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Acids and Bases. Acid/Base Definitions  Arrhenius Model  Acids produce hydrogen ions in aqueous solutions  Bases produce hydroxide ions in aqueous.

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Presentation on theme: "Acids and Bases. Acid/Base Definitions  Arrhenius Model  Acids produce hydrogen ions in aqueous solutions  Bases produce hydroxide ions in aqueous."— Presentation transcript:

1 Acids and Bases

2 Acid/Base Definitions  Arrhenius Model  Acids produce hydrogen ions in aqueous solutions  Bases produce hydroxide ions in aqueous solutions  Bronsted-Lowry Model  Acids are proton donors  Bases are proton acceptors  Lewis Acid Model  Acids are electron pair acceptors  Bases are electron pair donors

3 Who Theory: Acid= When Arrheniusincreases H + 1880’s Brønstedproton donor1923 Lowry ditto1923 Lewis Electron-pair acceptor 1923 Three definitions of acid

4 Some Definitions Arrhenius acids and bases –Acid:Substance that, when dissolved in water, increases the concentration of hydrogen ions (protons, H + ). –Base:Substance that, when dissolved in water, increases the concentration of hydroxide ions.

5 Some Definitions Brønsted–Lowry: must have both 1. an Acid:Proton donor and 2. a Base:Proton acceptor

6 If it can be either…...it is amphiprotic. HCO 3 – HSO 4 – H2OH2O

7 What Happens When an Acid Dissolves in Water? Water acts as a Brønsted–Lowry base and abstracts a proton (H + ) from the acid. As a result, the conjugate base of the acid and a hydronium ion are formed. Movies…

8 Conjugate Acids and Bases: From the Latin word conjugare, meaning “to join together.” Reactions between acids and bases always yield their conjugate bases and acids.

9 Acid and Base Strength Strong acids are completely dissociated in water. –Their conjugate bases are quite weak. Weak acids only dissociate partially in water. –Their conjugate bases are weak bases.

10 Acid and Base Strength Substances with negligible acidity do not dissociate in water. –Their conjugate bases are exceedingly strong.

11 Acid Dissociation HA  H + + A - Acid Proton Conjugate base Alternately, H + may be written in its hydrated form, H 3 O + (hydronium ion) Ka > 1 = product favored

12 Dissociation of Strong Acids Strong acids are assumed to dissociate completely in solution. Large K a or small K a ? Reactant favored or product favored?

13 Dissociation Constants: Strong Acids AcidFormula Conjugate Base KaKa Perchloric HClO 4 ClO 4 - Very large Hydriodic HI I - Very large Hydrobromic HBr Br - Very large Hydrochloric HCl Cl - Very large Nitric HNO 3 NO 3 - Very large Sulfuric H 2 SO 4 HSO 4 - Very large Hydronium ion H 3 O + H 2 O 1.0

14 Dissociation of Weak Acids Weak acids are assumed to dissociate only slightly (less than 5%) in solution. Large K a or small K a ? Reactant favored or product favored?

15 Dissociation Constants: Weak Acids AcidFormula Conjugate Base KaKa Iodic HIO 3 IO 3 - 1.7 x 10 -1 Oxalic H 2 C 2 O 4 HC 2 O 4 - 5.9 x 10 -2 Sulfurous H 2 SO 3 HSO 3 - 1.5 x 10 -2 Phosphoric H 3 PO 4 H 2 PO 4 - 7.5 x 10 -3 Citric H 3 C 6 H 5 O 7 H 2 C 6 H 5 O 7 - 7.1 x 10 -4 Nitrous HNO 2 NO 2 - 4.6 x 10 -4 Hydrofluoric HF F - 3.5 x 10 -4 Formic HCOOH HCOO - 1.8 x 10 -4 Benzoic C 6 H 5 COOH C 6 H 5 COO - 6.5 x 10 -5 Acetic CH 3 COOH CH 3 COO - 1.8 x 10 -5 Carbonic H 2 CO 3 HCO 3 - 4.3 x 10 -7 Hypochlorous HClO ClO - 3.0 x 10 -8 Hydrocyanic HCN CN - 4.9 x 10 -10

