2Acid-Base Equilibria 18.1 Acids and Bases in Water 18.2 Autoionization of Water and the pH Scale18.3 Proton Transfer and the Brønsted-Lowry Acid-Base Definition18.4 Solving Problems Involving Weak-Acid Equilibria18.5 Weak Bases and Their Relations to Weak Acids18.6 Molecular Properties and Acid Strength18.7 Acid-Base Properties of Salt Solutions18.8 Generalizing the Brønsted-Lowry Concept: The Leveling Effect18.9 Electron-Pair Donation and the Lewis Acid-Base Definition
3The Nature of Acids and Bases: Acids: Acids taste sour.React with metals and produce H gas.turns blue litmus redpH < 7* Bases: Bases taste bitter. They are slippery.turns red litmus blue.pH >7
7Acid Dissociation Constant (Ka) Write Ka expression for strong and weak acids.
8The extent of dissociation for strong acids. Figure 18.1The extent of dissociation for strong acids.Strong acid: HA(g or l) + H2O(l) H2O+(aq) + A-(aq)
9The extent of dissociation for weak acids. Figure 18.2The extent of dissociation for weak acids.Weak acid: HA(aq) + H2O(l) H2O+(aq) + A-(aq)
10Reaction of zinc with a strong and a weak acid. Figure 18.3Reaction of zinc with a strong and a weak acid.1M HCl(aq)1M CH3COOH(aq)
11The Acid-Dissociation Constant Strong acids dissociate completely into ions in water.HA(g or l) + H2O(l) H3O+(aq) + A-(aq)Kc >> 1Weak acids dissociate very slightly into ions in water.HA(aq) + H2O(l) H3O+(aq) + A-(aq)Kc << 1The Acid-Dissociation ConstantKc =[H3O+][A-][H2O][HA]stronger acid higher [H3O+]larger KaKc[H2O] = Ka =[H3O+][A-][HA]smaller Ka lower [H3O+]weaker acid
12Classifying the Relative Strengths of acids and Bases: Strong acids:HCl, HBr, HIOxo acids. HNO3, H2SO4, HClO4Weak acids:HFHCN , H2S (H not bonded to O or halogen)Oxo acids. HClO, HNO2, H3PO4Carboxylic acids. CH3COOH
13Classifying the Relative Strengths of acids and Bases: Strong bases:M2O, MOH M= (group 1A metal)MO , M(OH)2 M=group 2A metalWeak bases: (N atom and lone pair of electrons)NH3Amines.
15SAMPLE PROBLEM 18.1:Classifying Acid and Base Strength from theChemical FormulaPROBLEM:Classify each of the following compounds as a strong acid, weak acid, strong base, or weak base.(a) H2SeO4(b) (CH3)2CHCOOH(c) KOH(d) (CH3)2CHNH2PLAN: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.SOLUTION:(a) Strong acid - H2SeO4 - the number of O atoms exceeds the number of ionizable protons by 2.(b) Weak acid - (CH3)2CHCOOH is an organic acid having a -COOH group.(c) Strong base - KOH is a Group 1A(1) hydroxide.(d) Weak base - (CH3)2CHNH2 has a lone pair of electrons on the N and is an amine.
16Autoionization of water and the pH scale Water dissociates into its ions.
17Autoionization of Water and the pH Scale H2O(l)H2O(l)+OH-(aq)H3O+(aq)+
18H2O(l) + H2O(l) H3O+(aq) + OH-(aq) Kc =[H3O+][OH-][H2O]2The Ion-Product Constant for WaterKc[H2O]2 =Kw =[H3O+][OH-]= 1.0 x at 250CA change in [H3O+] causes an inverse change in [OH-].In an acidic solution, [H3O+] > [OH-]In a basic solution, [H3O+] < [OH-]In a neutral solution, [H3O+] = [OH-]
19The relationship between [H3O+] and [OH-] and the relative acidity of solutions.Figure 18.4Divide into Kw[H3O+][OH-][H3O+] > [OH-][H3O+] = [OH-][H3O+] < [OH-]ACIDIC SOLUTIONNEUTRAL SOLUTIONBASIC SOLUTION
20SAMPLE PROBLEM 18.2:Calculating [H3O+] and [OH-] in an AqueousSolutionPROBLEM:A research chemist adds a measured amount of HCl gas to pure water at 250C and obtains a solution with [H3O+] = 3.0x10-4M. Calculate [OH-]. Is the solution neutral, acidic, or basic?PLAN:Use the Kw at 250C and the [H3O+] to find the corresponding [OH-].SOLUTION:Kw = 1.0x10-14 = [H3O+] [OH-] so[OH-] = Kw/ [H3O+] = 1.0x10-14/3.0x10-4 =3.3x10-11M[H3O+] is > [OH-] and the solution is acidic.
