The `pK_a` of a weak acid (HA) is 4.5 The pOH of an aqueous buffered solution of HA in which 50% of it is ionised is

Options

A. `4.5`
B. ` 2.5`
C. ` 9.5`
D. ` 7.0`

Explanation

To solve the problem, we need to calculate the pOH of an aqueous buffered solution of the weak acid HA, given that the pK_a of HA is 4.5 and that 50% of it is ionized.

### Step-by-Step Solution:

1. **Understanding the Ionization of the Weak Acid:**
- The weak acid HA ionizes in water to form H⁺ and its conjugate base A⁻.
- When 50% of HA is ionized, it means that the concentration of HA that remains is equal to the concentration of A⁻ formed.

2. **Setting Up the Concentrations:**
- Let’s denote the initial concentration of HA as [HA]. Since 50% is ionized, we have:
- [HA] remaining = 0.5[HA]
- [A⁻] formed = 0.5[HA]
- Therefore, [HA] = [A⁻].

3. **Using the Henderson-Hasselbalch Equation:**
- The pH of a buffer solution can be calculated using the Henderson-Hasselbalch equation:
\[ \text{pH} = \text{pK}_a + \log\left(\frac{[\text{A}^-]}{[\text{HA}]}\right) \]
- Since [A⁻] = [HA], we can substitute into the equation:
\[ \text{pH} = \text{pK}_a + \log\left(\frac{[HA]}{[HA]}\right) \]
- This simplifies to:
\[ \text{pH} = \text{pK}_a + \log(1) \]
- Since \(\log(1) = 0\), we have:
\[ \text{pH} = \text{pK}_a \]

4. **Calculating the pH:**
- Given that \(\text{pK}_a = 4.5\):
\[ \text{pH} = 4.5 \]

5. **Calculating the pOH:**
- The relationship between pH and pOH is given by:
\[ \text{pH} + \text{pOH} = 14 \]
- Therefore, we can calculate pOH as follows:
\[ \text{pOH} = 14 - \text{pH} = 14 - 4.5 = 9.5 \]

### Final Answer:
The pOH of the aqueous buffered solution of HA in which 50% of it is ionized is **9.5**.