50 mL of 2N acetic acid mixed with 10 mL of 1N sodium acetate solution will have an approximate pH of `(K_(a)=10^(-5))`:

Options

A. 4
B. 6
C. 5
D. 7

Answer

To solve the problem of finding the approximate pH of a solution formed by mixing 50 mL of 2N acetic acid with 10 mL of 1N sodium acetate, we will use the Henderson-Hasselbalch equation, which is suitable for buffer solutions. Here are the steps to arrive at the solution:

### Step 1: Identify the components and their concentrations
- **Acetic Acid (CH₃COOH)**: Given as 2N (normal), which is equivalent to 2M (molar) since the n-factor for acetic acid is 1.
- **Volume of Acetic Acid**: 50 mL
- **Sodium Acetate (CH₃COONa)**: Given as 1N, which is also equivalent to 1M (molar) since the n-factor for sodium acetate is also 1.
- **Volume of Sodium Acetate**: 10 mL

### Step 2: Calculate the millimoles of each component
- **Millimoles of Acetic Acid**:
\[ \text{Millimoles of Acetic Acid} = \text{Molarity} \times \text{Volume (mL)} = 2 \, \text{mol/L} \times 50 \, \text{mL} = 100 \, \text{mmol} \]

- **Millimoles of Sodium Acetate**:
\[ \text{Millimoles of Sodium Acetate} = \text{Molarity} \times \text{Volume (mL)} = 1 \, \text{mol/L} \times 10 \, \text{mL} = 10 \, \text{mmol} \]

### Step 3: Calculate the pKa of acetic acid
Given \( K_a = 10^{-5} \):
\[ pK_a = -\log(K_a) = -\log(10^{-5}) = 5 \]

### Step 4: Use the Henderson-Hasselbalch equation
The Henderson-Hasselbalch equation for a buffer solution is:
\[ pH = pK_a + \log\left(\frac{[\text{Salt}]}{[\text{Acid}]}\right) \]
Here, the salt is sodium acetate and the acid is acetic acid.

### Step 5: Calculate the concentrations of the components in the final solution
- **Total Volume of the Solution**:
\[ \text{Total Volume} = 50 \, \text{mL} + 10 \, \text{mL} = 60 \, \text{mL} \]

- **Concentration of Sodium Acetate**:
\[ [\text{Sodium Acetate}] = \frac{10 \, \text{mmol}}{60 \, \text{mL}} = \frac{10}{60} = \frac{1}{6} \, \text{M} \]

- **Concentration of Acetic Acid**:
\[ [\text{Acetic Acid}] = \frac{100 \, \text{mmol}}{60 \, \text{mL}} = \frac{100}{60} = \frac{5}{3} \, \text{M} \]

### Step 6: Substitute the values into the Henderson-Hasselbalch equation
\[ pH = 5 + \log\left(\frac{10 \, \text{mmol}}{100 \, \text{mmol}}\right) = 5 + \log\left(\frac{10}{100}\right) = 5 + \log\left(\frac{1}{10}\right) \]
Using the property of logarithms:
\[ \log\left(\frac{1}{10}\right) = -1 \]
Thus,
\[ pH = 5 - 1 = 4 \]

### Final Answer
The approximate pH of the solution is **4**.

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