Acid & Base Equilibrium. Chapter 16

Transcription

1 Acid & Base Equilibrium Chapter 16 1

2 base 10 Log Facts log( ) 10 log( ) 3 log( ) 1 log( ) 0 log( ) -1 log( ) -3 log( ) -10 OK, but what if the # in front is not 1?? 2

3 Most of the time you would not be asked to think about logging numbers like this, unless the MC answers were far enough apart to make your choice easier. take note of the pattern of the exponent on the 10 in relation to the # before decimal on the ph. log( ) = 0.82 log( ) = 2.60 log( ) = 4.46 log( ) = 5.35 log( ) = 8.26 log( ) =

4 ...and remember sig figs. In a logged number only the numbers after the decimal are significant. Mantissa (aka significand) You see, the 4.6 is what causes the.34 The number in front of the.34 (abscissa) is a result of the magnitude of the number that is being logged, not a part of its significant figures, Thus only the numbers after the decimal are the significant figure part of a ph (or poh) log( ) = 0.34 log( ) = 2.34 log( ) = 6.34 log( ) = 9.34 log( ) = Abscissa 4

5 Let s Define Acids & Bases In 1880 s Arrhenius (same guy who studied reaction rates) defined: Acid: a substance that dissolves in water to form H + ü HCl H + + Cl Base: a substance that dissolves in water to form OH ü NaOH Na + + OH 1923 Bronsted & Lowry defined: Acids as a proton (H + ) donor Bases as a proton (H + ) acceptor 5

6 H + the proton - what is it? H + results from the dissociation of acids in solution. HCl(aq) H + (aq) + Cl (aq) The electronegativity of chlorine hogs the electron and polarizes the bond, combined with the weak strength of the bond causes the HCl molecule separate to as ions. Essentially H + is a naked proton. ü But is it really naked???? 6

7 Hydrogen ion = Hydronium Ion The H + (the naked proton) is so small it immediately attaches to one of the unshared pair of electrons on a water molecule forming H3O + + HCl(aq) + H2O H3O + (aq) + Cl (aq) As far as we re concerned, we ll symbolize the hydronium ion H + or H3O + We ll use these interchangeably. 7

8 Bronsted & Lowry Definition as applied to a strong acid HCl(g) + H2O(L) H3O + + Cl HCl is acting as the proton donor. H2O is acting as the proton acceptor. But for a strong acid, this B & L definition is not necessary, it s just as convenient and accurate to use the Arrhenius definition that defines an acid as a substance that dissolves to produce H + HCl(aq) H + + Cl 8

9 Bronsted & Lowry Definition as applied to a strong base NaOH (s) + H 2 O (L) Na + + OH + H 2 O (L) This of course fits the Arrhenius definition by dissolving to produce OH So what about the proton donor and acceptor? We could say that water donates a proton, and the OH accepts a proton. OH + H 2 O (L) H 2 O (L) + OH H 2 O (L) is acting as the proton donor. OH is acting as the proton acceptor. But of course it seems silly to even write this out because the exchange of proton turns into exactly the same thing. So for a strong base, this B & L definition is also not necessary, its just as convenient and accurate to use the Arrhenius definition that defines a base as a substance that dissolves to produce OH. 9

10 Let s Define Acids & Bases In 1880 s Arrhenius (same guy who studied reaction rates) defined: Acid: a substance that dissolves in water to form H + ü HCl H + + Cl Base: a substance that dissolves in water to form OH ü NaOH Na + + OH 1923 Bronsted & Lowry defined: Acids as a proton (H + ) donor Bases as a proton (H + ) acceptor Also in 1920 s Gilbert Lewis (same guy who invented {Electron Dot} Lewis Structures) expanded the definition: Acids as electron pair acceptors Bases as electron pair donors We have discussed this in reference to ions solvated in aqueous solution more on this later in the unit 10

11 Bronsted & Lowry Definition Most reactions we will study will be in solution, but gases can also act as acids and bases. You may recall... We tested the speed of gases with the following reaction: NH 4 Cl HCl (g) + NH 3(g) NH 4 Cl (s) NH4Cl forms Why is the white ring NOT in the middle? Why is it closer to the HCl end? 11

12 Bronsted & Lowry Definition HCl (g) + NH 3(g) NH 4 Cl (s) Draw the Lewis structure for HCl, NH3, and NH 4 Cl The HCl donates the proton and the NH 3 accepts the proton then the two ions stick together to form a white solid. NH 4 Cl HCl NH 3 12

13 Weak Acids & Weak Bases When an acid or base does not ionize completely in solution, the reaction does not go to completion and will reach equilibrium. This acid or base is said to be weak. concentrated vs dilute and strong vs weak You can have a concentrated strong acid lots dissolved, completely ionized (broken apart into ions) You can have a dilute strong acid not much dissolved, completely ionized You can have a concentrated weak acid lots dissolved, partially ionized You can have a dilute weak acid not much dissolved, partially ionized 13

14 Reaction of a Weak Acid HF (aq) + H 2 O (L) F (aq) + H 3 O + (aq) proton proton proton proton donor acceptor acceptor donor acid base base acid HF is the proton donor, the acid. H 2 O is the proton acceptor, the base This reaction happens to be one that doesn t go to completion. It s reversible and therefore reaches equilibrium. This reversibility allows us to label acids and bases for the reverse reaction. 14

15 Reaction of a Weak Base NH 3(g) + H 2 O (L) NH 4 + (aq) + OH (aq) proton proton proton proton acceptor donor donor acceptor base acid acid base NH 3 is the proton acceptor, the base. H 2 O is the proton donor, the acid. This reaction happens to be one that doesn t go to completion. It s reversible and therefore reaches equilibrium. This reversibility allows us to label acids and bases for the reverse reaction. 15

16 Amphoterism You may have noticed that in the last two reactions water has behaved both as an acid and as a base. NH 3(g) + H 2 O (L) NH 4 + (aq) + OH (aq) acid HF (aq) + H 2 O (L) base H 3 O + (aq) + F (aq) A substance that is capable of acting as either an acid or a base is called amphoteric (or amphoprotic). 16

17 Conjugate Acid & Base Pairs We can write a generic reaction to represent any weak acid HA + H 2 O A + H 3 O + acid base base acid You can think of this in reverse... You could just as easily write the reverse of this A + H 3 O + HA + H 2 O conjugate base conjugate acid conjugate acid conjugate base We call the corresponding acid/base pairs, conjugate pairs 17

18 K eq for Acids & Bases Write a generic reaction to represent any weak acid, HA: HA(aq) + H 2 O A + H 3 O + K a = A H 3 O + [ HA] HA(aq) A + H + K a = A H + [ HA] Now, write an equilibrium expression for this reaction Note what is NOT in the expression 18

19 Generic Rx s for Bases Write a generic reaction to represent any weak base, B: B + H 2 O HB + + OH Now, write an equilibrium expression for this reaction. (note: what is not in the expression) K b = HB+ OH [ B] 19

20 Generic Rx s for Bases Since every weak acid produces a conjugate weak base, we can also represent a weak base as A We can get A without getting it as an acid? by dissolving NaF into water: NaF(s) Na + + F F is a conjugate weak base (For now, ignore F.) Write an equation to represent the reaction of F in water. (Remember bases are proton acceptors.) F + H 2 O HF(aq) + OH Write the equilibrium expression for this equation [ K = HF ] OH b F 20

21 K eq for Acids & Bases Remember that conjugate acids form from weak bases, NH4 + We can get NH4 + without forming it from its base by dissolving an ammonium ionic compound? ü Dissolve NH4NO3 into water: NH4NO3(s) NH4 + + NO3 ü NH4 + is a conjugate weak acid (for now ignore NO3.) Write an equation for the acidic reaction of NH4 + in water (Remember acids are proton donors). NH4 + + H 2 O NH3(aq) + H 3 O + NH4 + NH3(aq) + H + ü write an equilibrium expression for the reaction of NH4 + in water [ ] H + K a = NH 3 [ ] H 3 O + K a = NH 3 NH 4 + NH

22 Just Who Are the WA & WB? Molecular WA generally have the formula HA Molecular WB generally have the formula: NH 3 stuff-nh 2 stuff-nh stuff-n Be sure and observe the conjugates. The accepted proton locates on the N 22

23 Consider the weak base, ethylamine C2H5NH2 Write an equation that represents the hydrolysis of ethylamine. aka: Write an equation that represents the ionization of ethylamine aka: Write an equation that represents ethylamine reacting with water C2H5NH2(aq) + H2O C2H5NH3 + (aq) + OH Then write the equilibrium expression. K = C 2H NH b OH [ ] C 2 H 5 NH 2 23

24 Consider the weak acid, lactic acid HC3H5O3 Write an equation that represents the hydrolysis of lactic acid. aka: Write an equation that represents the ionization of lactic acid aka: Write an equation that represents lactic acid reacting with water HC3H5O3(aq) + H2O C3H5O3 + H + Then write the equilibrium expression. K = H + a C 3 H 5 O 3 [ HC H O ]

25 Compare the inverse proportionality of the strengths of conjugate Acids ~ Bases HCl Cl H2SO4 HSO4 or SO4 2 HNO3 NO3 acid/base pairs. Acid strength increases H3O + HF HC2H3O2 C5H5NH + H2CO3 HOCl HOBr NH4 + H2O F C2H3O2 C5H5N HCO3 OCl OBr NH3 Base strength increases HCO3 CO3 2 C2H5NH3 + C2H5NH2 H2O OH Na + NaOH Ba 2+ Ba(OH)2 25

