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Chapter 18 Acid-Base Equilibria. 18.1 Acids and Bases in Water 18.2 Autoionization of Water and the pH Scale 18.3 Proton Transfer and the Brønsted-Lowry.

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Presentation on theme: "Chapter 18 Acid-Base Equilibria. 18.1 Acids and Bases in Water 18.2 Autoionization of Water and the pH Scale 18.3 Proton Transfer and the Brønsted-Lowry."— Presentation transcript:

1 Chapter 18 Acid-Base Equilibria

2 18.1 Acids and Bases in Water 18.2 Autoionization of Water and the pH Scale 18.3 Proton Transfer and the Brønsted-Lowry Acid-Base Definition 18.4 Solving Problems Involving Weak-Acid Equilibria 18.5 Weak Bases and Their Relations to Weak Acids 18.6 Molecular Properties and Acid Strength 18.7 Acid-Base Properties of Salt Solutions 18.8 Generalizing the Brønsted-Lowry Concept: The Leveling Effect 18.9 Electron-Pair Donation and the Lewis Acid-Base Definition

3 The Nature of Acids and Bases:  Acids: Acids taste sour. React with metals and produce H gas. React with metals and produce H gas. turns blue litmus red turns blue litmus red pH < 7 pH < 7  * Bases: Bases taste bitter. They are slippery. turns red litmus blue. turns red litmus blue. pH >7 pH >7

4

5 Arrhenius concept  Acids produce H+ ions. Bases produce OH- ions.  This concept is limited because it applies to only aqueous solution and defines only OH containing bases.

6 Neutralization:  acid + base _______salt + water.

7 Acid Dissociation Constant (Ka)  Write Ka expression for strong and weak acids.

8 Figure 18.1 Strong acid: HA( g or l ) + H 2 O( l ) H 2 O + ( aq ) + A - ( aq ) The extent of dissociation for strong acids.

9 Figure 18.2 The extent of dissociation for weak acids. Weak acid: HA( aq ) + H 2 O( l ) H 2 O + ( aq ) + A - ( aq )

10 Figure 18.3 Reaction of zinc with a strong and a weak acid. 1M HCl( aq ) 1M CH 3 COOH( aq )

11 The Acid-Dissociation Constant Weak acids dissociate very slightly into ions in water. Strong acids dissociate completely into ions in water. HA( g or l ) + H 2 O( l ) H 3 O + ( aq ) + A - ( aq ) HA( aq ) + H 2 O( l ) H 3 O + ( aq ) + A - ( aq ) K c >> 1 K c << 1 K c = [H 3 O + ][A - ] [H 2 O][HA] K c [H 2 O] = K a = [H 3 O + ][A - ] [HA] stronger acid higher [H 3 O + ] larger K a smaller K a lower [H 3 O + ] weaker acid

12 Classifying the Relative Strengths of acids and Bases:  Strong acids: –HCl, HBr, HI –Oxo acids. HNO3, H2SO4, HClO4  Weak acids: –HF –HCN, H2S (H not bonded to O or halogen) –Oxo acids. HClO, HNO2, H3PO4  Carboxylic acids. CH3COOH

13 Classifying the Relative Strengths of acids and Bases:  Strong bases: –M2O, MOH M= (group 1A metal) –MO, M(OH)2 M=group 2A metal  Weak bases: (N atom and lone pair of electrons) –NH3 –Amines.

14 ACID STRENGTH

15 SAMPLE PROBLEM 18.1: SOLUTION: Classifying Acid and Base Strength from the Chemical Formula PROBLEM:Classify each of the following compounds as a strong acid, weak acid, strong base, or weak base. (a) H 2 SeO 4 (b) (CH 3 ) 2 CHCOOH(c) KOH(d) (CH 3 ) 2 CHNH 2 PLAN:Pay attention to the text definitions of acids and bases. Look at O for acids as well as the -COOH group; watch for amine groups and cations in bases. (a) Strong acid - H 2 SeO 4 - the number of O atoms exceeds the number of ionizable protons by 2. (b) Weak acid - (CH 3 ) 2 CHCOOH is an organic acid having a -COOH group. (c) Strong base - KOH is a Group 1A(1) hydroxide. (d) Weak base - (CH 3 ) 2 CHNH 2 has a lone pair of electrons on the N and is an amine.

