Calculate the pH of a 0.47 M solution of sodium hydrogen sulfite, NaHSO3 (K, = 6.0 x 10). pH = Submit Answer Try Another Version 1 item attempt remaining
Calculate the pH of a 0.47 M solution of sodium hydrogen sulfite, NaHSO3 (K, = 6.0 x 10). pH = Submit Answer Try Another Version 1 item attempt remaining
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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![### pH Calculation Problem
**Problem Statement:**
Calculate the pH of a 0.47 M solution of sodium hydrogen sulfite, \( \text{NaHSO}_3 \) ( \( K_a = 6.0 \times 10^{-8} \) ).
**Field for Answer:**
`pH = __________`
**Buttons:**
- Submit Answer
- Try Another Version (1 item attempt remaining)
**Explanation:**
You are required to calculate the pH of a 0.47 M solution of sodium hydrogen sulfite (NaHSO₃) based on its given acid dissociation constant (\( K_a \)).
1. **Identify the species involved:**
Sodium hydrogen sulfite (\( \text{NaHSO}_3 \)) dissociates into sodium ions (\( \text{Na}^+ \)) and hydrogen sulfite ions (\( \text{HSO}_3^- \)) in solution.
2. **Establish the equilibrium expression:**
\( \text{HSO}_3^- \) can further dissociate:
\[
\text{HSO}_3^- \leftrightarrow \text{H}^+ + \text{SO}_3^{2-}
\]
3. **Expression for \( K_a \):**
\[
K_a = 6.0 \times 10^{-8} = \frac{[\text{H}^+][\text{SO}_3^{2-}]}{[\text{HSO}_3^-]}
\]
4. **Approach for calculation:**
- Assume the initial concentration of \( \text{HSO}_3^- \) is 0.47 M.
- Let the change in concentration \( [\text{H}^+] = x \).
5. **Solve the equation:**
Set up the equilibrium expression using the initial concentration and change:
\[
6.0 \times 10^{-8} \approx \frac{x^2}{0.47}
\]
Solve for \( x \) (representing \( [\text{H}^+] \)), then compute the pH using \( \text{pH} = -\log[\text{H}^+] \).
**Tips:**
- Ensure to check assumptions made during the simplification.
- Provide intermediate steps](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F848b3b37-0e47-4626-8c62-5eaa520ec29b%2F52da3da3-2da7-42e9-a154-98a9c2de58c9%2F3dla3nh.jpeg&w=3840&q=75)
Transcribed Image Text:### pH Calculation Problem
**Problem Statement:**
Calculate the pH of a 0.47 M solution of sodium hydrogen sulfite, \( \text{NaHSO}_3 \) ( \( K_a = 6.0 \times 10^{-8} \) ).
**Field for Answer:**
`pH = __________`
**Buttons:**
- Submit Answer
- Try Another Version (1 item attempt remaining)
**Explanation:**
You are required to calculate the pH of a 0.47 M solution of sodium hydrogen sulfite (NaHSO₃) based on its given acid dissociation constant (\( K_a \)).
1. **Identify the species involved:**
Sodium hydrogen sulfite (\( \text{NaHSO}_3 \)) dissociates into sodium ions (\( \text{Na}^+ \)) and hydrogen sulfite ions (\( \text{HSO}_3^- \)) in solution.
2. **Establish the equilibrium expression:**
\( \text{HSO}_3^- \) can further dissociate:
\[
\text{HSO}_3^- \leftrightarrow \text{H}^+ + \text{SO}_3^{2-}
\]
3. **Expression for \( K_a \):**
\[
K_a = 6.0 \times 10^{-8} = \frac{[\text{H}^+][\text{SO}_3^{2-}]}{[\text{HSO}_3^-]}
\]
4. **Approach for calculation:**
- Assume the initial concentration of \( \text{HSO}_3^- \) is 0.47 M.
- Let the change in concentration \( [\text{H}^+] = x \).
5. **Solve the equation:**
Set up the equilibrium expression using the initial concentration and change:
\[
6.0 \times 10^{-8} \approx \frac{x^2}{0.47}
\]
Solve for \( x \) (representing \( [\text{H}^+] \)), then compute the pH using \( \text{pH} = -\log[\text{H}^+] \).
**Tips:**
- Ensure to check assumptions made during the simplification.
- Provide intermediate steps
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