CHAPTER 8: ACID/BASE EQUILIBRIUMExamples of B.L. acids and basesConjugate Acids and BasesStrong acids and basesDetermining pHDetermining the pH of waterThe pH scaleStrength of acidska and pkaStrength of basesRelationship of ka to kbCompetition between acids and basesCompetition between acids and bases (cont.)Competition between acids and bases (cont.)Indicatorska’s of IndicatorsEquilibrium of acids and bases exampleExample continuedExample ContinuedExample 2Example 2 (cont.)Example 2HydrolysisIons and HydrolysisBuffer SolutionsExample of a buffer solutionExample of buffer solutionExample of a buffer solutionExample of a buffer solution – determining the new pHWhy did the pH change so little?Henderson-Hassellbalch EquationDesigning a bufferBuffered vs. Unbuffered SolutionsTitrations and pHCase 1: Titration of a strong acid by a strong baseCase 1: Titration of a strong acid by a strong baseCase 1: Titration of a strong acid by a strong baseCase 1 SummaryCase 2: Weak acid with strong baseCase 2: Weak acid with strong basePolyprotic AcidsPolyprotic acid examplePolyprotic acid examplePolyprotic example continuedLewis acids and basesExamples of Lewis acids and basesCHAPTER 8: ACID/BASE EQUILIBRIUM• Already mentioned acid-base reactions in Chapter 6 when discussing reaction types. • One way to define acids and bases is using the Brønsted-Lowry definitions.• A Brønsted-Lowry acid donates hydrogen ions; a Brønsted-Lowry base accepts hydrogen ions.CHEM 1310 A/B Fall 2006Examples of B.L. acids and basesH3O+(aq)+ OH-(aq)↔ 2H2O(l)B.L. acid B.L base both acid & baseCH3COOH(aq)+ H2O(l)↔ H3O+(aq)+ CH3COO-(aq)B.L. acid B.L. base B.L. acid B.L. baseCHEM 1310 A/B Fall 2006Conjugate Acids and Bases• The reaction determines what’s an acid and what’s a base.H2SO4(l) + CH3COOH ↔CH3COOH2++ HSO4-•CH3COOH2+is the conjugate acid of the base CH3COOH. •HSO4-is the conjugate base of H2SO4.• Add H+to get a conjugate acid; subtract H+to get a conjugate base.CHEM 1310 A/B Fall 2006Strong acids and bases• A strong acid is one which reacts almost completely with water to produce H+. This product is the hydronium ion, H3O+.– Examples: H2SO4, HCl, etc.• A strong base is one which reacts almost completely with water to produce OH-ions.– Examples: KOH, NaNH2CHEM 1310 A/B Fall 2006Determining pH • The pH measures the amount of H+(or H3O+) ions.•pH = -log10[H3O+]• What is the pH of pure water?2H2O ↔ H3O+(aq) + OH-(aq)“Self ionization” of waterkw= 1.0 x 10-14at 25ºC = [H3O+][OH-] CHEM 1310 A/B Fall 2006Determining the pH of water• Pure water must be neutral, therefore [H3O+] = [OH-].So, kw= 10-14= [H3O+]2[H3O+] = √(10-14) = 10-7MTherefore,pH = -log10(10-7) = 7CHEM 1310 A/B Fall 2006The pH scale• pH < 7 is acidic[H3O+] > [OH-]• pH = 7 is neutral [H3O+] = [OH-]• pH > 7 is basic[H3O+] < [OH-]It is also possible to compute a pOH scale.pOH = -log10[OH-]CHEM 1310 A/B Fall 2006Strength of acids• Stronger acids dissociate more than weaker acids (usually measured in water at 25ºC).• A general acid, HA, would dissociate according to the equation:HA (aq) + H2O (ℓ) ↔ H3O++ A-(aq)ka= ([H3O+][A-])/[HA]*1 CHEM 1310 A/B Fall 2006kaand pka• The bigger the constant ka, the more the acid dissociates.