EqulibriumPractice Solutions

EqulibriumPractice Solutions - pKa 10 Equilibrium Favors:...

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Equilibrium Practice Problems SOLUTIONS P A R T : 1. OH N O N H pKa 10 pKa 10 Equilibrium Favors: neither side (i.e. equal amts of reactants + products) – pKa’s equal K eq g167 10 (10-10) = 10 0 = 1 2. OH 2 OH pKa g167 0 pKa 50 Equilibrium Favors: right side K eq g167 10 (50-0) = 10 50 3. O O H 2 O O O H 3 O pKa 10 pKa 0 Equilibrium Favors: left side K eq g167 10 (0-10) = 10 -10
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4. H pKa 40 pKa 25 Equilibrium Favors: left side K eq g167 10 (25-40) = 10 -15 5. OH N O N H pKa 16
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Unformatted text preview: pKa 10 Equilibrium Favors: left side K eq g167 10 (10-16) = 10-6 6. O O O O O O pKa 10 pKa 20 Equilibrium Favors: right side K eq g167 10 (20-10) = 10 10 P A R T B : 1. O N O H N 2. O OH O OH O O 3. OH NH 2 N O NH 2 H N 4. O O CH 3 O O O CH 3 OH CENTRAL THEME: If you have a strong enough base, the MOST acidic proton on the acid gets pulled off FIRST...
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EqulibriumPractice Solutions - pKa 10 Equilibrium Favors:...

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