16 Autoionization of water and the pH scale  Water dissociates into its ions.

17 Autoionization of Water and the pH Scale H 2 O( l ) H 3 O + ( aq )OH - ( aq ) + +

18 K c = [H 3 O + ][OH - ] [H 2 O] 2 K c [H 2 O] 2 =[H 3 O + ][OH - ] The Ion-Product Constant for Water K w = A change in [H 3 O + ] causes an inverse change in [OH - ]. = 1.0 x at 25 0 C H 2 O( l ) + H 2 O( l ) H 3 O + ( aq ) + OH - ( aq ) In an acidic solution, [H 3 O + ] > [OH - ] In a basic solution, [H 3 O + ] < [OH - ] In a neutral solution, [H 3 O + ] = [OH - ]

19 Figure 18.4 The relationship between [H 3 O + ] and [OH - ] and the relative acidity of solutions. [H 3 O + ][OH - ] Divide into K w ACIDIC SOLUTION BASIC SOLUTION [H 3 O + ] > [OH - ] [H 3 O + ] = [OH - ][H 3 O + ] < [OH - ] NEUTRAL SOLUTION

20 SAMPLE PROBLEM 18.2:Calculating [H 3 O + ] and [OH - ] in an Aqueous Solution PROBLEM:A research chemist adds a measured amount of HCl gas to pure water at 25 0 C and obtains a solution with [H 3 O + ] = 3.0x10 -4 M. Calculate [OH - ]. Is the solution neutral, acidic, or basic? SOLUTION: PLAN:Use the K w at 25 0 C and the [H 3 O + ] to find the corresponding [OH - ]. K w = 1.0x = [H 3 O + ] [OH - ] so [OH - ] = K w / [H 3 O + ] = 1.0x /3.0x10 -4 = [H 3 O + ] is > [OH - ] and the solution is acidic. 3.3x M

21 pH scale  pH  log[H+]  pH in water ranges from 0 to 14.  Kw = 1.00  1014 = [H+] [OH]  pKw = = pH + pOH  As pH rises, pOH falls (sum = 14.00).  pH=7-neutral, pH>7-basic, pH 7-basic, pH<7- acidic

22 pH scale  Acidic solns have a higher pOH.  pK=-log K  equation reaches equ, mostly products present, low pK (high K)  Reverse of the above statement is also true.  pH is measure using pH meter, pH paper or acid-base indicator.

23 Figure 18.7 Methods for measuring the pH of an aqueous solution pH (indicator) paper pH meter

24 Figure 18.5 The pH values of some familiar aqueous solutions pH = -log [H 3 O + ]

25 Table 18.3 The Relationship Between K a and pK a Acid Name (Formula)K a at 25 0 CpK a Hydrogen sulfate ion (HSO 4 - ) 1.02x10 -2 Nitrous acid (HNO 2 ) Acetic acid (CH 3 COOH) Hypobromous acid (HBrO) Phenol (C 6 H 5 OH) 7.1x x x x

26 Figure 18.6 The relations among [H 3 O + ], pH, [OH - ], and pOH.

27 Problems  P.3.Calculate pH and pOH at 25 C for: 1.0 M H+  P.4.pH=6.88, Calculate [ H+] and [ OH-] for this sample.  P.5.Calculate pH of 0.10 M HNO3  P.6.Calculate pH of 1.0 x HCl.

