Chem 113 1st Edition Lecture 34 Outline of Last Lecture I. Acid-Base TitrationsII. Curve for a strong acid-strong base titrationsIII. Calculating pH during a strong acid-strong base titrationOutline of Current Lecture IV. Strong acid-strong base titrationsV. Curve for a weak acid-strong base titrationVI. Calculating the pH during a weak acid-strong base titrationCurrent LectureI. Strong acid-strong base titrationsa. K=1/Kw= 1.0x1014b. This implies that the reaction goes to completion and the limiting reactant is completely consumedc. pH=7.00 at equivalence point because there is no longer excess H3O+ or OH-II. Curve for a weak acid-strong base titrationa. The initial pH is higher than for the strong acid solutionb. The pH at the equivalence point is >7.00 due to the reaction of the conjugate base with waterc. The pH increases slowly beyond the equivalence pointd. The curve rises gradually in the buffer or common-ion region; the weak acid and conjugate base are both present in the solutionIII. Calculating the pH during a weak acid-strong base titrationa. Initial pHi. Ka=([H3O+][A-])/[HA]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. [H3O+]=√Ka×[ HA ]initiii. pH=-log[H3O+]b. pH before equivalence pointi. [H3O+]=Ka([HA]/[A]) or pH= pKa+ log([base]/[acid])1. You can use the Henderson-Hasselbalch equation as long as the [base]/[acid] ratio is <1000c. pH at the equivalence pointi. A-(aq)+H2O (l) HA(aq)+OH-(aq)ii. [OH-]=−¿A¿Kb׿√¿1. [A-]= (mol HAinit)/ (Vacid+Vbase) and Kb=Kw/Kaiii. [H3O+]= Kw/−¿A¿Kb׿√¿ and pH=-log[H3O+]d. pH beyond the equivalence pointi. [OH-]=(mol OH-excess)/(Vacid+Vbase)ii.
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