CHEM 102 1nd Edition Lecture 28Outline of Last Lecture I. TitrationsII. Acid-Base IndicatorsIII. IndicatorsIV. Titration CurvesOutline of Current Lecture I. Features of Strong Acid-Base pH CurvesII. Calculate of Solution pHIII. StepsCurrent LectureI. Features of Strong Acid-Base pH Curves- Initial pH from [H+] of strong acid- pH changes gradually until titration approaches equivalence point- excess strong acid determines the solution pH- Near the equivalence point, there is a dramatic pH increase- Past the equivalence point, there is a gradual pH rise againII. Calculation of Solution pH- After addition of 20.0 mL of 0.100 M NaOH to 100.0 mL of 0.100 M HCl Set up equation: HCl + NaOH NaCl + H20HCl NaOH NaCl H2OInitial 0.0100 0.0020 0Reaction -0.0020 -0.0020 +0.0020After Reaction 0.0080 0 0.0020 The pH will be determined by the excess HCl. M HCl = 0.0080/0.120L = 0.067 M HCl = [H3O+] pH = -log 0.067 = 1.17- When 100.0 mL of 0.100 M HCl have reacted with 100.0 mL of 0.100 M NaOH At equivalence point, all HCl has reacted with NaOHHCl NaOH NaCl H2OInitial 0.0100 0.0100 0Reaction -0.0100 -0.0100 +0.0100These 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.After Reaction 0 0 0.0100 Since NaCl is a neutral salt, the resulting solution will be neutral pH = 7III. Steps - React A & B, in moles- The reaction will be placed in a table such as those above- Change moles on the final line to molarity by dividing by the total volume- Get pH from the final line If buffer is present, use Henderson If SA or SB exist, get pH from the SA or SB If only the salt is left, Neutral salt = 7 If the salt is acidic or basic, get the pH from an ICF table and use the WA or WB, ion and water Use the Ka or Kb for the
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