26 Weak Conjugates Acids ~ Bases Pathetic Conjugates NaF Na + + F why is F a base? HCl H2SO4 HNO3 Cl HSO4 or SO4 2 NO3 pathetic H3O + H2O F + H2O HF + OH NH4Cl NH4 + + Cl why is NH4 + an acid? NH4 + NH3 + H + Acid strength increases HF HC2H3O2 C5H5NH + H2CO3 HOCl HOBr NH4 + F C2H3O2 C5H5N HCO3 OCl OBr NH3 Base strength increases NaCl Na + + Cl why is Na + pathetic? What part of water is Na + most attracted to? HCO3 C2H5NH3 + CO3 2 C2H5NH2 Na + + H2O NaOH + H + why is Cl pathetic? What part of water is Cl - most attracted to? pathetic H2O Na + Ba 2+ OH NaOH Ba(OH)2 Cl + H2O HCl + OH 26

27 Know your SA & SB... Strong Acids HCl HI HBr HNO3 H2SO4 HClO4 HClO3 Strong Bases LiOH NaOH KOH RbOH CsOH Ba(OH)2 Sr(OH)2 Ca(OH)2...and you know the pathetics. all other +ions are acids, and nearly all other ions are bases. 27

28 Stong, weak, pathetic What does it all mean? Acids ~ Bases A strong acid has a HCl Cl pathetic conjugate base H2SO4 HSO4 or SO4 2 (which we do not refer to as bases at all). A stronger-ish weak acid has a very weak conjugate weak base. Whereas a weaker weak acid has a stronger-ish Acid strength increases HNO3 H3O + HF HC2H3O2 C5H5NH + H2CO3 HOCl HOBr NH4 + NO3 H2O F C2H3O2 C5H5N HCO3 OCl OBr NH3 Base strength increases conjugate weak base HCO3 CO3 2 The strong bases have pathetic conjugate acids (which we do not refer to as acids at all). C2H5NH3 + H2O Na + Ba 2+ C2H5NH2 OH NaOH Ba(OH)2 28

29 Size of K a and K b Just as in the last unit, the equilibrium position depends on the degree to which the forward reaction dominates. We call this the strength of the acid or base. The weaker the acid, the smaller the K a ü Of course a weaker weak acid will result in a stronger-ish weak conjugate base. (but NOT a strong conjugate base) The weaker the base, the smaller the K b ü Likewise, a weaker weak base will result in a stronger-ish weak conjugate acid. (but NOT a strong conjugate acid) 29

30 Whaaat?? Acids ~ Bases Why is HCO3 in both the acid and base column? HCO3 is amphoteric. Depending on what substance HCO3 is near, HCO3 can act as either and acid or a base. Acid strength increases H2CO3 HCO3 Base strength increases HCO3 CO3 2 In water, HCO3 is a better base than it is acid. Ka = 5.6 x Kb = 1.8 x

31 In a 0.01 M aqueous solution of HCl, select the particles below that are in the beaker. Select all that apply. 1. H3O + 2. Cl 3. HCl 4. H2O 5. OH 31

32 Autoionization of water The extra equilibrium that happens in ALL aqueous solutions. 32

33 Autoionization As you saw, water can be amphoteric depending on what species it is with. What about when it is with itself? H 2 O + H 2 O H 3 O + + OH Or we could pretend it just decomposes: H 2 O H + + OH Write a Keq for these reactions above: Keq = [H3O + ] [OH ] or [H + ] [OH ] We call this equilibrium, Kw The amount of autoionization is small. How small...? 33

34 Ion Product In pure water, [H + ] = [OH ] = 1 x 10 7 M (at 25ºC) Thus, K w = [H + ] [OH ] = 1 x (at 25ºC) Does this mean that when you drink water you are drinking acid and base? Well, yes and no. In water, [H + ] = [OH ] = 1 x 10 7 M Since the amounts are equal and small, the water is considered neutral. Here s the punch line The ion product, K w is true not just for water, but for all aqueous solutions. 34

35 The ph scale Since in most aqueous solutions, [H + ] and [OH ] are such small quantities, and we would be writing out such small numbers, it is convenient to convert these small numbers to base 10 logs. We call this ph = log [H + ] (or -log [H 3 O + ]) So since in pure water [H + ] = 1 x 10 7 The ph of water = 7 In an acidic solution, [H + ] might = 1 x 10 4 M, so ph = -log [1 x 10 4 M] = 4 35

36 Other p scales Since its useful to report [H + ] as a log, it can also be useful to report [OH ] as a log. We call this poh poh = log [OH ] So for the ph = 4 solution, [H + ] = 1 x 10 4 M Since K w = [H + ] [OH ] = 1 x For this same solution [OH ] = 1 x 10 10, so poh = 10 Let s compare [H + ] and [OH ] concentrations, ph and poh of various common solutions on the next slide. 36

37 Do you notice a relationship between ph and poh of a solution? 37

38 Putting ph Values into Perspective Because the ph scale is a logarithmic scale, changes by one ph unit are actually changes by a factor of 10, and are thus relatively large changes. Consider this comparative scale. Start by assuming that a solution of ph=1, in which the [H + ] = 1 x 10 1 is similar in size as the length of a sports field. As ph increases, the [H + ] decreases by the same factor as the comparative scale. Let s take a look at the UColorado ph simmulation ph [H + ] Comparison 1 1 x yards 2 1 x yards 3 1 x yard 4 1 x inches 5 1 x inch 6 1 x inch 7 1 x inch 38

39 ph related to poh You see that ph + poh = 14 for a solution. Since [H + ] [OH ] = K w = 1 x So ph + poh = pk w = 14 On the next slide we ll take a closer look at the relationship between K a and K b 39

40 K a x K b = K w K a = [A ][H 3 O + ] HA + H 2 O A + H 3 O + A + H 2 O HA + OH [HA] K b = [HA][OH ] [A ] Take a moment to notice that the pink and blue equations are not opposites of each other. Add the two equations: 2 H 2 O H 3 O + + OH When adding reactions...multiply K s [A ][H 3 O + ] [HA] [HA][OH ] [A ] Thus K a x K b = [H 3 O + ] [OH ] = K w 40

41 Why do we care about ph? ph is incredibly important to aqueous solutions and the reactions that take place in those solutions. Body fluids must maintain a tight ph range because many metabolic reactions work only at certain [H + ] and [OH ] concentrations. Pool chemicals work best within a certain ph range. Fish are sensitive to changes in concentration of [H + ] and [OH ]. Soil ph is important for solubility and retention of particular plant nutrients. So as chemists, understanding and calculating [H + ] and [OH ] is very important. Thus the ph scale. 41

42 Acid & Base Clicker Questions Percent Ionization Polyprotic Acids Strength of Acids Chapter 16 Part 2 42

43 In a 0.01 M aqueous solution of HCl, select the particles below that are in the beaker. Select all that apply. 1. H3O + 2. Cl 3. HCl 4. H2O 5. OH 43

44 In a 0.01 M aqueous solution of HCl, select the particles below that are in the beaker. Select all that apply. 1. H3O + 2. Cl 3. HCl because HCl is a strong acid, all of the HCl molecules have dissociated. 4. H2O 5. OH Every aqueous solution has both OH and H + whether it is acidic, basic, or neutral. 44

45 In a 0.01 M solution of HC2H3O2, select the particles below that are in solution. Select all that apply. 1. H3O + 2. C2H3O2 3. HC2H3O2 4. H2O 5. OH 45

46 In a 0.01 M solution of HC2H3O2, select the particles below that are in solution. Select all that apply. 1. H3O + 2. C2H3O2 conjugate weak base 3. HC2H3O2 because HC2H3O2 is a weak acid, not all of the HC2H3O2 molecules have dissociated. 4. H2O 5. OH Every solution has both OH and H + whether it is acidic, basic, or neutral. 46

47 If you had an aqueous solution in which [OH ] = 1 x 10 4 M, Is this solution 1. acidic 2. basic No calculator but be prepared to justify your answer 3. neutral 4. cannot be determined without more info 47

48 If you had an aqueous solution in which [OH ] = 1 x 10 4 M, Is this solution No calculator 1. acidic 2. basic The poh of this solution would be 4, and the ph would be neutral 4. cannot be determined without more info If [OH ] > [H + ] the solution is basic. If [OH ] < [H + ] the solution is acidic. If [H + ] = [OH ] the solution is neutral. 48

49 Which reaction(s) below accurately represent the action of HCl or HF in water. Select all that apply. 1. HCl + H O H O + + Cl HF H + + F 3. HCl + H O H O + + Cl HCl H + + Cl 5. HF + H O H O + + F HF + 4H O H O + + F

50 Which reaction(s) below accurately represent the action of HCl or HF in water. 1. HCl + H 2 O H 3 O + + Cl there should not be any double arrow since HCl is a strong acid. 2. HF H + + F 3. HCl + H 2 O H 3 O + + Cl 4. HCl H + + Cl 5. HF + H 2 O H 3 O + + F there must be a double arrow since HF is a weak acid. 6. HF + 4H 2 O H 9 O F AP is never going to ask you about H 9 O 4 + however, it s probably a better representation of what s really going on. 50

51 In the reaction below, which of the species are acids? NH 3 + H 2 O NH OH 1. NH 3 2. H 2 O 3. NH There are no acids here, this is a reaction for a weak base. 51

52 In the reaction below, which species is an acid? NH 3 + H 2 O NH OH 1.NH 3 2.H 2 O 3.NH There are no acids here, this is a reaction for a weak base. 52

53 What would be the conjugate base for H2PO 4 (aq)? 1. H 3 O + 2. H 3 PO 4 3. HPO PO OH 6. There is no base, H2PO 4 is not an acid, negative ions are always bases. 53

