiluted H 2 O 2 solution is unsuitable for treating paintings a) i) Any one of the following: Add aqueous solution of silver nitrate. A yellow precipitate forms. Add aqueous chlorine and then an organic solvent. A purple organic layer forms. Suggested answers to in-text activities and unit-end exercises 9 Jing Kung. All rights reserved.

10 ii) Element Initial oxidation number Final oxidation number Iodine +5 1 Sulphur iii) The total change in oxidation number of S = 3 x (+2) = +6 The total change in oxidation number of I = 6 b) i) Pipette ii) Starch solution Dark blue to colourless iii) mol dm 3 x 24.0 dm 3 = 2.40 x 10 4 mol iv) According to the equation, 1 mole of iodine reacts with 2 moles of Na 2 S 2 O x 10 4 Number of moles of iodine = mol 2 = 1.20 x 10 4 mol mol v) Concentration of iodine solution = dm = mol dm Place 25.0 cm 3 of the water sample in a conical flask using a pipette. Add 5 cm 3 of dilute sulphuric acid and mix. Place the conical flask in a boiling water bath for 10 minutes. Add 5.00 cm 3 of the aqueous solution of potassium permanganate to the water sample using a pipette. After 10 minutes, add 5.00 cm 3 of the aqueous solution of sodium oxalate using a pipette. Wait until the solution becomes colourless. Fill a burette with the aqueous solution of potassium permanganate. Run the aqueous solution of potassium permanganate into the hot mixture until the first appearance of a persistent pale Suggested answers to in-text activities and unit-end exercises 10 Jing Kung. All rights reserved.

11 pink colour. Initially, 5.00 cm 3 of KMnO 4 (aq) and 5.00 cm 3 of Na 2 C 2 O 4 (aq) are added to the water sample. Number of moles of KMnO 4 in 5.00 cm 3 solution = mol dm = 1.00 x 10 5 mol Number of moles of Na 2 C 2 O 4 in 5.00 cm 3 solution = mol dm = 2.50 x 10 5 mol Under acidic conditions, the oxalate ions are converted to oxalic acid. Potassium permanganate reacts with the oxalic acid according to the following equation: 5H 2 C 2 O 4 (aq) + 2MnO 4 (aq) + 6H + (aq) 10CO 2 (g) + 2Mn 2+ (aq) + 8H 2 O(l) According to the equation, 5 moles of C 2 O 4 2 ions react with 2 moles of MnO 4 ions. If none of the permanganate ion is consumed by the oxidizable matter in the water sample, all the oxalate ions will react with the permanganate ions and no oxalate ion will remain in the mixture. Suppose V cm 3 of aqueous solution of permanganate are required to react with the oxalate ions remaining in the mixture. It can be deduced that this amount of permanganate ions is consumed in oxidizing the oxidizable matter in the water sample. Number of moles of MnO 4 ions consumed in oxidizing matter in water sample = mol dm 3 V = 2V x 10 6 mol = 2V x 10 3 mmol Number of millimoles of O 2 required to oxidize the same amount of matter = 1.2 x 2V x 10 3 mmol Mass of O 2 required to oxidize the same amount of matter = 2.4V x 10 3 mmol x 32.0 mg mmol 1 = 7.68V x 10 2 mg mg Permanganate index of water sample = 25.0 L = 3.07V mg of O 2 per litre Suggested answers to in-text activities and unit-end exercises 11 Jing Kung. All rights reserved.

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