16 Acid and Base Strength Acetate is a stronger base than H 2 O, so the equilibrium favors the left side (K<1). The stronger base “wins” the proton. HC 2 H 3 O 2 (aq) + H 2 OH 3 O + (aq) + C 2 H 3 O 2 – (aq)

17 Self-Ionization of Water aka “autoionization” H 2 O + H 2 O  H 3 O + + OH - At 25 , [H 3 O + ] = [OH - ] = 1 x 10 -7 The ion-product constant for water at 25  C: K w = [H 3 O + ][OH - ] K w = (1 x 10 -7 )(1 x 10 -7 ) = 1 x 10 -14

18 Calculating pH, pOH pH = -log (H 3 O + ) pOH = -log (OH - ) Relationship between pH and pOH pH + pOH = 14 Finding [H 3 O + ], [OH - ] from pH, pOH [H 3 O + ] = 10 -pH [OH - ] = 10 -pOH

19 Acids and Bases pH In pure water, K w = [H 3 O + ] [OH – ] = 1.0  10 -14 Because in pure water [H 3 O + ] = [OH - ], [H 3 O + ] = (1.0  10 -14 ) 1/2 = 1.0  10 -7

20 Acids and Bases pH Therefore, in pure water, pH = –log [H 3 O + ] = –log (1.0  10 -7 ) = 7.00 An acid has a higher [H 3 O + ] than pure water, so its pH is <7 A base has a lower [H 3 O + ] than pure water, so its pH is >7.

21 Other “p” Scales The “p” in pH tells us to take the negative log of the quantity (in this case, hydronium ions). Some similar examples are –pOH –log [OH - ] –pK w –log K w If you know one, you know them all: [H + ] [OH - ] pH pOH

22 pH and pOH Calculations

23 pH Scale pH Scale You need to know the acidity of common substances!

24 Acids and Bases How Do We Measure pH?  Litmus paper “Red” paper turns blue above ~pH = 8 “Blue” paper turns red below ~pH = 5  An indicator Compound that changes color in solution.

25 Acids and Bases Strong Acids You will recall that the seven strong acids are HCl, HBr, HI, HNO 3, H 2 SO 4, HClO 3, and HClO 4. These are strong electrolytes and exist totally as ions in aqueous solution. For the monoprotic strong acids, [H 3 O + ] = [acid].

26 Acids and Bases Strong Bases Strong bases are the soluble hydroxides, which are the alkali metal (NaOH, KOH)and heavier alkaline earth metal hydroxides (Ca(OH) 2, Sr(OH) 2, and Ba(OH) 2 ). Again, these substances dissociate completely in aqueous solution. [OH - ] = [hydroxide added].

27 Acids and Bases Dissociation Constants For a generalized acid dissociation, the equilibrium expression is This equilibrium constant is called the acid-dissociation constant, K a.

28 Acids and Bases Calculating K a from the pH The pH of a 0.10 M solution of formic acid, HCOOH, at 25°C is 2.38. Calculate K a for formic acid at this temperature. To calculate K a, we need all equilibrium concentrations. We can find [H 3 O + ], which is the same as [HCOO − ], from the pH.

29 Acids and Bases Calculating K a from the pH pH = –log [H 3 O + ] – 2.38 = log [H 3 O + ] 10 -2.38 = 10 log [H 3 O + ] = [H 3 O + ] 4.2  10 -3 = [H 3 O + ] = [HCOO – ]

30 Acids and Bases Calculating K a from pH In table form: [HCOOH], M[H 3 O + ], M[HCOO − ], M Initially0.1000 Change –4.2  10 -3 +4.2  10 -3 At Equilibrium 0.10 – 4.2  10 -3 = 0.0958 = 0.10 4.2  10 -3

31 Acids and Bases Calculating K a from pH [4.2  10 -3 ] [0.10] K a = = 1.8  10 -4

32 A Weak Acid Equilibrium Problem What is the pH of a 0.50 M solution of acetic acid, HC 2 H 3 O 2, K a = 1.8 x 10 -5 ? Step #1: Write the dissociation equation HC 2 H 3 O 2  C 2 H 3 O 2 - + H +

33 A Weak Acid Equilibrium Problem What is the pH of a 0.50 M solution of acetic acid, HC 2 H 3 O 2, K a = 1.8 x 10 -5 ? Step #2: ICE it! HC 2 H 3 O 2  C 2 H 3 O 2 - + H + I C E 0.50 0 0 - x +x 0.50 - x xx

