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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Calculate the pH of a 0.53 M solution of sodium hydrogen sulfite, NaHSO3(Ka = 6.0 x 10^-8)
![**Calculating the pH of a Sodium Hydrogen Sulfite Solution**
**Problem Statement:**
Calculate the pH of a 0.53 M solution of sodium hydrogen sulfite, NaHSO₃ (\(K_a = 6.0 \times 10^{-8}\)).
\[ \text{pH} = \boxed{\phantom{Insert Your Answer Here}} \]
**Explanation:**
To calculate the pH of the solution, you will need to use the provided concentration and the \(K_a\) value for hydrogen sulfite. The steps generally involve setting up an equilibrium expression based on the dissociation of \(HSO_3^-\), solving for the hydrogen ion concentration \([H^+]\), and then calculating the pH as follows:
1. Write the equilibrium expression for the dissociation of \(HSO_3^-\):
\[HSO_3^- \leftrightarrow H^+ + SO_3^{2-}\]
2. Set up the \(K_a\) expression:
\[K_a = \frac{[H^+][SO_3^{2-}]}{[HSO_3^-]}\]
3. Use the initial concentration of the solution and the \(K_a\) value to find \([H^+]\):
\[6.0 \times 10^{-8} = \frac{[H^+]^2}{0.53 - [H^+]}\]
4. Make any necessary approximations to simplify the mathematical solution.
5. Solve for \([H^+]\), then calculate the pH:
\[\text{pH} = -\log[H^+]\]
This explanation will help you solve for the pH using the given concentration and dissociation constant. Enjoy learning!](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fb91a8d0c-ccba-43d9-b564-e4e05a79bb72%2F87db8959-84cd-4d94-8fee-9391ccad29be%2Fww6g0zo_processed.png&w=3840&q=75)
Transcribed Image Text:**Calculating the pH of a Sodium Hydrogen Sulfite Solution**
**Problem Statement:**
Calculate the pH of a 0.53 M solution of sodium hydrogen sulfite, NaHSO₃ (\(K_a = 6.0 \times 10^{-8}\)).
\[ \text{pH} = \boxed{\phantom{Insert Your Answer Here}} \]
**Explanation:**
To calculate the pH of the solution, you will need to use the provided concentration and the \(K_a\) value for hydrogen sulfite. The steps generally involve setting up an equilibrium expression based on the dissociation of \(HSO_3^-\), solving for the hydrogen ion concentration \([H^+]\), and then calculating the pH as follows:
1. Write the equilibrium expression for the dissociation of \(HSO_3^-\):
\[HSO_3^- \leftrightarrow H^+ + SO_3^{2-}\]
2. Set up the \(K_a\) expression:
\[K_a = \frac{[H^+][SO_3^{2-}]}{[HSO_3^-]}\]
3. Use the initial concentration of the solution and the \(K_a\) value to find \([H^+]\):
\[6.0 \times 10^{-8} = \frac{[H^+]^2}{0.53 - [H^+]}\]
4. Make any necessary approximations to simplify the mathematical solution.
5. Solve for \([H^+]\), then calculate the pH:
\[\text{pH} = -\log[H^+]\]
This explanation will help you solve for the pH using the given concentration and dissociation constant. Enjoy learning!
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