Calculate the pH of a 0.53 M solution of sodium hydrogen sulfite, NaHSO3 (Ka = 6.0 x 10-8). pH =

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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!
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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