28 Bronsted-Lowry model:  An acid is a proton (H+ ) donor. A base is a proton acceptor.  * A conjugate acid-base pair only differs by one H.  HCl + H2O _________ H3O+ Cl- Acid base conj acid conj base  HA(aq) + H2O (l)  H3O+ (aq) +A-(aq) CA1 CB2 CA2 CB1

29 Bronsted-Lowry model:  conjugate base: everything that remains of the acid molecule after a proton is lost.  Has one H less and one more minus charge than the acid.  conjugate acid: formed when the proton is transferred to the base.  Has one more H and one less charge than the base.  Do Follow up problem-18.4-Page.779

30 Brønsted-Lowry Acid-Base Definition An acid-base reaction can now be viewed from the standpoint of the reactants AND the products. An acid reactant will produce a base product and the two will constitute an acid-base conjugate pair. An acid is a proton donor, any species which donates a H +. A base is a proton acceptor, any species which accepts a H +.

31 Table 18.4 The Conjugate Pairs in Some Acid-Base Reactions BaseAcid+ Base+ Conjugate Pair Reaction 4H 2 PO 4 - OH - + Reaction 5H 2 SO 4 N2H5+N2H5+ + Reaction 6HPO 4 2- SO Reaction 1HFH2OH2O+F-F- H3O+H3O+ + Reaction 3NH 4 + CO Reaction 2HCOOHCN - +HCOO - HCN+ NH 3 HCO HPO 4 2- H2OH2O+ HSO 4 - N 2 H PO 4 3- HSO 3 - +

32 Figure 18.8 Proton transfer as the essential feature of a Brønsted- Lowry acid-base reaction. (acid, H + donor)(base, H + acceptor) HClH2OH2O + Cl - H3O+H3O+ + Lone pair binds H + (base, H + acceptor)(acid, H + donor) NH 3 H2OH2O + NH 4 + OH - + Lone pair binds H +

33 SAMPLE PROBLEM 18.4:Identifying Conjugate Acid-Base Pairs PROBLEM:The following reactions are important environmental processes. Identify the conjugate acid-base pairs. (a) H 2 PO 4 - ( aq ) + CO 3 2- ( aq ) HPO 4 2- ( aq ) + HCO 3 - ( aq ) (b) H 2 O( l ) + SO 3 2- ( aq ) OH - ( aq ) + HSO 3 - ( aq ) SOLUTION: PLAN:Identify proton donors (acids) and proton acceptors (bases). (a) H 2 PO 4 - ( aq ) + CO 3 2- ( aq ) HPO 4 2- ( aq ) + HCO 3 - ( aq ) proton donor proton acceptor proton donor conjugate pair 1 conjugate pair 2 (b) H 2 O( l ) + SO 3 2- ( aq ) OH - ( aq ) + HSO 3 - ( aq ) conjugate pair 2 conjugate pair 1 proton donor proton acceptor proton donor

34 SAMPLE PROBLEM 18.5:Predicting the Net Direction of an Acid-Base Reaction PROBLEM:Predict the net direction and whether K a is greater or less than 1 for each of the following reactions (assume equal initial concentrations of all species): (b) H 2 O( l ) + HS - ( aq ) OH - ( aq ) + H 2 S( aq ) (a) H 2 PO 4 - ( aq ) + NH 3 ( aq ) HPO 4 2- ( aq ) + NH 4 + ( aq ) SOLUTION: PLAN:Identify the conjugate acid-base pairs and then consult Figure (button) to determine the relative strength of each. The stronger the species, the more preponderant its conjugate. (a) H 2 PO 4 - ( aq ) + NH 3 ( aq ) HPO 4 2- ( aq ) + NH 4 + ( aq ) stronger acidweaker acidstronger baseweaker base Net direction is to the right with K c > 1. (b) H 2 O( l ) + HS - ( aq ) OH - ( aq ) + H 2 S( aq ) stronger baseweaker basestronger acidweaker acid Net direction is to the left with K c < 1.

35 Figure 18.9 Strengths of conjugate acid- base pairs

36 Solving Problems Involving Weak-Acid Equilibria.  Write balanced equation and Ka expression.  Make an ICE table.  Make required assumptions.  Substitute values and solve for x.  Verify assumptions by calculating % error.

