Chem 113 1st Edition Lecture 31Outline of Last Lecture I. Predicting the Net Direction of ReactionsII. Comparison of acid-base reactionsIII. Leveling Effect in H2OOutline of Current Lecture IV. Solving Problems involving weak-acid/base equilibriaV. The relationship between Ka and pKaVI. The Henderson-Hasselbalch EquationVII. Acid-Base TitrationsVIII. Curve for a strong acid-base titrationIX. Calculating the pH during a strong acid-strong base titrationCurrent LectureI. Solving Problems involving weak-acid/base equilibriaa. Problem-solving approachi. Write a balanced equationii. Write an expression for Ka or Kbiii. Define x as the change in concentration that occurs during the reactioniv. Construct a reaction table in terms of xv. Make assumptions the simplify the calculationvi. Substitute values into the Ka/Kb expression and solve for xvii. Check that the assumptions are justifiedb. The notation systemi. Molar concentrations are indicated by []These notes represent a detailed interpretation of the professor’s lecture. GradeBuddy is best used as a supplement to your own notes, not as a substitute.ii. A bracketed formula with no subscript indicates an equilibrium concentrationc. The assumptionsi. [H3O+] from the autoionization of H2O is negligibleii. A weak acid/base has a small Ka/b and its dissociation is negligible. [HA]≈[HA]initII. The relationship between Ka and pKaa. pKa/b=-logKa/bb. pKa+pKb=14c. A low pKa/b corresponds to a high Ka/bIII. The Henderson-Hasselbalch Equationa.pH= p Ka+log([base ][acid])i. Base= A- or a weak base like NH3ii. Acid= proton donorIV. Acid-Base Titrationsa. In an acid-base titration, the concentration of an acid (or a base) is determined by neutralizing the acid (or base) with a solution of base (or acid) of known concentrationb. The equivalence point of the reaction occurs when the number of moles of OH- added equals the number of moles of H3O+ originally present, or vice versac. The end point occurs when the indicator changes colori. The indicator should be selected so that its color change occurs at a pH close to that of the equivalence pointV. Curve for a strong acid-base titrationa. The pH increases gradually when excess base has been addedb. The pH rises very rapidly at the equivalence point, which occurs at pH=7.00c. The initial pH is lowVI. Calculating the pH during a strong acid-strong base titrationa. Initial pHi. [H3O+]= [HA]initii. pH=-log[H3O+]b. pH before equivalence pointi. Initial mole H3O+= VacidxMacidii. Mole OH- added= VbasexMbaseiii. Mole H3O+ remaining= (mole H3O+init)-(mole OH-added)iv.+¿H3O¿¿mol H3Oremaining+¿Vacid+Vbase¿c. pH at the equivalence pointi. pH=7.00 for a strong acid-strong base titrationd. pH beyond the equivalence pointi. Initial mole H3O+= VacidxMacidii. Mole OH- added= VbasexMbaseiii. Mole OH-excess remaining= (mole OH-added)-(mole H3O+init)iv.+¿H3O¿¿molOHexcess−¿Vacid+Vbase¿v. pOH=-log[OH-] and
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