34 A Weak Acid Equilibrium Problem What is the pH of a 0.50 M solution of acetic acid, HC 2 H 3 O 2, K a = 1.8 x 10 -5 ? Step #3: Set up the law of mass action HC 2 H 3 O 2  C 2 H 3 O 2 - + H + 0.50 - xxx E

35 A Weak Acid Equilibrium Problem What is the pH of a 0.50 M solution of acetic acid, HC 2 H 3 O 2, K a = 1.8 x 10 -5 ? Step #4: Solve for x, which is also [H + ] HC 2 H 3 O 2  C 2 H 3 O 2 - + H + 0.50 - xxx E [H + ] = 3.0 x 10 -3 M

36 A Weak Acid Equilibrium Problem What is the pH of a 0.50 M solution of acetic acid, HC 2 H 3 O 2, K a = 1.8 x 10 -5 ? Step #5: Convert [H + ] to pH HC 2 H 3 O 2  C 2 H 3 O 2 - + H + 0.50 - xxx E

37 Acids and Bases Calculating Percent Ionization In the example: [A - ] eq = [H 3 O + ] eq = 4.2  10 -3 M [A - ] eq + [HCOOH] eq = [HCOOH] initial = 0.10 M

38 Acids and Bases Calculating pH from K a Calculate the pH of a 0.30 M solution of acetic acid, C 2 H 3 O 2 H, at 25°C. K a for acetic acid at 25°C is 1.8  10 -5. Is acetic acid more or less ionized than formic acid (K a =1.8 x 10 -4 ) ?

39 Acids and Bases Calculating pH from K a The equilibrium constant expression is:

40 Acids and Bases Calculating pH from K a Use the ICE table: [C 2 H 3 O 2 ], M[H 3 O + ], M[C 2 H 3 O 2 − ], M Initial0.3000 Change–x–x+x+x+x+x Equilibrium0.30 – xxx

41 Acids and Bases Calculating pH from K a Use the ICE table: [C 2 H 3 O 2 ], M[H 3 O + ], M[C 2 H 3 O 2 − ], M Initial0.3000 Change–x–x+x+x+x+x Equilibrium0.30 – xxx Simplify: how big is x relative to 0.30?

42 Acids and Bases Calculating pH from K a Use the ICE table: [C 2 H 3 O 2 ], M[H 3 O + ], M[C 2 H 3 O 2 − ], M Initial0.3000 Change–x–x+x+x+x+x Equilibrium0.30 – x ≈ 0.30xx Simplify: how big is x relative to 0.30?

43 Acids and Bases Calculating pH from K a Now, (1.8  10 -5 ) (0.30) = x 2 5.4  10 -6 = x 2 2.3  10 -3 = x Check: is approximation ok?

44 Acids and Bases Calculating pH from K a pH = –log [H 3 O + ] pH = – log (2.3  10 −3 ) pH = 2.64

45 Acids and Bases Polyprotic Acids Have more than one acidic proton. If the difference between the K a for the first dissociation and subsequent K a values is 10 3 or more, the pH generally depends only on the first dissociation.

46 Dissociation of Strong Bases  Strong bases are metallic hydroxides  Group I hydroxides (NaOH, KOH) are very soluble  Group II hydroxides (Ca, Ba, Mg, Sr) are less soluble  pH of strong bases is calculated directly from the concentration of the base in solution MOH(s)  M + (aq) + OH - (aq)

47 Reaction of Weak Bases with Water The generic reaction for a base reacting with water, producing its conjugate acid and hydroxide ion: B + H 2 O  BH + + OH - (all weak bases do this)

48 Weak Bases Bases react with water to produce hydroxide ion.

49 Weak Bases The equilibrium constant expression for this reaction is where K b is the base-dissociation constant.

50 Reaction of Weak Bases with Water The base reacts with water, producing its conjugate acid and hydroxide ion: CH 3 NH 2 + H 2 O  CH 3 NH 3 + + OH - K b = 4.38 x 10 -4