21pH scale pH log[H+] pH in water ranges from 0 to 14. Kw = 1.00 1014 = [H+] [OH]pKw = = pH + pOHAs pH rises, pOH falls (sum = 14.00).pH=7-neutral, pH>7-basic, pH<7-acidic
22pH 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.
23Methods for measuring the pH of an aqueous solution Figure 18.7Methods for measuring the pH of an aqueous solutionpH (indicator) paperpH meter
24The pH values of some familiar aqueous solutions Figure 18.5The pH values of some familiar aqueous solutionspH = -log [H3O+]
25Table 18.3 The Relationship Between Ka and pKa Acid Name (Formula)Ka at 250CpKaHydrogen sulfate ion (HSO4-)1.02x10-21.9913.15Nitrous acid (HNO2)7.1x10-44.74Acetic acid (CH3COOH)1.8x10-52.3x10-98.64Hypobromous acid (HBrO)Phenol (C6H5OH)1.0x10-1010.00
26The relations among [H3O+], pH, [OH-], and pOH. Figure 18.6The relations among [H3O+], pH, [OH-], and pOH.
27Problems 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 HNO3P.6.Calculate pH of 1.0 x HCl.
28Bronsted-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 baseHA(aq) + H2O (l) H3O+ (aq) +A-(aq)CA CB CA CB1
29Bronsted-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
30Brønsted-Lowry Acid-Base Definition An acid is a proton donor, any species which donates a H+.A base is a proton acceptor, any species which accepts a H+.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.
31Table 18.4 The Conjugate Pairs in Some Acid-Base Reactions +BaseAcid+Conjugate PairReaction 1HFH2O+F-H3O++Reaction 2HCOOHCN-+HCOO-HCN+Reaction 3NH4+CO32-+NH3HCO3-+Reaction 4H2PO4-OH-+HPO42-H2O+Reaction 5H2SO4N2H5++HSO4-N2H62++Reaction 6HPO42-SO32-+PO43-HSO3-+
32Proton transfer as the essential feature of a Brønsted- Lowry acid-base reaction.Figure 18.8HClH2O+Lone pair binds H+Cl-H3O++(acid, H+ donor)(base, H+ acceptor)NH3H2O+Lone pair binds H+NH4+OH-+(base, H+ acceptor)(acid, H+ donor)
33SAMPLE PROBLEM 18.4:Identifying Conjugate Acid-Base PairsPROBLEM:The following reactions are important environmental processes. Identify the conjugate acid-base pairs.(a) H2PO4-(aq) + CO32-(aq) HPO42-(aq) + HCO3-(aq)(b) H2O(l) + SO32-(aq) OH-(aq) + HSO3-(aq)PLAN:Identify proton donors (acids) and proton acceptors (bases).conjugate pair2conjugate pair1SOLUTION:(a) H2PO4-(aq) + CO32-(aq) HPO42-(aq) + HCO3-(aq)proton donorproton acceptorproton acceptorproton donorconjugate pair2conjugate pair1(b) H2O(l) + SO32-(aq) OH-(aq) + HSO3-(aq)proton donorproton acceptorproton acceptorproton donor
34SAMPLE PROBLEM 18.5:Predicting the Net Direction of an Acid-BaseReactionPROBLEM:Predict the net direction and whether Ka is greater or less than 1 for each of the following reactions (assume equal initial concentrations of all species):(a) H2PO4-(aq) + NH3(aq) HPO42-(aq) + NH4+(aq)(b) H2O(l) + HS-(aq) OH-(aq) + H2S(aq)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.SOLUTION:(a) H2PO4-(aq) + NH3(aq) HPO42-(aq) + NH4+(aq)stronger acidweaker acidstronger baseweaker baseNet direction is to the right with Kc > 1.(b) H2O(l) + HS-(aq) OH-(aq) + H2S(aq)stronger acidweaker acidstronger baseweaker baseNet direction is to the left with Kc < 1.
