1. Acids, Bases, and Salts Properties of Acids, Bases and Salts Calculating pH, pOH, [H 3 O + ], [OH - ] Calculating pH, pOH, [H 3 O + ], [OH - ] Acid-Base Theories, Strengths of Acids and Bases Acid-Base Theories, Strengths of Acids and Bases Calculating K a, K b, and percent dissociated Calculating K a, K b, and percent dissociated Salt Hydrolysis Buffers Titrations

2. Properties of Acids 1.Aqueous solutions of acids have a sour taste. For example, the sour taste of lemons and other citrus fruits is due to citric acid. 2.Acids are electrolytes; they conduct electricity when dissolved in water. 3.Acids change the color of acid-base indicators. Acids turn blue litmus red. 4.Some acids react with active metals to produce hydrogen gas. 5.Acids can be solids, liquids or gases in their pure state. 6.Nonmetals and nonmetal oxides tend to form acids in water.

3. Naming Binary Acids × Binary acids are named by using the prefix hydro followed by the root and the –ic suffix. Name the following binary acids. 1.HCl 2.HF 3.H 2 S 4.HBr hydrochloric acid hydrofluoric acid hydrosulfuric acid hydrobromic acid

4. Writing Formulas for Binary Acids × Remember to balance the charges when writing the formulas for binary acids. Write formulas for the following binary acids. 1.hydroiodic acid 2.hydroselenic acid H +, I - HI H +, Se 2- H 2 Se

5. Naming and Writing Formulas for Oxyacids × Oxyacids are named based upon the polyatomic ion making up the acid. Polyatomic Ion Ending Acid Ending ExampleName per –ateper - icHClO 4 perchloric acid -ate-icHClO 3 chloric acid -ite-ousHClO 2 chlorous acid hypo – ite hypo - ous HClO hypochlorous acid

6. Naming Writing Formulas for Oxyacids × Remember to balance the charges when writing the formulas for oxyacids. Write the name or formula for each of the following acids. 1.HIO 3 2.HNO 2 3. sulfuric acid 4. phosphorous acid IO 3 - = iodate iodic acid NO 2 - = nitrite nitrous acid H +, SO 4 2- H 2 SO 4 H +, PO 3 3- H 3 PO 3

7. Properties of Bases 1.Aqueous solutions of bases taste bitter. Have you ever had to wash out your mouth with soap? 2.Dilute aqueous solutions of bases feel slippery. 3.Bases change the color of acid-base indicators. Bases turn red litmus blue. 4.Bases are electrolytes; they conduct electricity when dissolved in water. 5.Most common bases are solids. An exception is ammonia (NH 3 ) which is a gas at room temperature. 6.Active metals and metal oxides tend to form bases in water.

8. Naming and Writing Formulas for Bases Bases are named by writing the name of the metal ion followed by the word hydroxide. The charges are balanced when writing the formulas for bases. Name or write the formula for the following bases 1.LiOH 2. Magnesium hydroxide × Lithium hydroxide Mg 2+, OH - Mg(OH) 2

9. Properties of Salts 1.A salt is an ionic compound composed of a cation (positive ion) and an anion (negative ion). 2.Salts are crystalline solids that have high melting points. 3.Salts are generally soluble in water. 4.Salts are electrolytes.

10. Naming and Writing Formulas for Salts Salts are named by writing the name of the cation followed by the name of the anion. The charges are balanced when writing the formulas for salts. Name or write the formula for the following salts. 1.CaSO 4 2. Magnesium nitrate 3.LiBr 4. Copper(II) chloride × Calcium sulfate Mg 2+, NO 3 - Mg(NO 3 ) 2 Lithium bromide Cu 2+, Cl - CuCl 2

11. Neutralization Reactions The reaction of an acid with a base is called a neutralization reaction. The products of a neutralization reaction are a salt and water. Neutralization reactions are double replacement reactions. Examples: HCl + NaOH  HNO 3 + KOH  H 2 SO 4 + LiOH  H2OH2O + NaCl H2OH2O + KNO 3 H2OH2O + Li 2 SO 4 2 2