54 What would be the conjugate base for H2PO 4 (aq)? 1. H 3 O + 2. H 3 PO 4 H2PO 4 + H 2 O H 3 O + + HPO 4 2 Acid base conjugate conjugate acid base 3. HPO PO OH 6. There is no base, H2PO 4 is not an acid, negative ions are always bases. 54

55 Which of the following particles are amphoteric? (can behave as either acid or base) 1. H 3 O + 2. H 3 PO 4 3. H2PO 4 4. HPO PO OH 7. H2O 55

56 Which of the following particles are amphoteric? 1. H 3 O + H2PO 4 + H 2 O H 3 O + + HPO 4 2 Acid Base Acid Base 2. H 3 PO 4 H2PO 4 + H 2 O OH + H3PO 4 3. H2PO 4 4. HPO PO OH Base Acid Base Acid HPO H 2 O H 3 O + + PO 3 4 Acid Base acid Base HPO H 2 O OH + H2PO 4 Base Acid Base Acid 7. H2O 56

57 Which base is the weakest? 1. F 2. NH3 3. OH 4. SO C2H3O2 57

58 Which base is the weakest? 1. F 2. NH3 3. OH 4. SO4 2 the base will be the weaker if its conjugate acid is stronger. 5. C2H3O2 58

59 Which acid is the strongest? 1. H O 2 2. H O HF 4. HC H O NH

60 Which acid is the strongest? 1. H 2 O 2. H O + 3 the acid will be the stronger if its conjugate base is weaker. 3. HF 4. HC 2 H 3 O 2 5. NH

61 Which base is the strongest? 1. NO 3 2. C H O HCO 3 4. CO 2 3 pka, what is this? Acid K a pk a HNO 3 Strong acid - HF 6.8 x HC 2 H 3 O x H 2 CO x NH 3 NH x HCO x Na + Pathetic acidity NA 61

62 Which base is the strongest? 1. NO 3 2. C 2 H 3 O 2 3. HCO 3 4. CO 2 3 pka, is -logka If pka is bigger, Ka is smaller which means weaker acid. Acid K a pk a HNO 3 Strong acid - The conjugate base of the weakest acid will be the strongest base HF 6.8 x HC 2 H 3 O x H 2 CO x NH x HCO x OH Pathetic acidity NA 5. NH 3 62

63 If the ph = 2 for an HNO 3 solution, what is the concentration of HNO 3? No calculator

64 If the ph = 2 for an HNO 3 solution, what is the concentration of HNO 3? = No calculator 64

65 If the ph = 10 for a Ca(OH) 2 solution, what is the concentration of Ca(OH) 2? No calculator x x x x x x x x x x

66 If the ph = 10 for a Ca(OH) 2 solution, what is the concentration of Ca(OH) 2? x x x x x 10 3 No calculator x x x x x poh = 4, log(1x10 4 ), thus [OH ] = 1x10 4, since Ca(OH)2 Ca OH, thus the concentration of dissolved Ca(OH)2 = 0.5 x 10 4 = 5 x

67 What is the approximate ph of an aqueous solution of M HCl? 1. 1 No calculator None of the above 67

68 What is the approximate ph of an aqueous solution of M HCl? 1. 1 No calculator = -log(1x10 3 ) = None of the above 68

69 What is the approximate ph of a M HF solution? No calculator Surely you re joking I can t do this without a calculator. 69

70 What is the ph of a M HF solution? No calculator We know a 0.01 M of SA would be ph = 2, thus WA would be lower ph Not able to be determined 70

71 What is the approximate ph of an aqueous solution of 1 x M HCl? 1. 2 No calculator None of the above 71

72 What is the approximate ph of an aqueous solution of M HCl? 1. 2 No calculator None of the above You can t make a solution basic by diluting an acid. The HCl is so dilute that the auto ionization water itself provides a larger source of H + ions,

73 The [H + ] in a M solution of HNO2 is 3.0 x 10 3 M. What is the Ka of HNO2? x 10 4 No calculator x x x None of the above x 10 5

74 The [H + ] in a M solution of HNO2 is 3.0 x 10 3 M. What is the Ka of HNO2? x 10 4 No calculator x x x 10 1 K = a [ 0.02] 5. None of the above x 10 5

75 If the cover were left off of a flask of saturated solution of magnesium hydroxide and the volume of the solution went down as some of the solution evaporated, 1. the ph would remain the same because as the solution evaporated, the ion concentrations would remain the same. 2. the ph would increase as the solution began to evaporate because the ion concentrations would increase. 3. the ph would decrease as the solution began to evaporate because the ion concentrations would decrease. 4. the ph would increase as the solution began to evaporate because the ion concentrations would decrease. 5. the ph would decrease as the solution began to evaporate because the ion concentrations would increase. 75

76 If the cover were left off of a flask of saturated solution of magnesium hydroxide and the volume of the solution went down as some of the solution evaporated, 1. the ph would remain the same because as the solution evaporated, the ion concentrations would remain the same. 2. the ph would increase as the solution began to evaporate because the ion concentrations would increase. 3. the ph would decrease as the solution began to evaporate because the ion concentrations would decrease. 4. the ph would increase as the solution began to evaporate because the ion concentrations would decrease. 5. the ph would decrease as the solution began to evaporate because the ion concentrations would increase. 76

77 After 250 ml of distilled water were added to a flask of saturated solution of magnesium hydroxide and solid precipitate remained on the bottom of the flask, 1. the ph would remain the same as some solid on the bottom of the flask dissolved, and the ion concentrations would remain the same. 2. the ph would increase as the increased volume of water would cause the ion concentrations to increase, and the solid on the bottom to decrease. 3. the ph would decrease as the increased volume of water would cause the ion concentrations to decrease, and the solid on the bottom to increase. 4. the ph would remain as some solid on the bottom of the flask dissolved, and the ion concentrations would increase. 5. the ph would decrease as the increased volume of water would cause the ion concentrations to increase, and the solid on the bottom to decrease. 77

78 After 250 ml of distilled water were added to a flask of saturated solution of magnesium hydroxide and solid precipitate remained on the bottom of the flask, 1. the ph would remain the same as some solid on the bottom of the flask dissolved, and the ion concentrations would remain the same. 2. the ph would increase as the increased volume of water would cause the ion concentrations to increase, and the solid on the bottom to decrease. 3. the ph would decrease as the increased volume of water would cause the ion concentrations to decrease, and the solid on the bottom to increase. 4. the ph would remain as some solid on the bottom of the flask dissolved, and the ion concentrations would increase. 5. the ph would decrease as the increased volume of water would cause the ion concentrations to increase, and the solid on the bottom to decrease. 78

79 Strong Acids and Bases Who are the strong acids? HCl, HBr, HI (not HF, it s a weak acid) HNO 3, HClO 3, HClO 4 H 2 SO 4 Who are the strong bases? Alkali hydroxides: KOH and NaOH ü LiOH, RbOH, and CsOH are not as common. Heavy alkaline hydroxides: Sr(OH) 2, and Ba(OH) 2 are fairly soluble and are completely ionized. ü Ca(OH) 2 is semi-soluble, Mg(OH) 2 is not so soluble ü The other hydroxides are very insoluble. 79

80 Problem Types Ka, Kb, ph, poh Percent Ionization neutralization 80

81 Consider a M hydrochloric acid solution. Calculate [H + ], [OH ], ph, poh Yes, calculator 81

82 Consider a M hydrochloric acid solution. Calculate [H + ], [OH ], ph, poh Because HCl is a strong acid, you can assume all of the M of HCl has ionized into M H + Calculate ph = log[0.035] = 1.46 It s an easy calculation from ph to poh Since ph + poh = 14, thus poh = = Since [OH ] = 10 poh thus [OH ] = 2.86 x Alternatively you could calculate the [OH ] from the [H + ] [H + ] [OH ] = 1 x Thus [OH ] = = M And poh = log[oh ] = log(2.86 x ) =

83 Calculate the molarity of a nitric acid solution with a ph of 2.34 What is the [OH ]? 83

84 Calculate the molarity of a nitric acid solution with a ph of 2.34 What is the [OH ]? Calculate [H + ] from ph = [H + ] = M Remember HNO3 is a strong acid, you can assume the M of H + represents completely ionized HNO3 Thus the concentration of the HNO3 is the same, M Calculate [OH - ] from poh = [OH + ] = 2.2 x M 84

85 Calculate the [OH ] for a NaOH solution with a ph of Calculate the mass of NaOH that was dissolved to produce 250 ml of this solution. Calculate the hydronium ion of this solution. 85

86 Calculate the [OH ] for a NaOH solution with a ph of Calculate the mass of NaOH that was dissolved to produce 250 ml of this solution. Calculate the hydronium ion of this solution. Change the ph to poh and undo it. poh = = 2.11 log[oh ] = 2.11 Solve to get [OH ] = 7.8 x 10 3 M Alternatively, you could undo the ph, solve for [H + ] and then use [OH ] [H + ] = 1 x to solve for [OH ] log[h + ] = Solve to get [H + ] = 1.3 x substitute into Kw = [OH ] [H + ] 1 x = [OH ] [1.3 x ] = [OH ] = 7.8 x 10 3 M 86

87 Calculate the [OH ] for a NaOH solution with a ph of Calculate the mass of NaOH that was dissolved to produce 250 ml of this solution. Calculate the hydronium ion of this solution. since [OH ] = 7.8 x 10 3 M Thus [NaOH] = 7.8 x 10 3 M x 0.25 L = mol mo x 40 g/mol = g NaOH dissolved Remember, hydronium is H3O + or H + From the ph, solve to get [H + ] = [H3O + ] = 1.3 x Or substitute into Kw = [OH ] [H + ] 1 x = [7.8 x 10 3 ] [H + ] [H + ] = 1.3 x

