What is the pH of a solution of 0.300 M HNO2 containing 0.210 M NANO2? (Ka of HNO2 is 4.5 x 10-4)

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### Determining the pH of a Solution Containing HNO₂ and NaNO₂

**Problem Statement:**
What is the pH of a solution of 0.300 M HNO₂ containing 0.210 M NaNO₂? (Ka of HNO₂ is 4.5 × 10⁻⁴)

**Solution Steps:**

1. **Identify the Components:**
   - HNO₂ (Nitrous Acid) is a weak acid with a known dissociation constant (Ka = 4.5 × 10⁻⁴).
   - NaNO₂ (Sodium Nitrite) dissociates completely in water to give Na⁺ and NO₂⁻ ions.
   
2. **Set Up the ICE Table:**

   | Species  | Initial Concentration | Change in Concentration | Equilibrium Concentration |
   |----------|-----------------------|-------------------------|---------------------------|
   | HNO₂     | 0.300 M               | -x                      | 0.300 - x                 |
   | H⁺       | 0 (negligible)        | +x                      | x                         |
   | NO₂⁻     | 0.210 M               | +x                      | 0.210 + x                 |

3. **Write the Equilibrium Expression:**
   \[
   Ka = \frac{[H⁺][NO₂⁻]}{[HNO₂]}
   \]
   \[
   4.5 × 10⁻⁴ = \frac{x(0.210 + x)}{0.300 - x}
   \]

4. **Simplify the Expression:**
   Assuming \( x \) is very small compared to 0.300 M and 0.210 M:
   \[
   4.5 × 10⁻⁴ \approx \frac{x (0.210)}{0.300}
   \]
   
5. **Solve for \( x \):**
   \[
   x = \frac{4.5 × 10⁻⁴ \times 0.300}{0.210}
   \]
   \[
   x = 6.42857 × 10⁻⁴ \approx 6.43 × 10⁻⁴
   \]

6. **Calculate the pH:**
   \[
   \text{
Transcribed Image Text:### Determining the pH of a Solution Containing HNO₂ and NaNO₂ **Problem Statement:** What is the pH of a solution of 0.300 M HNO₂ containing 0.210 M NaNO₂? (Ka of HNO₂ is 4.5 × 10⁻⁴) **Solution Steps:** 1. **Identify the Components:** - HNO₂ (Nitrous Acid) is a weak acid with a known dissociation constant (Ka = 4.5 × 10⁻⁴). - NaNO₂ (Sodium Nitrite) dissociates completely in water to give Na⁺ and NO₂⁻ ions. 2. **Set Up the ICE Table:** | Species | Initial Concentration | Change in Concentration | Equilibrium Concentration | |----------|-----------------------|-------------------------|---------------------------| | HNO₂ | 0.300 M | -x | 0.300 - x | | H⁺ | 0 (negligible) | +x | x | | NO₂⁻ | 0.210 M | +x | 0.210 + x | 3. **Write the Equilibrium Expression:** \[ Ka = \frac{[H⁺][NO₂⁻]}{[HNO₂]} \] \[ 4.5 × 10⁻⁴ = \frac{x(0.210 + x)}{0.300 - x} \] 4. **Simplify the Expression:** Assuming \( x \) is very small compared to 0.300 M and 0.210 M: \[ 4.5 × 10⁻⁴ \approx \frac{x (0.210)}{0.300} \] 5. **Solve for \( x \):** \[ x = \frac{4.5 × 10⁻⁴ \times 0.300}{0.210} \] \[ x = 6.42857 × 10⁻⁴ \approx 6.43 × 10⁻⁴ \] 6. **Calculate the pH:** \[ \text{
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