35Strengths of conjugate acid-base pairs Figure 18.9Strengths of conjugate acid-base pairs
36Solving 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.
37ProblemsP.7. Calculate the pH of 1.00 M aqueous solution of HF. Ka=7.2 x 10-4P.8 Calculate the pH of M aq solution of HOCl. Ka=3.5 x 10-8
38SAMPLE PROBLEM 18.6:Finding the Ka of a Weak Acid from the pH ofIts SolutionPROBLEM:Phenylacetic acid (C6H5CH2COOH, 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 Ka of phenylacetic acid?PLAN:Write out the dissociation equation. Use pH and solution concentration to find the Ka.Assumptions:With a pH of 2.62, the [H3O+]HPAc >> [H3O+]water.[PAc-] ≈ [H3O+]; since HPAc is weak, [HPAc]initial ≈ [HPAc]initial - [HPAc]dissociationSOLUTION:HPAc(aq) + H2O(l) H3O+(aq) + PAc-(aq)Ka =[H3O+][PAc-][HPAc]
39SAMPLE PROBLEM 18.6:Finding the Ka of a Weak Acid from the pH ofIts SolutioncontinuedConcentration(M)HPAc(aq) + H2O(l) H3O+(aq) + PAc-(aq)Initial0.12-1x10-7Change--x+xEquilibrium-0.12-xxx +(<1x10-7)[H3O+] = 10-pH = 2.4x10-3 M which is >> 10-7 (the [H3O+] from water)x ≈ 2.4x10-3 M ≈ [H3O+] ≈ [PAc-][HPAc]equilibrium = 0.12-x ≈ 0.12 M(2.4x10-3) (2.4x10-3)0.12So Ka == 4.8 x 10-5[H3O+]from water;1x10-7M2.4x10-3Mx100Be sure to check for % error.= 4x10-3 %x100[HPAc]dissn;2.4x10-3M0.12M= 2.0 %
40SAMPLE PROBLEM 18.7:Determining Concentrations from Ka andInitial [HA]PROBLEM:Propanoic acid (CH3CH2COOH, which we simplify and HPr) is an organic acid whose salts are used to retard mold growth in foods. What is the [H3O+] of 0.10M HPr (Ka = 1.3x10-5)?PLAN:Write out the dissociation equation and expression; make whatever assumptions about concentration which are necessary; substitute.Assumptions:For HPr(aq) + H2O(l) H3O+(aq) + Pr-(aq)x = [HPr]diss = [H3O+]from HPr= [Pr-]Ka =[H3O+][Pr-][HPr]SOLUTION:Concentration(M)HPr(aq) + H2O(l) H3O+(aq) + Pr-(aq)Initial0.10-Change--x+xEquilibrium-0.10-xxSince Ka is small, we will assume that x << 0.10
41SAMPLE PROBLEM 18.7:Determining Concentrations from Ka andInitial [HA]continued1.3x10-5 =[H3O+][Pr-][HPr]=(x)(x)0.10= 1.1x10-3 M = [H3O+]Check: [HPr]diss = 1.1x10-3M/0.10 M x 100 = 1.1%
42Percent 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.
43ProblemsP.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.
44Percent HA dissociation = [HA]dissociated [HA]initialx 100Polyprotic acidsacids with more than more ionizable protonKa1 =[H3O+][H2PO4-][H3PO4]H3PO4(aq) + H2O(l) H2PO4-(aq) + H3O+(aq)= 7.2x10-3Ka2 =[H3O+][HPO42-][H2PO4-]H2PO4-(aq) + H2O(l) HPO42-(aq) + H3O+(aq)= 6.3x10-8Ka3 =[H3O+][PO43-][HPO42-]HPO42-(aq) + H2O(l) PO43-(aq) + H3O+(aq)= 4.2x10-13Ka1 > Ka2 > Ka3
46Polyprotic AcidsThey 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.