88 Calculate the ph of a M barium hydroxide solution. First write the formula for barium hydroxide 88

89 Calculate the ph of a M barium hydroxide solution. First write the formula for barium hydroxide Ba(OH)2 Consider the stoichiometry of the dissociation Rx. Ba(OH)2 Ba OH Since Ba(OH)2 and thus [Ba 2+ ] is 0.05 M the concentration of OH = 0.10M Calculate poh = log[0.10 M] = 1 Since ph = 14 poh = Alternatively, use Kw = [OH ] [H + ] to solve for, [H + ] then calculate ph. 89

90 Calculate the ph of a 0.20 M solution of hypochlorous acid. Write a reaction for the ionization of hypochlorous acid. HClO H + + ClO Write the Ka expression. Construct an ricebox put in the concentrations we already know...next slide please

91 Calculate the ph of a 0.20 M solution of hypochlorous acid. We can assume that the initial [H + ] = 0, because the amount of H + in pure water is too small compared to any H + due to the presence of the hypochlorous acid. Since we don t know what the changes are... call them x And thus the equilibrium concentrations? HClO H + + ClO I 0.20 ~ 0 ~0 C E K a = [H + ][ClO ] [HClO]

92 Calculate the ph of a 0.20 M solution of hypochlorous acid. Consider how significant figures allow us to avoid the quadratic formula. We can consider the x to be insignificant compared to the 0.20M for HClO, thus the equilibrium [ClO ] = ~0.20 M Look up the Ka = 3.0 x 10 8 then fill in the equilibrium expression and solve for x x = [H + ] = 7.7 x 10 5 M Use the [H + ] to calculate ph if you did use the quadratic, x = x 10 5 ph = 4.11 HClO H + + ClO I 0.20 ~ 0 0 C -x +x +x E x x x ~ = [x][x] [0.20] In AP we will ALWAYS be able to ignore the -x

93 Calculate the ph of a 0.20 M solution of ammonia, pkb = Write a reaction for the ionization of ammonia. Write the Kb expression. Construct a ricebox put in the concentrations we know, calculate the others, ignore insignificant x s

94 Calculate the ph of a 0.20 M solution of ammonia, pkb = Write a reaction for the ionization of ammonia. NH3 + H2O NH4 + + OH Write the Kb expression. Construct a ricebox put in the concentrations we know, calculate the others, ignore insignificant x s

95 Calculate the ph of a 0.20 M solution of ammonia. Write a reaction for the ionization of ammonia. NH3 + H2O NH4 + + OH Write the Kb expression, pkb = 4.74 Kb = = 1.8 x 10 5 K b = [NH 4 + ][OH ] [NH 3 ]

96 Calculate the ph of a 0.20 M solution of ammonia. Write a reaction for the ionization of ammonia. NH3 + H2O NH4 + + OH Write the Kb expression, pkb = 4.74 Kb = = 1.8 x = [x][x] [0.20] x = [OH ] = if you did use the quadratic, x = x 10 3, poh = 2.72, ph = K = [NH + ][OH ] 4 b [NH ] 3 poh = 2.72 ph = This is an x 2 problem!

97 After preparing a 0.10 M solution of nitrous acid, the ph was 2.17, calculate the Ka for this acid. 97

98 After preparing a 0.10 M solution of nitrous acid, the ph was 2.17, calculate the Ka for this acid. Write the reaction for the equilibrium dissociation of nitrous acid in water. HNO2 H + + NO2 Write the Ka expression. Un-log the ph to determine the equilibrium [H + ] = 6.8 x 10 3 M Construct a ricebox put in the concentrations we already know... 98

99 After preparing a 0.10 M solution of nitrous acid, the ph was 2.17, calculate the Ka for this acid. What s the equilibrium concentration of NO2? same as H + What do you know about the initial [H + ] and [NO2 ]? Assume there is none in the container to start. Why do we assume the initial [H + ] = 0? Isn t there 1x10 7 M in water? 1x10 7 M is so much smaller than any amounts of that will form due to the presence of the acid so sig figs allow the 1x10 7 M to be considered ~0 HNO2 H + + NO2 I 0.10 ~ 0 0 C E 6.8x10 3 M K a = [H + ][NO 2 ] [HNO 2 ] 99

100 100 After preparing a 0.10 M solution of nitrous acid, the ph was 2.17, calculate the Ka for this acid. Now you can calculate the change row. Then the equilibrium concentration of HNO2 = M Fill in the equilibrium expression and calculate Ka R HNO2 H + + NO2 I C 6.8x10 3 M +6.8x10 3 M +6.8x10 3 M E x10 3 M 6.8x10 3 M 6.8x10 3 M = M K a = [H + ][NO 2 ] [HNO 2 ] In this case, we know the x so we might as well put it in to calculate = [0.0068][0.0068] [0.0932] =

101 After preparing a 0.10 M solution of nitrous acid, the ph was 2.17, calculate the Ka for this acid. What percentage of the acid is ionized in this 0.10M solution? Remember that % is part out of total. What part represents the part that is ionized? What part represents total acid? %ionization = [H 3 O+ ] [HA] 100 HNO2 H + + NO2 I C 6.8x10 3 M +6.8x10 3 M +6.8x10 3 M E M 6.8x10 3 M 6.8x10 3 M

102 After preparing a 0.10 M solution of nitrous acid, the ph was 2.17, calculate the Ka for this acid. What percentage of the acid is ionized in this 0.10M solution? Remember that % is part out of total. What part represents the part that is ionized? What part represents total acid? Calculate = 6.8% HNO2 H + + NO2 I C 6.8x10 3 M +6.8x10 3 M +6.8x10 3 M E M 6.8x10 3 M 6.8x10 3 M

103 Calculate the K a of lactic acid, HC 3 H 5 O 3. The acid is 3.7% dissociated in a M solution 103

104 Calculate the K a of lactic acid, HC 3 H 5 O 3. The acid is 3.7% dissociated in a M solution Write the equilibrium reaction and K a expression. Set up a RICE box Since we are told in the problem that there is 3.7% dissociation, we can calculate [H + ] R HC 3 H 5 O 3 H + + C 3 H 5 O 3 I C E 104

105 Calculate the K a of lactic acid, HC 3 H 5 O 3. The acid is 3.7% dissociated in a M solution Write the equilibrium reaction and K a expression. HC 3 H 5 O 3 H + + C 3 H 5 O 3 K a = [H + ][C 3 H 5 O 3 ] [HC 3 H 5 O 3 ] Set up a RICE box Since we are told in the problem that there is 3.7% dissociation, we know that the x is 3.7% of M x = This allows you to fill in the remainder of the RICE box. Using the E line of the RICE box allows you to calculate K a R HC 3 H 5 O 3 H + + C 3 H 5 O 3 I ~ 0 0 C E K a = [0.0037][0.0037] [0.0963] =

106 The ph of a M solution of pyridine, C5H5N is Calculate the Kb for this base. What is the percent dissociation of this solution? 106

107 The ph of a M solution of pyridine, C5H5N is Calculate the Kb for this base. What is the percent dissociation of this solution? Write the ionization reaction for pyridine in water. Write the K b expression, Make a rice box. 107

108 The ph of a M solution of pyridine, C5H5N is Calculate the Kb for this base. What is the percent dissociation of this solution? C 5 H 5 N + H 2 O C 5 H 5 NH + + OH Set up an RICE box ph = 9, poh = 5, thus [OH ] = 1 x 10 5 K b = [C 5H 5 NH + ][OH ] [C 5 H 5 N] 108

109 The ph of a M solution of pyridine, C5H5N is Calculate the Kb for this base. What is the percent dissociation of this solution? C 5 H 5 N + H 2 O C 5 H 5 NH + + OH Set up an RICE box K b = [C 5H 5 NH + ][OH ] [C 5 H 5 N] ph = 9, poh = 5, thus [OH ] = 1 x 10 5 R C 5 H 5 N C 5 H 5 NH + + OH I ~ 0 C - 1 x x x 10 5 K b = [ ][ ] [0.06] = E x x x 10 5 ~ = 0.017% Why is [OH ] ~ 0? You could argue that since it is an aqueous solution, that there [OH ] = 1 x 10 7 OH, but since the base will contribute more than that, we can ignore the OH contribution from the water. 109

110 Calculate the volume of M potassium hydroxide required to neutralize 25.0 ml of M nitric acid. Write the overall neutralization reaction. Write the net ionic reaction 110

111 Calculate the volume of M potassium hydroxide required to neutralize 25.0 ml of M nitric acid. KOH + HNO3 H2O + KNO3 OH + H + H2O M a V a = M b V b ( 0.05M )( 25ml) = ( 0.034M )V b V b = 36.8ml KOH 111

112 Calculate the volume of M potassium hydroxide required to neutralize 25.0 ml of M nitrous acid. KOH + HNO2 H2O + KNO2 OH + HNO2 H2O + NO2 Though the net ionic equation is different, solving for the quantity of base required for the neutralization of a weak acid would be exactly the same procedure. M a V a = M b V b ( 0.05M )( 25ml) = ( 0.034M )V b V b = 36.8ml KOH 112

113 Calculate the volume of M perchloric acid required to neutralize 65.0 ml of M barium hydroxide. Write the overall neutralization reaction. Write the net ionic reaction M a V a = M b V b 113

114 Calculate the volume of M perchloric acid required to neutralize 65.0 ml of M barium hydroxide. Ba(OH)2 + 2 HClO4 2 H2O + BaClO4 OH + H + H2O M a V a = M b V b 2 ( M )( V ) = ( M )( 65ml) 2 a V a = 208ml KOH 114

115 Calculate the ph after mixing 35 ml of M Acetic acid with 40 ml of M NaOH Write the overall neutralization reaction. 115

