Acid-Base Equilibria (Pt. 5)

Weak Acid Equilibria and Ka- Calculating the pH of a Weak

Acid 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 Acids

Strong acids dissociate completely in solution.

Weak acids only partially dissociate in solution.

Recall: Weak Acid SolutionsSuppose a 1.0 M solution of HF (a weak acid) is prepared.What is the concentration of hydronium (H3O+) in solution? We can’t do this by inspection. An equilibrium between the weak acid and the products exists.

We need to find [H3O+] by calculating the equilibrium concentrations of HF, H3O+, and F-

H3O+

F

HF

Calculating the pH of a Weak Acid Solution

An equilibrium exists between the weak acid and its products.

We can use the relationship between the value of the equilibrium constant K and the initial concentration of weak acid in solution. HOW?

The Equilibrium Constant Ka for Weak Acids

An equilibrium exists between the weak acid (HA) and its products.

conjugate base

weak acid

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.

𝐊 𝐚=¿¿ Recall heterogeneous equilibria… the activity for pure liquids and solids is “1”

Example: The Equilibrium Constant Ka for HF

Ka is called the “acid dissociation constant.”The value of Ka for HF is 3.5 10-4

𝐊 𝐚=¿¿

ICE Tables, Ka, and Calculating pH for a Weak Acid Solution

Use Ka and an ICE table to determine the [H3O+] at equilibrium.

Calculate the pH using the equilibrium [H3O+]

𝐊 𝐚=¿¿

A 0.25 M HNO2 solution is prepared.

The Ka for HNO2 is 4.6 10-4.

Calculate the pH of this solution.

Example Problem: Calculate the pH of a Weak Acid Solution

A 0.25 M HNO2 solution is prepared. The Ka for HNO2 is 4.6 10-4. Calculate the pH of this solution.

The first step… Write the chemical equation for the weak acid equilibrium.

Example Problem: Calculate the pH of a Weak Acid Solution

𝐇𝐍𝐎𝟐 (𝐚𝐪 )+𝐇𝟐𝐎 (𝐥 )⇌𝐇𝟑𝐎+¿ (𝐚𝐪 )+𝐍𝐎𝟐− (𝐚𝐪 ) ¿

A 0.25 M HNO2 solution is prepared. The Ka for HNO2 is 4.6 10-4. Calculate the pH of this solution.

ICE

Example Problem: Calculate the pH of a Weak Acid Solution

𝐇𝐍𝐎𝟐 (𝐚𝐪 )+𝐇𝟐𝐎 (𝐥 )⇌𝐇𝟑𝐎+¿ (𝐚𝐪 )+𝐍𝐎𝟐− (𝐚𝐪 ) ¿

A 0.25 M HNO2 solution is prepared. The Ka for HNO2 is 4.6 10-4. Calculate the pH of this solution.

Example Problem: Calculate the pH of a Weak Acid Solution

ICE

0.25 0 0𝐇𝐍𝐎𝟐 (𝐚𝐪 )+𝐇𝟐𝐎 (𝐥 )⇌𝐇𝟑𝐎+¿ (𝐚𝐪 )+𝐍𝐎𝟐− (𝐚𝐪 ) ¿

A 0.25 M HNO2 solution is prepared. The Ka for HNO2 is 4.6 10-4. Calculate the pH of this solution.

Example Problem: Calculate the pH of a Weak Acid Solution

ICE

+ x

0 00.25 x +

x

𝐇𝐍𝐎𝟐 (𝐚𝐪 )+𝐇𝟐𝐎 (𝐥 )⇌𝐇𝟑𝐎+¿ (𝐚𝐪 )+𝐍𝐎𝟐− (𝐚𝐪 ) ¿

A 0.25 M HNO2 solution is prepared. The Ka for HNO2 is 4.6 10-4. Calculate the pH of this solution.

Example Problem: Calculate the pH of a Weak Acid Solution

ICE

+ x

0 00.25 x +

x0.25 x xx

𝐇𝐍𝐎𝟐 (𝐚𝐪 )+𝐇𝟐𝐎 (𝐥 )⇌𝐇𝟑𝐎+¿ (𝐚𝐪 )+𝐍𝐎𝟐− (𝐚𝐪 ) ¿

A 0.25 M HNO2 solution is prepared. The Ka for HNO2 is 4.6 10-4. Calculate the pH of this solution.

Example Problem: Calculate the pH of a Weak Acid Solution

E 0.25 x xx

𝐊 𝐚=¿¿

𝐇𝐍𝐎𝟐 (𝐚𝐪 )+𝐇𝟐𝐎 (𝐥 )⇌𝐇𝟑𝐎+¿ (𝐚𝐪 )+𝐍𝐎𝟐− (𝐚𝐪 ) ¿

A 0.25 M HNO2 solution is prepared. The Ka for HNO2 is 4.6 10-4. Calculate the pH of this solution.

