1. Unit 11: Acid-Base Equilibrium Chapter 16 and 17 Problem Set Chapter 16: 17, 21, 37, 43, 45, 61, 65, 69, 77, 79, 101, 107 Chapter 17: 19, 23, 27, 31, 41,

2. WarmUp 1)nitric acid 2)hydrofluoric acid 3)hydrobromic acid 4)chloric acid 5)acetic acid 6)carbonic acid 7)tetraboric acid (tetraborate = B 4 O 7 2- ) 1)HCl 2)H 2 S 3)H 3 PO 4 4)HI 5)HNO 2 6)H 3 P 7)H 2 SO 4 Write the Formula Name the Acid

3. WarmUp 1)nitric acid 2)hydrofluoric acid 3)hydrobromic acid 4)chloric acid 5)acetic acid 6)carbonic acid 7)tetraboric acid (tetraborate = B 4 O 7 2- ) 1)HCl 2)H 2 S 3)H 3 PO 4 4)HI 5)HNO 2 6)H 3 P 7)H 2 SO 4 Write the Formula Name the Acid

4. Arrhenius Definition Acids produce hydrogen ions in aqueous solution. Bases produce hydroxide ions when dissolved in water. Limits to aqueous solutions. Only one kind of base. NH 3 ammonia could not be an Arrhenius base.

5. Bronsted-Lowry Definitions And acid is an proton (H + ) donor and a base is a proton acceptor. Acids and bases always come in pairs. HCl is an acid.. When it dissolves in water it gives its proton to water. HCl(g) + H 2 O(l) H 3 O + + Cl - Water is a base makes hydronium ion

6. Neutralization Reactions Remember: Acid + Base → Water + Salt The salt’s name and formula will be based off of the cation of the base and the anion of the acid Example: HCl + NaOH ↔ H 2 O + NaCl Neutralization reactions are one way that is used to produce pure salts. Amphoteric Substances: can act as either an acid or a base

7. Titration Method used to perform neutralization and to determine concentrations of an unknown acid or base. Titrant – added standard solution End Point – indicator color change Equivalence Point – [H + ] = [OH - ]

8. Acid Base Calculations In neutral solutions, [H + ] and [OH - ] are equal [H + ] = [OH - ] = 1 x 10 -7 M For other solutions, when [H + ] increases, then [OH - ] will decrease and vise versa [H + ] x [OH - ] = 1.0 x 10 -14 K w = [H + ] x [OH - ] = 1.0 x 10 -14 This is the Ion-Product Constant for Water Acidic Solutions: [H + ] > [OH - ] Basic Solutions: [H + ] < [OH - ]

9. Using pH Scale pH: Power of Hydrogen pH = 7 Neutral [H + ] = 1 x 10 -7 pH 1 x 10 -7 pH > 7 Basic [H + ] = 1 x 10 -7 pH = -log [H + ] Example Calculate pH if [H + ] = 1 x 10 -7 M pH = -log (1 x 10 -7 M) = -(log 1 + log 10 -7 ) = -(0 +(-7)) pH = 7 * The exponent shows you the pH!*

10. The pOH Scale pOH : Power of Hydroxide pOH = 7 Neutral[OH - ] = 1 x 10 -7 pOH 1 x 10 -7 pOH > 7 Acidic [OH - ] < 1 x 10 -7 pOH = -log [OH - ] *Same process as before, only now you are dealing with [OH - ] Other Key Equations pH + pOH = 14 pH = 14 – pOH pOH = 14 – pH

11. Calculations: Complete the following chart pHpOH[H][OH]Acid- Base? 2.3 4.5 3.4x10 -5 4.6x10 -9 7

12. How do we know if something is a strong or weak acid/base? We must calculate a value for the ionic dissociation for that substance. ACID K a = acid dissociation constant: the ratio of the concentration of the dissociated form of the acid to the concentration of the undissociated form. BASE K b = base dissociation constant: the ratio of the product of the conjugate acid and OH concentrations to the concentration of the conjugate base

13. Strong and Weak Acids and Bases Dependent on how they dissolve in water Strong Acids completely ionize in aqueous solutions Weak acids only partially ionize in aqueous solutions Strong bases dissociate completely into metal ions and hydroxide ions in aqueous solutions Weak bases react with water to form OH and the conjugate acid of the base **Memorize Strong Acids and Bases**

