Calculate the pH and the equilibrium concentrations of HSeO3 and SeO32- in a 0.0769 M selenious acid solution, H₂SO3 (aq). For H₂SeO3, Ka1 = 2.4x10-³ and K₂2 = 4.8x10-⁹
Calculate the pH and the equilibrium concentrations of HSeO3 and SeO32- in a 0.0769 M selenious acid solution, H₂SO3 (aq). For H₂SeO3, Ka1 = 2.4x10-³ and K₂2 = 4.8x10-⁹
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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![**Title: Calculating pH and Equilibrium Concentrations in Selenious Acid Solutions**
**Objective:**
To calculate the pH and equilibrium concentrations of \( \text{HSeO}_3^- \) and \( \text{SeO}_3^{2-} \) in a 0.0769 M selenious acid (\( \text{H}_2\text{SeO}_3 \)) aqueous solution.
**Given Data:**
- Initial concentration of \( \text{H}_2\text{SeO}_3 \): 0.0769 M
- \( \text{K}_a1 \) for \( \text{H}_2\text{SeO}_3 \): \( 2.4 \times 10^{-3} \)
- \( \text{K}_a2 \) for \( \text{H}_2\text{SeO}_3 \): \( 4.8 \times 10^{-9} \)
**Tasks:**
1. Calculate the pH of the solution.
2. Determine the equilibrium concentration of \( [\text{HSeO}_3^-] \) in M.
3. Determine the equilibrium concentration of \( [\text{SeO}_3^{2-}] \) in M.
**Blank Sections for Solutions:**
- **pH =**
- \( [\text{HSeO}_3^-] = \) M
- \( [\text{SeO}_3^{2-}] = \) M
---
**Instructions:**
Using the provided \( \text{K}_a \) values and the initial concentration of selenious acid, apply the principles of chemical equilibrium to solve for the missing values. Consider each dissociation step separately and account for the contribution of hydrogen ions to the solution’s pH.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fb6d50391-92c3-428f-a052-8385218d41f0%2Fdf3f7991-16c0-450c-bde1-58b587f634a9%2Fyx7j8ua_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Title: Calculating pH and Equilibrium Concentrations in Selenious Acid Solutions**
**Objective:**
To calculate the pH and equilibrium concentrations of \( \text{HSeO}_3^- \) and \( \text{SeO}_3^{2-} \) in a 0.0769 M selenious acid (\( \text{H}_2\text{SeO}_3 \)) aqueous solution.
**Given Data:**
- Initial concentration of \( \text{H}_2\text{SeO}_3 \): 0.0769 M
- \( \text{K}_a1 \) for \( \text{H}_2\text{SeO}_3 \): \( 2.4 \times 10^{-3} \)
- \( \text{K}_a2 \) for \( \text{H}_2\text{SeO}_3 \): \( 4.8 \times 10^{-9} \)
**Tasks:**
1. Calculate the pH of the solution.
2. Determine the equilibrium concentration of \( [\text{HSeO}_3^-] \) in M.
3. Determine the equilibrium concentration of \( [\text{SeO}_3^{2-}] \) in M.
**Blank Sections for Solutions:**
- **pH =**
- \( [\text{HSeO}_3^-] = \) M
- \( [\text{SeO}_3^{2-}] = \) M
---
**Instructions:**
Using the provided \( \text{K}_a \) values and the initial concentration of selenious acid, apply the principles of chemical equilibrium to solve for the missing values. Consider each dissociation step separately and account for the contribution of hydrogen ions to the solution’s pH.
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