chemistry graduate student is given 450. mL of a 1.10M nitrous acid (HNO,) solution. Nitrous acid is a weak acid with = 4.5 × 10 ". What mass of NaNO, should the student dissolve in the HNO, solution to turn it into a buffer with = 3.18? u may assume that the volume of the solution doesn't change when the NaNO, is dissolved in it. Be sure your answer has unit symbol, and round it to 2 significant digits. Ox10 O

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### Problem Statement: Buffer Solution Preparation

**Objective:**  
Calculate the mass of sodium nitrite (NaNO₂) that should be dissolved in a nitrous acid (HNO₂) solution to achieve a desired pH.

**Given Data:**

- Volume of HNO₂ solution: 450 mL
- Initial concentration of HNO₂ solution: 1.10 M
- Acid dissociation constant (Kₐ) for HNO₂: 4.5 × 10⁻⁴
- Desired pH of the buffer solution: 3.18

**Assumptions:**

- The volume of the solution does not change when NaNO₂ is dissolved.

**Steps for Calculation:**

1. **Henderson-Hasselbalch Equation:**
   To find the required concentration of the conjugate base (NO₂⁻), we use the Henderson-Hasselbalch equation:
   \[
   \text{pH} = \text{p}K_a + \log \left( \frac{[\text{A}^-]}{[\text{HA}]} \right)
   \]
   where:
   - \(\text{pH} = 3.18\)
   - \(\text{p}K_a = - \log (4.5 \times 10^{-4}) = 3.35\)

2. **Rearrange the Equation:**
   Solving for the ratio of conjugate base (NO₂⁻) to acid (HNO₂):
   \[
   3.18 = 3.35 + \log \left( \frac{[NO_2^-]}{[HNO_2]} \right)
   \]
   \[
   \log \left( \frac{[NO_2^-]}{[HNO_2]} \right) = 3.18 - 3.35 = -0.17
   \]
   \[
   \frac{[NO_2^-]}{[HNO_2]} = 10^{-0.17} \approx 0.68
   \]

3. **Find the Concentration of NO₂⁻:**
    \[
    [NO_2^-] = 0.68 \times [HNO_2]
    \]
    Since the initial concentration of HNO₂ is 1.
Transcribed Image Text:### Problem Statement: Buffer Solution Preparation **Objective:** Calculate the mass of sodium nitrite (NaNO₂) that should be dissolved in a nitrous acid (HNO₂) solution to achieve a desired pH. **Given Data:** - Volume of HNO₂ solution: 450 mL - Initial concentration of HNO₂ solution: 1.10 M - Acid dissociation constant (Kₐ) for HNO₂: 4.5 × 10⁻⁴ - Desired pH of the buffer solution: 3.18 **Assumptions:** - The volume of the solution does not change when NaNO₂ is dissolved. **Steps for Calculation:** 1. **Henderson-Hasselbalch Equation:** To find the required concentration of the conjugate base (NO₂⁻), we use the Henderson-Hasselbalch equation: \[ \text{pH} = \text{p}K_a + \log \left( \frac{[\text{A}^-]}{[\text{HA}]} \right) \] where: - \(\text{pH} = 3.18\) - \(\text{p}K_a = - \log (4.5 \times 10^{-4}) = 3.35\) 2. **Rearrange the Equation:** Solving for the ratio of conjugate base (NO₂⁻) to acid (HNO₂): \[ 3.18 = 3.35 + \log \left( \frac{[NO_2^-]}{[HNO_2]} \right) \] \[ \log \left( \frac{[NO_2^-]}{[HNO_2]} \right) = 3.18 - 3.35 = -0.17 \] \[ \frac{[NO_2^-]}{[HNO_2]} = 10^{-0.17} \approx 0.68 \] 3. **Find the Concentration of NO₂⁻:** \[ [NO_2^-] = 0.68 \times [HNO_2] \] Since the initial concentration of HNO₂ is 1.
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