Section 10.3: Acid Base Stoichiometry Tutorial 1 Questions, page 481 1. It is necessary to keep the volume of indicator used to a minimum because many acid base indicators are weak acids. Some of the base used in a titration reacts with the indicator. 2. It is necessary to rinse with titrant after rinsing with water so that the inner walls of the burette are lined with titrant rather than with water. If water were present, it could dilute the titrant. 3. The air bubble occupies a certain volume in the burette. The presence of an air bubble would make the volume of titrant used seem larger than it actually is. 4. Rinsing the sides of a titration flask near the end of the titration washes any unreacted titrant into the bottom of the flask. 5. The solution turned pink momentarily because the sodium hydroxide solution was briefly concentrated in one spot. Then, as the flask is swirled, the sodium hydroxide spreads throughout the flask where it reacts with excess acid, changing the indicator to colourless. Tutorial 2 Practice, page 484 1. (a) Given: three volumes of acid Required: average volume of acid used, V HNO3 Calculate the average of titrant (acid) used by subtracting the final from the initial burette volume for each trial. 8.00 m + 8.06 m + 8.12 m V HNO3 (average) 3 8.06 m V HNO3 (average) Statement: The average volume of nitric acid used is 8.06 m. (b) Given: concentration of acid, c HNO3 0.500 mol/ volume of base, 25.00 m average volume of acid, V HNO3 8.06 m (average) Required: amount concentration of base, c NaOH Analysis: c n V Step 1. Convert the volume of solution to litres. V HNO3 8.06 m! (average) V HNO3 (average) 8.06! 10"3 25.00 m! 2.500! 10 "2 Copyright 2011 Nelson Education td. Chapter 10: Acids and Bases 10.3-1
Step 2. Write a balanced equation for the reaction listing the given values. NaOH(aq) + HNO 3 (aq) NaNO 3 + H 2 O 2.5 10 2 8.06 10 3 c NaOH 0.500 mol/ Step 3. Use the concentration equation to determine the amount of acid. n HNO3 c HNO3 V HNO3 (average) n HNO3 0.500 mol 4.03! 10 "3 mol! 8.06! 10 "3 Step 4. Use the amount of acid and the mole ratio in the balanced equation to determine the amount of base. 4.03! 10 "3 mol HNO3! 1 mol NaOH 1mol HNO3 4.03! 10 "3 mol NaOH Step 5. Use the concentration equation to determine the concentration of base. c NaOH 4.03! 10"3 mol 2.500! 10 "2 c NaOH 0.161 mol Statement: The amount concentration of the sodium hydroxide solution is 0.161 mol/. 2. Given: mass of potassium carbonate, m K2 CO 3 1.00 g volume of acid, V HNO3 24.20 m Required: amount concentration of acid, c HNO3 Analysis: c n V Step 1. Convert the volume of titrant used to litres. V HNO3 24.20 m! V HNO3 2.420! 10 "2 Step 2. Write the balanced equation listing the given values. K 2 CO 3 (s) + 2 HNO 3 (aq) 2 KNO 3 + H 2 CO 3 1.00 g 2.420 10 2 138.21 g/mol c HNO3 Copyright 2011 Nelson Education td. Chapter 10: Acids and Bases 10.3-2
Step 3. Convert the mass of potassium carbonate to the amount. n K2 CO 3 1.00 g! 1 mol 138.21 g n K2 CO 3 7.2354! 10 "3 mol [2 extra digits carried] Step 4. Use the amount and the mole ratio in the balanced equation to determine the amount of the acid. n HNO3 7.2354! 10 "3 mol K2 CO 3! 2 mol HNO 3 1mol K2 CO 3 n HNO3 1.4471! 10 "2 mol HNO3 [2 extra digits carried] Step 5. Use the concentration equation to determine the concentration of acid. c HNO3 n HNO 3 V HNO3 1.4471! 10"2 mol 2.420! 10 "2 c HNO3 0.598 mol Statement: The amount concentration of nitric acid is 0.598 mol/. 3. Given: concentration of acid, c HCl 0.100 mol/ volume of acid, 15.52 m volume of base, V Ba(OH)2 25.00 m Required: amount concentration of base, c Ba(OH)2 Analysis: c n V Step 1. Convert the volume of the solutions to litres. 15.52 m! 1.552! 10 "2 V Ba(OH)2 25.00 m! V Ba(OH)2 2.500! 10 "2 Copyright 2011 Nelson Education td. Chapter 10: Acids and Bases 10.3-3
