Chemistry 132 NTPowerPoint PresentationAcid-Base EquilibriaReviewThe Common Ion EffectSlide 6Slide 7Slide 8A Problem To ConsiderSlide 10Slide 11Slide 12Slide 13Slide 14Slide 15Slide 16BuffersSlide 18Slide 19Slide 20The pH of a BufferSlide 22Slide 23Slide 24Slide 25Slide 26Slide 27Slide 28Slide 29Slide 30Adding Acid or Base to a BufferSlide 32Slide 33Slide 34Slide 35Slide 36Slide 37Slide 38Slide 39Slide 40Slide 41Slide 42Slide 43Slide 44Slide 45Slide 46Slide 47The Henderson-Hasselbalch EquationSlide 49Slide 50Slide 51Slide 52Slide 53Slide 54Slide 55Slide 56Slide 57Acid-Ionization Titration CurvesTitration of a Strong Acid by a Strong BaseFigure 16.11Slide 61Slide 62Slide 63Slide 64Slide 65Slide 66Slide 67Titration of a Weak Acid by a Strong BaseSlide 69Figure 16.12Slide 71Slide 72Slide 73Slide 74Slide 75Titration of a Weak Base by a Strong AcidFigure 16.13A Little “Trick”Slide 79Operational SkillsHomeworkSlide 8211111Chemistry 132 NTInstead of having “answers” on a math test, they should just call them “ impressions”, and if you got a different “impression”, so what, can’t we all be brothers?Jack Handey2222233333Acid-Base EquilibriaChapter 16Module 3Sections 16.5, 16.6, and 16.7Reaction of zinc metal with hydrochloric acid.44444ReviewCalculating the concentration of a species in a weak base solution using KbPredicting whether a salt solution is acidic, basic, or neutralObtaining Ka from Kb or Kb from KaCalculating concentrations of species in a salt solution55555The Common Ion EffectThe common-ion effect is the shift in an ionic equilibrium caused by the addition of a solute that provides an ion common to the equilibrium.Consider a solution of acetic acid (HC2H3O2), in which you have the equilibrium above. (aq)OHC(aq)OH2323 )l(OH)aq(OHHC223266666The Common Ion EffectThe common-ion effect is the shift in an ionic equilibrium caused by the addition of a solute that provides an ion common to the equilibrium.If we were to add NaC2H3O2 to this solution, it would provide C2H3O2- ions which are present on the right side of the equilibrium. (aq)OHC(aq)OH2323 )l(OH)aq(OHHC223277777The Common Ion EffectThe common-ion effect is the shift in an ionic equilibrium caused by the addition of a solute that provides an ion common to the equilibrium.The equilibrium composition would shift to the left and the degree of ionization of the acetic acid is decreased. (aq)OHC(aq)OH2323 )l(OH)aq(OHHC223288888The Common Ion EffectThe common-ion effect is the shift in an ionic equilibrium caused by the addition of a solute that provides an ion common to the equilibrium.This repression of the ionization of acetic acid by sodium acetate is an example of the common-ion effect. (aq)OHC(aq)OH2323 )l(OH)aq(OHHC223299999A Problem To ConsiderAn aqueous solution is 0.025 M in formic acid, HCH2O and 0.018 M in sodium formate, NaCH2O. What is the pH of the solution. The Ka for formic acid is 1.7 x 10-4.Consider the equilibrium below. (aq)OCH(aq)OH23 )l(OH)aq(OHCH22Starting0.025 0 0.018Change-x +x +xEquilibrium0.025-x x 0.018+x1010101010A Problem To ConsiderAn aqueous solution is 0.025 M in formic acid, HCH2O and 0.018 M in sodium formate, NaCH2O. What is the pH of the solution. The Ka for formic acid is 1.7 x 10-4.The equilibrium constant expression is:a223K]OHCH[]OCH][OH[1111111111A Problem To ConsiderAn aqueous solution is 0.025 M in formic acid, HCH2O and 0.018 M in sodium formate, NaCH2O. What is the pH of the solution. The Ka for formic acid is 1.7 x 10-4.Substituting into this equation gives:4107.1)x025.0()x018.0(x1212121212A Problem To ConsiderAn aqueous solution is 0.025 M in formic acid, HCH2O and 0.018 M in sodium formate, NaCH2O. What is the pH of the solution. The Ka for formic acid is 1.7 x 10-4.Assume that x is small compared with 0.018 and 0.025. Then025.0)x025.0(018.0)x018.0(1313131313A Problem To ConsiderAn aqueous solution is 0.025 M in formic acid, HCH2O and 0.018 M in sodium formate, NaCH2O. What is the pH of the solution. The Ka for formic acid is 1.7 x 10-4.The equilibrium equation becomes4107.1)025.0()018.0(x1414141414A Problem To ConsiderAn aqueous solution is 0.025 M in formic acid, HCH2O and 0.018 M in sodium formate, NaCH2O. What is the pH of the solution. The Ka for formic acid is 1.7 x 10-4.Hence,44104.2018.0025.0)107.1(x1515151515A Problem To ConsiderAn aqueous solution is 0.025 M in formic acid, HCH2O and 0.018 M in sodium formate, NaCH2O. What is the pH of the solution. The Ka for formic acid is 1.7 x 10-4.Note that x was much smaller than 0.018 or 0.025.63.3)104.2log(pH4For comparison, the pH of 0.025 M formic acid alone (without the sodium formate) is 2.69.1616161616A Problem To ConsiderAn aqueous solution is 0.025 M in formic acid, HCH2O and 0.018 M in sodium formate, NaCH2O. What is the pH of the solution. The Ka for formic acid is 1.7 x 10-4.Note that x was much smaller than 0.018 or 0.025.63.3)104.2log(pH4(see Example 16.10 and Problem 16.61)1717171717BuffersA buffer is a solution characterized by the ability to resist changes in pH when limited amounts of acid or base are added to it.Buffers contain either a weak acid and its conjugate base or a weak base and its conjugate acid.Thus, a buffer contains both an acid species and a base species in equilibrium.1818181818BuffersA buffer is a solution characterized by the ability to resist changes in pH when limited amounts of acid or base are added to it.Consider a buffer with equal molar amounts of HA and its conjugate base A-.When H3O+ is added to the buffer it reacts with the base A-.)l(OH)aq(HA)aq(A)aq(OH231919191919BuffersA buffer is a solution characterized by the ability to resist changes in pH when limited amounts of acid or base are added to it.Consider a buffer with equal molar amounts of HA and its conjugate base A-.When OH- is added to the buffer it reacts with the acid HA.)aq(A)l(OH)aq(HA)aq(OH22020202020BuffersA buffer is a solution characterized by the ability to resist changes in pH when limited amounts of acid or base are added to it.Two important characteristics of a buffer are its buffer capacity and its pH.Buffer capacity depends on the amount of acid and conjugate base present in the solution.The next example illustrates how to calculate the pH of a