116 Calculate the [H + ] after mixing 100. ml of 0.62 M hydrofluoric acid with 250. ml of M KOH Write out any relevant reactions that you should be thinking about when doing this problem to this problem 116

117 Calculate the ph after mixing 25 ml of 0.25 M ammonia with 40. ml of 0.11 M HCl Write out any relevant reactions that you should be thinking about when doing this problem to this problem 117

118 Clicker Questions Weak Acids 118

119 For a certain weak acid, K a = 2.0 x What is K b for its conjugate base? x 10 3 No calculator x x x x

120 For a certain weak acid, K a = 2.0 x What is K b for its conjugate base? x x x x x

121 Which acid is the weakest? 1. chlorous 2. nitrous 3. acetic 4. hypobromous 5. they are all equally weak acid pka chlorous 1.96 nitrous 3.35 acetic 4.74 hypobromous there is not enough information to distinguish 121

122 Which acid is the weakest? 1. chlorous 2. nitrous 3. acetic 4. hypobromous 5. they are all equally weak acid pka chlorous 1.96 nitrous 3.35 acetic 4.74 hypobromous there is not enough information to distinguish The largest pka is the weakest acid 122

123 Dissolving Salts in Water ph above 7? ph 7? ph below 7? 123

124 Anions of a Strong Acid HCl, HBr, HI, HNO 3, H 2 SO 4, HClO 4, HClO 3 What would happen if any of the anions (the pathetic conjugate base) of a SA were put into water? Putting XCl in water will dissolve to produce Cl ions. XCl X + + Cl If the Cl could interact with water as a acid, this is what would occur Cl + H2O HCl + OH however, these HCl particles would immediately dissociate (They are after all a SA) to give this HCl + OH H + + Cl + OH these H + and OH ions would immediately find each other H + + Cl + OH Cl + H2O which is right back where we started, thus no change to ph. 124

125 Cations of a Strong Base NaOH, KOH, LiOH, Ca(OH) 2, Sr(OH) 2, and Ba(OH) 2 What would happen if any of the cations (the pathetic conjugate acid) of a SB were put into water? Putting KX into water would dissolve to produce K + ions. KX K + + X If the K + could interact with water, this is what might occur. K + + H 2 O KOH + H + however these KOH particles would immediately dissociate (They are after all a SB.) KOH + H + K + + OH + H + these H + and OH ions would immediately find each other K + + OH + H + K + + H 2 O which is right back where we started, thus no change to ph 125

126 Up until now, when a soluble salt dissolved, we thought that s all there was to it, and for NaCl, that s all there is to it. But for many salts, there s more to the story... When salts such as sodium fluoride or ammonium nitrate dissolve, we must consider how the dissolved ions will interact with the water NaF Na + + F NaF NH4 + + NO3 Now, knowing what you know about conjugate acids and bases, what will happen after the salt breaks into ions? H2O + F HF + OH NH4 + + H2O H3O + + NH3 and this affects the ph of the solution. 126

127 Anions of a Weak Acid HF, HNO2, HC2H3O2, HC7H5O2, HClO, HCN What would happen if any of the anions (attached to a metal, X instead of H) of a WA were put into water? Putting XNO2 in water will dissolve to produce NO2 ions. XNO 2 X + + NO2 The NO2 can interact with water as a weak base NO2 + H2O HNO2 + OH This interaction of the NO2 ion with water causes the formation of OH thus the ph rises above

128 Cations of a Weak Base NH3 NH4 +, C5H5N C5H5NH +, H2NOH H3NOH +, NH 2CH3 NH3CH3 + What happens when a conjugate acid of a WB is put into water? Putting NH 4 X in water will dissolve to produce NH4 + ions. NH4X NH4 + + X The NH 4 + can interact with water as a weak acid NH4 + + H2O NH3+ H3O + This interaction NH4 + ion with water causes the formation of H + thus the ph drops below

129 In Summary, Salts: Acidic, Basic, or Neutral? Anions (the pathetic conjugate base) of a strong acid Cation (the pathetic conjugate acid) of a strong base The X ions that are the pathetic conjugate base of a strong acid, and the M + ions that are the pathetic conjugate acid of a strong base, will not be able to hydrolyze in water and thus will hang around as spectator ions having no effect on ph. Anions (the conjugate base) of a weak acid A ions that are the conjugate bases of weak acids will hydrolyze in water causing the ph to rise. Cations of a weak base positive ions (usually contianing N) that are conjugate acids of weak bases will hydrolyze in water causing the ph to drop. Metallic ions (more on the next three slides ) Fe 3+, Zn 2+, Cr 3+, Al 3+ attract the unshared pairs of e- on solvating water molecules, weakening the bonds, causing an H + to ionize causing the ph to drop. 129

130 Hydration Shells All ions are hydrated in water - surrounded by H2O molecules. hydrated cations can be represented as Fe(H 2 O) 6 3+ Zn(H 2 O) 5 2+ The amount of waters represented around the ion vary from ion to ion. You would not be expected to predict the amount. Sometimes a salt solution may have higher entropy than the solid salt and water since the water molecules become more ordered. anion 130

131 Hydrated Metallic Ions For metals other than column I and II Ions such as; Fe 3+, Zn 2+, Cr 3+, Al 3+ The cation attracts the electron pairs on the water towards the cation causing the O H bond in the water molecule to become even more polarized, weakening it enough that one H + will be released into solution. [Fe(H 2 O) 6 ] 3+ [Fe(H 2 O) 5 (OH)] 2+ + H + The strength of the metal ion as an acid is affected by charge higher charge = stronger acid size smaller size = stronger acid 131

132 Hydrated Metallic Ions A = Angstrom = meter Note the size of the hydration shell, ~6 A compared to the length of the ion-dipole bond, 2.56 A Al 3+ compared to the length of the O H bond in water, 0.9 A Al

133 Know the SA & SB... Strong Acids HCl HI HBr HNO3 H2SO4 HClO4 HClO3 Strong Bases LiOH NaOH KOH RbOH CsOH Ba(OH)2 Sr(OH)2 Ca(OH)2...and you the the pathetics. all other +ions are acids, and nearly all other ions are bases. 133

134 Salt Hydrolysis Negative ions are bases (or pathetic) Positive ions are acids (or pathetic)

135 Which chemicals below will produce a ph below 7 when dissolved in water? Select all that apply. Be ready to comment on any increase in ph, and those that still = 7 1. CsI 6. CaSO3 2. HBr 7. NH4NO3 3. FeCl3 8. NaOH 4. KCN 9. NH3 5. HCN 135

136 Which chemicals below will produce a ph below 7 when dissolved in water? Select all that apply. Be ready to comment on any increase in ph 1. CsI 2. HCl SA 3. FeCl3 hydrated ion 4. KCN 5. HCN WA 6. CaSO3 7. NH4NO3 conjwa 8. NaOH 9. NH3 136

137 Which chemicals below will produce a ph above 7 when dissolved in water? Select all that apply. Be ready to comment on any decrease in ph 1. NaC2H3O2 6. NaClO 2. Au(NO3)3 7. Ca(OH)2 3. Ca3(PO4)2 8. HC2H3O2 4. KF 9. (C2H3)2NH 5. NH4ClO4 137

138 Which chemicals below will produce a ph above 7 when dissolved in water? Select all that apply. Be ready to comment on any decrease in ph 1. NaC2H3O2 conjwb 2. Al(NO3)3 3. Ca3(PO4)2 Although if calcium phosphate is so insoluble it may not put out any PO4 ions 4. KF 5. NH4ClO4 6. NaClO conjwb 7. Ca(OH)2 SB 8. HC2H3O2 9. (C2H3)2NH I doubt AP would expect you to recognize this as a WB without telling you. 138

139 Which chemicals below will produce a solution with ph = 7 when dissolved in water? Select all that apply. 1. Ca(CN)2 5. RbI 2. AlCl3 6. NaNO2 3. NaCl 7. HF 4. Ba(ClO3)2 139

140 Which chemicals below will produce a solution with ph = 7 when dissolved in water? Select all that apply. You are looking for pathetic/pathetic 1. Ca(CN)2 5. RbI 2. AlCl3 6. NaNO2 3. NaCl 7. HF 4. Ba(ClO3)2 140

141 Percent Ionization The amount of ionization as related to molarity 141

142 Percent Ionization and Concentration HA + H2O H3O + + A The amount of H3O + and A per original amount of original amount HA present. %ionization = [H 3 O+ ] [HA] 100 For example: Calculate % ionization for 0.10 M HA, Ka = 4.0 x 10 5 First solve an x 2 problem, then calculate % = x2 [0.1] x = [H 3 O + ] = %ionization = [ ] [0.1] 100 = 2% 142

143 Percent Ionization and Concentration For varying concentrations of acetic acid HA + H2O H3O + + A Consider the varying concentration, and resulting % ionization and ph values. Conc in M % ionization ph Percent Ionization Percent Ionization Concentration

144 Percent Ionization and Concentration Percent ionization varies with concentration changes. What is the correlation? As concentration decreases, percent ionization increases. At lower concentration, the % ionization is greater, but the [H + ] is still smaller (higher ph) Although more of the molecular acid is ionized, there is still much less of it available to provide [H + ] Why? When diluting the solution, there will be less concentration of both HA and of H + and A in the solution, thus the equilibrium will shift toward products to minimize the dilution disturbance, and because one acid molecule can produce two ions, increasing the overall concentration. 144

145 Percent Ionization and Concentration HA H + + A Imagine the ridiculous scenario of molar a weak acid concentration in which all were 1 M K a = 1 [ ][ 1] [ 1] = 1 Then diluting the solution to double the volume and all the concentrations would halved. Q would now be less than Ka, Q = 0.5 [ ][ 0.5] [ 0.5] = 0.5 < K a the system would shift to the right, thus higher percentage ionized 145