•pka= -log10kaAcid kapkaHCl ~ 107~ -7H2SO4~ 102~ -2CH3COOH ~ 1.8 x 10-5~ 4.74CHEM 1310 A/B Fall 2006Strength of bases• Measures how strongly a substance wants to accept H+.H2O(ℓ) + NH3(aq) ↔ NH4+(aq) + OH-(aq)kb= ([NH4+][OH-])/[NH3]•kbis the “basicity constant” analogous to kaCHEM 1310 A/B Fall 2006Relationship of kato kb• Since, the autoionization reaction of water tells us that,[OH-] = (kw/[H3O+])thenkb= ([NH4+][NH3]) x (kw/[H3O+]) = kw/kawhere kais acidity constant of NH4+, the conjugate acid of NH3.• In general, kakb= kwand pka+ pkb= pkw= 14CHEM 1310 A/B Fall 2006Competition between acids and bases• In equilibrium reactions, you can determine if reactants or products are more favored.• For an arbitrary reactions, equilibrium constants are not usually tabulated, but can be determined from corresponding kaand kbvalues.CHEM 1310 A/B Fall 2006Competition between acids and bases (cont.)• Consider the reactionHF(aq) + CN-(aq) ↔ HCN(aq) + F-(aq) 1• The corresponding reactions areHF(aq) + H2O(ℓ) ↔ H3O+(aq) + F-(aq) 2Ka= 6.6 x 10-4= ([H3O+][F-])/[HF]H3O+(aq) + CN-(aq) ↔ HCN(aq) + H2O(ℓ) 3(1/ka’) = (1/6.2 x 10-10) = [HCN]/([H3O+][CN-])CHEM 1310 A/B Fall 2006Competition between acids and bases (cont.)• Reaction 1 is the sum of reactions 2 and 3.• When you add reactions, you multiply equilibrium constants.•krx= ka* (1/ka’) = (ka/ka’) = 1.1 x 106• This means products are favored because HF is a stronger acid (or because CN-is a stronger base).CHEM 1310 A/B Fall 2006Indicators• Indicators are molecules which change colors when pH changes over a certain range.• Phenolphthalein turns from colorless to pink as pH gets up to ~7 and higher• Other indicators change color at different pH’s.• Indicators are not ultra-precise because the color change occurs in a pH window.CHEM 1310 A/B Fall 2006ka’s of Indicators• An indicator in water could be represented by the reactionHIn(aq) + H2O ↔ H3O+(aq) + In-(aq)which has a kaofka= [H3O+][In-]/[HIn]• The color starts to change as ([In-]/[HIn]) = (ka/[H3O+]) approaches 1, that is when there are almost equal amounts of In-and HIn (each of which has a different color).• Color change happens when ka≈ [H3O+], or when pH ≈ pkaof the indicator.CHEM 1310 A/B Fall 2006Equilibrium of acids and bases example• Propionic acid CH3CH2COOH has a ka= 1.3 x 10-5at 25ºC. What is the pH of a 0.65 M solution? What fraction of the acid ionizes?CHEM 1310 A/B Fall 2006Example continued[CH3CH2COOH] [CH3CH2COO-][H3O+]Start 0.65 M ~0 ~0FinalkaCHEM 1310 A/B Fall 2006Example ContinuedCHEM 1310 A/B Fall 2006Example 2• 0.100 mol NaCH3COO is dissolved in water to make 1.00 L of solution. What is the solution’s pH? CHEM 1310 A/B Fall 2006Example 2 (cont.)[CH3COO-][CH3COOH] [OH-]Initial 0.100 0 0FinalCHEM 1310 A/B Fall 2006Example 2CHEM 1310 A/B Fall 2006Hydrolysis• The previous example demonstrated hydrolysis – when aqueous ions change the pH of a solution (the NaCH3COO increased the pH).• Not all ions do this; notice that the CH3COO-ions grabbed protons (releasing OH-which raised the pH) while