Chem 2 - Acid-Base Equilibria V: Weak Acid Equilibria and Calculating the pH of a Weak Acid Solution

1. Acid-Base Equilibria (Pt. 5)
Weak Acid Equilibria and Ka-
Calculating the pH of a Weak
Acid 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 Acids
Strong acids dissociate completely in
solution.
𝐇𝐂𝐥 𝐚𝐪 + 𝐇 𝟐 𝐎 𝐥 → 𝐇 𝟑 𝐎+
𝐚𝐪 + 𝐂𝐥−
(𝐚𝐪)
Weak acids only partially dissociate in
solution.
𝐇𝐅 𝐚𝐪 + 𝐇 𝟐 𝐎 𝐥 ⇌ 𝐇 𝟑 𝐎+
𝐚𝐪 + 𝐅−
(𝐚𝐪)

3. Recall: Weak Acid Solutions
Suppose 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

4. 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?

5. 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)
conjugate baseweak acid

6. 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 𝟏
Recall heterogeneous
equilibria… the
activity for pure
liquids and solids is
“1”

7. 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
𝐇𝐅 𝐚𝐪 + 𝐇 𝟐 𝐎 𝐥 ⇌ 𝐇 𝟑 𝐎+
𝐚𝐪 + 𝐅−
𝐚𝐪
𝐊 𝐚 =
𝐇 𝟑 𝐎+
𝐅−
𝐇𝐀
= 𝟑. 𝟓 × 𝟏𝟎−𝟒

8. 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+]
𝐇𝐀 𝐚𝐪 + 𝐇 𝟐 𝐎 𝐥 ⇌ 𝐇 𝟑 𝐎+
𝐚𝐪 + 𝐀−
𝐚𝐪
𝐊 𝐚 =
𝐇 𝟑 𝐎+
𝐀−
𝐇𝐀

9. 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

10. 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
𝐇𝐍𝐎 𝟐 𝐚𝐪 + 𝐇 𝟐 𝐎 𝐥 ⇌ 𝐇 𝟑 𝐎+
𝐚𝐪 + 𝐍𝐎 𝟐
−
𝐚𝐪

11. A 0.25 M HNO2 solution is prepared. The Ka for
HNO2 is 4.6  10-4. Calculate the pH of this
solution.
I
C
E
Example Problem: Calculate the pH of a
Weak Acid Solution
𝐇𝐍𝐎 𝟐 𝐚𝐪 + 𝐇 𝟐 𝐎 𝐥 ⇌ 𝐇 𝟑 𝐎+
𝐚𝐪 + 𝐍𝐎 𝟐
−
𝐚𝐪

12. 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
I
C
E
0.25 0 0
𝐇𝐍𝐎 𝟐 𝐚𝐪 + 𝐇 𝟐 𝐎 𝐥 ⇌ 𝐇 𝟑 𝐎+
𝐚𝐪 + 𝐍𝐎 𝟐
−
𝐚𝐪

13. 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
I
C
E
+ x
0 00.25
 x + x
𝐇𝐍𝐎 𝟐 𝐚𝐪 + 𝐇 𝟐 𝐎 𝐥 ⇌ 𝐇 𝟑 𝐎+
𝐚𝐪 + 𝐍𝐎 𝟐
−
𝐚𝐪

14. 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
I
C
E
+ x
0 00.25
 x + x
0.25  x xx
𝐇𝐍𝐎 𝟐 𝐚𝐪 + 𝐇 𝟐 𝐎 𝐥 ⇌ 𝐇 𝟑 𝐎+
𝐚𝐪 + 𝐍𝐎 𝟐
−
𝐚𝐪

15. 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 =
𝐇 𝟑 𝐎+
𝐍𝐎 𝟐
−
𝐇𝐍𝐎 𝟐
=
𝐱 ∙ 𝐱
𝟎. 𝟐𝟓 − 𝐱
=
𝐱 𝟐
𝟎. 𝟐𝟓 − 𝐱
𝐇𝐍𝐎 𝟐 𝐚𝐪 + 𝐇 𝟐 𝐎 𝐥 ⇌ 𝐇 𝟑 𝐎+
𝐚𝐪 + 𝐍𝐎 𝟐
−
𝐚𝐪

16. 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
=
𝐱 𝟐
𝟎. 𝟐𝟓 − 𝐱
𝐇𝐍𝐎 𝟐 𝐚𝐪 + 𝐇 𝟐 𝐎 𝐥 ⇌ 𝐇 𝟑 𝐎+
𝐚𝐪 + 𝐍𝐎 𝟐
−
𝐚𝐪
Solve for x