47ProblemsP.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.
48SAMPLE PROBLEM 18.8:Calculating Equilibrium Concentrations for aPolyprotic AcidPROBLEM:Ascorbic acid (H2C6H6O6; H2Asc for this problem), known as vitamin C, is a diprotic acid (Ka1 = 1.0x10-5 and Ka2 = 5x10-12) found in citrus fruit. Calculate [H2Asc], [HAsc-], [Asc2-], and the pH of 0.050M H2Asc.PLAN:Write out expressions for both dissociations and make assumptions.Ka1 >> Ka2 so the first dissociation produces virtually all of the H3O+.Ka1 is small so [H2Asc]initial ≈ [H2Asc]dissAfter finding the concentrations of various species for the first dissociation, we can use them as initial concentrations for the second dissociation.SOLUTION:Ka1 =[HAsc-][H3O+][H2Asc]H2Asc(aq) + H2O(l) HAsc-(aq) + H3O+(aq)= 1.0x10-5Ka2 =[Asc2-][H3O+][HAsc-]HAsc-(aq) + H2O(l) Asc2-(aq) + H3O+(aq)= 5x10-12
50Bases“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)
52Problems 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 )
53Abstraction of a proton from water by methylamine. Figure 18.11Abstraction of a proton from water by methylamine.Lone pair binds H++CH3NH2H2OMethylamineBase in water+CH3NH3+OH-methylammonium ion
54SAMPLE PROBLEM 18.9:Determining pH from Kb and Initial [B]PROBLEM:Dimethylamine, (CH3)2NH, a key intermediate in detergent manufacture, has a Kb of 5.9x What is the pH of 1.5M (CH3)2NH?PLAN:Perform this calculation as you did those for acids. Keep in mind that you are working with Kb and a base.(CH3)2NH(aq) + H2O(l) (CH3)2NH2+(aq) + OH-(aq)Assumptions:Kb >> Kw so [OH-]from water is neglible[(CH3)2NH2+] = [OH-] = x ; [(CH3)2NH2+] - x ≈ [(CH3)2NH]initialSOLUTION:(CH3)2NH(aq) + H2O(l) (CH3)2NH2+(aq) + OH-(aq)ConcentrationInitial1.50M-Change- x-+ xEquilibriumx-x
55SAMPLE PROBLEM 18.9:Determining pH from Kb and Initial [B]continuedKb = 5.9x10-4 =[(CH3)2NH2+][OH-][(CH3)2NH]5.9x10-4 =(x) (x)1.5Mx = 3.0x10-2M = [OH-]Check assumption:3.0x10-2M/1.5M x 100 = 2%[H3O+] = Kw/[OH-] = 1.0x10-14/3.0x10-2 = 3.3x10-13MpH = -log 3.3x10-13 = 12.48
56SAMPLE PROBLEM 18.10:Determining the pH of a Solution of A-PROBLEM:Sodium acetate (CH3COONa, or NaAc for this problem) has applications in photographic development and textile dyeing. What is the pH of 0.25M NaAc? Ka of acetic acid (HAc) is 1.8x10-5.PLAN:Sodium salts are soluble in water so [Ac-] = 0.25M.Write the association equation for acetic acid; use the Ka to find the Kb.SOLUTION:Ac-(aq) + H2O(l) HAc(aq) + OH-(aq)ConcentrationInitial0.25M-Change-x+x-Equilibrium-0.25M-xxKb =[HAc][OH-][Ac-]=KwKaKb =1.0x10-141.8x10-5= 5.6x10-10M
57SAMPLE PROBLEM 18.10:Determining the pH of a Solution of A-continued[Ac-] = 0.25M-x ≈ 0.25MKb =[HAc][OH-][Ac-]5.6x10-10 = x2/0.25Mx = 1.2x10-5M = [OH-]Check assumption:1.2x10-5M/0.25M x 100 = 4.8x10-3 %[H3O+] = Kw/[OH-]= 1.0x10-14/1.2x10-5= 8.3x10-10MpH = -log 8.3x10-10M = 9.08
58Molecular 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.
