The document discusses acid-base equilibria, focusing on weak base equilibria and the calculation of pH and pOH for weak base solutions. It explains the differences between strong and weak bases, the Brønsted-Lowry definition, and how to use the base dissociation constant (Kb) to find the equilibrium concentrations in solutions. An illustrative example is provided using a 0.25 M NO2− solution with Kb = 2.2 × 10^-11, including step-by-step calculations of pH and pOH.

1. Acid-Base Equilibria (Pt. 6)
Weak Base Equilibria and Kb-
Calculating pH and pOH for a
Weak Base Solution
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: Strong versus Weak Bases
Strong bases dissociate completely in
solution to produce OH.
𝐍𝐚𝐎𝐇 𝐚𝐪 → 𝐎𝐇−
𝐚𝐪 + 𝐍𝐚+
(𝐚𝐪)
Weak bases only partially react in solution.
𝐍𝐇 𝟑 𝐚𝐪 + 𝐇 𝟐 𝐎 𝐥 ⇌ 𝐍𝐇 𝟒
+
𝐚𝐪 + 𝐎𝐇−
(𝐚𝐪)

3. Recall: Brønsted-Lowry Bases
Brønsted-Lowry bases accept protons H+.
𝐍𝐇 𝟑 𝐚𝐪 + 𝐇 𝟐 𝐎 𝐥 ⇌ 𝐍𝐇 𝟒
+
𝐚𝐪 + 𝐎𝐇−
(𝐚𝐪)
NH3 accepts H+ from water (H2O)
(NH3 acts as a base)
Ammonium (NH4
+) is the conjugate
acid for NH3

4. Calculating the pH of a Weak Base Solution
An equilibrium exists between the weak
base and its products.
We can use the relationship between the
value of the equilibrium constant K and
the initial concentration of weak base in
solution as we did for weak acids.

5. The Equilibrium Constant Kb for Weak Bases
An equilibrium exists between the
weak base (B) and its products.
𝐁−
𝐚𝐪 + 𝐇 𝟐 𝐎 𝐥 ⇌ 𝐎𝐇−
𝐚𝐪 + 𝐇𝐁(𝐚𝐪)
conjugate acid
of the weak
base
weak base

6. The Equilibrium Constant Kb for Weak Bases
An equilibrium exists between the weak base (B)
and its products.
𝐁−
𝐚𝐪 + 𝐇 𝟐 𝐎 𝐥 ⇌ 𝐎𝐇−
𝐚𝐪 + 𝐇𝐁(𝐚𝐪)
The equilibrium
constant K is
“renamed” for
bases to Kb
𝐊 𝐛 =
𝐎𝐇−
𝐇𝐁
𝐁− 𝟏
Recall heterogeneous
equilibria… the
activity for pure
liquids and solids is
“1”

7. Example:
The Equilibrium Constant Kb for NO2

Kb is called the “base dissociation constant.”
The value of Kb for NO2
 is 2.2  10-11
𝐍𝐎 𝟐
−
𝐚𝐪 + 𝐇 𝟐 𝐎 𝐥 ⇌ 𝐎𝐇−
𝐚𝐪 + 𝐇𝐍𝐎 𝟐(𝐚𝐪)
𝐊 𝐛 =
𝐎𝐇−
𝐇𝐍𝐎 𝟐
𝐍𝐎 𝟐
− = 𝟐. 𝟐 × 𝟏𝟎−𝟏𝟏

8. ICE Tables, Kb, and Calculating pH for
a Weak Base Solution
Use Kb and an ICE table to determine the
[OH] at equilibrium. Next, calculate pOH
and use this value to calculate pH.
Calculate the pOH using the equilibrium [OH]
𝐍𝐎 𝟐
−
𝐚𝐪 + 𝐇 𝟐 𝐎 𝐥 ⇌ 𝐎𝐇−
𝐚𝐪 + 𝐇𝐍𝐎 𝟐(𝐚𝐪)
𝐊 𝐛 =
𝐎𝐇−
𝐇𝐍𝐎 𝟐
𝐍𝐎 𝟐
− = 𝟐. 𝟐 × 𝟏𝟎−𝟏𝟏

9. A 0.25 M NO2
 solution is prepared.
The Kb for NO2
 is 2.2  10-11.
Calculate the pH of this solution.
Example Problem: Calculate the pH of a
Weak Base Solution

10. A 0.25 M NO2
 solution is prepared. The Kb for
NO2
 is 2.2  10-11. Calculate the pH of this
solution.
The first step… Write the chemical equation
for the weak base equilibrium.
Example Problem: Calculate the pH of a
Weak Base Solution
𝐍𝐎 𝟐
−
𝐚𝐪 + 𝐇 𝟐 𝐎 𝐥 ⇌ 𝐎𝐇−
𝐚𝐪 + 𝐇𝐍𝐎 𝟐(𝐚𝐪)

11. A 0.25 M NO2
 solution is prepared. The Kb for
NO2
 is 2.2  10-11. Calculate the pH of this
solution.
I
C
E
Example Problem: Calculate the pH of a
Weak Base Solution
𝐍𝐎 𝟐
−
𝐚𝐪 + 𝐇 𝟐 𝐎 𝐥 ⇌ 𝐎𝐇−
𝐚𝐪 + 𝐇𝐍𝐎 𝟐(𝐚𝐪)

12. A 0.25 M NO2
 solution is prepared. The Kb for
NO2
 is 2.2  10-11. Calculate the pH of this
solution.
Example Problem: Calculate the pH of a
Weak Base Solution
I
C
E
0.25 0 0
𝐍𝐎 𝟐
−
𝐚𝐪 + 𝐇 𝟐 𝐎 𝐥 ⇌ 𝐎𝐇−
𝐚𝐪 + 𝐇𝐍𝐎 𝟐(𝐚𝐪)

13. A 0.25 M NO2
 solution is prepared. The Kb for
NO2
 is 2.2  10-11. Calculate the pH of this
solution.
Example Problem: Calculate the pH of a
Weak Base Solution
I
C
E
+ x
0 00.25
 x + x
𝐍𝐎 𝟐
−
𝐚𝐪 + 𝐇 𝟐 𝐎 𝐥 ⇌ 𝐎𝐇−
𝐚𝐪 + 𝐇𝐍𝐎 𝟐(𝐚𝐪)

14. A 0.25 M NO2
 solution is prepared. The Kb for
NO2
 is 2.2  10-11. Calculate the pH of this
solution.
Example Problem: Calculate the pH of a
Weak Base Solution
I
C
E
+ x
0 00.25
 x + x
0.25  x xx
𝐍𝐎 𝟐
−
𝐚𝐪 + 𝐇 𝟐 𝐎 𝐥 ⇌ 𝐎𝐇−
𝐚𝐪 + 𝐇𝐍𝐎 𝟐(𝐚𝐪)

15. A 0.25 M NO2
 solution is prepared. The Kb for
NO2
 is 2.2  10-11. Calculate the pH of this
solution.
Example Problem: Calculate the pH of a
Weak Base Solution
E 0.25  x xx
𝐊 𝐛 =
𝐎𝐇−
𝐇𝐍𝐎 𝟐
𝐍𝐎 𝟐
− =
𝐱 ∙ 𝐱
𝟎. 𝟐𝟓 − 𝐱
=
𝐱 𝟐
𝟎. 𝟐𝟓 − 𝐱
𝐍𝐎 𝟐
−
𝐚𝐪 + 𝐇 𝟐 𝐎 𝐥 ⇌ 𝐎𝐇−
𝐚𝐪 + 𝐇𝐍𝐎 𝟐(𝐚𝐪)

16. A 0.25 M NO2
 solution is prepared. The Kb for
NO2
 is 2.2  10-11. Calculate the pH of this
solution.
Example Problem: Calculate the pH of a
Weak Base Solution
E 0.25  x xx
𝟐. 𝟐 × 𝟏𝟎−𝟏𝟏
=
𝐱 𝟐
𝟎. 𝟐𝟓 − 𝐱
𝐍𝐎 𝟐
−
𝐚𝐪 + 𝐇 𝟐 𝐎 𝐥 ⇌ 𝐎𝐇−
𝐚𝐪 + 𝐇𝐍𝐎 𝟐(𝐚𝐪)
Solve for x

17. Solving for x (assuming x is negligible)
𝟐. 𝟐 × 𝟏𝟎−𝟏𝟏
=
𝐱 𝟐
𝟎. 𝟐𝟓 − 𝐱
Because Kb is small, x is
very small.
Assume x is zero to
simplify the calculation.
𝟐. 𝟐 × 𝟏𝟎−𝟏𝟏
=
𝐱 𝟐
𝟎. 𝟐𝟓 − 𝟎
=
𝐱 𝟐
𝟎. 𝟐𝟓
𝟐. 𝟐 × 𝟏𝟎−𝟏𝟏
=
𝐱 𝟐
𝟎. 𝟐𝟓
𝟐. 𝟐 × 𝟏𝟎−𝟏𝟏
𝟎. 𝟐𝟓 = 𝐱 𝟐

18. Solving for x (assuming x is negligible)
𝟐. 𝟐 × 𝟏𝟎−𝟏𝟏
𝟎. 𝟐𝟓 = 𝐱 𝟐
𝟓. 𝟓 × 𝟏𝟎−𝟏𝟐
= 𝐱 𝟐
𝟓. 𝟓 × 𝟏𝟎−𝟏𝟐
𝟏
𝟐 = 𝐱 𝟐
𝟏
𝟐
𝟐. 𝟑𝟓 × 𝟏𝟎−𝟔
= 𝐱
x is the [OH]

19. Calculate the pOH of the Weak Base
Solution
A 0.25 M NO2
 solution is prepared. The Kb for NO2

is 2.2  10-11. Calculate the pH of this solution.
pOH =  log [OH] =  log [2.3510-6 ] = 5.63
We still need the pH!
0.25  2.3510-6
~ 0.25 M
2.3510-6 M
𝐍𝐎 𝟐
−
𝐚𝐪 + 𝐇 𝟐 𝐎 𝐥 ⇌ 𝐎𝐇− 𝐚𝐪 + 𝐇𝐍𝐎 𝟐(𝐚𝐪)
2.3510-6 M

20. Recall: Relationships Between pH,
pOH, and pKw
pH + pOH = pKw
pKw =  log [1.01014] = 14 (at 25C)
pH + pOH = 14 (at 25C)

21. For the weak base solution
pOH = 5.63
Use the relationship:
Calculate the pH of a Weak Base Solution
using the pOH
𝐩𝐇 + 𝐩𝐎𝐇 = 𝟏𝟒
A basic solution! (pH greater than 7)
𝐩𝐇 = 𝟏𝟒 − 𝐩𝐎𝐇 = 𝟏𝟒 − 𝟓. 𝟔𝟑 = 𝟖. 𝟑𝟕

22. Next up,
Conjugate Acid/Base Pairs and
Relationships Between Ka, Kb,
and Kw.
(Pt 7)