February 18, 2009

February 18, 2009 - • Acid Molecule that donates an H •...

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February 18, 2009
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Multiple Equilibria E + S ES K Step 1 Assign E = 1, K[S] to ES 1 K[S] E ES • f E = 1/{1 +K[S]} f ES = K[S]/{1+k[S]} • [E] = E t f e [ES] = E t f ES
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. For example if K[S] = 2, E t = 2M • f E = 1/{1+2} = 1/3 [E] = 2(1/3) = 2/3 f ES = 2/{1+2} = 2/3 [ES] = 2(2/3) = 4/3 M
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K[S] ES 1 • E ES 2 K[S] K[S]K[S] SE • f E = 1/{1 + 2K[S] + K[S]K[S]} = 1/Sum • F es,SE = 2K[S]/Sum • f ES2 = K[S]K[S]/Sum
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. If K[S] = 4 f E = 1/{1 +2x4 + 4 2 } = 1/25 very little free f ES2 = 16/25 = 64%
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Isomers K 1 K 2 A B C Just equilibrium constant 1 K 1 K 1 K 2 A B C f A = 1/{1+ K 1 + K 1 K 2+ } = 1/Sum’ f B = K 1 /Sum’ f C = K 1 K 2 /Sum
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Dissociation K d1 K d2 H 2 A HA - + H + A 2- + 2H + Becomes 1 2 K d1 /[H + ] K d1 /[H + ] K d2 /[H + ] H 2 A HA - A 2- Clicker problem Determine f H2A when K d1 /[H] = K d2 /[H] = 1
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Arrhenius Acids and Bases Arrhenius acid dissociates in solution to produce H + Arrhenius base dissociates in solution to produce OH - strong acid, base complete dissociation HCl + H 2 O H + (aq) + Cl - (aq) NaOH + H 2 O OH - (aq) + Na + (aq)
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Bronsted Lowry
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Unformatted text preview: • Acid Molecule that donates an H + • Base Molecule that accepts H + • HCl + NH 3 ↔ Cl-+ NH 4 + • Acid Base Base Acid • HCl + NaOH ↔ HOH + Cl-+ Na + • Acid Base Acid Base .Water • HCl + H 2 O ↔ H 3 O + + Cl-• Acid Base Acid Base • Autoionization • H 2 O + H 2 O ↔ H 3 O + + OH-• Acid Base Acid Base Equilibrium • HOH + HOH ↔ H 3 O + + OH-• or for convenience • HOH ↔ H + + OH-• K w = [H + ][OH-] = 10-14 at 25C • [H + ] = 0 + 1x • [OH-] = 0 + 1x 1 – 1 stoichiometry • K w = [H + ][OH-] = x 2 • x = 10-7 M = [H + ] = [OH-] • CO 2 anhydride H 2 CO 3 • Add strong acid value 0.2M • [H + ] = 10-7 + 0.2 ≈ 0.2M • K w = 10-14 = [0.2][OH-] • [OH-] = 10-14 /0.02 = 50x10-14 5x10-13 Dilution warning • Add 1mL of 0.4M HCl to 3mL of water • MV = MV 0.4M(1mL) = xM(3+1) • x = 0.4(1/4) = 0.1M = [H + ]...
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February 18, 2009 - • Acid Molecule that donates an H •...

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