12. Self-Ionization of Water The origination of the idea of pH is the self- ionization of water. In the self-ionization of water, two water molecules collide producing a hydronium ion and a hydroxide ion by transfer of a proton. H 2 O(l) + H 2 O(l) ⇄ H 3 O + (aq) + OH - (aq) also written as H 2 O(l) ⇄ H + (aq) + OH - (aq)

13. Ion Product Constant for Water (K w ) For aqueous solutions, the product of the hydronium ion concentration and the hydroxide ion concentration equals 1.0 × 10 -14 M 2 (at 25°C). [H 3 O + ][OH - ] = 1.0 × 10 -14 M 2 also written as [H + ] [OH - ] = 1.0 × 10 -14 M 2 The product of the concentrations of the hydronium ions and the hydroxide ions in water is called the Ion Product Constant for water (K w ). K w = [H 3 O + ][OH - ] = 1 × 10 -14 M 2 also written as K w = [H + ][OH - ] = 1 × 10 -14 M 2

14. [H 3 O + ] versus [OH - ] All aqueous solutions have H 3 O + and OH - present. A solution can be classified as acidic, basic or neutral by comparing the number of hydronium ions in solution to the number of hydroxide ions. Neutral: [H 3 O + ] = [OH - ] = 1 × 10 -7 M Acidic: [H 3 O + ] > [OH - ] [H 3 O + ] > 1 × 10 -7 M Basic (alkaline): [H 3 O + ] < [OH - ] [H 3 O + ] < 1 × 10 -7 M

15. Example Problems. Calculate the [OH - ] if the [H 3 O + ] is 1.0 x 10 -5 M. Is the solution acidic, basic or neutral? [H 3 O + ] > [OH - ] acidic

16. Example Problems. Calculate the [OH - ] if the [H 3 O + ] is 1.8 × 10 -8 M. Is the solution acidic, basic or neutral? [H 3 O + ] < [OH - ] basic

17. pH An easier way to determine whether a solution is acidic, basic or neutral is by determining the pH of the solution. The pH scale was developed by Sorenson in 1909. The pH scale ranges from 0 to 14 at 25°C. acidic solution: neutral solution: basic solution: pH = -log [H 3 O + ][H 3 O + ] = 10 x (-pH) pOH = -log [OH - ][OH - ] = 10 x (-pOH) pH + pOH = 14 pH < 7 pH = 7 pH > 7

18. Example Problems Calculate the pH of a solution that has a [H 3 O + ] of 3.4 × 10 -5 M. pH = -log [H 3 O + ] pH = 4.47 Calculate the pOH of a solution that has a [OH - ] of 2.5 × 10 -2 M. pOH = -log [OH - ] pOH = 1.60

19. Example Problems Calculate the [H 3 O + ] of a solution that has a pH of 3.5. [H 3 O + ] = 10 x (-pH) [H 3 O + ] = 3.16 × 10 -4 Calculate the [OH - ] of a solution that has a pOH of 2.3. [OH - ] = 10 x (-pOH) [OH - ] = 5.01 × 10 -3 Calculate the pH of a solution that has a pOH of 12. pH + pOH =14 pH = 14-12 = 2

20. Complete the following table. pHpOH[H 3 O + ] [OH - ] Acidic, Basic or Neutral 4.001.0 × 10 -10 Acidic 8.001.0 × 10 -6 Acidic 7.001.0 × 10 -7 Neutral 5.001.0 × 10 -5 1.0 × 10 -9 3.503.2 × 10 -4 Acidic 13.900.791.26 × 10 -14 4.5 × 10 -11 2.51 × 10 -4 Basic 9.151.4 × 10 -5 Basic pH = -log [H 3 O + ][H 3 O + ] = 10 x (-pH) pOH = -log [OH - ][OH - ] = 10 x (-pOH) pH + pOH = 14 10.00 6.00 3.6010.40 0.10 10.50 9.00 7.00 4.85 1.0 × 10 -4 1.0 × 10 -8 1.0 × 10 -7 3.2 × 10 -11 7.08 × 10 -10 Acidic

