Lab Report #3-2-3: 0.1 M NaC2H3O2 solution, using the given pH data, write expression for equ constant (Ka or Kb): Kb= [C2H3O2'] [H*] / [HC2H3O2] Ka= [C2H3O2] [H*] / [HC2H3O2] Ka = [HC2H3O2] [OH]/ [C2H3O2] Kb = [HC2H3O2] [OH¯] / [C2H3O2]
Lab Report #3-2-3: 0.1 M NaC2H3O2 solution, using the given pH data, write expression for equ constant (Ka or Kb): Kb= [C2H3O2'] [H*] / [HC2H3O2] Ka= [C2H3O2] [H*] / [HC2H3O2] Ka = [HC2H3O2] [OH]/ [C2H3O2] Kb = [HC2H3O2] [OH¯] / [C2H3O2]
Chemistry
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Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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Question
Transcribed Image Text:### Table of pH Values for Various Solutions
#### pH Measurements
The table below shows the pH values of various solutions, indicating the acidity or basicity of each solution. The pH scale ranges from 0 to 14, with lower values indicating higher acidity, 7 being neutral, and higher values indicating higher alkalinity (basicity).
| Solutions | pH Values |
|-------------------|-----------|
| H₂O (unboiled) | 3.5 |
| H₂O (boiled) | 7.0 |
| NaCl | 7.0 |
| NaC₂H₃O₂ | 9.1 |
| NH₄Cl | 4.5 |
| NaHCO₃ | 9.5 |
| Na₃PO₄ | 11.9 |
| Na₂CO₃ | 11.0 |
#### Explanation
1. **H₂O (unboiled)**: The pH of unboiled water is 3.5, indicating slight acidity. This may be due to the presence of dissolved carbon dioxide forming carbonic acid.
2. **H₂O (boiled)**: The pH of boiled water is neutral (7.0). Boiling removes dissolved gases, neutralizing the water.
3. **NaCl**: Sodium chloride solution has a neutral pH of 7.0.
4. **NaC₂H₃O₂**: The pH of sodium acetate solution is 9.1, indicating it is basic (alkaline).
5. **NH₄Cl**: Ammonium chloride solution has a pH of 4.5, indicating it is acidic.
6. **NaHCO₃**: Sodium bicarbonate (baking soda) has a pH of 9.5, indicating it is basic.
7. **Na₃PO₄**: Sodium phosphate solution is strongly basic with a pH of 11.9.
8. **Na₂CO₃**: Sodium carbonate solution also shows strong alkalinity with a pH of 11.0.
Using this table, students can understand the pH levels of different solutions and how different compounds affect the acidity or basicity of solutions.
![**Lab Report #3-2-3:**
**0.1 M NaC₂H₃O₂ solution, using the given pH data, write expression for equilibrium constant (Ka or Kb):**
1. \( \text{Kb} = \frac{[\text{C₂H₃O₂}^-][\text{H}^+]}{[\text{HC₂H₃O₂}]} \)
2. \( \text{Ka} = \frac{[\text{C₂H₃O₂}^-][\text{H}^+]}{[\text{HC₂H₃O₂}]} \)
3. \( \text{Ka} = \frac{[\text{HC₂H₃O₂}][\text{OH}^-]}{[\text{C₂H₃O₂}^-]} \)
4. \( \text{Kb} = \frac{[\text{HC₂H₃O₂}][\text{OH}^-]}{[\text{C₂H₃O₂}^-]} \)
Select the correct expression for the equilibrium constant based on the given pH data and the dissociation of the 0.1 M NaC₂H₃O₂ solution.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F6a70e2d2-d641-435a-b212-d4266cff05fc%2Fb69d1a87-2c4f-41b6-9bbd-bd612b8366b7%2F62g5tk_processed.png&w=3840&q=75)
Transcribed Image Text:**Lab Report #3-2-3:**
**0.1 M NaC₂H₃O₂ solution, using the given pH data, write expression for equilibrium constant (Ka or Kb):**
1. \( \text{Kb} = \frac{[\text{C₂H₃O₂}^-][\text{H}^+]}{[\text{HC₂H₃O₂}]} \)
2. \( \text{Ka} = \frac{[\text{C₂H₃O₂}^-][\text{H}^+]}{[\text{HC₂H₃O₂}]} \)
3. \( \text{Ka} = \frac{[\text{HC₂H₃O₂}][\text{OH}^-]}{[\text{C₂H₃O₂}^-]} \)
4. \( \text{Kb} = \frac{[\text{HC₂H₃O₂}][\text{OH}^-]}{[\text{C₂H₃O₂}^-]} \)
Select the correct expression for the equilibrium constant based on the given pH data and the dissociation of the 0.1 M NaC₂H₃O₂ solution.
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