https://file.aiquickdraw.com/mathbot/file/1729973827054dpp4hhug.jpg

To solve for the pH of a 0.30 M solution of sodium propionate (\( \text{NaC}_3\text{H}_5\text{O}_2 \)) at 25°C, given:

- **Propionic acid (\( \text{HC}_3\text{H}_5\text{O}_2 \)) is monoprotic** with a dissociation constant \( K_a = 1.3 \times 10^{-5} \).
- **Water dissociation constant** \( K_w = 1.01 \times 10^{-14} \).

### Solution:

1. **Determine the \( K_b \) of sodium propionate:**
   Sodium propionate is the conjugate base of propionic acid. So,
   \[
   K_b = \frac{K_w}{K_a} = \frac{1.01 \times 10^{-14}}{1.3 \times 10^{-5}} = 7.77 \times 10^{-10}
   \]

2. **Set up the equilibrium expression:**
   For a weak base, we assume the reaction in water as:
   \[
   \text{C}_3\text{H}_5\text{O}_2^- + \text{H}_2\text{O} \leftrightarrow \text{HC}_3\text{H}_5\text{O}_2 + \text{OH}^-
   \]
   The equilibrium expression is:
   \[
   K_b = \frac{[ \text{OH}^- ][ \text{HC}_3\text{H}_5\text{O}_2 ]}{[ \text{C}_3\text{H}_5\text{O}_2^- ]}
   \]

3. **Calculate \( [\text{OH}^-] \):**
   Since the initial concentration of \( \text{C}_3\text{H}_5\text{O}_2^- \) is 0.30 M and assuming \( x \) is the concentration of \( \text{OH}^- \) produced, we get:
   \[
   K_b = \frac{x^2}{0.30 - x} \approx \frac{x^2}{0.30}
   \]
   Solving for \( x \):
   \[
   x = \sqrt{K_b \times 0.30} = \sqrt{7.77 \times 10^{-10} \times 0.30} \approx 1.53 \times 10^{-5} \, \text{M}
   \]

4. **Find the pOH and then pH:**
   \[
   \text{pOH} = -\log[ \text{OH}^- ] = -\log(1.53 \times 10^{-5}) \approx 4.82
   \]
   \[
   \text{pH} = 14 - \text{pOH} = 14 - 4.82 = 9.18
   \]

### Answer:
The correct answer is **(e) 9.18**.

Would you like further details or explanations on this solution?

---

Here are 5 related questions for practice:

1. What would the pH be if the concentration of sodium propionate were doubled?
2. How would the pH change if the temperature increased, thus increasing \( K_w \)?
3. Explain the relationship between \( K_a \) and \( K_b \) for a conjugate acid-base pair.
4. Calculate the pH of a solution containing 0.30 M of a strong base.
5. How would the pH of a weak acid solution compare to that of its conjugate base?

**Tip:** Remember, for conjugate acid-base pairs, \( K_a \times K_b = K_w \) always holds true.

What is the pH of a 0.30 M solution of sodium propionate, NaC3H5O2, at 25°C (propionic acid, HC3H5O2, is monoprotic and has a Ka = 1.3 × 10^−5 at 25°C. Kw = 1.01 × 10^−14)?

Find the pH of a 0.30 M sodium propionate solution at 25°C, given propionic acid's Ka of 1.3 × 10^−5 and Kw of 1.01 × 10^−14. This problem covers key concepts in acid-base equilibrium and pH calculation, ideal for college-level chemistry students.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation