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Lecture15_S08

# Lecture15_S08 - Feb 22 2008 Lecture 15 Chapter 16 Acids and...

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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) I 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 0 M 0 M K b = [H(caffeine) + ][OH - ] [caffeine] C E 0.15 M - x (0.15-x) M + x x 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 3 - , ClO 4 - , HSO 4 - Ions of strong acids or bases have zero tendency to reform acid or base conjugate acid cations or high oxidation state metals cations Make solutions Acidic : NH 4 + , CH 3 NH 3 + , Al 3+ , Fe 3+ , Cr 3+ Anions, conjugate bases
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