Step 2. Write a balanced equation for the reaction listing the given values. 2 HCl(aq) + Ba(OH) 2 (aq) BaCl 2 + 2 H 2 O 1.552 10 2 2.500 10 2 0.100 mol/ c Ba(OH)2 Step 3. Use the concentration equation to determine the amount of acid. c HCl 0.100 mol 1.552! 10 "3 mol! 1.552! 10 "2 Step 4. Use the amount of acid and the mole ratio in the balanced equation to determine the amount of base. n Ba(OH)2 1.552! 10 "3 mol HCl! 1 mol Ba(OH) 2 2 mol HCl n Ba(OH)2 7.760! 10 "4 mol Ba(OH)2 Step 5. Use the concentration equation to determine the concentration of base. c Ba(OH)2 n Ba(OH) 2 V Ba(OH)2 7.760! 10"4 mol 2.500! 10 "2 c Ba(OH)2 3.10! 10 "2 mol Statement: The amount concentration of the barium hydroxide solution is 3.10 10 2 mol/. Section 10.3 Questions, page 485 1. The purpose of an acid base indicator in a titration is to signify when the endpoint of the titration is reached. 2. The equivalence point is the point during a titration when neutralization is complete. The endpoint of a titration is a sudden change in a property of the solution such as a change in ph or conductivity. Ideally, the endpoint occurs at a ph very close to that of the equivalence point. 3. (a) Given: mass of sodium hydroxide, m NaOH 4.00 g volume of solution, 100.0 m Required: amount concentration of base, c NaOH Analysis: c NaOH Copyright 2011 Nelson Education td. Chapter 10: Acids and Bases 10.3-4
Step 1. Convert the volume of solution to litres. 100.0 m! 0.1000 Step 2. Convert mass of the sodium hydroxide to the amount. 4.00 g! 1 mol 40.00 g 0.100 mol Step 3. Use the concentration equation to determine the concentration of the solution. c NaOH 0.100 mol 0.100 c NaOH 1.00 mol Statement: The amount concentration of sodium hydroxide solution is 1.00 mol/. (b) The actual concentration differs from the predicted concentration because, over time, contaminants may enter the solution and react with sodium hydroxide. Carbon dioxide, for example, is known to react with sodium hydroxide to form sodium carbonate. This decreases the amount of sodium hydroxide present. Assuming the volume of solution remains constant, carbon dioxide contamination decreases the concentration of the solution. 4. It is necessary to re-standardize a solution after it has been stored for some time because the concentration of the solution may vary over time. This may be caused by several factors including reactions with contaminants and changes in the volume of the solution. 5. Storage bottles of solutions are often filled right to the top to eliminate the amount of air that would otherwise occupy the space above the solution. Components in air, such as oxygen and carbon dioxide, may react with the solution. 6. Drying the samples in an oven before using them to standardize a base is necessary to remove any water that may have been absorbed by the solid. If the mass were not dried, the mass of a sample contaminated with water would be larger than a dried sample. As a result, the mass of potassium hydrogen phthalate recorded would be larger than it actually is, and the calculated amount and concentration of base would be incorrect. 