146 The ph of a M weak acid is what is the percentage ionization? % No calculator % % % % %

147 The ph of a M weak acid is what is the percentage ionization? % % % % % No calculator Since ph = 3.00, [H + ] = 1.0 x 10 3 M, so % ionization = [H+ ]eq [HA]o % %ionization =

148 Titration Curves 148

149 What is the ph of an aqueous solution to which 51.0 ml of 0.10 M NaOH have been added to 50.0 ml 0.10 M HCl?

150 What is the poh of an aqueous solution to which 51.0 ml of 0.10 M NaOH have been added to 50.0 ml 0.10 M HCl? ml of base beyond the equivalence point. (0.1 M)(1 ml) = M(101 ml) [OH -1 ] = M Then poh then ph 150

151 What is the ph of an aqueous solution to which 26.5 ml of 0.10 M NaOH have been added to 50.0 ml M HCl?

152 What is the ph of an aqueous solution to which 26.5 ml of 0.10 M NaOH have been added to 50.0 ml M HCl? M * 26.5 ml = 2.65 mmoles OH M * 50 ml = 2.5 mmoles of H +1 Subtract to get mmoles of excess OH mmoles OH -1 Then calculate molarity of OH moles / 76 ml = M Then poh and then ph 152

153 Calculate the amount of 0.10 M KOH needed to completely neutralize 50 ml of 0.10 M H2SO ml ml ml 4. H2SO4 can not be completely neutralized 5. This can not be determined without knowing the Ka1 and Ka2 of H2SO4 153

154 Calculate the amount of 0.10 M KOH needed to completely neutralize 50 ml of 0.10 M H2SO ml ml ml 4. H2SO4 can not be completely neutralized 5. This can not be determined without knowing the Ka1 and Ka2 of H2SO4 MbVb = MaVa but remember that the acid is buy 1 get 2 so 0.1 M * Vb = 2 * 0.1 M * 50 ml 154

155 Comparing Titration Curves pka=5 Equivalence point of WA + SA titration is above 7 due to the presence of the conj WB Equivalence point of SB + SA titration = 7 Weak Base Strong Base poh = pkb Equivalence point of SB + SA titration = 7 Equivalence point of WB + SA titration is below 7 due to the presence of the conj WA ph at equivalence point SA + SB = 7 WA + SB > 7 WB + SA < 7 Halfway to equivalence point is the easiest way to determine the Ka or Kb. 155

156 Titration Curves Acids with different Ka values 156

157 This titration curve was obtained in the titration of an unknown 0.10 M acid with 0.10 M NaOH. What is the Ka of the acid? 1. 1 x x x x x Not enough info 7. 1 x two Ka values 157

158 This titration curve was obtained in the titration of an unknown acid with 0.10 M NaOH. What is the Ka of the acid? 1. 1 x x x x x 10-8 Halfway to the equivalence point [H +1 ] = K a Thus ph = pk a 158

159 The equivalence point of this titration curve occurs at what ph? 1.~ 5 2.~ 7 3.~ 9 4.~ 11 5.~ 12.5 ph Weak Acid vs Strong Base Volume of Base Added 159

160 The equivalence point of this titration curve occurs at what ph? 1.~ 5 2.~ 7 3.~ 9 4.~ 11 5.~ 12.5 ph Weak Acid vs Strong Base Volume of Base Added 160

161 This is the titration curve of a titrated by a 1. WA by a SB 2. SB by a WA 3. WB by a SA 4. WA by a SA 5. SA by a SB 6. Can not be determined without more information. ph Weak Acid vs Strong Base Volume of Base Added 161

162 This is the titration curve of a titrated by a 1. WA by a SB 2. SB by a WA 3. WA by a SA 4. SA by a SB ph Can not be determined without more information. The equivalence point has a ph above 7 Weak Acid vs Strong Base Volume of Base Added 162

163 What is the Ka of this acid? 1.cannot be determined without more info x x x x ph Weak Acid vs Strong Base Volume of Base Added 163

164 What is the Ka of this acid? 1.cannot be determined without more info x 10-4 ph halfway = 5.4 = pka un-logs for the Ka 3.1 x x 10-6 ph Weak Acid vs Strong Base Volume of Base Added 164

165 Determine the molar mass for this acid g of it was titrated with 0.15 M strong base? 1.~100 g/mol 2.~200 g/mol 3.~300 g/mol 4.~400 g/mol 5.~500 g/mol 6.cannot be determined without more ph Weak Acid vs Strong Base Volume of Base Added info 165

166 Determine the molar mass for this acid g of it was titrated with 0.15 M strong base? 1.~100 g/mol 2.~200 g/mol 3.~300 g/mol 4.~400 g/mol 5.~500 g/mol 6.cannot be determined without more info ph Weak Acid vs Strong Base Volume of Base Added 0.15 M * L = moles of base, which = moles of acid at equivalence point g/ moles = 211 g/mole 166

167 Why is the choice of indicator more crucial for a WA-SB titration than for a SA-SB titration? 1. The nearly vertical equivalence point portion of the titration curve is large for a weak acid-strong base titration, and fewer indicators undergo their color change so quickly it s difficult to monitor. 2. The nearly vertical equivalence point portion of the titration curve is smaller for a weak acid-strong base titration, and fewer indicators undergo their color change within this narrow range. 3. Many indicators do not change colors at the equivalence points of weak acid-strong base titrations. 4. Equivalence points at ph s other than 7.00 are difficult to determine. 167

168 Why is the choice of indicator more crucial for a WA-SB titration than for a SA-SB titration? 1. The nearly vertical equivalence point portion of the titration curve is large for a weak acid-strong base titration, and fewer indicators undergo their color change so quickly it s difficult to monitor. 2. The nearly vertical equivalence point portion of the titration curve is smaller for a weak acid-strong base titration, and fewer indicators undergo their color change within this narrow range. 3. Many indicators do not change colors at the equivalence points of weak acid-strong base titrations. 4. Equivalence points at ph s other than 7.00 are difficult to determine. 168

169 The ph at position II is 1.at 7 2.above 7 3.below 7 169

170 The ph at position II is 1.at 7 2.above 7 3.below 7 170

171 Which titration curve represents the acid with the largest Ka? 1. Blue 2. Green 3. Yellow 4. Orange ph 5. Red 171

172 Which titration curve represents the acid with the largest K a? 1. Blue 2. Green 3. Yellow 4. Orange 5. Red You could argue that the red curve is a SA that does not have a Ka. All acids have Ka values, but its Ka is just too large to worry about equillibrium. 172

173 Measuring ph during titration curves ph can be accurately measured using a ph meter. ph can also be indicated using dyes that are different colors at different ph values and change color over a very particular ph range. These are called acid base indicators. They indicate when we reach a particular ph and if chosen correctly for the titration, they will accurately indicate the endpoint of a titration. 173

174 Indicators Phenolphthalein aka HIn 174

175 Indicators are molecules that are one color (or lack of) in molecular form, and a different color in ionized form. We can represent phenolphthalein s reaction as HPh H + + Ph Which form of phenolphthalein is pink, the molecular form, or the ionized form? 1. HPh 2. Ph 175

176 phenolphthalein It is a very complex weak acid molecule that changes structure dramatically in neutral vs ionized form. HIn H + + In Adding base forces the weak acid to ionize into its pink form. HIn In 176

177 Thus the ph of the end point should be near the pka of the indicator. 177 At what ph does the color change? We can simplify nearly any indicator as weak acid molecule that is two different colors in its neutral form and ionized form. HIn H + + In H + = K [ a HIn ] In The Ka of phenolphthalein = x more or less of either neutral or ionized form is considered the range of the color change. [ ] [ ] H + = K 10 a 1 ph = 8 = H + = K a 1 10 ph = 10 [ ] [ ] =

178 HIn H + + In Methyl Red Which hydrogen ionizes? Ka = ph between ph below and 6.2 ph above 6.2 O + H + 178

179 Choosing a Indicator 179

180 Choosing a Indicator 180

181 Choosing a Indicator 181

182 Which indicator would best determine the equivalence point of the titration of a solution of NaC2H3O2 with HNO3? 1. Methyl violet 2. Methyl orange 4. Phenolphthalein 5. Alizarin yellow R 3. Bromthymol blue 182

183 Which indicator would best determine the equivalence point of the titration of NaC2H3O2 with HNO3? 1. Methyl violet Weak Acid vs Strong Base 2. Methyl orange 3. Bromthymol blue 4. Phenolphthalein ph Alizarin yellow R Vol of Base Added (ml) The endpoint of the titration of WB by a SA is likely to have a ph below 7 because of the presence of a conjugate acid. 183

184 Buffers Made Easy 184

185 What is a Buffer? An aqueous solution that has a highly stable ph even upon addition of small amounts of H + or OH. The solution is able to absorb and neutralize incoming acid or incoming base. How?? Because of the presence of significant quantities of both a conjugate weak acid/base pair. HA H + + A X-NH2 + H 2 O X-NH3 + + OH 185

186 There are two methods to construct a buffer Dissolve a conjugate WA-salt (or WB-salt) into a solution of WA or WB. Titrate a WA (or WB) partway along. HA H + + A NH3 + H 2 O NH4 + + OH 186

187 The ph of an acetic acid solution is ~ 4.7 If you added some sodium acetate to acetic acid, 1. The ph would go up 2. The ph would go down 3. The ph would stay the same 187

188 The ph of an acetic acid solution is ~ 4.7 HA H + + A If you added some sodium acetate, A to acetic acid, 1. The ph would go up HA H + + A Adding A will shift the equilibrium to the left, reducing the amount of H +. This is the common ion effect 2. The ph would go down 3. The ph would stay the same 188