37 Problems  P.7. Calculate the pH of 1.00 M aqueous solution of HF. Ka=7.2 x 10-4  P.8 Calculate the pH of M aq solution of HOCl. Ka=3.5 x 10-8

38 SAMPLE PROBLEM 18.6:Finding the K a of a Weak Acid from the pH of Its Solution PROBLEM:Phenylacetic acid (C 6 H 5 CH 2 COOH, simplified here as HPAc) builds up in the blood of persons with phenylketonuria, an inherited disorder that, if untreated, causes mental retardation and death. A study of the acid shows that the pH of 0.12M HPAc is What is the K a of phenylacetic acid? PLAN:Write out the dissociation equation. Use pH and solution concentration to find the K a. K a = [H 3 O + ][PAc - ] [HPAc] Assumptions: With a pH of 2.62, the [H 3 O + ] HPAc >> [H 3 O + ] water. [PAc - ] ≈ [H 3 O + ]; since HPAc is weak, [HPAc] initial ≈ [HPAc] initial - [HPAc] dissociation SOLUTION:HPAc( aq ) + H 2 O( l ) H 3 O + ( aq ) + PAc - ( aq )

39 SAMPLE PROBLEM 18.6:Finding the K a of a Weak Acid from the pH of Its Solution continued Concentration(M)HPAc( aq ) + H 2 O( l ) H 3 O + ( aq ) + PAc - ( aq ) Initial0.12-1x Change--x-x+x+x+x+x Equilibrium-0.12-xxx +(<1x10 -7 ) [H 3 O + ] = 10 -pH = 2.4x10 -3 M which is >> (the [H 3 O + ] from water) x ≈ 2.4x10 -3 M ≈ [H 3 O + ] ≈ [PAc - ][HPAc] equilibrium = 0.12-x ≈ 0.12 M So K a = (2.4x10 -3 ) 0.12 = 4.8 x Be sure to check for % error.= 4x10 -3 % x100 [HPAc] dissn ; 2.4x10 -3 M 0.12M [H 3 O + ] from water ; 1x10 -7 M 2.4x10 -3 M x100 = 2.0 %

40 SAMPLE PROBLEM 18.7:Determining Concentrations from K a and Initial [HA] PROBLEM:Propanoic acid (CH 3 CH 2 COOH, which we simplify and HPr) is an organic acid whose salts are used to retard mold growth in foods. What is the [H 3 O + ] of 0.10M HPr (K a = 1.3x10 -5 )? SOLUTION: PLAN:Write out the dissociation equation and expression; make whatever assumptions about concentration which are necessary; substitute. x = [HPr] diss = [H 3 O + ] from HPr = [Pr - ] Assumptions:For HPr( aq ) + H 2 O( l ) H 3 O + ( aq ) + Pr - ( aq ) K a =[H 3 O + ][Pr - ] [HPr] HPr( aq ) + H 2 O( l ) H 3 O + ( aq ) + Pr - ( aq )Concentration(M) Initial Change--x-x+x+x+x+x Equilibrium-0.10-xxx Since K a is small, we will assume that x << 0.10

41 SAMPLE PROBLEM 18.7:Determining Concentrations from K a and Initial [HA] continued (x)(x) x10 -5 = [H 3 O + ][Pr - ] [HPr] = = 1.1x10 -3 M = [H 3 O + ] Check: [HPr] diss = 1.1x10 -3 M/0.10 M x 100 = 1.1%

42 Percent dissociation:  For a weak acid % dissociation increases as the acid becomes more dilute.  For solution of any weak acid, [ H+] decreases as [HA]0 decreases, but % dissociation increases as [HA]0 increases.

43 Problems P.11.Calculate the % dissociation of acetic acid (Ka= 1.8 x 10-5 ) 1.00 M and M solutions. P.12.In a M HC3H5O3 aqueous solution, lactic acid is 3.7 % dissociated. Calculate ka for the acid.