7. A ph meter can be used to monitor the ph changes during a titration. Near the equivalence point, a sharp rise or drop should be observed. For an acid base indicator to be suitable for a titration, the colour change should occur when the equivalence point is reached. 8. (a) The two acids have different initial ph because Acid 1 is stronger than Acid 2 and, thus, it has a lower ph. (b) The volume that is required to neutralize these acids is 50 m. The same volume is required for both acids because the acids are of the same concentration. This also assumes that both acids contain the same number of hydrogen atoms that ionize. (c) An acid base indicator that could be used for one of the acids, but not the other is methyl red. Copyright 2011 Nelson Education td. Chapter 10: Acids and Bases 10.3-5
9. (a) CaCO 3 (s) + 2 HCl(aq) CO 2 (g) + 2 H 2 O(l) + CaCl 2 (aq) (b) Given: volume of hydrochloric acid, 25.20 m amount concentration of hydrochloric acid, c HCl 0.50 mol/ Required: mass of calcium carbonate, m CaCO3 Analysis: c n V Step 1. Convert the volume of solution to litres. 25.20 m! 0.025 20 Step 2. Rearrange the concentration equation to determine the amount of hydrochloric acid. c HCl! 0.50 mol! 0.025 20 0.0126 mol Step 3. Use the amount of acid and the mole ratio in the balanced equation to determine the amount of calcium carbonate. CaCO 3 (s) + 2 HCl(aq) CO 2 (g) + 2 H 2 O(l) + CaCl 2 (aq) n CaCO3 0.0126 mol HCl! 1 mol CaCO 3 2 mol HCl n CaCO3 0.006 30 mol CaCO3 Step 4. Calculate the mass of calcium carbonate. m CaCO3 0.006 30 mol! 100.09 g 1 mol m CaCO3 0.63 g Statement: The mass of calcium carbonate required is 0.63 g. 10. (a) 2 NH 4 OH(aq) + H 2 SO 4 (aq) 2 H 2 O(l) + (NH 4 ) 2 SO 4 (aq) (b) Given: three volumes of acid Required: average volume of acid used, V H2 SO 4 Calculate the average of titrant (acid) used by subtracting the final from the initial burette volume for each trial. 8.46 m + 8.50 m + 8.66 m V H2 SO 4 (average) 3 8.54 m V H2 SO 4 (average) Statement: The average volume of sulfuric acid used is 8.54 m. Copyright 2011 Nelson Education td. Chapter 10: Acids and Bases 10.3-6
(c) Given: concentration of acid, c H2 0.25 mol/ SO 4 volume of base, V NH4 10.00 m OH average volume of acid, V H2 8.54 m SO 4 (average) Required: amount concentration of base, c NH4 OH Analysis: c n V Step 1. Convert the volume of solution to litres. V NH4 10.00 m! OH V NH4 OH 0.010 00 V H2 SO 4 (average) 8.54 m! V H2 SO 4 (average) 0.00854 Step 2. Write a balanced equation for the reaction listing the given values. 2 NH 4 OH(aq) + H 2 SO 4 (aq) 2 H 2 O(l) + (NH 4 ) 2 SO 4 (aq) 0.010 00 0.008 54 0.25 mol/ c NH4 OH Step 3. Use the concentration equation to determine the amount of acid. n H2 c SO 4 H2 V SO 4 H2 SO 4 (average) n H2 SO 4 0.25 mol! 0.008 54 0.002 135 mol Step 4. Use the amount of acid and the mole ratio in the balanced equation to determine the amount of base. n NH4 OH 0.002 135 mol H 2 SO 4! 2 mol NH 4 OH 1mol H2 SO 4 n NH4 OH 0.004 27 mol NH 4 OH Copyright 2011 Nelson Education td. Chapter 10: Acids and Bases 10.3-7
