Downloaded 124 times
![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](https://image.slidesharecdn.com/unit16weakacidequilpt5final-160601181512/75/Chem-2-Acid-Base-Equilibria-V-Weak-Acid-Equilibria-and-Calculating-the-pH-of-a-Weak-Acid-Solution-3-2048.jpg)
![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](https://image.slidesharecdn.com/unit16weakacidequilpt5final-160601181512/75/Chem-2-Acid-Base-Equilibria-V-Weak-Acid-Equilibria-and-Calculating-the-pH-of-a-Weak-Acid-Solution-8-2048.jpg)
![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+]](https://image.slidesharecdn.com/unit16weakacidequilpt5final-160601181512/75/Chem-2-Acid-Base-Equilibria-V-Weak-Acid-Equilibria-and-Calculating-the-pH-of-a-Weak-Acid-Solution-18-2048.jpg)
![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
𝐇𝐍𝐎 𝟐 𝐚𝐪 + 𝐇 𝟐 𝐎 𝐥 ⇌ 𝐇 𝟑 𝐎+ 𝐚𝐪 + 𝐍𝐎 𝟐
−
𝐚𝐪
1.0710-2 M](https://image.slidesharecdn.com/unit16weakacidequilpt5final-160601181512/75/Chem-2-Acid-Base-Equilibria-V-Weak-Acid-Equilibria-and-Calculating-the-pH-of-a-Weak-Acid-Solution-19-2048.jpg)
![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%](https://image.slidesharecdn.com/unit16weakacidequilpt5final-160601181512/75/Chem-2-Acid-Base-Equilibria-V-Weak-Acid-Equilibria-and-Calculating-the-pH-of-a-Weak-Acid-Solution-21-2048.jpg)
Uploaded byLumen Learning
Chem 2 - Acid-Base Equilibria V: Weak Acid Equilibria and Calculating the pH of a Weak Acid Solution
This document discusses calculating the pH of weak acid solutions. It explains that weak acids only partially dissociate into hydronium ions and conjugate base ions according to an equilibrium reaction. The equilibrium constant Ka relates the concentrations of these species. An example problem calculates the pH of a 0.25 M nitric acid solution using its Ka value, finding the hydronium ion concentration to be 1.07 × 10-2 M and the pH to be 1.97. It also calculates the percent dissociation of the weak acid.
In this document
Powered by AI
Introduction to the concept of weak acids vs strong acids, equilibrium in weak acid solutions, and basic calculation methods.
Detailed discussion on the equilibrium constant Ka for weak acids, including formulation and relation.
Example problem demonstrating the calculation of pH for a weak acid solution using HF.
Continuous example calculating pH for a 0.25 M HNO2 solution, detailing ICE table setup and calculations.
Introduction to percent dissociation involving calculations for weak acids, based on HNO2 example.
Brief mention of the upcoming topic on calculating pH and Kb for weak bases.
More Related Content
Similar to Chem 2 - Acid-Base Equilibria V: Weak Acid Equilibria and Calculating the pH of a Weak Acid Solution
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 versusWeak 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 AcidSolutions 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 pHof 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 ConstantKa 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 Kis “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 ConstantKa 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 MHNO2 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 MHNO2 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 MHNO2 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 MHNO2 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 MHNO2 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 MHNO2 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 MHNO2 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 MHNO2 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 pHof 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 𝐇𝐍𝐎 𝟐 𝐚𝐪 + 𝐇 𝟐 𝐎 𝐥 ⇌ 𝐇 𝟑 𝐎+ 𝐚𝐪 + 𝐍𝐎 𝟐 − 𝐚𝐪 1.0710-2 M
- 20. Percent Dissociation What percentof 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 percentof 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 thepH, Kb and Weak Base Equilibria (Pt 6)