lec20_whc_a - Exam Feb. 26; Room Assignments Room 1051 SLB...

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Exam Feb. 26; Room Assignments ALL OTHER SECTIONS ( B01, B03, B05, B08, B10, B11, B13, B14, B18, B19, BR0 ) Meet in Sciences Lecture Hall Edna B06 2076 SLB Edna B02 2068 SLB James B15 2059 SLB James B09 2051 SLB Mateo B16 1076 SLB Mateo B12 1068 SLB Troy B20 1075 SLB Troy B04 1067 SLB Kelly B17 1059 SLB Kelly B07 1051 SLB TA Section Room
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Announcements • Review Session--Wednesday, Feb. 24, 5:-6 pm; 1100 SocSci (Death Star)
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The Common-Ion Effect • This is a manifestation of Le Chatelier's Principle • Affects the pH buffers we covered in the previous lecture. • Also affects equilibrium with a soluble solid.
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Common-Ion Effect The effect on an equilibrium by a second substance that produces ions that can participate in that equilibrium. The added ions are said to be common to the equilibrium. CH 3 CO 2 H + H 2 O H 3 O + + CH 3 CO 2 - HCl + H 2 O H 3 O + + Cl - Common ion Le Chatelier’s Principle
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Add a small amount of strong acid (H 3 O + is the common ion) to an solution with acetate: CH 3 CO 2 + H 3 O + Æ CH 3 CO 2 H + H 2 O drives the reaction to relieve the chemical stress. Acetate concentration decreases, acetic acid increases, but the H 3 O + concentration will have hardly changed .
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Q: Why must a buffer contain weak acid/conj base? A1: Need a lot of available acid to react with added base (or base to react with added acid). p K a + log [HA] pH = [A - ] Weak-acid equilibrium buffers pHand prevents large excusions if a common ion, in this case H 3 O + , is added: [H 3 O + ] [A - ] [HA] K a = HA + H 2 O A - + H 3 O + 1M±HCl Æ pH=0 1 M HOAc / OAc- Æ pH~4.7 0.01 M HCl Æ pH=2 0.01 M HOAc / OAc- Æ pH~4.7 0.0001 M HCl Æ pH=4 0.0001 M HOAc / OAc- Æ pH~4.7
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Henderson-Hasselbalch Equation p K a + log [acid] pH = [conjugate base] For this to be convenient, need to be able to use initial molarities and not have to set up an equilibrium table! Æ Need [acid] and [conj base] to be both much larger than K a , (i.e. “x” will be small Æ ignore changes)! 0.1 < [HA] < 10 [A - ] [A - ] > 100 × K a and [HA] > 100 × K a Æ To keep pH within ±1: For the buffer to work acid and conjugate base must have similar molarities!:
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What is the pH of 500 mL of an acetate buffer solution that contains 0.5 M of NaOAc and 1.0 M HOAc when 2 g of NaOH are added (pK a = 4.75). Example – pH in Buffer Solution Strong base NaOH: 2g / (40.0 g/mol) / 0.5L = 0.1 M HOAc + OH- Æ OAc - + H 2 O Initial 1.0 M 0 0.5 M Strong base: 0.1 M (acid in excess, OH - is limiting) Change: -0.1M -0.1M +0.1M Final: 0.9 M 0 0.6 M p K a + log = 4.75 + log 0.6/0.9 = 4.6 [HOAc] pH = [OAc-]
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Add to acetate buffer: 0.04 M NaOAc and 0.08 M HOAc (K a = 1.8 x 10 –5 ).
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This note was uploaded on 06/13/2010 for the course CHEM 995940767 taught by Professor Topadakis during the Spring '10 term at UC Davis.

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lec20_whc_a - Exam Feb. 26; Room Assignments Room 1051 SLB...

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