AcidBaseEquil - Acid Base Equilibria Acid Chapter 17 Topics...

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Acid Base Equilibria Acid Base Equilibria Chapter 17
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11/01/11 CHEM 106, Acid-Base Equilibria 2 Topics of Discussion Topics of Discussion Solutions of a weak acid and base Acid-Ionization Equilibria Polyprotic Acids Base-Ionization Equilibria Acid-base Properties of Salt Solutions Solutions of Weak Acid/Base with Another Solute Common-Ion Effect Buffers Acid-Base Titration Curves
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11/01/11 CHEM 106, Acid-Base Equilibria 3 Weak acid Weak acid HA + H 2 O H 3 O + + A - K a is acid dissociation constant HA H + + A - a C 2 3 C 2 3 K 0]K [H [HA] ] ][A 0 [H K 0] [HA][H ] ][A 0 [H = = = - + - + a K [HA] ] ][A [H = - +
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11/01/11 CHEM 106, Acid-Base Equilibria 4 Some weak acids Some weak acids N a m e F o r m u l a K a p K a I o d i c a c i d H I O 3 0 . 1 7 0 . 2 3 N i t r o u s a c i d H N O 2 7 . 1 x 1 0 - 4 3 . 1 5 H y d r o f l u o r i c a c i d H F 6 . 8 x 1 0 - 4 3 . 1 7 F o r m i c a c i d H C O O H 1 . 8 x 1 0 - 4 3 . 7 4 A c e t i c a c i d C H 3 C O O H 1 . 8 x 1 0 - 5 4 . 7 4 H y p o c h l o r o u s a c i d H O C l 3 . 0 x 1 0 - 8 7 . 5 2
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11/01/11 CHEM 106, Acid-Base Equilibria 5 Weak Base Weak Base B + H 2 0 BH + + OH - NH 3 + H 2 O NH 4 + + OH - b K [B] ] ][OH [BH = - + b 3 4 K ] [NH ] ][OH [NH = - +
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11/01/11 CHEM 106, Acid-Base Equilibria 6 Some weak bases Some weak bases N a m e F o r m u l a K b p K b M e t h y l a m i n e C H 3 N H 2 4 . 4 x 1 0 - 4 3 . 3 6 H y d r a z i n e N 2 H 4 1 . 7 x 1 0 - 6 5 . 7 7 A m m o n i a N H 3 1 . 8 x 1 0 - 5 4 . 7 4 H y d r o x i l a m i n e H O N H 2 6 . 6 x 1 0 - 9 8 . 1 8 A n i l i n e C 6 H 5 N H 2 4 . 4 x 1 0 - 1 0 9 . 3 6
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11/01/11 CHEM 106, Acid-Base Equilibria 7 Equilibrium calculations Equilibrium calculations K a and K b from percentage ionization %ionization = (amount ionized)/(amount available)x100 Example In a 0.01 M solution of butyric acid the acid is 4 % ionized at 20 0 C. Calculate K a and pK a for butyric acid at this temperature.
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11/01/11 CHEM 106, Acid-Base Equilibria 8 Solution Solution HBu + H 2 O H 3 O + Bu - H 2 O + H B u H 3 O + + B u - I n i t i a l 0 . 0 1 0 0 C h a n g e - 0 . 0 1 x 0 . 0 4 + 0 . 0 0 0 4 + 0 . 0 0 0 4 E q u i l i b r i u m 0 . 0 0 9 9 6 0 . 0 0 0 4 0 . 0 0 0 4 K a = (0.0004) 2 /(0.00996) = 1.6x10 -5 pK a = 4.8
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11/01/11 CHEM 106, Acid-Base Equilibria 9 Equilibrium calculations Equilibrium calculations K a and K b from initial concentrations and pH Example In a 0.1 M solution of formic acid, the pH is 2.38 at 25 0 C. Calculate the K a and pK a for formic acid at this temperature. HCOOH + H 2 O HCOO - + H 3 O +
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11/01/11 CHEM 106, Acid-Base Equilibria 10 Solution Solution [H 3 O + ] = 10 -2.38 = 0.0042 M H 2 O + H C O O H H 3 O + + H C O O - I n i t i a l 0 . 1 0 0 C h a n g e - 0 . 0 0 4 2 + 0 . 0 0 4 2 + 0 . 0 0 4 2 e q u i l i b r i u m 0 . 0 9 6 0 . 0 0 4 2 0 . 0 0 4 2 [HCOOH] ] O ][H [HCOO K 3 a + - = K a = 1.84x10 -4
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11/01/11 CHEM 106, Acid-Base Equilibria 11 Simplifying assumption Simplifying assumption HA H + + A - Accuracy of calculations may be set within 5% [H + ]/[HA] initial < 0.05 [H + ] 2 /[HA] initial = K a [HA] initial > K a 400 [HA] equilibrium = [HA] initial - x (x is negligible) [HA] equilibrium ~ [HA] initial
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11/01/11 CHEM 106, Acid-Base Equilibria 12 Example Example What is the pH of 0.010 M solution of dimethyl amine, (CH 3 ) 2 NH? K b = 9.6x10 -4 (CH 3 ) 2 NH + H 2 O (CH 3 ) 2 NH 2 + + OH - Is simplification possible?
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