44 Percent HA dissociation = [HA] dissociated [HA] initial x 100 Polyprotic acids acids with more than more ionizable proton H 3 PO 4 ( aq ) + H 2 O( l ) H 2 PO 4 - ( aq ) + H 3 O + ( aq ) H 2 PO 4 - ( aq ) + H 2 O( l ) HPO 4 2- ( aq ) + H 3 O + ( aq ) HPO 4 2- ( aq ) + H 2 O( l ) PO 4 3- ( aq ) + H 3 O + ( aq ) K a1 = [H 3 O + ][H 2 PO 4 - ] [H 3 PO 4 ] K a2 = [H 3 O + ][HPO 4 2- ] [H 2 PO 4 - ] K a3 = [H 3 O + ][PO 4 3- ] [HPO 4 2- ] K a1 > K a2 > K a3 = 7.2x10 -3 = 6.3x10 -8 = 4.2x10 -13

45 ACID STRENGTH

46 Polyprotic Acids  They can furnish more than one proton (H+) to the solution.  Characteristics: 1. Ka values are much smaller than the first value, so only the first dissociation step makes a significant contribution to equilibrium concentration of H+.  2. For sulfuric acid, it behaves as a strong acid in the first dissociation step and a weak acid in the second step, where its contribution of H+ ions can be neglected. Use quadratic equation to solve such kind of a problem.

47 Problems P.13.Calculate the pH of a 5.0 M H3PO4 solution and the equilibrium concentration of the species H3PO4, H2PO4-, HPO42-, PO43- P.14.Calculate the pH of 1.0 M sulfuric acid solution.

48 SAMPLE PROBLEM 18.8:Calculating Equilibrium Concentrations for a Polyprotic Acid PROBLEM:Ascorbic acid (H 2 C 6 H 6 O 6 ; H 2 Asc for this problem), known as vitamin C, is a diprotic acid (K a1 = 1.0x10 -5 and K a2 = 5x ) found in citrus fruit. Calculate [H 2 Asc], [HAsc - ], [Asc 2- ], and the pH of 0.050M H 2 Asc. SOLUTION: PLAN:Write out expressions for both dissociations and make assumptions. K a1 >> K a2 so the first dissociation produces virtually all of the H 3 O +. K a1 is small so [H 2 Asc] initial ≈ [H 2 Asc] diss After finding the concentrations of various species for the first dissociation, we can use them as initial concentrations for the second dissociation. K a1 = [HAsc - ][H 3 O + ] [H 2 Asc] = 1.0x10 -5 K a2 = [Asc 2- ][H 3 O + ] [HAsc - ] = 5x H 2 Asc( aq ) + H 2 O (l ) HAsc - ( aq ) + H 3 O + ( aq ) HAsc - ( aq ) + H 2 O (l ) Asc 2- ( aq ) + H 3 O + ( aq )

49 - x-+ x SAMPLE PROBLEM 18.8:Calculating Equilibrium Concentrations for a Polyprotic Acid continued H 2 Asc( aq ) + H 2 O (l ) HAsc - ( aq ) + H 3 O + ( aq )Concentration(M) Initial Equilibrium x-xx K a1 = [HAsc - ][H 3 O + ]/[H 2 Asc] = 1.0x10 -5 = (x)(x)/0.050 M pH = -log(7.1x10 -4 ) = x10 -4 M-00 Change- x-+ x 7.1x x-xx Equilibrium Change Initial x x = 7.1x10 -4 M HAsc - ( aq ) + H 2 O (l ) Asc 2- ( aq ) + H 3 O + ( aq )Concentration(M) x = 6x10 -8 M

50 Bases  “Strong” and “weak” are used in the same sense for bases as for acids.  strong = complete dissociation (hydroxide ion supplied to solution)  NaOH(s)  Na+(aq) + OH(aq)  weak = very little dissociation (or reaction with water)  H3CNH2(aq) + H2O(l)  H3CNH3+(aq) + OH(aq)

51 BASE STRENGTH K b = [BH + ][OH - ] [B]

52 Problems P.15.Calculate pH of 5.0 x 10-2 M NaOH solution. P.16.Calculate the pH for a 1.50 M solution of ammonia (Kb=1.8 x 10-5 )

53 + CH 3 NH 3 + OH - methylammonium ion Figure Abstraction of a proton from water by methylamine. + CH 3 NH 2 H2OH2O Methylamine Base in water Lone pair binds H +