14. Memorize the 8 strong acids… all others are weak HCl hydrochloric acid HNO3 nitric acid HBr hydrobromic acid HIO4 periodic acid HI hydroiodic acid H2SO4 sulfuric acid HClO4 perchloric acid HClO3 chloric acid

15. Writing the K a Expression Example HCl + H 2 O ↔ H 3 O + + Cl - First you need to write a K eq expression Second, water can be eliminated because its concentration is a constant. Write the K a expression for this acid: CH 3 COOH + H 2 O ↔ H 3 O + + CH 3 COO -

16. Significance of K a K a indicated the fraction of acid in the ionized form Weak acids have small K a values Strong acids have large K a values because their ionization is more complete

17. Writing the K b Expression Example NH 3 + H 2 O ↔ NH 4 + + OH - Remember: K b tells us how weak bases compete for OH - from strong bases The smaller the value for K b, the weaker the base

18. Relationships K W = [H + ][OH - ] -log K W = -log([H + ][OH - ]) -log K W = -log[H + ]+ -log[OH - ] pK W = pH + pOH K W = 1.0 x10 -14 14.00 = pH + pOH [H + ],[OH - ],pH and pOH Given any one of these we can find the other three.

19. Example Calculate the pH of 2.0 M acetic acid HC 2 H 3 O 2 with a Ka 1.8 x10 -5 Calculate pOH, [OH - ], [H + ]

20. A mixture of Weak Acids The process is the same. Determine the major species. The stronger will predominate. Bigger Ka if concentrations are comparable Calculate the pH of a mixture 1.20 M HF (Ka = 7.2 x 10 -4 ) and 3.4 M HOC 6 H 5 (Ka = 1.6 x 10 -10 )

21. Salt Hydrolysis Most salt solutions are neutral, but some can be acidic or basic Salts like sodium chloride and potassium sulfate are neutral WHY? Neutral salts are made from strong acids and strong bases Salt Hydrolysis – the cations or anions of the dissolved salt remove H + from, or donate H + to water A solution is either acidic or basic based on H + transfer.

22. Salt Cations/Anions Effect on pH In water: If cation and anion do not react, then pH = neutral If cation does not react, but anion does, then hydroxide is produced, pH = basic If cation reacts but anion does not, then H are produced, pH = acidic

23. If both react: OH - and H 3 O + both produced 1.Anions from strong acids will not change pH (ex: Br - ) 2.Anions from a weak acid will increase pH (ex: CN - ) 3.Cations from weak bases will decrease pH (ex: CH 3 NH 3 + ) 4.Group 1A ions will increase pH 5.Group 2A ions will decrease pH

24. What is the pH of a 0.140M solution of sodium acetate? C 2 H 3 O 2 - + HOH↔HC 2 H 3 O 2 + OH - But what is K b ? KbxKa = Kw Kb = Kw/Ka = 1x10 -14 /1.8x10 -5 Solve for x x = 8.77x10 -6 =[OH - ] pOH=5.06, pH = 8.94 so the solution is basic. I0.140 M00 C-x+x E0.140 – xxx

25. The Common Ion Effect When the salt with the anion of a weak acid is added to that acid, It reverses the dissociation of the acid. Lowers the percent dissociation of the acid. The same principle applies to salts with the cation of a weak base.. The calculations are the same as last chapter.

26. Types of Acids Monoprotic Acied – one acidic hydrogen Polyprotic Acids- more than 1 acidic hydrogen (diprotic, triprotic). Oxyacids - Proton is attached to the oxygen of an ion. Organic acids contain the Carboxyl group - COOH with the H attached to O Generally very weak. H 2 SO 3 ↔ H + + HSO 3 - K a1 = 1.7 x 10 -2 HSO 3 - ↔ H + + SO 3 2- K a2 = 6.4 x 10 -8 Why is K smaller? It is always easier to remove the first proton

27. Structure and Acid-Base Properties Any molecule with an H in it is a potential acid. The stronger the X-H bond the less acidic (compare bond dissociation energies). The more polar the X-H bond the stronger the acid (use electronegativities). The more polar H-O-X bond -stronger acid.

