Lecture15_S08

Lecture15_S08 - Feb. 22, 2008 Lecture 15, Chapter 16: Acids...

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Feb. 22, 2008 Lecture 15, Chapter 16: Acids and Bases Dissolving weak base in pure water only partly ionize H 2 O to give OH - gives basic solution: use K b Base (aq) + H 2 O (l) H(Base) + (aq) + OH- (aq) K b = [H(base) + ][OH - ] [Base] Caffeine is a weak base. A 0.15 M solution has pH = 8.45, find K b caffeine (aq) + H 2 O (l) H(caffeine) + (aq) + OH- (aq) Find pOH from pH, find [OH - ] eq from pOH pOH = 14-pH = 14.00-8.45 = 5.55; [OH-] eq = 10 -pOH = 10 -5.55 = 2.82x10 -6 K b = [H(caffeine) + ][OH - ] [caffeine] I C E 0.15 M - x (0.15-x) M 0 M + x x M 0 M + x x M [OH-] eq = 2.82x10 -6 M = [H(caffeine) + ] eq [caffeine] eq = (0.15 – 2.82x10-6)M = 0.15 M K b = [H(caffeine) + ][OH - ] [caffeine] = (2.86x10 -5 ) 2 (0.15) = 5.30x10 -11
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Most remaining questions will ask to find pH of salt solutions This depends on your ability to recognize whether ions in the salt are acids, bases, or spectators Spectator ions do not affect pH : Li + , Na + , K + , Mg 2+ , Ca 2+ , Ba 2+ Cl - , Br - , I - , NO 3 - , ClO
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This note was uploaded on 04/17/2008 for the course CHEM 2 taught by Professor Ryan during the Spring '08 term at Marquette.

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Lecture15_S08 - Feb. 22, 2008 Lecture 15, Chapter 16: Acids...

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