54 SAMPLE PROBLEM 18.9:Determining pH from K b and Initial [B] PROBLEM:Dimethylamine, (CH 3 ) 2 NH, a key intermediate in detergent manufacture, has a K b of 5.9x What is the pH of 1.5M (CH 3 ) 2 NH? SOLUTION: PLAN:Perform this calculation as you did those for acids. Keep in mind that you are working with K b and a base. (CH 3 ) 2 NH( aq ) + H 2 O( l ) (CH 3 ) 2 NH 2 + ( aq ) + OH - ( aq ) Assumptions: [(CH 3 ) 2 NH 2 + ] = [OH - ] = x ; [(CH 3 ) 2 NH 2 + ] - x ≈ [(CH 3 ) 2 NH] initial K b >> K w so [OH - ] from water is neglible Initial1.50M00- Change- x-+ x Equilibrium x-xx (CH 3 ) 2 NH( aq ) + H 2 O( l ) (CH 3 ) 2 NH 2 + ( aq ) + OH - ( aq )Concentration

55 SAMPLE PROBLEM 18.9:Determining pH from K b and Initial [B] continued K b = 5.9x10 -4 = [(CH 3 ) 2 NH 2 + ][OH - ] [(CH 3 ) 2 NH] 5.9x10 -4 = (x) 1.5M x = 3.0x10 -2 M = [OH - ] Check assumption:3.0x10 -2 M/1.5M x 100 = 2% [H 3 O + ] = K w /[OH - ] = 1.0x /3.0x10 -2 = 3.3x M pH = -log 3.3x = 12.48

56 SAMPLE PROBLEM 18.10:Determining the pH of a Solution of A - PROBLEM:Sodium acetate (CH 3 COONa, or NaAc for this problem) has applications in photographic development and textile dyeing. What is the pH of 0.25M NaAc? K a of acetic acid (HAc) is 1.8x SOLUTION: PLAN:Sodium salts are soluble in water so [Ac - ] = 0.25M. Write the association equation for acetic acid; use the K a to find the K b. Initial0.25M-00 Change-x-x+x+x+x- Equilibrium-0.25M-xxx Ac - ( aq ) + H 2 O( l ) HAc( aq ) + OH - ( aq )Concentration K b = [HAc][OH - ] [Ac - ] = KwKw KaKa = 5.6x M K b = 1.0x x10 -5

57 SAMPLE PROBLEM 18.10:Determining the pH of a Solution of A - continued K b = [HAc][OH - ] [Ac - ] [Ac-] = 0.25M-x ≈ 0.25M 5.6x = x 2 /0.25M x = 1.2x10 -5 M = [OH - ] Check assumption:1.2x10 -5 M/0.25M x 100 = 4.8x10 -3 % [H 3 O + ] = K w /[OH - ]= 1.0x /1.2x10 -5 = 8.3x M pH = -log 8.3x M = 9.08

58 Molecular Properties and Acid strength:  Trends in acid strength of Nonmetal Hydrides:  Across a period nonmetal hydride acid strength increases.  Down a group nonmetal hydride acid strength increases.

59 Figure The effect of atomic and molecular properties on nonmetal hydride acidity. 6A(16) H2OH2O H2SH2S H 2 Se H 2 Te 7A(17) HF HCl HBr HIHI Electronegativity increases, acidity increases Bond strength decreases, acidity increases

60 Trends in Acid strength of Oxoacids 1. For oxoacids with same number of O around E, acid strength increases with electronegativity of E. 2. For oxoacids with different number of O around E, acid strength increases with number of O atoms.

