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Unformatted text preview: Monday, February 23, 2009 • 1M Ac in flask • Add 0.25M H + • Ac + H + ↔ HAC H + absorbed strongly • 1 0.25 0+ 0.25 • K c = 106 = {[H + ][Ac ]/[HAc] = • [H + ]{[Ac ]/[HAc] = [H + ] {0.75/.25} [H + ]3 • [H + ] = 1/3 x106 • 0.25M acid in but [H + ] changes in 106 range • Add 0.5M acid • initial final • Ac x • 1 0.5 0.5 • HAc • 0 +0.5 0.5 • 106 = [H + ] {0.5/0.5} • [ H + ] = 106 increased from 1/3 x106 to 1x106 by adding 0.25 more H + . • Add 0.75M H + • Ac 1  0.75 = 0.25M • HAc 0 + 0.75 = 0.75M • 106 = [H + ] {0.25/0.75} = 1/3 [H + ] • [H + ] = 3x106 • small H + change in 106 M (pH6) range • 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 • 106 = [H + ]{ [Ac ]/[HAc]} = [H + ] [0.5/1.5] • [H + ] = 3x106 • Remember at start 2M HAC → xH + + xAc • X = 1.4x103 = [H + ] = [Ac ] base pulled it down quickly at start • Alternative...
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This note was uploaded on 02/26/2012 for the course CHEM 108 taught by Professor Dr.brennan during the Spring '09 term at Binghamton University.
 Spring '09
 DR.BRENNAN

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