Chem_208_Lecture_17_BW - 40.0 12.36 50.0 12.52 Buffer...

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1 Today: Finish Titrations (17-4) Lewis Acids/Bases (16.9) Thurs, 3/27: Lewis Acids/Bases Solubility Equilibria (Chapter 18) Question #5(b): . . . 300 mL of KOH (not HCl) has been added PS7 due by 2:00 Fri, 3/28 PS8 available Thurs, 3/27
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2 Titration of 25.0 mL of 0.10 M HOCl with 0.10 M NaOH (K a = 2.9 x 10 –8 for HOCl) In any titration of an acid by 0.10 M NaOH, each 1.0 mL of the added NaOH will deliver 0.10 mmol of OH . Each 0.10 mmol of OH will neutralize (100%) 0.10 mmol of acid, be it H 3 O + or HX
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3 Titration of 25.0 mL of 0.10 M HOCl with 0.10 M NaOH (K a = 2.9 x 10 –8 for HOCl) mL NaOH added pH 0.0 4.27 1.0 6.16 5.0 6.94 10.0 7.36 12.5 7.54 15.0 7.72 20.0 8.14 22.0 8.41 24.0 8.92 25.0 10.12 26.0 11.29 30.0 11.96 35.0 12.22
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Unformatted text preview: 40.0 12.36 50.0 12.52 Buffer Region Weak acid Excess Base Equivalence Point 4 Titration of a weak acid using NaOH pH > 7 at equivalence point (moles base = moles acid) Good indicator choice (Equiv. pt. = Endpoint) Poor indicator choice (Endpoint reached BEFORE equiv. pt) Titration of 25.0 ml of 0.100 M Acetic Acid 5 Titration of Polyprotic acids Consider the titration of 20.0 mL of 0.10 M H 3 PO 4 with 0.10 M NaOH Recall K a 1 = 7.1 x 10 3 ; pK a 1 = 2.15 K a 2 = 6.3 x 10 6 ; pK a 2 = 7.20 K a 3 = 4.2 x 10 13 ; pK a 3 = 12.38 Again, each 1.0 mL of 0.10 M NaOH added will neutralize 0.1 mmol of acid in solution....
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Chem_208_Lecture_17_BW - 40.0 12.36 50.0 12.52 Buffer...

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