### Table of pH Values for Various Solutions#### pH MeasurementsThe 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 |#### Explanation1. **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.

The question is based on the concept of salt hydrolysis. we have to write expression of equilibrium…

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

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

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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