Chem 2 - Acid-Base Equilibria VII: Conjugate Acid/Base Pairs and Relationships Between Ka, Kb, and Kw

1. Acid-Base Equilibria (Pt. 7)
Conjugate Acid/Base Pairs
and Relationships Between
Ka, Kb, and Kw
By Shawn P. Shields, Ph.D.
This work is licensed by Dr. Shawn P. Shields-Maxwell under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0
International License.

2. Recall: Conjugate Acid-Base Pairs
Conjugate acid-base pairs are two
related species differing only by a
proton (H+).
𝐇𝐅 𝐚𝐪 + 𝐇 𝟐 𝐎 𝐥 ⇌ 𝐇 𝟑 𝐎+
𝐚𝐪 + 𝐅−
(𝐚𝐪)
𝐍𝐇 𝟑 𝐚𝐪 + 𝐇 𝟐 𝐎 𝐥 ⇌ 𝐍𝐇 𝟒
+
𝐚𝐪 + 𝐎𝐇−
(𝐚𝐪)
base conjugate acid for NH3
acid conjugate
base for HF

3. The equilibrium
constant K is
“renamed” for
acids to Ka
The Equilibrium Constant Ka for Weak Acids
An equilibrium exists between the weak
acid (HA) and its products.
𝐇𝐀 𝐚𝐪 + 𝐇 𝟐 𝐎 𝐥 ⇌ 𝐇 𝟑 𝐎+
𝐚𝐪 + 𝐀−
(𝐚𝐪)
𝐊 𝐚 =
𝐇 𝟑 𝐎+
𝐀−
𝐇𝐀

4. Ka and Kb for Conjugate Acid-Base Pairs
Suppose acetic acid (CH3COOH) is dissolved in
water…
𝐂𝐇 𝟑 𝐂𝐎𝐎𝐇 𝐚𝐪 + 𝐇 𝟐 𝐎 𝐥 ⇌ 𝐂𝐇 𝟑 𝐂𝐎𝐎−
𝐚𝐪 + 𝐇 𝟑 𝐎+
𝐚𝐪
acetic acid conjugate base
for acetic acid
“loseable” H+

5. Ka and Kb for Conjugate Acid-Base Pairs
Suppose acetate (CH3COO-) is dissolved in
water…
𝐂𝐇 𝟑 𝐂𝐎𝐎−
+ 𝐇 𝟐 𝐎 𝐥 ⇌ 𝐂𝐇 𝟑 𝐂𝐎𝐎𝐇 𝐚𝐪 + 𝐎𝐇−
(𝐚𝐪)
acetate conjugate acid
for acetate
acetate gained H+
from water

6. Ka and Kb for Conjugate Acid-Base Pairs
CH3COOH and CH3COO- are conjugate
acid-base pairs
acetic acid acetate (base)
Ka = 1.76  10-5 Kb = 5.68  10-10

7. Multiply Ka and Kb for Conjugate
Acid-Base Pairs
𝐊 𝐚 ∙ 𝐊 𝐛= 𝟏. 𝟕𝟔 × 𝟏𝟎−𝟓
[𝟓. 𝟔𝟖 × 𝟏𝟎−𝟏𝟎
]
𝐊 𝐚 ∙ 𝐊 𝐛= 𝟏 × 𝟏𝟎−𝟏𝟒
= 𝐊 𝐖
𝐊 𝐚 ∙ 𝐊 𝐛 = 𝐊 𝐖
Use this relationship to interconvert
between Ka and Kb.

8. Calculating (Converting Between)
Ka and Kb using Kw
𝐊 𝐚 ∙ 𝐊 𝐛 = 𝐊 𝐖
𝐊 𝐚 ∙ 𝐊 𝐛= 𝟏. 𝟎 × 𝟏𝟎−𝟏𝟒
= 𝐊 𝐖
𝐊 𝐚 =
𝐊 𝐖
𝐊 𝐛
𝐊 𝐛 =
𝐊 𝐖
𝐊 𝐚
Use KW to interconvert between Ka and Kb.

9. Calculating (Converting Between)
Ka and Kb using Kw
𝐊 𝐚 ∙ 𝐊 𝐛= 𝟏. 𝟕𝟔 × 𝟏𝟎−𝟓
[𝟓. 𝟔𝟖 × 𝟏𝟎−𝟏𝟎
]
𝐊 𝐚 ∙ 𝐊 𝐛= 𝟏. 𝟎 × 𝟏𝟎−𝟏𝟒
= 𝐊 𝐖
𝐊 𝐚 =
𝟏.𝟎×𝟏𝟎−𝟏𝟒
𝟓.𝟔𝟖×𝟏𝟎−𝟏𝟎 𝐊 𝐛 =
𝟏.𝟎×𝟏𝟎−𝟏𝟒
𝟏.𝟕𝟔×𝟏𝟎−𝟓
Use KW to interconvert between Ka and Kb.

10. pKa and pKb
Calculate the pKa for an acid as
pKa =  log Ka
Calculate the pKb for a base as
pKb =  log Kb

11. Calculating pKa and pKb
Calculate the pKa pKa =  log Ka
𝐩𝐊 𝐚 = −𝐥𝐨𝐠 𝟏. 𝟕𝟔 × 𝟏𝟎−𝟓
= 𝟒. 𝟕𝟓
Calculate the pKb pKb =  log Kb
𝐩𝐊 𝐛 = −𝐥𝐨𝐠 𝟓. 𝟔𝟖 × 𝟏𝟎−𝟏𝟎
= 𝟗. 𝟐𝟓

12. Inverse logs for pKa, pKa, and pKw
pKa =  log Ka 10pKa = Ka
pKb =  log Kb 10pKb = Kb
pKw =  log Kw = 14 10pKw = Kw
Recall: Kw (and pKw) will have different
values at temperatures other than 25C

13. A Few More Relationships Between
pKa, pKb, and pKw
pKa + pKb = pKw
pKw =  log [Kw] = 14 (at 25C)
pKa + pKb = 14 (at 25C)

14. Next up,
The Conjugate See-Saw and
Analyzing Ka and Kb for Acid or
Base Relative Strength
(Pt. 8)