21. Acid-Base Theories- Arrhenius Acids and Bases Arrhenius theorized that acids and bases must produce ions in solution. acid – produces hydrogen ions in water solutions. Examples: HCl, H 2 SO 4 base – produces hydroxide ions in water solutions. Examples: NaOH, Ba(OH) 2

22. Acid-Base Theories- Brønsted-Lowry Acids and Bases Brønsted-Lowry defined an acid as a proton donor Brønsted-Lowry defined a base as a proton acceptor

23. Acid + Base  conjugate base + conjugate acid A conjugate base is the remaining part of an acid after it has released a proton. A conjugate acid is the acid formed when a base accepts a proton. NH 3 + H 2 O ⇄ NH 4 + + OH - base acid CA CB

24. Label the acid, base, CA, and CB HNO 2 + H 2 O ⇄ H 3 O + + NO 2 - H 2 O + C 2 H 3 O 2 - ⇄ HC 2 H 3 O 2 + OH - acid base CACB acid base CACB

25. Strengths of Acids and Bases When you refer to the strength of an acid or a base, you are talking about the degree to which it is ionized in aqueous solutions. Strong acids are completely ionized into aqueous solution. This makes them strong electrolytes. Examples: HCl HBr, HI, HClO 4, HClO 3, HNO 3, and H 2 SO 4 Example: HCl + H 2 O → H 3 O + + Cl - The concentration of H 3 O + present after the HCl ionizes is equal to the original concentration of HCl.

26. Weak Acids Weak acids ionize only slightly in aqueous solution. Weak acids are weak electrolytes. Ex. HCN, HF, HC 2 H 3 O 2 HC 2 H 3 O 2 + H 2 O ⇄ H 3 O + + C 2 H 3 O 2 - The initial concentration of HC 2 H 3 O 2 is much greater than the concentration of H 3 O + at equilibrium. The acid dissociation (ionization) constant, K a, can be written for a weak acid. The acid dissociation constant is a ratio of the dissociated form of an acid to the undissociated form.

27. Weak Acids Write the K a expression for acetic acid. HC 2 H 3 O 2 + H 2 O ⇄ H 3 O + + C 2 H 3 O 2 -

28. Comparing K a values The smaller the value of K a, the weaker the acid. Which of the following acids is the weakest? carbonic acid K a = 4.2 × 10 -7 formic acidK a = 1.8 × 10 -4 benzoic acid K a = 6.3 × 10 -5 Carbonic Acid

29. Bases Strong bases ionize completely in aqueous solution. Strong bases are strong electrolytes. Examples: Ca(OH) 2, NaOH, KOH, LiOH, Ba(OH) 2, Sr(OH) 2 Weak bases partially ionize in aqueous solution. They are weak electrolytes. Examples NH 3, Al(OH) 3

30. K b values The base dissociation constant, K b, can be written for a weak base. The base dissociation, K b, is the ratio of the dissociated form of a base to the undissociated form.

31. K b Expressions Write an ionization equation and a K b expression for hydrazine, N 2 H 4. N 2 H 4 + H 2 O ⇄ N 2 H 5 + + OH -

32. Calculating the pH of Solutions of Strong Acids and Strong Bases 1.Calculate the pH of a 1.00 M HI solution. 2.Calculate the pH of a 1.0 M KOH solution. Since it is a strong acid it completely dissociates in water. [HI] = [H 3 O + ] pH = -log [H 3 O + ] pH= -log[1.00] = 0 Since it is a strong base it completely dissociates in water. [KOH] = [OH - ] pOH = -log [OH - ] pOH= -log[1.00] = 0 pH = 14 – 0 = 14

33. Calculations Involving Weak Acids and Bases 1.A 0.100 M solution of acetic acid (HC 2 H 3 O 2 ) is only partially ionized. The K a of acetic acid is 1.8×10 -5. a.Write a dissociation reaction for acetic acid. HC 2 H 3 O 2 + H 2 O ⇄ C 2 H 3 O 2 - + H 3 O +

34. Calculations Involving Weak Acids and Bases 1.A 0.100 M solution of acetic acid (HC 2 H 3 O 2 ) is only partially ionized. The K a of acetic acid is 1.8×10 -5. b. Write a K a expression for acetic acid.