Example Problem: Calculate the pH of a Weak Acid Solution

E 0.25 x xx

𝟒 .𝟔×𝟏𝟎−𝟒= 𝐱𝟐𝟎 .𝟐𝟓−𝐱

𝐇𝐍𝐎𝟐 (𝐚𝐪 )+𝐇𝟐𝐎 (𝐥 )⇌𝐇𝟑𝐎+¿ (𝐚𝐪 )+𝐍𝐎𝟐− (𝐚𝐪 ) ¿

Solve for x

Solving for x (assuming x is negligible)

𝟒 .𝟔×𝟏𝟎−𝟒= 𝐱𝟐𝟎 .𝟐𝟓−𝐱

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

𝟒 .𝟔×𝟏𝟎−𝟒= 𝐱𝟐𝟎 .𝟐𝟓−𝟎=

𝐱𝟐𝟎 .𝟐𝟓

𝟒 .𝟔×𝟏𝟎−𝟒= 𝐱𝟐𝟎 .𝟐𝟓 (𝟒 .𝟔×𝟏𝟎−𝟒 )𝟎 .𝟐𝟓=𝐱𝟐

Solving for x (assuming x is negligible)

(𝟒 .𝟔×𝟏𝟎−𝟒 )𝟎 .𝟐𝟓=𝐱𝟐

𝟏 .𝟏𝟓×𝟏𝟎−𝟒=𝐱𝟐

(𝟏 .𝟏𝟓×𝟏𝟎−𝟒 )𝟏𝟐= (𝐱𝟐 )

𝟏𝟐

𝟏 .𝟎𝟕×𝟏𝟎−𝟐=𝐱x is the [H3O+]

Calculate the pH of the Weak Acid Solution

A 0.25 M HNO2 solution is prepared. The Ka for HNO2 is 4.6 10-4. Calculate the pH of this solution.

pH = log [H3O+] = log [1.0710-2 ] = 1.97

0.25 1.0710-

2

= 0.2393 M

1.0710-2 M

𝐇𝐍𝐎𝟐 (𝐚𝐪 )+𝐇𝟐𝐎 (𝐥 )⇌𝐇𝟑𝐎+¿ (𝐚𝐪 )+𝐍𝐎𝟐− (𝐚𝐪 ) ¿

1.0710-2 M

Percent Dissociation

What percent of the weak acid is dissociated?

𝐇𝐍𝐎𝟐 (𝐚𝐪 )+𝐇𝟐𝐎 (𝐥 )⇌𝐇𝟑𝐎+¿ (𝐚𝐪 )+𝐍𝐎𝟐− (𝐚𝐪 ) ¿

% 𝐃𝐢𝐬𝐬𝐨𝐜𝐢𝐚𝐭𝐢𝐨𝐧=𝐂𝐨𝐧𝐜𝐞𝐧𝐭𝐫𝐚𝐭𝐢𝐨𝐧𝐨𝐟 𝐡𝐲𝐝𝐫𝐨𝐧𝐢𝐮𝐦𝐂𝐨𝐧𝐜𝐞𝐧𝐭𝐫𝐚𝐭𝐢𝐨𝐧𝐨𝐟 𝐭𝐡𝐞𝐰𝐞𝐚𝐤 𝐚𝐜𝐢𝐝 ×𝟏𝟎𝟎

Percent Dissociation

What percent of the weak acid is dissociated?

𝐇𝐍𝐎𝟐 (𝐚𝐪 )+𝐇𝟐𝐎 (𝐥 )⇌𝐇𝟑𝐎+¿ (𝐚𝐪 )+𝐍𝐎𝟐− (𝐚𝐪 ) ¿

% 𝐃𝐢𝐬𝐬𝐨𝐜𝐢𝐚𝐭𝐢𝐨𝐧=𝐂𝐨𝐧𝐜𝐞𝐧𝐭𝐫𝐚𝐭𝐢𝐨𝐧𝐨𝐟 𝐡𝐲𝐝𝐫𝐨𝐧𝐢𝐮𝐦𝐂𝐨𝐧𝐜𝐞𝐧𝐭𝐫𝐚𝐭𝐢𝐨𝐧𝐨𝐟 𝐭𝐡𝐞𝐰𝐞𝐚𝐤 𝐚𝐜𝐢𝐝 ×𝟏𝟎𝟎

% 𝐃𝐢𝐬𝐬𝐨𝐜=¿¿

Next up, Calculating the pH, Kb and Weak Base Equilibria (Pt 6)