Acid-Base Equilibria (Pt. 6)

Weak Base Equilibria and Kb- Calculating pH and pOH for a

Weak Base SolutionBy 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.

Weak bases only partially react in solution.

Recall: Brønsted-Lowry Bases

Brønsted-Lowry bases accept protons H+.

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.

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.

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

𝐊 𝐛=[𝐎𝐇− ] [𝐇𝐍𝐎𝟐 ]

[𝐍𝐎𝟐− ]

=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]

𝐊 𝐛=[𝐎𝐇− ] [𝐇𝐍𝐎𝟐 ]

[𝐍𝐎𝟐− ]

=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

𝐍𝐎𝟐− (𝐚𝐪 )+𝐇𝟐𝐎 (𝐥 )⇌𝐎𝐇− (𝐚𝐪 )+𝐇𝐍𝐎𝟐(𝐚𝐪)

A 0.25 M NO2 solution is prepared. The Kb for

NO2 is 2.2 10-11. Calculate the pH of this

solution.

ICE

Example Problem: Calculate the pH of a Weak Base Solution

𝐍𝐎𝟐− (𝐚𝐪 )+𝐇𝟐𝐎 (𝐥 )⇌𝐎𝐇− (𝐚𝐪 )+𝐇𝐍𝐎𝟐(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

ICE

0.25 0 0𝐍𝐎𝟐

− (𝐚𝐪 )+𝐇𝟐𝐎 (𝐥 )⇌𝐎𝐇− (𝐚𝐪 )+𝐇𝐍𝐎𝟐(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

ICE

+ x

0 00.25 x +

x

𝐍𝐎𝟐− (𝐚𝐪 )+𝐇𝟐𝐎 (𝐥 )⇌𝐎𝐇− (𝐚𝐪 )+𝐇𝐍𝐎𝟐(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

ICE

+ x

0 00.25 x +

x0.25 x xx

𝐍𝐎𝟐− (𝐚𝐪 )+𝐇𝟐𝐎 (𝐥 )⇌𝐎𝐇− (𝐚𝐪 )+𝐇𝐍𝐎𝟐(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− ] [𝐇𝐍𝐎𝟐 ]

[𝐍𝐎𝟐− ]

= 𝐱 ∙𝐱𝟎 .𝟐𝟓−𝐱= 𝐱𝟐

𝟎 .𝟐𝟓−𝐱

𝐍𝐎𝟐− (𝐚𝐪 )+𝐇𝟐𝐎 (𝐥 )⇌𝐎𝐇− (𝐚𝐪 )+𝐇𝐍𝐎𝟐(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= 𝐱𝟐

𝟎 .𝟐𝟓−𝐱

𝐍𝐎𝟐− (𝐚𝐪 )+𝐇𝟐𝐎 (𝐥 )⇌𝐎𝐇− (𝐚𝐪 )+𝐇𝐍𝐎𝟐(𝐚𝐪)

Solve for x

Solving for x (assuming x is negligible)

𝟐 .𝟐×𝟏𝟎−𝟏𝟏= 𝐱𝟐

𝟎 .𝟐𝟓−𝐱

Because Kb is small, x is very small. Assume x is zero to simplify the calculation.

𝟐 .𝟐×𝟏𝟎−𝟏𝟏= 𝐱𝟐

𝟎 .𝟐𝟓−𝟎=𝐱𝟐

𝟎 .𝟐𝟓

𝟐 .𝟐×𝟏𝟎−𝟏𝟏= 𝐱𝟐

𝟎 .𝟐𝟓 (𝟐 .𝟐×𝟏𝟎−𝟏𝟏)𝟎 .𝟐𝟓=𝐱𝟐

Solving for x (assuming x is negligible)

(𝟐 .𝟐×𝟏𝟎−𝟏𝟏)𝟎 .𝟐𝟓=𝐱𝟐

𝟓 .𝟓×𝟏𝟎−𝟏𝟐=𝐱𝟐

(𝟓 .𝟓×𝟏𝟎−𝟏𝟐)𝟏𝟐=(𝐱𝟐 )

𝟏𝟐

𝟐 .𝟑𝟓×𝟏𝟎−𝟔=𝐱x 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 [2.3510-6 ] = 5.63We still need the pH!

0.25 2.3510-

6

~ 0.25 M

2.3510-6 M

𝐍𝐎𝟐− (𝐚𝐪 )+𝐇𝟐𝐎 (𝐥 )⇌𝐎𝐇− (𝐚𝐪 )+𝐇𝐍𝐎𝟐(𝐚𝐪)

2.3510-6 M

Recall: Relationships Between pH, pOH, and pKw

pH + pOH = pKw

pKw = log [1.01014] = 14 (at 25C)

pH + pOH = 14 (at 25C)

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)