35. Calculations Involving Weak Acids and Bases c. Calculate the pH of a 0.15 M solution of HC 2 H 3 O 2. [HC 2 H 3 O 2 ][H 3 O + ][C 2 H 3 O 2 - ] Initial Change Equilibrium 0.15 M0.0 M -x +x 0.15 M-x ≈ 0.15 M* *The K a value is so small that it is assumed that the amount the [HC 2 H 3 O 2 ] changes is negligible. xx x 2 =2.73×10 -6, x = 0.00165 M pH = -log[H 3 O + ] = -log(0.00165) = 2.78

36. Calculations Involving Weak Acids and Bases d.Another way of expression how much of a weak acid (or base) is in ionic form is to give the percent dissociated (also called percent ionized). HA + H 2 O ⇄ H 3 O + + A -

37. Calculations Involving Weak Acids and Bases e.Find the percent of dissociation of the 0.15 M HC 2 H 3 O 2 solution. HC 2 H 3 O 2 + H 2 O ⇄ H 3 O + + C 2 H 3 O 2 -

38. 2.Nitrous acid (HNO 2 ) is a weak acid with a K a of 4.6×10 -4 at 25°C. a.Write a dissociation reaction for nitrous acid. HNO 2 + H 2 O ⇄ NO 2 - + H 3 O + b.Write a K a expression for nitrous acid.

39. c. Calculate the pH of a 0.0450 M nitrous acid solution. [HNO 2 ][H 3 O + ][NO 2 - ] Initial Change Equilibrium 0.0450 M0.0 M -x+x 0.0450 M-x ≈ 0.0450 M xx x 2 = 2.07×10 -5 ; x = 0.00455 pH = -log(0.00455) = 2.34

40. 3.Lactic acid (HC 3 H 5 O 3 ) is a waste product that accumulates in muscle tissue during exertion, leading to pain and a feeling of fatigue. a.Write a dissociation reaction for lactic acid. HC 3 H 5 O 3 + H 2 O ⇄ C 3 H 5 O 3 - + H 3 O + b.Write a K a expression for lactic acid. c.In a 0.100 M aquous solution, lactic acid is 3.7% dissociated. Calculate [H 3 O + ].

41. d. Calculate the value of K a for lactic acid. [HC 3 H 5 O 3 ][H 3 O + ][C 3 H 5 O 3 - ] Initial Change Equilibrium 0.100 M0.0 M -3.7×10 -3 * +3.7×10 -3 0.0963 M 3.7×10 -3 * In this case, the [H 3 O + ] (we calculated in part C) represents the amount the initial concentrations change.

42. Salt Hydrolysis A salt is made by neutralizing an acid with a base. When a salt dissolves in water, it releases ions having an equal number of positive and negative charges. Thus a solution of a salt should be neither acidic nor basic. Some salts do form neutral solutions, but other react with water (hydrolyze) to form acidic or basic solutions.

43. Types of Salt Solutions 1.A neutral solution results when the salt produced from a strong acid and strong base is dissolved in water. ex. HCl + NaOH  NaCl + H 2 O 2.An acidic solution results when the salt produced from a strong acid and a weak base is dissolved in water. ex. 3HCl + Al(OH) 3  AlCl 3 + 3H 2 O 3.A basic solution results when the salt produced from a weak acid and a strong base is dissolved in water. ex. H 2 CO 3 + 2NaOH  Na 2 CO 3 + 2H 2 O 4.The salt produced from a weak acid and a weak base may form an acidic, basic, or neutral solution. (You would have to compare the K a and K b values to determine whether or not the salt formed from a weak acid and base was acidic, basic, or neutral. We will not be doing that in this class.)

