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03II26 20.0 cm3 of 2.0 M aqueous ammonia required 16.0 cm3 of sulphuric acid for complete neutralization. What is the concentration of the sulphuric acid in g dm-3? (Relative atomic masses : H = 1.0, O = 16.0, S = 32.1) A. 61.3 g dm-3 B. 122.6 g dm-3 C. 183.9 g dm-3 D. 245.2 g dm-3 B 2NH3(aq) + H2SO4(aq)  (NH4)2SO4(aq) No. of moles of NH3 = 2.0 x (20.0 / 1000) = 0.04 mol Mole ratio of NH3 : H2SO4 = 2 : 1 No. of moles of H2SO4 = 0.04 / 2 = 0.02 mol Mass of H2SO4 = 0.02 x (2 x 1 + 32.1 + 16 x 4) = 1.962 g 1.962 Concentration of H2SO4 = 16.0 / 1000 = 122.6 g dm-3 01II6 When potassium carbonate solution and calcium chloride solution are mixed, calcium carbonate is precipitated. Which of the following mixtures would produce the greatest amount of precipitate? A. 5 cm3 of 1 M K2CO3(aq) + 15 cm3 of 1 M CaCl2(aq) B. 10 cm3 of 1 M K2CO3(aq) + 10 cm3 of 1 M CaCl2(aq) C. 15 cm3 of 1 M K2CO3(aq) + 8 cm3 of 1 M CaCl2(aq) D. 18 cm3 of 1 M K2CO3(aq) + 5 cm3 of 1 M CaCl2(aq) B 01II15 A mixture consists of one mole of sodium carbonate and one mole of sodium hydrogencarbonate. What is the least number of moles of hydrochloric acid required to liberate all the available carbon dioxide from the mixture? A. 1.5 B. 2.0 C. 3.0 D. 4.0 C Na2CO3(aq) + 2HCl(aq)  2NaCl(aq) + CO2(g) + H2O(l) NaHCO3(aq) + HCl(aq)  NaCl(aq) + CO2(g) + H2O(l)

1 mole of Na2CO3 reacts with 2 moles of HCl and 1 moles of NaHCO3 reacts with 1 mole of HCl. All together 3 moles of HCl is required.

99II25 In an experiment, 1.00 M sodium hydroxide solution was added to 25.0 cm3 of 1.00 M sulphuric acid until the acid was completely neutralized. What is the concentration of sodium sulphate (correct to two decimal places) in the resulting solution? A. 1.00 M B. 0.50 M C. 0.33 M D. 0.25 M C 2NaOH(aq) + H2SO4(aq)  Na2SO4(aq) + 2H2O(l) No. of moles of H2SO4 = 1.00 x (25.0 / 1000) = 0.025 mol Mole ratio of NaOH : H2SO4 = 2 : 1 No. of moles of NaOH = 0.025 x 2 = 0.05 mol Volume of NaOH added = 0.05 / 1 = 0.05 dm3 Mole ratio of Na2SO4 : H2SO4 = 1 : 1 No. of moles of Na2SO4 = 0.025 Concentration of Na2SO4 = 0.025 / (0.05 + 0.025) = 0.33 M 99II6 The concentration of an aqueous solution of an acid is 1.0 M. 25.0 cm3 of this acid solution requires 37.5 cm3 of 2.0 M sodium hydroxide solution for complete neutralization. What is the basicity of the acid? A. 1 B. 2 C. 3 D. 4 C H+ + OH-  H2O 1 mole of H+ ion reacts with 1 mole of OH- ion. No. of moles of OH- ion = 2.0 x (37.5 / 1000) = 0.075 mol No. of moles of the acid = 1.0 x (25.0 / 1000) = 0.025 mol 0.025 moles of the acid can provide 0.075 mol of H+ ion to react with 0.075 moles of OH- ion.

