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

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Chem 2 - Acid-Base Equilibria VI: Weak Base Equilibria and Kb - Calculating pH and pOH for a Weak Base Solution

  • 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)