What is the pH of a 0.680 M solution of Ca(NO2)2 (Ka of HNO, is 4.5 x 104)?

Chemistry
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Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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Chapter1: Chemical Foundations
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**Question:**
What is the pH of a 0.680 M solution of Ca(NO₂)₂ (Ka of HNO₂ is 4.5 × 10⁻⁴)?

**Explanation:**
This question involves calculating the pH of a solution containing calcium nitrite, which is a salt derived from a weak acid (HNO₂) and a strong base (Ca(OH)₂). The dissociation constant (Ka) is provided to understand the extent to which the weak acid dissociates in solution, which will help determine the pH.

**Key Concepts to Understand:**
1. **pH Calculation:** This involves using the concentration of the solution and the dissociation constant to find the concentration of hydrogen ions, and then using this to calculate the pH.
2. **Hydrolysis of Salts:** Ca(NO₂)₂ will undergo hydrolysis in water, and the nitrite ion (NO₂⁻) will react with water to produce OH⁻, making the solution basic.
3. **Equilibrium Expressions:** Setting up an equilibrium expression using the known values and solving for the unknowns.

This question requires knowledge of chemical equilibrium and acid-base reactions to approach the solution methodically.
Transcribed Image Text:**Question:** What is the pH of a 0.680 M solution of Ca(NO₂)₂ (Ka of HNO₂ is 4.5 × 10⁻⁴)? **Explanation:** This question involves calculating the pH of a solution containing calcium nitrite, which is a salt derived from a weak acid (HNO₂) and a strong base (Ca(OH)₂). The dissociation constant (Ka) is provided to understand the extent to which the weak acid dissociates in solution, which will help determine the pH. **Key Concepts to Understand:** 1. **pH Calculation:** This involves using the concentration of the solution and the dissociation constant to find the concentration of hydrogen ions, and then using this to calculate the pH. 2. **Hydrolysis of Salts:** Ca(NO₂)₂ will undergo hydrolysis in water, and the nitrite ion (NO₂⁻) will react with water to produce OH⁻, making the solution basic. 3. **Equilibrium Expressions:** Setting up an equilibrium expression using the known values and solving for the unknowns. This question requires knowledge of chemical equilibrium and acid-base reactions to approach the solution methodically.
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