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The problem here involves calculating the pH of a solution of sodium propionate (\( \text{NaC}_3\text{H}_5\text{O}_2 \)) given its concentration and the \( \text{p}K_a \) of its conjugate acid, propionic acid (\( \text{HC}_3\text{H}_5\text{O}_2 \)).

### Solution Outline:

1. **Identify Relevant Information:**
   - The concentration of sodium propionate is \( 0.85 \, \text{M} \).
   - The \( \text{p}K_a \) of propionic acid is \( 4.89 \).

2. **Relationship Between \( \text{p}K_a \) and \( K_b \):**
   Since sodium propionate is the conjugate base of propionic acid, we calculate the \( K_b \) for propionate:
   \[
   \text{p}K_a + \text{p}K_b = 14
   \]
   \[
   \text{p}K_b = 14 - 4.89 = 9.11
   \]
   Then, convert \( \text{p}K_b \) to \( K_b \):
   \[
   K_b = 10^{-\text{p}K_b} = 10^{-9.11}
   \]

3. **Use the Base Ionization Equation:**
   For the reaction:
   \[
   \text{C}_3\text{H}_5\text{O}_2^- + \text{H}_2\text{O} \rightleftharpoons \text{HC}_3\text{H}_5\text{O}_2 + \text{OH}^-
   \]
   The \( K_b \) expression is:
   \[
   K_b = \frac{[ \text{HC}_3\text{H}_5\text{O}_2 ][ \text{OH}^- ]}{[ \text{C}_3\text{H}_5\text{O}_2^- ]}
   \]
   Assuming \( x \) is the concentration of \( \text{OH}^- \) formed:
   \[
   K_b = \frac{x^2}{0.85 - x} \approx \frac{x^2}{0.85}
   \]
   Solve for \( x \), then find \( \text{pOH} \) and finally \( \text{pH} = 14 - \text{pOH} \).

Would you like me to go through the calculations step-by-step? 

### Related Questions:

1. How does \( \text{p}K_a \) relate to \( K_a \) in acid-base equilibrium calculations?
2. What is the significance of the \( \text{p}K_b \) in calculating the pH of basic solutions?
3. How do we calculate the pH of weak acid and weak base salt solutions in general?
4. Why do we assume \( x \) is small when simplifying equilibrium expressions?
5. What are the roles of conjugate acid-base pairs in buffer solutions?

### Tip:

Always verify if \( x \) is small enough by checking if it’s less than 5% of the initial concentration to validate the approximation.

Calculate the pH at 25°C of a 0.85 M solution of sodium propionate (NaC3H5O2). Note that propionic acid (HC3H5O2) is a weak acid with a pKa of 4.89. Round your answer to 1 decimal place.

Learn how to calculate the pH of a 0.85 M sodium propionate solution at 25°C, using the pKa of propionic acid (4.89). This example covers key acid-base equilibrium concepts and involves pH and pOH calculations, Kb expression, and conjugate acid-base pairs. Ideal for undergraduate chemistry students.

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