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Let's solve this step-by-step.

We are asked to find the **pH** of a solution made by combining:
- 157 mL of **0.35 M NaC₂H₃O₂** (sodium acetate, the conjugate base of acetic acid)
- 139 mL of **0.46 M HC₂H₃O₂** (acetic acid, the weak acid)

### Step 1: Recognize the system as a buffer solution
Since we are combining a weak acid (acetic acid) with its conjugate base (sodium acetate), this forms a **buffer solution**. For buffer solutions, we can use the **Henderson-Hasselbalch equation**:

\[
\text{pH} = \text{p}K_a + \log \left( \frac{[\text{A}^-]}{[\text{HA}]} \right)
\]

- \( \text{p}K_a = -\log K_a \)
- \( [\text{A}^-] \) is the concentration of the conjugate base (acetate, NaC₂H₃O₂)
- \( [\text{HA}] \) is the concentration of the weak acid (acetic acid)

---

### Step 2: Calculate pKa
Given:  
\[
K_a = 1.75 \times 10^{-5}
\]

\[
\text{p}K_a = -\log(1.75 \times 10^{-5}) = 4.76
\]

---

### Step 3: Find the concentrations of acid and base after mixing
The total volume of the solution is:

\[
V_{\text{total}} = 157 \, \text{mL} + 139 \, \text{mL} = 296 \, \text{mL}
\]

#### Moles of each component:
\[
\text{Moles of NaC}_2\text{H}_3\text{O}_2 = 0.35 \, M \times 0.157 \, L = 0.05495 \, \text{mol}
\]

\[
\text{Moles of HC}_2\text{H}_3\text{O}_2 = 0.46 \, M \times 0.139 \, L = 0.06394 \, \text{mol}
\]

#### Concentrations after mixing:
\[
[\text{A}^-] = \frac{0.05495 \, \text{mol}}{0.296 \, L} = 0.1856 \, M
\]

\[
[\text{HA}] = \frac{0.06394 \, \text{mol}}{0.296 \, L} = 0.2160 \, M
\]

---

### Step 4: Apply the Henderson-Hasselbalch equation
\[
\text{pH} = 4.76 + \log \left( \frac{0.1856}{0.2160} \right)
\]

\[
\frac{0.1856}{0.2160} = 0.859
\]

\[
\log(0.859) = -0.066
\]

\[
\text{pH} = 4.76 - 0.066 = 4.69
\]

---

### Final Answer:
The **pH** of the solution is approximately **4.69**.

So the correct option is:
\[
\boxed{a) \, 4.69}
\]

---

Would you like any further details or clarifications?  

Here are five related questions:
1. What happens to the pH if more acetic acid is added to the buffer?
2. How does the buffer capacity depend on the concentrations of acid and base?
3. Can the Henderson-Hasselbalch equation be used for strong acids and bases?
4. What is the effect of dilution on the pH of a buffer solution?
5. How do you calculate pH if only acetic acid is present without its conjugate base?

**Tip:** Buffer solutions resist changes in pH upon small additions of acid or base, making them essential in biological systems.

What is the pH of a solution made by combining 157 mL of 0.35 M NaC₂H₃O₂ with 139 mL of 0.46 M HC₂H₃O₂? The Kₐ of acetic acid is 1.75 × 10⁻⁵.

Find the pH of a buffer solution created by combining 157 mL of 0.35 M NaC₂H₃O₂ and 139 mL of 0.46 M HC₂H₃O₂ using the Henderson-Hasselbalch equation. Includes step-by-step calculations and explanation for high school chemistry students.

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