1. Chapter Ten Acids, Bases, and Salts

2. Chapter 10 | Slide 2 of 66 Acid-Base Theories Arrhenius Acid-Base Theory Bronsted-Lowry Acid-Base Theory

3. Chapter 10 | Slide 3 of 66 Arrhenius Acids Arrhenius Acids: –substances that ionize to form H + in solution (e.g. HCl, HNO 3, CH 3 CO 2 H, lemon, lime, vitamin C). Ionization: the process in which individual positive and negative ions are produced from a compound that is dissolved in solution HCl (g)  H + (aq) + Cl - (aq) Characteristics –Sour taste –Change blue litmus to red –Quite corrosive

4. Chapter 10 | Slide 4 of 66 Arrhenius Bases –Hydroxide-containing compounds that, in water, produces hydroxide ions –NaOH (s)  Na + (aq) + OH - (aq) Characteristics –Bitter taste –Change red litmus blue –Slippery to the touch

5. Chapter 10 | Slide 5 of 66 The difference between the aqueous solution processes of ionization and dissociation. Acids, Bases, and Salts cont’d

6. Chapter 10 | Slide 6 of 66 Litmus is a vegetable dye obtained from certain lichens found principally in the Netherlands. Acids, Bases, and Salts cont’d

7. Chapter 10 | Slide 7 of 66 Bronsted-Lowry Acid-Base Theory Bronsted-Lowry Acids –A substance that can donate a proton (H + ion) to some other substance Bronsted-Lowry Base –A substance that can accept a proton (H + ion) from another substance Which is the acid? The base?

8. Chapter 10 | Slide 8 of 66 Conjugate Acid-Base Pairs Conjugate acid –The species formed when a Bronsted-Lowry base accepts a proton Conjugate base –The species formed when a Bronsted-Lowry acid loses a proton HF (aq) + H 2 O (l)  H 3 O + (aq) + F - (aq)

9. Chapter 10 | Slide 9 of 66 Practice: Conjugate Acids and Bases Write the chemical formula of each of the following: –The conjugate acid of ClO 3 - –The conjugate base of NH 4 + –The conjugate acid of PO 4 3- –The conjugate base of HS -

10. Chapter 10 | Slide 10 of 66 Amphoteric Substances Molecules that are able to function as either Bronsted-Lowry acids or bases –HNO 3 (aq) + H 2 O (l)  H 3 O + (aq) + NO 3 - (aq) –NH 3 (aq) + H 2 O (l)   NH 4 + (aq) + OH - (aq)

11. Chapter 10 | Slide 11 of 66 Mono-, Di-, and Polyprotic Acids –Acids with one acidic proton are called monoprotic (e.g., HCl). –Acids with two acidic protons are called diprotic (e.g., H 2 SO4). –Acids with many acidic protons are called polyprotic. (e.g., citric acid)

12. Chapter 10 | Slide 12 of 66 Strong v. Weak Acids and Bases Strong acids and bases completely ionize or dissociate (break into their ions) in solution. –100% of strong acid molecules break up into their ions Weak acids and bases partially ionizes in solution. –Usually, less than 5% of weak acid molecules break up into their ions. The majority of the molecules remain in the molecular form

13. Chapter 10 | Slide 13 of 66 Fig. 10.5 A comparison of the number of acidic species present in strong acid and weak acid solutions of the same concentration. Acids, Bases, and Salts cont’d

14. Chapter 10 | Slide 14 of 66 Strong Acids and Bases Memorize these!

15. Chapter 10 | Slide 15 of 66 Ionization Constants Acid ionization constant –The equilibrium constant for the reaction of a weak acid with water –HA (aq) + H 2 O (l)  H 3 O + (aq) + A - (aq) The higher the Ka, the stronger the acid.

16. Chapter 10 | Slide 16 of 66 Acids, Bases, and Salts cont’d

17. Chapter 10 | Slide 17 of 66 Ionization Constants Base ionization constant –The equilibrium constant for the reaction of a weak base with water –B (aq) + H 2 O (l)  BH + (aq) + OH - (aq) The higher the Kb, the stronger the base.

