Monday, February 23, 2009

Monday, February 23, 2009 - Monday, February 23, 2009 •...

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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 = 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 • 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 + . • 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 • 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 • 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.

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Monday, February 23, 2009 - Monday, February 23, 2009 •...

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