chem126--Acids II Using Ka

chem126--Acids II Using Ka - ACIDS WITH LARGE VALUES OF Ka...

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ACIDS WITH LARGE VALUES OF K a ARE STRONGER THAN ACIDS WITH SMALLER VALUES OF K a K a = 1.74x10 -5 M EXAMPLE 5 : (a) Find pH of 0.0500 M CH 3 COOH; (b) the concentrations of the species; and (c) % dissociation of acetic acid. (a) Init. 0.05 0 0 Change -x +x +x Eqm. 0.05-x x x 2 5 1.74 10 0.05 x x x = ( ) 25 1.74 10 0.05 x xx = Solve quadratic equation or use method of successive approximations
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METHOD OF SUCCESSIVE APPROXIMATION ( ) 25 1.74 10 0.05 x xx =− x is small with respect to 0.05. First approximation, x = 0. ( ) 1.74 10 0.05 = x = 9.33 x 10 -4 M For small K a this is sufficient. 2 nd approximation, ( ) 1.74 10 0.05 9.33 10 x −− 4 x = 9.24 x 10 -4 M 3 rd approximation, ( ) 1.74 10 0.05 9.24 10 x 4 x = 9.24 x 10 -4 M
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SHORTCUT METHOD OF SUCCESSIVE APPROXIMATION () 5 1.74 10 0.05 xx x ⇒= ( ) 25 1.74 10 0.05 x =− 0 ENTER ( ) ( ) E- 1.74 5 .05 ANS ENTER ENTER ENTER 0 9.33 E -4 9.24 E -4 9.24 E -4 TI-83 [H 3 O + ] = 9.24 x 10 -4 M Aside, –log(9.33 E -4) = 3.03 pH = -log [H 3 O + ] = -log(9.24 E - 4) = 3.03 [CH 3 COO - ] = 9.24 x 10 -4 M [CH 3 COOH] = 0.05 -x = 0.05 - 9.24 x 10 -4 = 0.0491 M (b) (c) % CH 3 COOH dissociated = ( x/[CH 3 COOH] 0 )(100%) = (9.24 x 10 -4 /0.05)(100%) = 1.8%
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% Dissociation of Acetic Acid as a Function of Its Initial Concentration Technique of example 5 works for [CH 3 COOH] 0 > 2x10 -5 M Why not for smaller concentrations? McQuarrie ( ) 55 5 1 5 1.74 10 2.0 10 1.9 1 ;1 . 1 8 1 0 0 x xM xxx x x x −− = = −→ = % CH 3 COOH dissociated = ( x/[CH 3 COOH] 0 )(100%) = (1.18 x 10 -5 /2x10 -5 )(100%) = 59 %
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p K a Defined p K a = -log K a = -log(1.74 E - 5) = 4.76 for acetic acid p K a = -log K a McQuarrie
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BASES WITH LARGE VALUES OF K b ARE STRONGER THAN BASES WITH SMALLER VALUES OF K b McQuarrie
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This note was uploaded on 04/07/2008 for the course CHEM 126 taught by Professor Biolsi during the Spring '08 term at NJIT.

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chem126--Acids II Using Ka - ACIDS WITH LARGE VALUES OF Ka...

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