CHEM 122 11th Edition Lecture 28 Outline of Last LectureI. Percent IonizationII. Calculations of Weak Acid and Base SolutionsIII. Polytropic acidsOutline of Current Lecture I. Calculating polyprotic AcidsII. Properties of Solutions of SaltsCurrent LectureCalculating Polyprotic Acids:If Ka1 is more that 10^3 greater than Ka2, then we can treat the acid as a simple monoprotic acid. 2 reasons why:- Due to the difference in the Ka’s the first ionization is a stronger acid than the second.- Due to le Chanteliers principle the H+ produced in the first acid dissociation forces the second acid dissociation to the left.- [H+]equil.= [H+]first + [H+]second- if [H+]first>>> [H+]second then [H+]equil.= [H+]firstProperties of Solutions of Salts:- Salt: an ionic compoundo Excludes anions with OH- or O2-o Excludes cations with H+- Acid Base or Neutral?o Find the ions of the dissolved saltso Alkali metal ions do not react with water therefore they do not effect the pHo For Cations, add an H+ and decide if it becomes a strong base or weak base. Weak base has a conjugate weak acid, creating an acidic solutiono For Anions, add an H+ ion and decide if it becomes a strong acid or weak acid.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. Strong base dissociates completely so it would be neutral.o If neither cation nor anion are acidic or basic then the solution is
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