CHEM 102 1nd Edition Lecture 30Outline of Last Lecture I. Characteristic Features of Weak Acid-Strong BaseII. Calculation of Solution pHOutline of Current Lecture I. Calculation of Solution pHII. Choosing an IndicatorsCurrent LectureI. Calculation of Solution pH- At the equivalence point when 100.0 mL of 0.100 M NaOH have reacted with 100.0 mL of 0.100 M CH3COOH.CH3COOH NaOH NaCH3COO H2OInitial0.0100 mol 0.0100 mol 0 molReaction-0.0100 mol -0.0100 mol +0.0100 molFinal0 mol 0 mol + 0.0100 mol The salt has basic properties Salt will hydrolyze to make the resulting solution basic M NacH3COO- = 0.0100 mol/0.200 L solution = 0.050 M- Now to determine pH determine the extent of salt hydrolysisCH3COO- H2O CH3COOH OH-Initial0.05 M ~0 M ~0MReaction- x M + x M + x MFinal(0.05 – x) M X M X M Kb = Kw/Ka = 1.1e-14/1.8e-5 = 5.56e-10 5.56e-10 = x2/0.05 – x; x = 5.27e-6 = [OH-] pOH = -log(5.27e-6) = 5.28 pH = 14 – pOH = 14 - 5.28 = 8.72- Past the equivalence point – after addition of 120.0 mL of 0.100 M NaOH to 100.0 mL of 0.100 M CH3COOH After the equivalence point, the excess NaOH added will determine the pH of the solutionCH3COOH NaOH NaCH3COO H2OThese 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.Initial0.0100 0.0120 0Reaction-0.0100 -0.0100 +0.0100Final0 0.0020 0.0100 MNaOH = 0.0020 mol NaOH/0.220 L = 0.00909 M NaOH [OH-] = 0.00909 M; pOH = 2.04; pH = 11.96II. How to Choose an Indicator- The pKa of the indicator acid is about in the middle of its color range- The pKa should be as close as possible to the pH of the equivalence point of the titration- The pKa of indicator should equal the pH at the equivalence point- Indicator should change colors at the equivalence
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