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MIT OpenCourseWare http://ocw.mit.edu 5.111 Principles of Chemical Science Fall 2008 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms .
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24.1 5.111 Lecture Summary #24 Topics: Oxidation States, and Balancing Oxidation/Reduction Reactions (Read Section K, Chapter 12) FROM MONDAY: Titrations Curves for Weak acid/Strong base and for Weak base/Strong acid V > V eq strong base in water V = V eq (salt) conj. base of weak acid A - A - A - A - V=V half-eq special category of buffer A H A A - A - = H 0 < V < V A H A A A - H H buffer eq Start A H A H A A H H weak acid in water V = 0 V = 0 weak base in water B B B B 0 < V < V buffer eq B + B B B H V=V half eq B + H B + H special category of buffer B B = V = V eq (salt) conj. acid of weak base B + H B + B + B + H H H V > V eq strong acid in water 14 12 10 8 6 7 4 2 0 0 10 20 30 Volume of base added (mL) pH Strong base Weak acid S equivalence point >7 buffering region half-eq point Figure by MIT OpenCourseWare. Figure by MIT OpenCourseWare. 14 12 10 8 6 7 4 2 0 0 10 20 30 Volume of acid added (mL) pH Strong acid Weak base S buffering region equivalence point pH <7
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- - 24.2 Example: Titration of weak acid with strong base 25.0 mL of 0.10 M HCOOH with 0.15 M NaOH (K a = 1.77 x 10 -4 for HCOOH) 1. Volume = 0 mL of NaOH added Before any NaOH is added, the problem is that of an ionization of a weak acid in water. - HCOOH (aq) + H 2 O (l) H 3 O + (aq) + HCO 2 (aq) - HCOOH (aq) H 3 O + (aq) + HCO 2 (aq) initial molarity 0.10 M 0 0 change in molarity -x +x +x__ equilibrium molarity 0.10-x x x K a = 1.77 x 10 -4 = (x) 2 /(0.10-x) ~= (x) 2 /0.10 x = 0.00421 (check 0.00421 is 4.2% of 0.10) okay pH = -log [0.00421] = 2.38 (to how many sig figs?) 2. 0 < V < V eq In this range, the acid has been partly ionized by the strong base (buffering region). Calculate the pH of the solution resulting from the addition of 5.0 mL of 0.15 M NaOH. Because OH
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lecnotes24 - MIT OpenCourseWare http:/ocw.mit.edu 5.111...

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