When a solution of benzoic acid was titrated with `NaOH` the `pH` of the solution when half the acid neutralized was `4.2`. Dissociation constant of the acid is

Options

A. `6.31xx10^(-5)`
B. `3.2xx10^(-5)`
C. `8.7xx10^(-8)`
D. `6.42xx10^(-4)`

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the Situation
When benzoic acid (C6H5COOH) is titrated with NaOH, it reacts to form sodium benzoate (C6H5COONa) and water. At the point of half-neutralization, the concentrations of benzoic acid and sodium benzoate are equal.

### Step 2: Use the pH at Half-Neutralization
We are given that the pH of the solution at half-neutralization is 4.2. At this point, the concentrations of the weak acid (benzoic acid) and its conjugate base (sodium benzoate) are equal.

### Step 3: Apply the Henderson-Hasselbalch Equation
The Henderson-Hasselbalch equation is given by:
\[ \text{pH} = \text{pKa} + \log\left(\frac{[\text{Salt}]}{[\text{Acid}]}\right) \]

Since at half-neutralization, the concentrations of salt and acid are equal:
\[ \text{pH} = \text{pKa} + \log(1) \]
Since \(\log(1) = 0\), we have:
\[ \text{pH} = \text{pKa} \]

### Step 4: Relate pKa to Ka
From the information given:
\[ \text{pKa} = 4.2 \]

We know that:
\[ \text{pKa} = -\log(K_a) \]
Thus:
\[ K_a = 10^{-\text{pKa}} = 10^{-4.2} \]

### Step 5: Calculate Ka
Calculating \(K_a\):
\[ K_a = 10^{-4.2} \]
Using a calculator:
\[ K_a \approx 6.31 \times 10^{-5} \]

### Step 6: Conclusion
The dissociation constant \(K_a\) of benzoic acid is approximately \(6.31 \times 10^{-5}\).

### Final Answer
The dissociation constant of the acid is \(6.31 \times 10^{-5}\) (Option A).

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