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Monday, February 23, 2009

Monday, February 23, 2009 - Monday 1M Ac in flask Add 0.25M...

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Monday, February 23, 2009
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1M Ac - in flask Add 0.25M H + Ac - + H + HAC H + absorbed strongly 1- 0.25 0+ 0.25 K c = 10 -6 = {[H + ][Ac - ]/[HAc] = [H + ]{[Ac - ]/[HAc] = [H + ] {0.75/.25} [H + ]3 [H + ] = 1/3 x10 -6 0.25M acid in but [H + ] changes in 10 -6 range
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Add 0.5M acid initial final Ac - x 1 -0.5 0.5 HAc 0 +0.5 0.5 10 -6 = [H + ] {0.5/0.5} [ H + ] = 10 -6 increased from 1/3 x10 -6 to 1x10 -6 by adding 0.25 more H + .
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Add 0.75M H + Ac - 1 - 0.75 = 0.25M HAc 0 + 0.75 = 0.75M 10 -6 = [H + ] {0.25/0.75} = 1/3 [H + ] [H + ] = 3x10 -6 small H + change in 10 -6 M (pH6) range
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Can do same for HAc 2M HAc in flask titrate with OH - HAc + OH - → Ac - +HOH K a way Add 0.5M OH - HAc 2 -0.5 = 1.5M Ac - 0 +0.5 = 0.5M 10 -6 = [H + ]{ [Ac - ]/[HAc]} = [H + ] [0.5/1.5] [H + ] = 3x10 -6 Remember at start 2M HAC → xH + + xAc - X = 1.4x10 -3 = [H + ] = [Ac - ] base pulled it down quickly at start
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• Alternative K b Ac - + HOH → HAc + OH - • K b = [HAc][OH - ]/[Ac - ] Notice • K a = [H + ][Ac - ]/ [HAc] K b = [HAc][OH - ]/[Ac - ] • K a K b = [H][Ac]/[HAC] [HAc][OH]/Ac =[H + ][OH - ] = 10 -14
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If K a = 10 -6 then 10 -6 K b = 10 -14 K b = 10 -8 pK a = -log 10 K a = 6 pK b = -log 10 K b = 8 • pK a + pK b = 14
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K b = 10 -8 Repeat problem 2M HAc becomes 1.5MHAc when 0.5M OH - added; 0M Ac - becomes 0.5M Ac -
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