**Determining the pH of a Sulfurous Acid Solution**Sulfurous acid (\(H_2SO_3\)) is a diprotic acid with dissociation constants \(K_{a1} = 1.39 \times 10^{-2}\) and \(K_{a2} = 6.73 \times 10^{-8}\). Determine the pH of a 0.232 M sulfurous acid (\(H_2SO_3\)) solution.\[ \text{pH} = \underline{\hspace{5cm}} \] **Explanation**:- **Diprotic Acid**: Sulfurous acid can donate two protons. The first dissociation is stronger and contributes more significantly to the pH.- **Dissociation Constants**: \(K_{a1}\) indicates the equilibrium constant for the first dissociation, while \(K_{a2}\) is for the second.- **Concentration**: The initial concentration of sulfurous acid is 0.232 M.To find the pH, the contribution from the first ionization is primarily considered because \(K_{a1} \gg K_{a2}\). Use the formula for calculating pH from the concentration of \(H^+\) ions derived from the dissociation equilibrium.

Answered: Sulfurous acid (H, SO,) is a diprotic…

Sulfurous acid (H, SO,) is a diprotic acid with Kal = 1.39 × 10-2 and K2 = 6.73 x 10–8. Determine the pH of a 0.232 M sulfurous acid (H, SO,) solution. pH =

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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Question

**Determining the pH of a Sulfurous Acid Solution**

Sulfurous acid (\(H_2SO_3\)) is a diprotic acid with dissociation constants \(K_{a1} = 1.39 \times 10^{-2}\) and \(K_{a2} = 6.73 \times 10^{-8}\). Determine the pH of a 0.232 M sulfurous acid (\(H_2SO_3\)) solution.

\[ \text{pH} = \underline{\hspace{5cm}} \]

**Explanation**:
- **Diprotic Acid**: Sulfurous acid can donate two protons. The first dissociation is stronger and contributes more significantly to the pH.
- **Dissociation Constants**: \(K_{a1}\) indicates the equilibrium constant for the first dissociation, while \(K_{a2}\) is for the second.
- **Concentration**: The initial concentration of sulfurous acid is 0.232 M.

To find the pH, the contribution from the first ionization is primarily considered because \(K_{a1} \gg K_{a2}\). Use the formula for calculating pH from the concentration of \(H^+\) ions derived from the dissociation equilibrium.

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