Chem 220 Sep26b - Monmouth College

[Pages:2]Chem 220 ? Lecture 9 (9/26) ? Chemical Equilibrium- Ka Kb .

Conjugate Acid/Conjugate Base Pair

CH3COOH + CH3NH2

CH3COO-

+ CH3NH3+

Aceitic acid is an Acid since it donates a proton; the acetate ion is a conjugate base because it

can accept a proton to become aceitic acid

CH3COOH + H2O

CH3COO- + H3O+ Ka = 1.75 x 10-5

CH3COO- + H2O

CH3COOH + OH- Kb = Kw/Ka = 5.71 x 10-10

Methylamine is a Base because it accepts a proton; methlyammonium ion is a conjugate acid

because it can donate a proton to become methylamine

CH3NH2 + H2O

CH3NH3+ + OH- Kb = 4.4 x 10-4

CH3NH3+ + H2O

CH3NH2 + H3O+ Ka = Kw/Kb = 2.3 x 10-11

Therefore, Kw = Ka*Kb

Equilibrium and Acid/Base Chemistry

1. Strong Acids and Strong Bases completely dissociate i.e. large K a. [H+] = [HA]before dissociation since HA completely dissociates b. pH = -log[H+]

2. We can write K for weak acids and bases a. [H+] not equal to [HA] b. need to find [H+] with K c. pH = -log[H+]

REMEMBER:

If you have [H+] and need [OH-] then [OH-] = 1.0 x 10-14/[H+]

If you have [OH-] and need [H+] then [H+] = 1.0 x 10-14/[OH-]

If you have pH and need [H+] then [H+] = 10-pH

If you have pOH and need [OH-] then [OH-] = 10-pOH

pH + pOH = 14

Kw = Ka x Kb

Ka = Kw/Kb Kb= Kw/Ka Kw = 1.0 x 10-14

Ka = weak acid; acetic acid; carboxylic acids

1. Weak acid

conjugate base + H+

2. HA + H2O

A- + H3O+

3. Ka = [A-][H3O+] ; [H+] ................
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