Step 5. Use the concentration equation to determine the concentration of base. c NH4 n NH 4 OH 2 OH 2 V NH4 OH 2 0.004 27 mol 0.0100 c NH4 0.43 mol OH 2 Statement: The amount concentration of ammonium hydroxide solution is 0.43 mol/. (d) It is necessary to conduct more than one trial when standardizing a solution to show that the data is reproducible and reliable. 11. Given: concentration of base, c NaOH 0.100 mol/ average volume of base, (average) 1.60 m volume of acid, V milk 10.00 m Required: amount concentration of lactic acid in milk, c HC3 H 5 O 2 Analysis: c n V Step 1. Convert the volume of solutions to litres. (average) 1.60 m! (average) 0.001 60 V milk 10.00 m! V milk 0.0100 Step 2. Write the balanced equation for the reaction listing the given values. HC 3 H 5 O 2 (aq) + NaOH(aq) H 2 O(l) + NaC 3 H 5 O 2 (aq) c HC3 0.00160 H 5 O 2 0.100 mol/ Step 3. Use the concentration equation to determine the amount of base. c NaOH (average) 0.100 mol 1.60! 10 4 mol! 0.00160 Copyright 2011 Nelson Education td. Chapter 10: Acids and Bases 10.3-8
Step 4. Use the amount of acid and the mole ratio in the balanced equation to determine the amount of base. n HC3 H 5 O 2 1.60! 10 4 mol NaOH! 1 mol HC 3 H 5 O 2 1mol NaOH n HC3 H 5 O 2 1.60! 10 4 mol HC3 H 5 O 2 Step 5. Use the concentration equation to determine the concentration of base. c HC3 H 5 O 2 n HC 3 H 5 O 2 V HC3 H 5 O 2 1.60! 10 4 0.0100 c HC3 0.0160 mol H 5 O 2 Statement: The amount concentration of lactic acid in milk is 0.0160 mol/. 12. (a) KH(IO 3 ) 2 (aq) + KOH(aq) H 2 O(l) + 2 KIO 3 (aq) (b) Given: mass of potassium hydrogen iodate, m KH(IO3 1.50 g ) 2 average volume of base, V KOH 25.42 m Required: amount concentration of potassium hydroxide, c KOH Step 1. Convert the volume of titrant used to litres. V KOH 25.42 m! V KOH 2.542! 10 "2 Step 2. Write the balanced equation for the reaction listing the given values. KH(IO 3 ) 2 (aq) + KOH(aq) H 2 O(l) + 2 KIO 3 (aq) 1.50 g 2.420 10 2 389.91 g/mol c KOH Step 3. Convert the mass of potassium hydrogen iodate to the amount. n KH(IO3 ) 2 1.50 g! 1 mol 389.91 g n KH(IO3 ) 2 3.847! 10 "3 mol Step 4. Use the amount and the mole ratio in the balanced equation to determine the amount of the base. n KOH 3.847! 10 "3 mol KH(IO3 ) 2! 1 mol KOH 1mol KH(IO3 ) 2 n KOH 3.847! 10 "3 mol KOH Copyright 2011 Nelson Education td. Chapter 10: Acids and Bases 10.3-9
Step 5. Use the concentration equation to determine the concentration of base. c KOH n KOH V KOH 3.847! 10"3 mol 2.542! 10 "2 c KOH 0.151 mol Statement: The amount concentration of potassium hydroxide is 0.151 mol/. 13. Given: mass of potassium carbonate, m K2 CO 3 1.00 g average volume of acid, (average) 24.20 m volume of base, V K2 CO 3 100 m Required: amount concentration of acid, c HCl Analysis: c n V Step 1. Convert the volume of solutions to litres. (average) 24.20 m! (average) 2.420! 10 "2 V K2 100 m! CO 3 V K2 CO 3 0.100 Step 2. Write a balanced equation for the reaction listing the values. K 2 CO 3 (aq) + 2 HCl(aq) 2 H 2 O(l) + CO 2 (g) + 2 KCl(aq) 1.00 g 2.420 10 2 138.21 g/mol c HCl Step 3. Convert the mass of potassium carbonate to the amount. n K2 CO 3 1.00 g! 1 mol 138.21 g n K2 CO 3 7.235! 10 "3 mol Step 4. Use the amount of potassium carbonate and the mole ratio in the balanced equation to determine the amount of acid. 7.235! 10 "3 mol K2 CO 3! 2 mol HCl 1mol K2 CO 3 1.447! 10 "2 mol HCl Copyright 2011 Nelson Education td. Chapter 10: Acids and Bases 10.3-10
Step 5. Use the concentration equation to determine the concentration of acid. c HCl 1.447! 10"2 mol 2.420! 10 "2 c HCl 0.598 mol Statement: The amount concentration of the hydrochloric acid solution is 0.598 mol/. Copyright 2011 Nelson Education td. Chapter 10: Acids and Bases 10.3-11