17. Solving for x (assuming x is negligible)
𝟒. 𝟔 × 𝟏𝟎−𝟒
=
𝐱 𝟐
𝟎. 𝟐𝟓 − 𝐱
Because Ka is small, x is
very small.
Assume x is zero to
simplify the calculation.
𝟒. 𝟔 × 𝟏𝟎−𝟒
=
𝐱 𝟐
𝟎. 𝟐𝟓 − 𝟎
=
𝐱 𝟐
𝟎. 𝟐𝟓
𝟒. 𝟔 × 𝟏𝟎−𝟒
=
𝐱 𝟐
𝟎. 𝟐𝟓
𝟒. 𝟔 × 𝟏𝟎−𝟒
𝟎. 𝟐𝟓 = 𝐱 𝟐

18. Solving for x (assuming x is negligible)
𝟒. 𝟔 × 𝟏𝟎−𝟒
𝟎. 𝟐𝟓 = 𝐱 𝟐
𝟏. 𝟏𝟓 × 𝟏𝟎−𝟒
= 𝐱 𝟐
𝟏. 𝟏𝟓 × 𝟏𝟎−𝟒
𝟏
𝟐 = 𝐱 𝟐
𝟏
𝟐
𝟏. 𝟎𝟕 × 𝟏𝟎−𝟐
= 𝐱
x is the [H3O+]

19. 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 [1.0710-2 ] = 1.97
0.25  1.0710-2
= 0.2393 M
1.0710-2 M
𝐇𝐍𝐎 𝟐 𝐚𝐪 + 𝐇 𝟐 𝐎 𝐥 ⇌ 𝐇 𝟑 𝐎+ 𝐚𝐪 + 𝐍𝐎 𝟐
−
𝐚𝐪
1.0710-2 M

20. Percent Dissociation
What percent of the weak acid is
dissociated?
𝐇𝐍𝐎 𝟐 𝐚𝐪 + 𝐇 𝟐 𝐎 𝐥 ⇌ 𝐇 𝟑 𝐎+
𝐚𝐪 + 𝐍𝐎 𝟐
−
𝐚𝐪
% 𝐃𝐢𝐬𝐬𝐨𝐜𝐢𝐚𝐭𝐢𝐨𝐧 =
𝐂𝐨𝐧𝐜𝐞𝐧𝐭𝐫𝐚𝐭𝐢𝐨𝐧 𝐨𝐟 𝐡𝐲𝐝𝐫𝐨𝐧𝐢𝐮𝐦
𝐂𝐨𝐧𝐜𝐞𝐧𝐭𝐫𝐚𝐭𝐢𝐨𝐧 𝐨𝐟 𝐭𝐡𝐞 𝐰𝐞𝐚𝐤 𝐚𝐜𝐢𝐝
× 𝟏𝟎𝟎

21. Percent Dissociation
What percent of the weak acid is
dissociated?
𝐇𝐍𝐎 𝟐 𝐚𝐪 + 𝐇 𝟐 𝐎 𝐥 ⇌ 𝐇 𝟑 𝐎+
𝐚𝐪 + 𝐍𝐎 𝟐
−
𝐚𝐪
% 𝐃𝐢𝐬𝐬𝐨𝐜𝐢𝐚𝐭𝐢𝐨𝐧 =
𝐂𝐨𝐧𝐜𝐞𝐧𝐭𝐫𝐚𝐭𝐢𝐨𝐧 𝐨𝐟 𝐡𝐲𝐝𝐫𝐨𝐧𝐢𝐮𝐦
𝐂𝐨𝐧𝐜𝐞𝐧𝐭𝐫𝐚𝐭𝐢𝐨𝐧 𝐨𝐟 𝐭𝐡𝐞 𝐰𝐞𝐚𝐤 𝐚𝐜𝐢𝐝
× 𝟏𝟎𝟎
% 𝐃𝐢𝐬𝐬𝐨𝐜 =
[𝐇 𝟑 𝐎+]
[𝐇𝐍𝐎 𝟐]
× 𝟏𝟎𝟎 =
[𝟏. 𝟎𝟕 × 𝟏𝟎−𝟐]
[𝟎. 𝟐𝟓]
× 𝟏𝟎𝟎 = 𝟒. 𝟐𝟖%

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