189 The ph of an acetic acid solution is ~ 4.7 If you added some sodium hydroxide to acetic acid, 1. The ph would go up 2. The ph would go down 3. The ph would stay the same 189

190 The ph of an acetic acid solution is ~ 4.7 HA H + + A If you added some sodium hydroxide, OH to acetic acid, 1. The ph would go up K = H + a A [ HA] HA H + + A + H2O LeChatelier s Principle, adding OH will combine with H + forming water, and shift the equilibrium to the right. As the A HA [ ] becomes larger, due to the shift right, the the [H+] to be smaller 2. The ph would go down 3. The ph would stay the same 190

191 What is the ph of a solution made with equal moles of HC2H3O2 and C2H3O2? Ka = ph = ph = ph = ph =

192 What is the ph of a solution made with equal moles of HC2H3O2 and C2H3O2? Ka = ph = ph = ph = 4.74 K = H + a A [ HA] When WA = WB in a buffer ph = pka 4. ph =

193 A mixture of 0.10 mol of NH4Cl and 0.12 mole of NH3 is added to enough water to make 1.0 L of solution. Which of the ion(s) is(are) a spectator ion(s) and not involved in the equilbrium? 1.There are no spectator ions. 2.Cl 3.Both NH + and Cl 4 4.NH

194 A mixture of 0.10 mol of NH4Cl and 0.12 mole of NH3 is added to enough water to make 1.0 L of solution. Which of the ions is a spectator ion? 1.There are no spectator ions. 2.Cl 3.Both NH + and Cl 4 4.NH

195 A mixture of 0.10 mol of NH4Cl and 0.12 mole of NH3 is added to enough water to make 1.0 L of solution. Which equilibrium equation(s) accurately describes the reaction to allow for ph calculation? 1. NH 3(aq) + H 2 O (L) ó NH 4 + (aq) + OH (aq) 2. NH 4 + (aq) + OH (aq) ó NH 3(aq) + H 2 O (L) 3. H 3 O + (aq) + OH (aq) ó 2H 2 O (l) 4. NH 4 + (aq) + H 2 O (L) ó H 3 O + (aq) + NH 3(aq) 5. H 3 O + (aq) + NH 3(aq) ó NH 4 + (aq) + H 2 O (L) 195

196 A mixture of 0.10 mol of NH4Cl and 0.12 mole of NH3 is added to enough water to make 1.0 L of solution. Which equilibrium equation(s) accurately describes the reaction to allow for ph calculation? K a = H + [ cjwb] [ WA] 1. NH 3(aq) + H 2 O (L) ó NH 4 + (aq) + OH (aq) 2. NH 4 + (aq) + OH (aq) ó NH 3(aq) + H 2 O (L) 3. H 3 O + (aq) + OH (aq) ó 2H 2 O (l) 4. NH 4 + (aq) + H 2 O (L) ó H 3 O + (aq) + NH 3(aq) 5. H 3 O + (aq) + NH 3(aq) ó NH 4 + (aq) + H 2 O (L) K b = OH [ cjwa] [ WB] 196

197 A mixture of 0.10 mol of NH4Cl and 0.12 mole of NH3 is added to enough water to make 1.0 L of solution. Which equilibrium equation(s) accurately describes the reaction to allow for ph calculation? NH 3(aq) + H 2 O (L) ó NH 4 + (aq) + OH (aq) NH 4 + (aq) + H 2 O (L) ó H 3 O + (aq) + NH 3(aq) [ ] K a = H + NH 3 + NH = H + [ 0.12] [ 0.1] H + = ph = = OH [ 0.10] [ 0.12] OH = poh = 4.67 ph =

198 A mixture of 0.10 mol of NH4Cl and 0.12 mole of NH3 is added to enough water to make 1.0 L of solution. Now, add 250 ml of 0.20 M HCl and calculate the ph. NH 3(aq) + H 2 O (L) ó NH + 4 (aq) + OH (aq) NH + 4 (aq) + H 2 O (L) ó H 3 O + (aq) + NH 3(aq) [ ] K a = H + NH 3 + NH = H + [ 0.12] [ 0.1] H + = ph = 9.33 K b = OH + NH 4 [ NH ] = OH [ 0.10] [ 0.12] OH = poh = 4.67 ph =

199 A mixture of 0.10 mol of NH4Cl and 0.12 mole of NH3 is added to enough water to make 1.0 L of solution. Now, add 250 ml of 0.20 M HCl and calculate the ph. NH 3(aq) + H 2 O (L) ó NH + 4 (aq) + OH (aq) NH + 4 (aq) + H 2 O (L) ó H 3 O + (aq) + NH 3(aq) [ ] K a = H + NH 3 + NH = H + [ ] [ ] H + = ph = 8.92 K b = OH + NH 4 [ NH ] = OH [ ] [ ] OH = poh = 5.08 ph =

200 Which of the following combinations will result in a buffer? Will the ph of the solution be 7, above 7, or below 7? ml of 1 M KNO3 and 20 ml of HNO3 2. A solution of calcium acetate and acetic acid ml of 0.5 M CH3NH2 with 5 g of dissolved CH3NH3NO ml of 1 M NaOH and 20 ml of NaCl 200

201 Which of the following combinations will result in a buffer? Will the ph of the solution be 7, above 7, or below 7? ml of 1 M KNO3 and 20 ml of HNO3 SA (HNO3) and it s pathetic (NO3 ) does NOT make a buffer 2. A solution of calcium acetate and acetic acid? WA (acetic acid)and it s conj WB (acetate ion) = buffer, ph < ml of 0.5 M CH3NH2 with 5 g of dissolved CH3NH3NO3 WB (CH3NH2) and it s conj WA (CH3NH3 + ) = buffer, ph > ml of 1 M NaOH and 20 ml of NaCl SB (NaOH) and it s pathetic (NO3 ) does NOT make a buffer 201

202 Which of the following combinations will result in a buffer? Will the ph of the solution be 7, above 7, or below 7? ml of 0.5 M NaOH with 25 ml of 0.5 M HF ml of 1 M HNO3 and 10 ml of 1 M NaOH ml of 0.5 M CH3NH2 with 100 ml of 0.5 M HNO ml of 0.5 M CH3NH2 with 60 ml of 0.5 M HNO3 202

203 Which of the following combinations will result in a buffer? Will the ph of the solution be 7, above 7, or below 7? ml of 0.5 M NaOH with 25 ml of 0.5 M HF WA (HF) titrated beyond equiv point = not buffer, ph > ml of 1 M HNO3 and 10 ml of 1 M NaOH SA (HNO3) and SB (NaOH) in any quantities does NOT make a buffer ml of 0.5 M CH3NH2 with 100 ml of 0.5 M HNO3 WB (CH3NH2) titrated to equiv point does not make a buffer ml of 0.5 M CH3NH2 with 60 ml of 0.5 M HNO3 WB (CH3NH2) titrated partway = buffer, likely ph > 7 203

204 Consider the weak acid ionization reaction represented below. HNO2 H + + NO2 Which of the actions below would shift the equilibrium position toward reactants? 1. adding HCl 2. adding KOH 3. adding H2O 4. adding NaNO2 204

205 Consider the weak acid ionization reaction represented below. HNO2 H + + NO2 Which of the actions below would shift the equilibrium position toward reactants? 1. adding HCl common ion H+ will shift equilibrium to the left 2. adding KOH 3. adding H2O 4. adding NaNO2 205

206 Calculate the ph of a M HNO2 solution. Ka = 4.5 x

207 Calculate the ph of a M HNO2 solution. Ka = 4.5 x 10 4 For the reaction: HA H + + A K = H + a A [ HA] this is an x 2 problem K a = [H + ][A ] [HA] = = [x][x] [0.025] x = [H + ] = ph =

208 Calculate the ph of a solution made by adding 2.1 g of sodium nitrite to 2 L of M solution. = 4.5 x 10 HNO2 Ka 4 (Assume no volume change when the salt is added.) K a = [H + ][A ] [HA] MM = 69 g/mol 208

209 Calculate the ph of a solution made by combining 0.20 mole of sodium nitrite with 0.20 mole of in a 2.0 L HNO2 solution. = 4.5 x 10 Ka 4 K = H + a A [ HA] = = H + ( 0.030) ( 0.050) x = H + = ph = 3.12 This is a buffer solution 209

210 Which of the following combinations can produce a buffer solution? Select all that apply. 1. HCl / NaCl 2. HC2H3O2 / NH3 3. H3PO4 / NaH2PO4 4. HNO3 / Ca(OH)2 5. HCN / NaOH 6. NH4NO3 / NH3 210

211 Which of the following will produce a buffer solution? 1. HCl / NaCl 2. HC 2 H 3 O 2 / NH 3 This solution can buffer, with a WA and WB, but since they are not conjugates...you will never see them together in a question. 3. H3PO4 / NaH2PO4 H3PO4 H + + H2PO HNO3 / Ca(OH)2 5. HCN / NaOH Acid with ~half base 6. NH4NO3 / NH3 Two ways to make a buffer put in a conjugate WA /WB pair put a WA with ~half of SB or WB with ~half of SA 211

212 Which of the following combinations can produce a buffer solution? Select all that apply. 1. KNO3 / NaOH 2. HCN / NaOH 3. NH3 / HCl 4. HCl / KOH 5. NH3NO3 / NH3 212

213 Which of the following will produce a buffer solution? 1. KNO3 / NaOH 2. HCN / NaOH 3. NH3 / HCl 4. HCl / KOH 5. NH4NO3 / NH3 Two ways to make a buffer put in a conjugate WA /WB pair put a WA with ~half of SB or WB with ~half of SA 213