59The effect of atomic and molecular properties on Figure 18.12The effect of atomic and molecular properties onnonmetal hydride acidity.6A(16)H2OH2SH2SeH2Te7A(17)HFHClHBrHIElectronegativity increases, acidity increasesBond strength decreases, acidity increases
60Trends 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<< The relative strengths of oxoacids. H O I Br Cl > Figure 18.13The relative strengths of oxoacids.HOIBrCl>HOClHOCl<<
62Table 18.7 Ka Values of Some Hydrated Metal Ions at 250C Free IonHydrated IonKaACID STRENGTHFe3+Fe(H2O)63+(aq)6 x 10-3Sn2+Sn(H2O)62+(aq)4 x 10-4Cr3+Cr(H2O)63+(aq)1 x 10-4Al3+Al(H2O)63+(aq)1 x 10-5Cu2+Cu(H2O)62+(aq)3 x 10-8Pb2+Pb(H2O)62+(aq)3 x 10-8Zn2+Zn(H2O)62+(aq)1 x 10-9Co2+Co(H2O)62+(aq)2 x 10-10Ni2+Ni(H2O)62+(aq)1 x 10-10
63The acidic behavior of the hydrated Al3+ ion. Figure 18.13The acidic behavior of the hydrated Al3+ ion.Electron density drawn toward Al3+Nearby H2O acts as baseAl(H2O)63+Al(H2O)5OH2+H3O+H2O
64Acid-Base properties of Salt: Salt is an ionic compound.Ka xKb=Kw
66SAMPLE PROBLEM 18.11:Predicting Relative Acidity of Salt SolutionsPROBLEM: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, KClO4(b) Sodium benzoate, C6H5COONa(c) Chromium trichloride, CrCl3(d) Sodium hydrogen sulfate, NaHSO4PLAN: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.SOLUTION:(a) The ions are K+ and ClO4- , both of which come from a strong base(KOH) and a strong acid(HClO4). Therefore the solution will be neutral.(b) Na+ comes from the strong base NaOH while C6H5COO- is the anion of a weak organic acid. The salt solution will be basic.(c) Cr3+ is a small cation with a large + charge, so it’s hydrated form will react with water to produce H3O+. Cl- comes from the strong acid HCl. Acidic solution.(d) Na+ comes from a strong base. HSO4- can react with water to form H3O+. So the salt solution will be acidic.
67SAMPLE PROBLEM 18.12:Predicting the Relative Acidity of SaltSolutions from Ka and Kb of the IonsPROBLEM:Determine whether an aqueous solution of zinc formate, Zn(HCOO)2, is acidic, basic, or neutral.PLAN:Both Zn2+ and HCOO- come from weak conjugates. In order to find the relatively acidity, write out the dissociation reactions and use the information in Tables and 18.7.SOLUTION:Zn(H2O)62+(aq) + H2O(l) Zn(H2O)5OH+(aq) + H3O+(aq)HCOO-(aq) + H2O(l) HCOOH(aq) + OH-(aq)Ka Zn(H2O)62+ = 1x10-9Ka HCOO- = 1.8x10-4 ; Kb = Kw/Ka = 1.0x10-14/1.8x10-4 = 5.6x10-11Ka for Zn(H2O)62+ >>> Kb HCOO-, therefore the solution is acidic.
68Molecules as Lewis Acids An acid is an electron-pair acceptor.A base is an electron-pair donor.acidbaseadductH2O(l)M(H2O)42+(aq)M2+adduct
69The Mg2+ ion as a Lewis acid in the chlorophyll molecule. Figure 18.15The Mg2+ ion as a Lewis acid in the chlorophyll molecule.
70SAMPLE PROBLEM 18.13:Identifying Lewis Acids and BasesPROBLEM:Identify the Lewis acids and Lewis bases in the following reactions:(a) H+ + OH H2O(b) Cl- + BCl BCl4-(c) K+ + 6H2O K(H2O)6+PLAN:Look for electron pair acceptors (acids) and donors (bases).SOLUTION:acceptor(a) H+ + OH H2Odonordonor(b) Cl- + BCl BCl4-acceptoracceptor(c) K+ + 6H2O K(H2O)6+donor