44. Identify the acid and base from which each of the following salts was formed and then classify the solution as acidic, basic, or neutral. a.0.5 M NH 4 ClO 4 NH 3 - weak base HClO 4 – strong acid acidic b.1.0 M BaSO 4 Ba(OH) 2 - strong base H 2 SO 4 – strong acid neutral c. 0.4 M K 2 CO 3 KOH - strong base H 2 CO 3 – weak acid basic

45. Buffers A buffer system is a solution that can absorb moderate amounts of acid or base without a significant change in its pH. A buffer provides ions that will react with H 3 O + or OH - ions if they are introduced into the solution. Because the added H 3 O + or OH - ions are thereby neutralized, the pH of the system remains nearly constant. Buffer solutions are prepared by using a weak acid with one of its salts or a weak base with one of its salts.

46. Example of a Buffer System Many of the fluids in your body must be maintained within a very narrow pH range if you are to remain healthy. Let’s look at the buffer system that is present in your blood. This buffer contains: HCO 3 - (from the salt) and H 2 CO 3 (carbonic acid) When excess hydronium ions enter the blood, the hydrogen carbonate ion undergoes the following reaction to reduce the [H 3 O + ]. HCO 3 - (aq) + H 3 O + (aq) ⇄ H 2 CO 3 (aq) + H 2 O(l) When excess hydroxide ions form in the blood the following reaction occurs. H 2 CO 3 (aq) + OH - (l) ⇄ HCO 3 - (aq) + H 2 O(l)

47. Titrations Sometimes we want to know the concentration of an unknown solution of an acid or base. The concentration of an acid (or base) in a solution is determined by carrying out a neutralization reaction.

48. Titration Steps 1.A measured amount of an acid of unknown concentration is added to an Erlenmeyer flask. 2.An appropriate indicator (such as phenolphthalein) is added to the solution. 3.Measured amounts of a base of known concentration are mixed into the acid. The solution of known concentration is called the standard solution. The addition of the base is carried out using a buret. This process is continued until the indicator indicates that the end point has been reached. 4.The point at which the two solutions used in a titration are present in chemically equivalent amounts is the equivalence point.

49. Titration Graphs A graph can be made of pH versus volume of standard solution. The equivalence point of the titration corresponds to the middle of that portion of the graph showing a very large change in pH with the addition of a small amount of the standard solutions.

50. Titration Graphs

51. The equivalence point is not always at a pH of 7. Why not? It will not be 7 if a weak acid and a strong base are combined or if a strong acid and a weak base are combined.

52. Titration Calculations When the acid and base used in a titration react in a 1:1 mole ratio, the following relationship can be used to determine the concentration of the unknown solution or the volume of known needed to neutralize the unknown solution. M a V a = M b V b

53. Titration Calculations When the acid and base do not react in a 1:1 mole ratio, a mole factor must be used. 1.H 2 SO 4 + 2NaOH  Na 2 SO 4 + 2H 2 O 2(M a V a ) = M b V b 2.2HCl + Ba(OH) 2  BaCl 2 + 2H 2 O M a V a = 2(M b V b )

54. Example problems How many mL of 0.50 M HCl must be added to 25.0 mL of 2.0 M KOH to make a neutral solution? HCl + KOH  KCl + H 2 O Since it is a 1:1 ratio you can use the relationship M a V a =M b V b

55. Example problems What is the molarity of a solution of HNO 3 if 30.0 mL of 1.5 M Ba(OH) 2 are required to neutralize 10 mL of the acid? 2HNO 3 + Ba(OH) 2  Ba(NO 3 ) 2 + 2H 2 O Since it is not a 1:1 ratio you must use a mole factor.M a V a =2M b V b