1. Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Chapter 8 Acids and Bases 8.5 Reactions of Acids and Bases 1

2. Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Acids and Metals Acids react with metals such as K, Na, Ca, Mg, Al, Zn, Fe, and Sn to produce hydrogen gas and the salt of the metal Molecular equations: 2K(s) + 2HCl(aq)  2KCl(aq) + H 2 (g) metal acid salt hydrogen gas Zn(s) + 2HCl(aq)  ZnCl 2 (aq) + H 2 (g) metal acid salt hydrogen gas 2

3. Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Learning Check Write a balanced equation for the reaction of magnesium metal with HCl(aq). Label the metal, acid, and salt. 3

4. Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Learning Check Write a balanced equation for the reaction of magnesium metal with HCl(aq). Mg(s) + 2HCl(aq)  MgCl 2 (aq) + H 2 (g) metal acid salt 4

5. Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Acids and Carbonates 5 Acids react with carbonates and hydrogen carbonates to produce carbon dioxide gas, a salt, and water 2HCl(aq) + CaCO 3 (s)  CO 2 (g) + CaCl 2 (aq) + H 2 O(l) acid carbonate carbon salt water dioxide HCl(aq) + NaHCO 3 (s)  CO 2 (g) + NaCl (aq) + H 2 O(l) acid bicarbonate carbon salt water dioxide

6. Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Learning Check Write a balanced equation for the following reactions: A. MgCO 3 (s) + HBr(aq)  B. HCl(aq) + NaHCO 3 (aq)  6

7. Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Solution Write a balanced equation for the following reactions: A. MgCO 3 (s) + 2HBr(aq)  MgBr 2 (aq) + CO 2 (g) + H 2 O(l) carbonate acid salt carbon water dioxide B. HCl(aq) + NaHCO 3 (aq)  NaCl(aq) + CO 2 (g) + H 2 O(l) acid bicarbonate salt carbon water dioxide 7

8. Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Neutralization Reactions In a neutralization reaction, an acid reacts with a base to produce salt and water the acid HCl reacts with NaOH to produce salt and water the salt formed is the anion from the acid and cation of the base HCl(aq) + NaOH(aq)  NaCl(aq) + H 2 O(l) acid base salt water 8

9. Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Neutralization Reactions In neutralization reactions, if we write the strong acid and strong base as ions, we see that H + reacts with OH − to form water, leaving the ions Na + and Cl  in solution: H + (aq) + Cl  (aq) + Na + (aq) + OH  (aq)  Na + (aq) + Cl  (aq) + H 2 O(l) the overall reaction is H 3 O + from the acid and OH  from the base form water: H + (aq) + OH  (aq)  H 2 O(l) 9

10. Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Guide for Balancing Neutralization Reactions 10

11. Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Balancing Neutralization Reactions Write the balanced equation for the neutralization of magnesium hydroxide and nitric acid. Step 1Write the reactants and products. Mg(OH) 2 + HNO 3 Step 2Balance the H + in the acid with the OH  in the base. Mg(OH) 2 + 2HNO 3 Step 3Balance the H 2 O with H + and the OH . Mg(OH) 2 + 2HNO 3  salt + 2H 2 O Step 4Write the salt from the remaining ions. Mg(OH) 2 + 2HNO 3  Mg(NO 3 ) 2 + 2H 2 O 11

12. Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Learning Check Select the correct group of coefficients for each of the following neutralization equations. 1. HCl(aq) + Al(OH) 3 (aq)  AlCl 3 (aq) + H 2 O(l) A. 1, 3, 3, 1 B. 3, 1, 1, 1 C. 3, 1, 1, 3 2. Ba(OH) 2 (aq) + H 3 PO 4 (aq)  Ba 3 (PO 4 ) 2 (s) + H 2 O(l) A. 3, 2, 2, 2 B. 3, 2, 1, 6 C. 2, 3, 1, 6 12

13. Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Solution 1. HCl(aq) + Al(OH) 3 (aq)  AlCl 3 (aq) + H 2 O(l) Step 1Write the reactants and products. HCl + Al(OH) 3 Step 2 Balance the H + in the acid with the OH  in the base. 3HCl + Al(OH) 3 Step 3Balance the H 2 O with H + and the OH . 3HCl + Al(OH) 3  salt + 3H 2 O Step 4Write the salt from the remaining ions. 3HCl(aq) + Al(OH) 3 (aq)  AlCl 3 (aq) + 3H 2 O(l) The answer is C. 3, 1, 1, 3. 13

14. Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Solution 2. Ba(OH) 2 (aq) + H 3 PO 4 (aq)  Ba 3 (PO 4 ) 2 (s) + H 2 O(l) Step 1Write the reactants and products. Ba(OH) 2 + H 3 PO 4 Step 2Balance the H + in the acid with the OH  in the base. 3Ba(OH) 2 + 2H 3 PO 4 Step 3Balance the H 2 O with H + and the OH . 3Ba(OH) 2 + 2H 3 PO 4  salt + 6H 2 O Step 4Write the salt from the remaining ions. 3Ba(OH) 2 (aq) + 2H 3 PO 4 (aq)  Ba 3 (PO 4 ) 2 (s) + 6H 2 O(l) The answer is B. 3, 2, 1, 6. 14

15. Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Basic Compounds in Antacids 15

16. Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Learning Check 16 Write the neutralization reactions for stomach acid, HCl, and the ingredients in Mylanta. Mylanta: Al(OH) 3 and Mg(OH) 2

17. Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Solution 17 Write the neutralization reactions for stomach acid, HCl, and the ingredients in Mylanta. Mylanta: For Al(OH) 3 : Step 1Write the reactants and products. Al(OH) 3 + HCl Step 2Balance the H + in the acid with the OH  in the base. Al(OH) 3 + 3HCl Step 3Balance the H 2 O with H + and the OH . Al(OH) 3 + 3HCl  salt + 3H 2 O Step 4Write the salt from the remaining ions. Al(OH) 3 (aq) + 3HCl(aq)  AlCl 3 (aq) + 3H 2 O(l)

18. Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Solution 18 Write the neutralization reactions for stomach acid, HCl, and the ingredients in Mylanta. Mylanta: For Mg(OH) 2 : Step 1Write the reactants and products. Mg(OH) 2 + HCl Step 2Balance the H + in the acid with the OH  in the base. Mg(OH) 2 + 2HCl Step 3Balance the H 2 O with H + and the OH . Mg(OH) 2 + 2HCl  salt + 2H 2 O Step 4Write the salt from the remaining ions. 2HCl(aq) + Mg(OH) 2 (aq)  MgCl 2 (aq) + 2H 2 O(l)

19. Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Acid–Base Titration 19 Titration is a laboratory procedure used to determine the molarity of an acid uses a base such as NaOH to neutralize a measured volume of an acid Base  NaOH Acid  Solution

20. Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Indicator An indicator is added to the acid in the flask causes the solution to change color when the acid is neutralized 20

21. Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Endpoint of Titration At the endpoint, the indicator gives the solution a permanent pink color the volume of the base used to reach the endpoint is measured the molarity of the acid is calculated using the neutralization equation for the reaction 21

22. Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Guide to Calculations for Acid–Base Titrations 22

23. Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Acid–Base Titration Calculations What is the molarity of an HCl solution if 18.5 mL of 0.225 M NaOH are required to neutralize 10.0 mL of HCl? HCl(aq) + NaOH(aq)  NaCl(aq) + H 2 O(l) Step 1 State given and needed quantities. Given: 18.5 mL of 0.225 M NaOH 10.0 mL of HCl Needed: Molarity of HCl 23

24. Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Acid–Base Titration Calculations What is the molarity of an HCl solution if 18.5 mL of 0.225 M NaOH are required to neutralize 10.0 mL of HCl? HCl(aq) + NaOH(aq)  NaCl(aq) + H 2 O(l) Step 2 Write a plan to calculate molarity or volume. molarity equation molarity NaOH coefficients HCl 18.5 mL  L  moles NaOH  moles HCl  L HCl 24

25. Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Acid–Base Titration Calculations What is the molarity of an HCl solution if 18.5 mL of 0.225 M NaOH are required to neutralize 10.0 mL of HCl? HCl(aq) + NaOH(aq)  NaCl(aq) + H 2 O(l) Step 3 State equalities and conversion factors including concentration.1 L = 1000 mL 0.225 mole NaOH and 1 mole HCl x 1 L NaOH 1 mole NaOH 25

26. Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Acid–Base Titration Calculations What is the molarity of an HCl solution if 18.5 mL of 0.225 M NaOH are required to neutralize 10.0 mL of HCl? HCl(aq) + NaOH(aq)  NaCl(aq) + H 2 O(l) Step 4 Set up the problem to calculate the needed quantity. 18.5 mL NaOH  1 L NaOH  0.225 mole NaOH 1000 mL NaOH 1 L NaOH  1 mole HCl = 0.00416 mole of HCl 1 mole NaOH M HCl = 0.00416 mole HCl = 0.416 M HCl 0.0100 L HCl 26

27. Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Learning Check Calculate the mL of 2.00 M H 2 SO 4 required to neutralize 50.0 mL of 1.00 M KOH. H 2 SO 4 (aq) + 2KOH(aq)  K 2 SO 4 (aq) + 2H 2 O(l) A. 12.5 mL B. 50.0 mL C. 200. mL 27

28. Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Solution Calculate the mL of 2.00 M H 2 SO 4 required to neutralize 50.0 mL of 1.00 M KOH. H 2 SO 4 (aq) + 2KOH(aq)  K 2 SO 4 (aq) + 2H 2 O(l) Step 1 State given and needed quantities. Given: 2.00 M H 2 SO 4 50.0 mL of 1.00 M KOH Needed: mL of H 2 SO 4 28

29. Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Solution Calculate the mL of 2.00 M H 2 SO 4 required to neutralize 50.0 mL of 1.00 M KOH. H 2 SO 4 (aq) + 2KOH(aq)  K 2 SO 4 (aq) + 2H 2 O(l) Step 2 Write a plan to calculate molarity or volume. molarity equation molarity KOH coefficients H 2 SO 4 50.0 mL  L  moles KOH  moles H 2 SO 4  L H 2 SO 4 29

30. Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Solution Calculate the mL of 2.00 M H 2 SO 4 required to neutralize 50.0 mL of 1.00 M KOH. H 2 SO 4 (aq) + 2KOH(aq)  K 2 SO 4 (aq) + 2H 2 O(l) Step 3 State equalities and conversion factors including concentration.1 L = 1000 mL 1 L H 2 SO 4 and 1 mole H 2 SO 4 2.00 mole H 2 SO 4 2 mole KOH 30

31. Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Solution Calculate the mL of 2.00 M H 2 SO 4 required to neutralize 50.0 mL of 1.00 M KOH. H 2 SO 4 (aq) + 2KOH(aq)  K 2 SO 4 (aq) + 2H 2 O(l) Step 4 Set up the problem to calculate the needed quantity. 50.0 mL  1 L = 0.0500 L 10 3 mL 0.0500 L KOH  1.00 mole KOH  1 mole H 2 SO 4 1 L KOH 2 mole KOH  1 L H 2 SO 4  1000 mL = 12.5 mL. 2.00 mole H 2 SO 4 1 L H 2 SO 4 The answer is A. 31