`pH` of a solution of `10 ml`. `1 N` sodium acetate and `50 ml 2N` acetic acid `(K_(a)=1.8xx10^(-5))` is approximately

Options

A. 3.74
B. 5
C. 6
D. 7

Answer

To find the pH of a solution containing sodium acetate and acetic acid, we can use the Henderson-Hasselbalch equation, which is suitable for buffer solutions. Here’s a step-by-step solution:

### Step 1: Identify the components of the solution
- We have **sodium acetate (CH₃COONa)** which is a salt of acetic acid and acts as a weak base.
- We have **acetic acid (CH₃COOH)** which is a weak acid.

### Step 2: Calculate the number of moles of sodium acetate and acetic acid
- **Sodium acetate**:
- Normality (N) = 1 N
- Volume (V) = 10 mL = 0.010 L
- Moles of sodium acetate = Normality × Volume = 1 N × 0.010 L = **0.010 moles**.

- **Acetic acid**:
- Normality (N) = 2 N
- Volume (V) = 50 mL = 0.050 L
- Moles of acetic acid = Normality × Volume = 2 N × 0.050 L = **0.100 moles**.

### Step 3: Calculate the total volume of the solution
- Total volume = Volume of sodium acetate + Volume of acetic acid = 10 mL + 50 mL = **60 mL = 0.060 L**.

### Step 4: Calculate the concentrations of sodium acetate and acetic acid
- **Concentration of sodium acetate**:
\[ \text{Concentration} = \frac{\text{Moles}}{\text{Total Volume}} = \frac{0.010 \text{ moles}}{0.060 \text{ L}} = \frac{10}{60} = \frac{1}{6} \approx 0.167 \text{ M} \]

- **Concentration of acetic acid**:
\[ \text{Concentration} = \frac{\text{Moles}}{\text{Total Volume}} = \frac{0.100 \text{ moles}}{0.060 \text{ L}} = \frac{100}{60} \approx 1.67 \text{ M} \]

### Step 5: Calculate pKa from Ka
- Given \( K_a = 1.8 \times 10^{-5} \):
\[ pK_a = -\log(K_a) = -\log(1.8 \times 10^{-5}) \approx 4.74 \]

### Step 6: Apply the Henderson-Hasselbalch equation
The Henderson-Hasselbalch equation is given by:
\[ pH = pK_a + \log\left(\frac{[\text{Salt}]}{[\text{Acid}]}\right) \]
Substituting the values:
\[ pH = 4.74 + \log\left(\frac{0.167}{1.67}\right) \]
Calculating the log term:
\[ \frac{0.167}{1.67} = 0.1 \quad \Rightarrow \quad \log(0.1) = -1 \]
Thus,
\[ pH = 4.74 - 1 = 3.74 \]

### Final Answer
The pH of the solution is approximately **3.74**.