A weak monobasic acid is half neutralized by a strong base. If the Ph of the solution is 5.4 its `pK_(a)` is

Options

A. 6.8
B. 2.7
C. 5.4
D. 10.8

Explanation

To find the \( pK_a \) of a weak monobasic acid that is half neutralized by a strong base, we can follow these steps:

### Step 1: Understand the Situation
When a weak acid is half neutralized by a strong base, it means that the concentration of the weak acid (\( HA \)) is equal to the concentration of its conjugate base (\( A^- \)). This is because half of the acid has reacted with the base to form the salt.

### Step 2: Use the Henderson-Hasselbalch Equation
The Henderson-Hasselbalch equation relates the pH of a buffer solution to the \( pK_a \) of the weak acid and the ratio of the concentrations of the conjugate base and the weak acid:
\[ \text{pH} = pK_a + \log\left(\frac{[A^-]}{[HA]}\right) \]

### Step 3: Substitute Values
Since the acid is half neutralized, the concentrations of the weak acid and its conjugate base are equal:
\[ [A^-] = [HA] \]
This means that the ratio \(\frac{[A^-]}{[HA]} = 1\). Therefore, the logarithm of 1 is 0:
\[ \log(1) = 0 \]
Thus, the equation simplifies to:
\[ \text{pH} = pK_a + 0 \]
or
\[ \text{pH} = pK_a \]

### Step 4: Solve for \( pK_a \)
Given that the pH of the solution is 5.4, we can directly conclude:
\[ pK_a = 5.4 \]

### Conclusion
The \( pK_a \) of the weak monobasic acid is **5.4**.

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