18. Chapter 10 | Slide 18 of 66 Acids, Bases, and Salts cont’d

19. Chapter 10 | Slide 19 of 66

20. Chapter 10 | Slide 20 of 66 →Fig. 10.6 The acid-base reaction between sulfuric acid and barium hydroxide produces the insoluble salt barium sulfate. Acids, Bases, and Salts cont’d

21. Chapter 10 | Slide 21 of 66 ←Fig. 10.3 A white cloud of finely divided solid NH 4 Cl is produced by the acid-base reaction that results when the colorless gases HCl and NH 3 mix. Acids, Bases, and Salts cont’d Ken O’Donoghue © Houghton Mifflin Company

22. Chapter 10 | Slide 22 of 66 Formation of water by the transfer of protons from H 3 O+ ion to OH- ions. A Neutralization or double replacement reaction. Acids, Bases, and Salts cont’d

23. Chapter 10 | Slide 23 of 66 Self-Ionization of Water A small percentage of water molecules interact with each other to form ions [H + ] = [OH - ] = 1 x 10 -7 M

24. Chapter 10 | Slide 24 of 66 Ion Product Constant for Water  We can use the ion product constant for water to calculate the concentration of hydroxide or hydronium ions in any solution

25. Chapter 10 | Slide 25 of 66 The relationship between (H 3 O + ) and (OH - ) in aqueous solution is an inverse proportion; when (H 3 O + ) is increased, (OH - ) decreases, and vice versa.

26. Chapter 10 | Slide 26 of 66 Relationship Between [H 3 O + ] and [OH - ] in Neutral, Acidic, and Basic Solutions.

27. Chapter 10 | Slide 27 of 66  CC 10.1 Acids, Bases, and Salts cont’d

28. Chapter 10 | Slide 28 of 66 pH Hydronium ion concentrations can range from very small numbers to very large numbers. It is very inconvenient to work with numbers extended over a large range pH scale was developed to give a practical way to deal with a large range of numbers pH = -log[H 3 O + ] “p” = “take the negative logarithm of” [H 3 O + ] = 10 -pH

29. Chapter 10 | Slide 29 of 66 As the hydronium ion concentration increases, the pH AND the hydroxide ion concentration decrease.

30. Chapter 10 | Slide 30 of 66 Practice Calculating pH Calculate the pH for the following solutions: [H 3 O + ] = 1.00 x 10 -11 [H 3 O + ] = 1.00 x 10 -5 [H 3 O + ] = 1.00 x 10 -6 Calculate the hydronium ion concentration of solutions with the following pH’s pH = 3 pH = 9

31. Chapter 10 | Slide 31 of 66 Interpreting pH Values Acids –Have pH less than 7 Bases –Have pH greater than 7 Neutral solutions –Have pH = 7

32. Chapter 10 | Slide 32 of 66 Interpreting pH Values A change of 1 unit in pH always corresponds to a tenfold change in hydronium ion concentration 0.1 M HCl10 -1 MpH = 1 0.01 M HCl10 -2 MpH = 2 10 fold difference difference of 1

33. Chapter 10 | Slide 33 of 66 →Fig. 10.10 Most fruits and vegetable are acidic. Acids, Bases, and Salts cont’d

34. Chapter 10 | Slide 34 of 66 →Fig. 10.12 pH values of selected common liquids. Acids, Bases, and Salts cont’d

35. Chapter 10 | Slide 35 of 66 Fig. 10.13 A pH meter gives an accurate measurement of pH values. Acids, Bases, and Salts cont’d

36. Chapter 10 | Slide 36 of 66 pKa and Acid Strength K a is the acid ionization constant –The higher the K a, the stronger the acid pK a = -log(K a ) –The lower the pK a, the stronger the acid Determine the pK a for hydrocyanic acid, HCN, given that the K a for this acid is 4.4 x 10 -10. -9.4

37. Chapter 10 | Slide 37 of 66 Acids, Bases, and Salts cont’d

38. Chapter 10 | Slide 38 of 66 Salts An ionic compound containing a metal or polyatomic ion as the positive ion and a nonmetal or polyatomic ion (except hydroxide) as the negative ion –Ionic compounds with hydroxides as the negative ions are bases All common soluble salts completely dissociate into ions in solution

39. Chapter 10 | Slide 39 of 66 “Neutralization” Reactions When solutions of an acid and a base are mixed, the products of the reaction have none of the characteristic properties of the acid or the base. –Typically, water and a salt are formed (a neutral solution) when an acid reacts with a metal hydroxide base. Neutralization reactions are double replacement reactions AX + BY  AY + BX

40. Chapter 10 | Slide 40 of 66 “Neutralization” Reaction Examples Write a balanced complete chemical equation for the reaction between aqueous solutions of HCl and KOH. Write a balanced complete chemical equation for the reaction between aqueous solutions of acetic acid (HC 2 H 3 O 2 ) and barium hydroxide [Ba(OH) 2 ]. Write a balanced complete chemical equation for the reaction between aqueous solutions of H 3 PO 4 and NaOH.

41. Chapter 10 | Slide 41 of 66 Acids, Bases, and Salts cont’d  CAG 10.1

42. Chapter 10 | Slide 42 of 66 Buffers An aqueous solution consisting of a weak acid and its conjugate base in approximately equal amounts that prevent major changes in solution pH when small amounts of acid or base are added to it –The weak acid can react with added base to neutralize it –The conjugate base can react with added acid to neutralize it

43. Chapter 10 | Slide 43 of 66 A Comparison of pH Changes in Buffered and Unbuffered Solutions.

44. Chapter 10 | Slide 44 of 66 ←CC 10.2 Acid Rain Acids, Bases, and Salts cont’d

45. Chapter 10 | Slide 45 of 66 Table 10.8 Acids, Bases, and Salts cont’d

46. Chapter 10 | Slide 46 of 66 Acids, Bases, and Salts cont’d

47. Chapter 10 | Slide 47 of 66 Fig. 10.14 (a) The buffered and unbuffered solutions have the same pH level. Acids, Bases, and Salts cont’d Fig. 10.14 (b) After adding 1mL of a 0.01 M HCl solution, the pH of the buffered solution has not perceptibly changed, but the unbuffered solution has become acidic. Ken O’Donoghue © Houghton Mifflin Company

48. Chapter 10 | Slide 48 of 66 Acids, Bases, and Salts cont’d CAG 10.2

49. Chapter 10 | Slide 49 of 66 Acids, Bases, and Salts cont’d CC 10.4

50. Chapter 10 | Slide 50 of 66 Buffers and Le Chatelier’s Principle HC 2 H 3 O 2 (aq) + H 2 O (l)  H 3 O +1 (aq) + C 2 H 3 O 2 -1 (aq) What happens if we add acid to the equilibrium? What happens if we add base to the equilibrium?

51. Chapter 10 | Slide 51 of 66 Buffers Practice Predict whether each of the following pairs of substances could function as a buffer system in aqueous solution –HCl and NaOH –HC 2 H 3 O 2 and KC 2 H 3 O 2 –NaCl and NaCN –HCN and HC 2 H 3 O 2

52. Chapter 10 | Slide 52 of 66 Strong v. Weak Electrolytes Electrolyte: A substance whose aqueous solution conducts electricity Strong electrolytes: completely dissociate in solution. –For example: Weak electrolytes: produce a small concentration of ions when they dissolve. –These ions exist in equilibrium with the unionized substance. –For example: Nonelectrolytes: do not ionize in solution

53. Chapter 10 | Slide 53 of 66 This simple device can be used to distinguish among strong electrolytes, weak electrolytes, and nonelectrolytes. Acids, Bases, and Salts cont’d

54. Chapter 10 | Slide 54 of 66 CC 10.5 Electrolyte and Body Fluids Acids, Bases, and Salts cont’d

55. Chapter 10 | Slide 55 of 66 Titration A laboratory procedure to determine the concentration of a compound in solution. If required, the total amount of the compound present can be determined from the experimentally determined concentration. –Must have a balanced chemical equation to describe the reaction. –Must know the volume of the solution for which the unknown concentration of the compound is to be determined. –Must have a means to indicate the endpoint of the reaction or when all of the compound of unknown concentration has completely reacted. –The volume of a solution of known concentration is determined that will completely react with the compound for which the concentration or amount is unknown using a titration equipment setup. –Use the concept of Volume in L X concentration in moles / L = moles

56. Chapter 10 | Slide 56 of 66 Titration This now becomes a stoichiometry calculation. grams X mole / gram = moles L X moles / L = moles Use stoichiometric mole ratios to determine the number of moles of the compound of unknown amount or concentration.

57. Chapter 10 | Slide 57 of 66 ←Fig. 10.16 Diagram showing setup for titration procedures. Acids, Bases, and Salts cont’d

58. Chapter 10 | Slide 58 of 66 Titrations: Reactions Used to Determine an Unknown Concentration

59. Chapter 10 | Slide 59 of 66 Fig. 10.17 An acid-base titration using an indicator that is yellow in acidic solution and red in basic solution. Acids, Bases, and Salts cont’d

60. Chapter 10 | Slide 60 of 66 Titration Calculations: An Example Suppose you are titrating an unknown sample of sulfuric acid (H 2 SO 4 ) with sodium hydroxide. It takes 25 mL of 0.24 M sodium hydroxide to titrate a 15 mL sample of the acid. What is the molarity of the acid? –2NaOH (aq) + H 2 SO 4 (aq)  2H 2 O (l) + Na 2 SO 4 (aq)

61. Chapter 10 | Slide 61 of 66 Titration Calculations: Another Example Determine the molarity of a NaOH solution when the following amount of acid neutralizes 25.0 mL of the NaOH solution: 23.76 mL of 1.00 M HCl.

62. Chapter 10 | Slide 62 of 66 Group Work: Titration Problems Suppose you are titrating an unknown sample of hydrochloric acid with magnesium hydroxide. It takes 14.28 mL of 1.35 M magnesium hydroxide to titrate a 23 mL sample of the acid. What is the molarity of the acid? 2HCl (aq) + Mg(OH) 2 (aq)  MgCl 2 (aq) + 2 H 2 O (l)