61 HO I HOBrHOCl>> HO O O O << Figure The relative strengths of oxoacids.  HOCl  

62 Table 18.7 K a Values of Some Hydrated Metal Ions at 25 0 C Free IonHydrated IonKaKa Fe 3+ Fe(H 2 O) 6 3+ ( aq ) 6 x Sn 2+ Sn(H 2 O) 6 2+ ( aq ) 4 x Cr 3+ Cr(H 2 O) 6 3+ ( aq ) 1 x Al 3+ Al(H 2 O) 6 3+ ( aq ) 1 x Cu 2+ Cu(H 2 O) 6 2+ ( aq ) 3 x Pb 2+ Pb(H 2 O) 6 2+ ( aq ) 3 x Zn 2+ Zn(H 2 O) 6 2+ ( aq ) 1 x Co 2+ Co(H 2 O) 6 2+ ( aq ) 2 x Ni 2+ Ni(H 2 O) 6 2+ ( aq ) 1 x ACID STRENGTH

63 Al(H 2 O) 5 OH 2+ Al(H 2 O) 6 3+ Figure The acidic behavior of the hydrated Al 3+ ion. H2OH2O H3O+H3O+ Electron density drawn toward Al 3+ Nearby H 2 O acts as base

64 Acid-Base properties of Salt:  Salt is an ionic compound.  Ka xKb=Kw

65

66 SAMPLE PROBLEM 18.11:Predicting Relative Acidity of Salt Solutions PROBLEM:Predict whether aqueous solutions of the following are acidic, basic, or neutral, and write an equation for the reaction of any ion with water: (a) Potassium perchlorate, KClO 4 (b) Sodium benzoate, C 6 H 5 COONa (c) Chromium trichloride, CrCl 3 (d) Sodium hydrogen sulfate, NaHSO 4 SOLUTION: PLAN:Consider the acid-base nature of the anions and cations. Strong acid-strong base combinations produce a neutral solution; strong acid-weak base, acidic; weak acid-strong base, basic. (a) The ions are K + and ClO 4 -, both of which come from a strong base(KOH) and a strong acid(HClO 4 ). Therefore the solution will be neutral. (b) Na + comes from the strong base NaOH while C 6 H 5 COO - is the anion of a weak organic acid. The salt solution will be basic. (c) Cr 3+ is a small cation with a large + charge, so it’s hydrated form will react with water to produce H 3 O +. Cl - comes from the strong acid HCl. Acidic solution. (d) Na + comes from a strong base. HSO 4 - can react with water to form H 3 O +. So the salt solution will be acidic.

67 SAMPLE PROBLEM 18.12:Predicting the Relative Acidity of Salt Solutions from K a and K b of the Ions PROBLEM:Determine whether an aqueous solution of zinc formate, Zn(HCOO) 2, is acidic, basic, or neutral. SOLUTION: PLAN:Both Zn 2+ and HCOO - come from weak conjugates. In order to find the relatively acidity, write out the dissociation reactions and use the information in Tables 18.2 and K a Zn(H 2 O) 6 2+ = 1x10 -9 K a HCOO - = 1.8x10 -4 ; K b = K w /K a = 1.0x /1.8x10 -4 = 5.6x K a for Zn(H 2 O) 6 2+ >>> K b HCOO -, therefore the solution is acidic. Zn(H 2 O) 6 2+ ( aq ) + H 2 O( l ) Zn(H 2 O) 5 OH + ( aq ) + H 3 O + ( aq ) HCOO - ( aq ) + H 2 O( l ) HCOOH( aq ) + OH - ( aq )

68 Molecules as Lewis Acids acidbaseadduct An acid is an electron-pair acceptor. A base is an electron-pair donor. M 2+ H 2 O( l ) M(H 2 O) 4 2+ ( aq ) adduct

69 Figure The Mg 2+ ion as a Lewis acid in the chlorophyll molecule.

70 SAMPLE PROBLEM 18.13:Identifying Lewis Acids and Bases PROBLEM:Identify the Lewis acids and Lewis bases in the following reactions: (a) H + + OH - H 2 O (b) Cl - + BCl 3 BCl 4 - (c) K + + 6H 2 O K(H 2 O) 6 + SOLUTION: PLAN:Look for electron pair acceptors (acids) and donors (bases). (a) H + + OH - H 2 O acceptor donor (b) Cl - + BCl 3 BCl 4 - donor acceptor (c) K + + 6H 2 O K(H 2 O) 6 + acceptor donor


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