So the basicity of the acid must be 3. 97II49 1st statement : When filling a pipette with a solution, a pipette filler is preferred to sucking with the mouth. 2nd statement : It is more accurate to fill a pipette with a solution by using a pipette filler than by sucking with the mouth. C 95II8 In order to prepare 250 cm3 of 0.10 M sodium hydroxide solution from 1.0 M sodium hydroxide solution, which of the following combinations of apparatus should be used ? A. burette, measuring cylinder, pipette B. conical flask, measuring cylinder, volumetric flask C. burette, conical flask, wash bottle D. pipette, volumetric flask, wash bottle 96II28 Which of the following apparatus would be most suitable for measuring the volume of soap solution (from 0.6 cm3 - 9.3 cm3)? A. 50 cm3 burette B. 50 cm3 measuring cylinder C. 25 cm3 pipette D. 10 cm3 beaker A 98I(6ai) A student prepared sodium nitrate solution by reacting 1 M sodium hydroxide solution with dilute nitric acid. The student carried out a titration to determine the amount of dilute nitric acid required to react with a known volume of 1 M sodium hydroxide solution. (1) Write the chemical equation for the reaction. (An ionic equation with NOT be accepted for this question.) (2) Draw a labeled diagram for the set-up of the titration. (3) Phenolphthalein can be used to determine the end point of the titration. State the colour change at the end point. (4) Suggest how the student can prepare a sodium nitrate solution using the titration

results. Answers: (1) NaOH + HNO3  NaNO3 + H2O 1 (2) (1 mark for a diagram showing the set-up for the titration experiment; 2 marks for labeling the apparatus and reagents: 4 labels) 1+2 (3) from pink/ purple/ red to colourless 1 (4) Add dilute nitric acid to 1 M sodium hydroxide solution in the same volume ratio as that in the titration result, without adding the indicator or repeat the titration procedure without adding the indicator. 1 02I(9b) In an experiment to determine the concentration of ammonia in a sample of glass cleaner, 25.0 cm3 of the sample was diluted to 250.0 cm3 in a volumetric flask. 25.0 cm3 of the diluted sample was transferred to a conical flask and was then titrated against 0.23 M hydrochloric acid. 28.7 cm3 of the acid was required to reach the endpoint. (i) State the liquid that should be used to rinse the following pieces of apparatus used in this experiment. (1) volumetric flask (2) conical flask (ii) Name the apparatus that should be used to transfer 25.0 cm3 of the diluted sample to the conical flask. (iii) Calculate the concentration, in mol dm-3, of ammonia in the sample of glass cleaner. (You may assume that ammonia is the only substance in the sample that reacts with hydrochloric acid.) (6 marks) Answer: (i) (1) distilled water / deionized water (2) distilled water / deionized water (ii) pipette (iii) NH3 + HCl  NH4+ + ClNo. of moles of HCl used = 0.23 x 28.7 x 10-3 = 6.60 x 10-3 Concentration of NH3 in glass cleaner 6.60 x10 −3 250 × = 25 25 x10 −3 = 2.64 mol dm-3

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03I(8b) An experiment was carried out to determine the concentration of a nickel(II) sulphate solution. The experiment consisted of the following three stages: Stage 1: 25.0 cm3 of 0.503 M sodium hydroxide solution was added to 25.0 cm3 of the nickel(II) sulphate solution to precipitate out nickel(II) hydroxide. Stage 2: The mixture obtained in Stage 1 was filtered and the residue was washed thoroughly with distilled water. Stage 3: The excess alkali in the filtrate was titrated against 0.251 M hydrochloric acid with methyl orange as indicator. 18.5 cm3 of the acid was required to reach the end-point. (i) Write the ionic equation for the reaction in Stage 1. (ii) Stage the colour change at the end-point of the titration in Stage 3. (iii) (1) Based on the titration result in Stage 3, calculate the number of moles of hydroxide ions present in the filtrate. (2) Calculate the number of moles of sodium hydroxide that was added in Stage 12. (3) Using your results in (1) and (2) above, calculate the molarity of the nickel(II) sulphate solution. (iv) Why was it necessary to wash the residue thoroughly in Stage 2? (9 marks) Answers: (i) Ni2+ + 2OH-  Ni(OH)2 1 (ii) yellow to orange 1 (iii) (1) H+ + OH-  H2O 1 + 1 mole of H ion reacts with 1 mole of OH ion (1) No. of moles of OH- in the filtrate = 0.251 x 18.5 x 10-3 = 4.64 x 10-3 1 (2) No. of moles of NaOH used = 0.503 x 25 x 10-3 = 0.0126 1 (3) No. of NaOH that has reacted with Ni2+ = 0.012575 – 4.6435 x 10-3 = 7.93 x 10-3 1 2+ Ni + 2OH  Ni(OH)2 Concentration of Ni2+ ions 7.9315 x10 −3 = 1 2 x 25 x10 −3

= 0.158 mol dm-3 1 (iv) To remove, as far as possible, OH- ions which have adhered to the surface of the residue/ to transfer OH- ions to the filtrate. 1