51 K b for Some Common Weak Bases BaseFormula Conjugate Acid KbKb Ammonia NH 3 NH 4 + 1.8 x 10 -5 Methylamine CH 3 NH 2 CH 3 NH 3 + 4.38 x 10 -4 Ethylamine C 2 H 5 NH 2 C 2 H 5 NH 3 + 5.6 x 10 -4 Diethylamine (C 2 H 5 ) 2 NH (C 2 H 5 ) 2 NH 2 + 1.3 x 10 -3 Triethylamine (C 2 H 5 ) 3 N (C 2 H 5 ) 3 NH + 4.0 x 10 -4 Hydroxylamine HONH 2 HONH 3 + 1.1 x 10 -8 HydrazineH 2 NNH 2 H 2 NNH 3 + 3.0 x 10 -6 Aniline C 6 H 5 NH 2 C 6 H 5 NH 3 + 3.8 x 10 -10 Pyridine C 5 H 5 N C 5 H 5 NH + 1.7 x 10 -9 Many students struggle with identifying weak bases and their conjugate acids.What patterns do you see that may help you?

52 A Weak Base Equilibrium Problem What is the pH of a 0.50 M solution of ammonia, NH 3, K b = 1.8 x 10 -5 ? Step #1: Write the equation for the reaction NH 3 + H 2 O  NH 4 + + OH -

53 A Weak Base Equilibrium Problem What is the pH of a 0.50 M solution of ammonia, NH 3, K b = 1.8 x 10 -5 ? Step #2: ICE it! I C E 0.50 0 0 - x +x 0.50 - x xx NH 3 + H 2 O  NH 4 + + OH -

54 A Weak Base Equilibrium Problem Step #3: Set up the law of mass action 0.50 - xxx E What is the pH of a 0.50 M solution of ammonia, NH 3, K b = 1.8 x 10 -5 ? NH 3 + H 2 O  NH 4 + + OH -

55 A Weak Base Equilibrium Problem Step #4: Solve for x, which is also [OH - ] 0.50 - xxx E [OH - ] = 3.0 x 10 -3 M NH 3 + H 2 O  NH 4 + + OH - What is the pH of a 0.50 M solution of ammonia, NH 3, K b = 1.8 x 10 -5 ?

56 A Weak Base Equilibrium Problem Step #5: Convert [OH - ] to pH 0.50 - xxx E What is the pH of a 0.50 M solution of ammonia, NH 3, K b = 1.8 x 10 -5 ? NH 3 + H 2 O  NH 4 + + OH -

57 Acids and Bases pH of Basic Solutions What is the pH of a 0.15 M solution of NH 3 ? [NH 4 + ] [OH − ] [NH 3 ] K b = = 1.8  10 -5

58 Acids and Bases pH of Basic Solutions Tabulate the data. [NH 3 ], M[NH 4 + ], M[OH − ], M Initial0.1500 Equilibrium 0.15 - x  0.15 xx Simplify: how big is x relative to 0.15?

59 Acids and Bases pH of Basic Solutions (1.8  10 -5 ) (0.15) = x 2 2.7  10 -6 = x 2 1.6  10 -3 = x 2 (x) 2 (0.15) 1.8  10 -5 = Check: is approximation ok?

60 Acids and Bases pH of Basic Solutions Therefore, [OH – ] = 1.6  10 -3 M pOH = –log (1.6  10 -3 ) pOH = 2.80 pH = 14.00 – 2.80 pH = 11.20

61 Acids and Bases K a and K b are linked: Combined reaction = ?

62 Acids and Bases K a and K b are linked: Combined reaction = ?

63 Acids and Bases K a and K b K a and K b are related in this way: K a  K b = K w Therefore, if you know one of them, you can calculate the other.