28. Strength of oxyacids The more oxygen hooked to the central atom, the more acidic the hydrogen. HClO 4 > HClO 3 > HClO 2 > HClO Remember that the H is attached to an oxygen atom. The oxygens are electronegative Pull electrons away from hydrogen

29. Strength of oxyacids Electron Density ClOH

30. Strength of oxyacids Electron Density ClOHO

31. Strength of oxyacids ClOH O O Electron Density

32. Strength of oxyacids ClOH O O O Electron Density

33. Calculate the Concentration Of all the ions in a solution of 1.00 M Arsenic acid H 3 AsO 4 n Ka 1 = 5.0 x 10 -3 n Ka 2 = 8.0 x 10 -8 n Ka 3 = 6.0 x 10 -10

34. Sulfuric acid is special In first step it is a strong acid. Ka 2 = 1.2 x 10 -2 Calculate the concentrations in a 2.0 M solution of H 2 SO 4 Calculate the concentrations in a 2.0 x 10 -3 M solution of H 2 SO 4

35. Hydrated metals Highly charged metal ions pull the electrons of surrounding water molecules toward them. Make it easier for H + to come off. Al +3 O H H

36. Acid-Base Properties of Oxides Non-metal oxides dissolved in water can make acids. SO 3 (g) + H 2 O(l) H 2 SO 4 (aq) Ionic oxides dissolve in water to produce bases. CaO(s) + H 2 O(l) Ca(OH) 2 (aq)

37. Lewis Acids and Bases Most general definition. Acids are electron pair acceptors. Bases are electron pair donors. BF F F :N:N H H H

38. Lewis Acids and Bases Boron triflouride wants more electrons. BF F F :N:N H H H

39. Lewis Acids and Bases Boron triflouride wants more electrons. BF 3 is Lewis base NH 3 is a Lewis Acid. B F F F N H H H

40. Lewis Acids and Bases Al +3 ( ) H H O Al ( ) 6 H H O + 6 +3

41. Buffers Solutions in which the pH remains relatively constant when small amounts of acid of base are added. Acidic Buffer – composed of a weak acid and one of its salts Basic Buffer – composed of a weak base and one of its salts Buffer Capacity – the amount of acid or base that can be added to a buffer solution before a significant change in pH occurs

42. Buffers have the ability to acquire or give a way a H +, which allows it to maintain a standard pH Example: Ethanoic Acid/Ethanoate system (Acetic Acid / Acetate) CH 3 COO - + H + ↔ CHCOOH CH 3 COOH + OH - ↔ CHCOO - + H 2 O Both reactions occur simultaneously to maintain pH

43. pH of Buffers To determine pH of Buffers, we use the Henderson-Hasselbalch Equation Calculate the pH of a solution that is.50 M HAc and.25 M NaAc (Ka = 1.8 x 10 -5 )

44. Calculate the pH of the following mixtures 1.0.75 M lactic acid (HC 3 H 5 O 3 ) and 0.25 M sodium lactate (Ka = 1.4 x 10 -4 ) 2.0.25 M NH 3 and 0.40 M NH 4 Cl (Kb = 1.8 x 10 -5 )

45. Titration Curves

46. pH mL of Base added 7 Strong acid with strong Base Equivalence at pH 7

47. pH mL of Base added >7 l Weak acid with strong Base l Equivalence at pH >7

48. pH mL of Base added 7 l Strong base with strong acid l Equivalence at pH 7

49. pH mL of Base added <7 l Weak base with strong acid l Equivalence at pH <7

50. Summary Strong acid and base just stoichiometry. Determine Ka, use for 0 mL base Weak acid before equivalence point –Stoichiometry first –Then Henderson-Hasselbach Weak acid at equivalence point Kb Weak base after equivalence - leftover strong base.

51. Summary Determine Ka, use for 0 mL acid. Weak base before equivalence point. –Stoichiometry first –Then Henderson-Hasselbach Weak base at equivalence point Ka. Weak base after equivalence - leftover strong acid.

52. Indicators Weak acids that change color when they become bases. weak acid written HIn Weak base HIn H + + In - clear red Equilibrium is controlled by pH End point - when the indicator changes color.

53. Indicators Since it is an equilibrium the color change is gradual. It is noticeable when the ratio of [In - ]/[HI] or [HI]/[In - ] is 1/10 Since the Indicator is a weak acid, it has a Ka. pH the indicator changes at is. pH=pKa +log([In - ]/[HI]) = pKa +log(1/10) pH=pKa - 1 on the way up

54. Indicators pH=pKa + log([HI]/[In - ]) = pKa + log(10) pH=pKa+1 on the way down Choose the indicator with a pKa one less than the pH at equivalence point if you are titrating with base. Choose the indicator with a pKa one greater than the pH at equivalence point if you are titrating with acid.