Chlorous acid, HC1O2, has a K. = 1.2 × 10-2.Calculate the pH when 0.15mol of HC1O2 is dissolved in 1 L of water to make an aqueous solution.

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Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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**Chlorous Acid and pH Calculation**

**Problem Statement:**

Chlorous acid, \( \text{HClO}_2 \), has a \( K_a = 1.2 \times 10^{-2} \). Calculate the pH when 0.15 mol of \( \text{HClO}_2 \) is dissolved in 1 L of water to make an aqueous solution.

**Solution Approach:**

1. **Identify the Chemical Equation:**
   - Dissociation of chlorous acid: 
     \[ \text{HClO}_2 \rightleftharpoons \text{H}^+ + \text{ClO}_2^- \]

2. **Use the Ionization Constant \( K_a \):**
   \[ K_a = \frac{[\text{H}^+][\text{ClO}_2^-]}{[\text{HClO}_2]} \]

3. **Initial Concentration:**
   - \([\text{HClO}_2] = 0.15 \, \text{M}\)
   - \([\text{H}^+] = [\text{ClO}_2^-] = 0\)

4. **Calculate Change in Concentration:**
   - Let \( x \) be the concentration of \( \text{H}^+ \) that dissociates:
   - \( [\text{HClO}_2] = 0.15 - x \)
   - \( [\text{H}^+] = [\text{ClO}_2^-] = x \)

5. **Set Up the Expression and Solve for \( x \):**
   \[ 1.2 \times 10^{-2} = \frac{x^2}{0.15 - x} \]
   
6. **Assume \( x \) is small compared to 0.15 M and solve:**
   \[ 1.2 \times 10^{-2} \approx \frac{x^2}{0.15} \]
   \[ x^2 = 1.2 \times 10^{-2} \times 0.15 \]
   \[ x^2 = 1.8 \times 10^{-3} \]
   \[ x = \sqrt{1.8 \times 10^{-3}} \
Transcribed Image Text:**Chlorous Acid and pH Calculation** **Problem Statement:** Chlorous acid, \( \text{HClO}_2 \), has a \( K_a = 1.2 \times 10^{-2} \). Calculate the pH when 0.15 mol of \( \text{HClO}_2 \) is dissolved in 1 L of water to make an aqueous solution. **Solution Approach:** 1. **Identify the Chemical Equation:** - Dissociation of chlorous acid: \[ \text{HClO}_2 \rightleftharpoons \text{H}^+ + \text{ClO}_2^- \] 2. **Use the Ionization Constant \( K_a \):** \[ K_a = \frac{[\text{H}^+][\text{ClO}_2^-]}{[\text{HClO}_2]} \] 3. **Initial Concentration:** - \([\text{HClO}_2] = 0.15 \, \text{M}\) - \([\text{H}^+] = [\text{ClO}_2^-] = 0\) 4. **Calculate Change in Concentration:** - Let \( x \) be the concentration of \( \text{H}^+ \) that dissociates: - \( [\text{HClO}_2] = 0.15 - x \) - \( [\text{H}^+] = [\text{ClO}_2^-] = x \) 5. **Set Up the Expression and Solve for \( x \):** \[ 1.2 \times 10^{-2} = \frac{x^2}{0.15 - x} \] 6. **Assume \( x \) is small compared to 0.15 M and solve:** \[ 1.2 \times 10^{-2} \approx \frac{x^2}{0.15} \] \[ x^2 = 1.2 \times 10^{-2} \times 0.15 \] \[ x^2 = 1.8 \times 10^{-3} \] \[ x = \sqrt{1.8 \times 10^{-3}} \
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