64 Acid-Base Properties of Salts Type of SaltExamplesCommentpH of solution Cation is from a strong base, anion from a strong acid KCl, KNO 3 NaCl NaNO 3 Both ions are neutral Neutral These salts simply dissociate in water: KCl(s)  K + (aq) + Cl - (aq)

65 Acid-Base Properties of Salts Type of SaltExamplesCommentpH of solution Cation is from a strong base, anion from a weak acid NaC 2 H 3 O 2 KCN, NaF Cation is neutral, Anion is basic Basic C 2 H 3 O 2 - + H 2 O  HC 2 H 3 O 2 + OH- base acid acid base The basic anion can accept a proton from water:

66 Acid-Base Properties of Salts Type of SaltExamplesCommentpH of solution Cation is the conjugate acid of a weak base, anion is from a strong acid NH 4 Cl, NH 4 NO 3 Cation is acidic, Anion is neutral Acidic NH 4 + (aq)  NH 3 (aq) + H + (aq) Acid Conjugate Proton base The acidic cation can act as a proton donor:

67 Acid-Base Properties of Salts Type of SaltExamplesCommentpH of solution Cation is the conjugate acid of a weak base, anion is conjugate base of a weak acid NH 4 C 2 H 3 O 2 NH 4 CN Cation is acidic, Anion is basic See below  IF K a for the acidic ion is greater than K b for the basic ion, the solution is acidic  IF K b for the basic ion is greater than K a for the acidic ion, the solution is basic  IF K b for the basic ion is equal to K a for the acidic ion, the solution is neutral

68 Acid-Base Properties of Salts Type of SaltExamplesCommentpH of solution Cation is a highly charged metal ion; Anion is from strong acid Al(NO 3 ) 2 FeCl 3 Hydrated cation acts as an acid; Anion is neutral Acidic Step #1: AlCl 3 (s) + 6H 2 O  Al(H 2 O) 6 3+ (aq) + Cl - (aq) Salt water Complex ion anion Step #2: Al(H 2 O) 6 3+ (aq)  Al(OH)(H 2 O) 5 2+ (aq) + H + (aq) Acid Conjugate base Proton

69 Reactions of Anions with Water Anions are bases. As such, they can react with water in a hydrolysis reaction to form OH – and the conjugate acid: X – (aq) + H 2 O(l)HX(aq) + OH – (aq)

70 Reactions of Cations with Water Cations with acidic protons (like NH 4 + ) lower the pH of a solution by releasing H +. Most metal cations (like Al 3+ ) that are hydrated in solution also lower the pH of the solution; they act by associating with H 2 O and making it release H +.

71 Reactions of Cations with Water Attraction between nonbonding electrons on oxygen and the metal causes a shift of the electron density in water. This makes the O-H bond more polar and the water more acidic. Greater charge and smaller size make a cation more acidic.

72 Effect of Cations and Anions 1.An anion that is the conjugate base of a strong acid will not affect the pH. 2.An anion that is the conjugate base of a weak acid will increase the pH. 3.A cation that is the conjugate acid of a weak base will decrease the pH.

73 Effect of Cations and Anions 4.Cations of the strong Arrhenius bases will not affect the pH. 5.Other metal ions will cause a decrease in pH. 6.When a solution contains both the conjugate base of a weak acid and the conjugate acid of a weak base, the affect on pH depends on the K a and K b values.

74 What effect on pH?Why? An anion that is the conjugate base of a strong acid does not affect pH. = very weak base An anion that is the conjugate base of a weak acid increases pH. = strong base A cation that is the conjugate acid of a weak base decreases pH. = strong acid Cations of the strong Arrhenius bases (Na +, Ca 2+ ) do not affect pH. = very weak acid (not really acidic at all) Other metal ions cause a decrease in pH. = moderate bases (cations) Weak acid + weak baseDepends on K a and K b

75 Factors Affecting Acid Strength The more polar the H-X bond and/or the weaker the H-X bond, the more acidic the compound. Acidity increases from left to right across a row and from top to bottom down a group.

76 Factors Affecting Acid Strength In oxyacids, in which an OH is bonded to another atom, Y, the more electronegative Y is, the more acidic the acid.

77 Factors Affecting Acid Strength For a series of oxyacids, acidity increases with the number of oxygens.

78 Factors Affecting Acid Strength Resonance in the conjugate bases of carboxylic acids stabilizes the base and makes the conjugate acid more acidic.

79 Lewis Acids Lewis acids are defined as electron-pair acceptors. Atoms with an empty valence orbital can be Lewis acids. A compound with no H’s can be a Lewis acid.

80 Lewis Bases Lewis bases are defined as electron-pair donors. Anything that is a Brønsted–Lowry base is also a Lewis base. (B-L bases also have a lone pair.) Lewis bases can interact with things other than protons.


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