214 What is [H + ] of a M HF solution dissolved with enough solid NaF to produce a solution 0.20 M NaF Ka (HF) = 6.8 x x 10-4 M x 10-4 M x 10-4 M x 10-3 M x 10-3 M 6. Cannot be determined without knowing volumes of each, or total volume 214

215 What is [H + ] of a M HF solution dissolved with enough solid NaF to produce a solution 0.20 M NaF Ka (HF) = 6.8 x M M K a = H + F [ HF] = = H + [ 0.20] [ 0.05] M x = H + = M M 6. Cannot be determined without knowing volumes of each, or total volume 215

216 Calculate the ph of a 0.50 M solution of sodium acetate (NaC2H3O2 or NaAc) in 0.50 M acetic acid (HC2H3O2 or HAc). pk a (HAc) =

217 Calculate the ph of a 0.50 M solution of sodium acetate (NaC2H3O2 or NaAc) in 0.50 M acetic acid (HC2H3O2 or HAc). pk a (HAc) =

218 Calculate the ph of a 0.50 M solution of sodium acetate (NaC2H3O2 or NaAc) in 0.50 M acetic acid (HC2H3O2 or HAc). pk a (HAc) = K = H + a A [ HA] For any buffer, when [WA] = [WB], the ph = pk a 218

219 Using equal volumes and concentrations of which of the following combinations would be best to make a buffer with a ph above 7? 1. NaCl and HCl 2. NH3 and NH4Cl 3. NaOH and HCl 4. HNO2 and NaNO2 5. NH3 and C5H5NHCl acid pk a HNO C5H5NH NH

220 Using equal volumes and concentrations of which of the following combinations would make a buffer with a ph above 7? 1. NaCl and HCl acid pk a 2. NH 3 and NH 4 Cl 3. NaOH and HCl 4. HNO 2 and NaNO 2 HNO C5H5NH NH NH 3 and C 5 H 5 NHCl When the conjugate pair [base] = [acid] in the above expression K a = [H + ], so pk a = ph K = H + a A [ HA] 220

221 Calculate the ph of a 0.50 M solution of sodium acetate (NaC2H3O2 or NaAc) in M acetic acid (HC2H3O2 or HAc). pk a (HAc) =

222 Calculate the ph of a 0.50 M solution of sodium acetate (NaC2H3O2 or NaAc) in M acetic acid (HC2H3O2 or HAc). pk a (HAc) = K = H + a A [ HA] Note that the ph was diffeent from pka by one unit ( ) = H ( 0.05) H + =

223 In an NH4 + /NH3 buffer, what concentration changes will occur if a small amount of OH is added? [NH4 + ] [NH3] ph 1 Increase Increase Increase 2 Decrease Increase Increase 3 Increase Decrease Increase 4 Increase Decrease Decrease 5 Decrease Increase Decrease 6 No Changes because OH is not part of the equilibrium reaction. 223

224 In an NH4 + /NH3 buffer, what concentration changes will occur if a small amount of OH - is added? [NH4 + ] [NH3] ph 1 Increase Increase Increase 2 Decrease Increase Increase 3 Increase Decrease Increase 4 Increase Decrease Decrease 5 Decrease Increase Decrease 6 No Changes because OH -1 is not part of the equilibrium reaction. 1. A shift will always occur, however it may not always make any appreciable effect on the ph value. 2. To think about this write the equilibrium reaction (from either perspective) NH3 + H2O OH + NH4 + adding OH shifts left OR NH4 + NH3 + H + adding OH shifts right 224

225 The pka of phenolphthalein is likely to be HIn H + + In 1. below 8 2. ~8 3. ~9 4. ~10 5. above there is not enough information to know 225

226 The pka of phenolphthalein is likely to be HIn H + + In 1. below 8 2. ~8 3. ~9 4. ~10 5. above there is not enough information to know 226

227 Calculate the ph of a buffer solution containing 0.20 moles sodium acetate and 0.30 moles acetic acid (HAc) pk a (HAc) = Yes, you may use a calculator not enough information, volume information must be known to calculate a ph. 227

228 Calculate the ph of a buffer solution containing 0.20 moles sodium acetate and 0.30 moles acetic acid (HAc) pk a (HAc) = K = H + a A [ HA] = H + ( 0.2) ( 0.3) H + = not enough information, volume information must be known to calculate a ph. 228

229 Which of the following conjugate acid-base pairs will not function as a buffer? HCHO2/CHO2 HCO3 /CO3-2 HNO3/NO3 1. All acid-base pairs will function as buffers. 2. HCHO 2 and CHO 2 will not work as a buffer because CHO 2 is a spectator ion. 3. HCO 3 and CO 3 2 will not work as a buffer because HCO 3 is a spectator ion. 4. HNO 3 and NO 3 will not work as a buffer because NO 3 is a spectator ion. 229

230 Which of the following conjugate acid-base pairs will not function as a buffer? HCHO2/CHO2-1 HCO3-1 /CO3-2 HNO3/NO All acid-base pairs will function as buffers. 2. HCHO 2 and CHO 2 will not work as a buffer because CHO 2 is a spectator ion. 3. HCO 3 and CO 3 2 will not work as a buffer because HCO 3 is a spectator ion. 4. HNO 3 and NO 3 will not work as a buffer because NO 3 is a pathetic spectator ion. 230

231 Calculate the ph of a buffer solution containing 0.25 moles sodium acetate and 0.30 moles acetic acid (HAc) to which 0.20 moles HCl are added. pk a (HAc) = not enough information, total volume must be known. 231

232 Calculate the ph of a buffer solution containing 0.25 moles sodium acetate and 0.30 moles acetic acid (HAc) to which 0.20 moles HCl are added. pk a (HAc) = K = H + a A [ HA] = H + ( ) ( ) = H + ( 0.05) ( 0.5) H + = not enough information, total volume must be known. 232

233 Calculate the ph of 2000 ml of a buffer solution containing 0.25 moles sodium acetate and 0.30 moles acetic acid (HAc) to which 0.35 moles HCl are added. pk a (HAc) = submit your answer. 233

234 Calculate the ph of 2000 ml of a buffer solution containing 0.25 moles sodium acetate and 0.30 moles acetic acid (HAc) to which 0.35 moles HCl are added. pk a (HAc) = This is a buffer which has been pushed beyond its buffer capacity. 2. The 0.25 moles of acetate ion would be able to neutralize 0.25 moles of incoming HCl, however that would leave 0.10 moles of excess HCl. 3. This would be more important than any H+ contributed by the weak acid. H + = 0.1mol 2L = 0.05M ph =

235 Predicting Strength of Acids Using the Chemical Formula to Predict Acid Strength 235

236 What Causes H + s to Fall Off? Acid is only ph changing when H + s fall off Not every H falls off Not every H attached to an O falls off When a H O bond become especially polarized, the H is more likely to fall off. What makes these bonds more polar? the electron drawing ability of nearby C=O bonds 236

237 What Causes H + s to Fall Off? Acid is only ph changing when H + s fall off Not every H falls off Not every H attached to an O falls off When a H O bond become especially polarized, the H is more likely to fall off. What makes these bonds more polar? the electron drawing ability of nearby C=C bonds 237

238 Three Types of Acids there are other types, but creating these 3 categories, lets us look for trends Oxyacids HNO 3, HNO 2 H2SO4, H2SO3 HClO... HClO4 Binary acids, H-X HF, HCl, HBr, HI H2O, H2S Carboxylic Acid (organic acid) CxHyOz COOH 238

239 3 Factors causing stronger acids 1. Polarizability of the H bond 2. Weakness of the X H bond 3. Stability of the conjugate base (the resulting anion). 239

240 Which acid is the weakest? 1.HBrO 2.HBrO 2 3.HBrO 3 4.HBrO 4

241 Which acid is the weakest? 1. HBrO 2. HBrO 2 3. HBrO 3 4. HBrO 4 Oxy acids with varying # of O atoms more oxygens = stronger acid Less oxygens have less electron drawing capability, thus the -H bond is less polarized, thus a weaker acid.

242 Oxyacids: HXOn (really XOnH) As additional oxygens attached to X, increases acidity. HClO < HClO 2 < HClO 3 < HClO 4 The additional oxygens draw the electrons in the H O bond causing the bond to become even more polarized. Also, the additional oxygens help to stabilize the conjugate base by providing more room to spread out its negative charge. 242

243 Which acid is the weakest? 1.HClO 2 2.HBrO 2 3.HIO 2

244 Which acid is the weakest? 1.HClO 2 2.HBrO 2 3.HIO 2 Oxy acids different central atom but same # oxygens I is less electronegative than Cl thus the H bond is less polarized, thus the H is more likely to hang on and a therefore a weaker acid.

245 Oxyacids: HOX As the electronegativity of X increases, the acidity increases HOI < HOBr < HOCl Because the H O bond becomes more polarized due to greater electron-drawing of the nearby electronegative atom. The resulting conjugate base ion will be more stable since the presence of the more electronegative Cl atom can help distribute and stablilize the negative charge. 245

246 Which molecule is the weakest acid? 1.H3P 2.H2S 3.HCl

247 Which acid is the weakest? 1.H3P 2.H2S 3.HCl Binary acids across a row P is less electronegative than Cl thus the H bond is less polarized, thus the H is more likely to hang on and a weaker acid results.

248 Which acid is the weakest? 1.H2S 2.H2Se 3.H2Te

249 Which acid is the weakest? 1.H2S 2.H2Se 3.H2Te Binary acids in a column S is smaller in size and thus the bond is stronger resulting in a weaker acid. (If you use the electronegativity argument, you will get to the wrong answer.)

250 Making Sense of Binary Acid Strength What is the explanation...? 250