`pK_(a)` of a base `(K_(b)=1 xx 10^(-5)` is

Options

A. 5
B. `-9`
C. `-5`
D. 9

Answer

To find the \( pK_a \) of a base given its \( K_b \), we can use the relationship between \( K_a \), \( K_b \), and the ion product of water \( K_w \). The relationship is given by:

\[ K_a \cdot K_b = K_w \]

Where \( K_w \) at 25°C is \( 1 \times 10^{-14} \).

### Step-by-Step Solution:

1. **Identify the given values**:
- \( K_b = 1 \times 10^{-5} \)
- \( K_w = 1 \times 10^{-14} \)

2. **Calculate \( K_a \)**:
Using the relationship \( K_a \cdot K_b = K_w \), we can rearrange this to find \( K_a \):
\[ K_a = \frac{K_w}{K_b} \]
Substituting the known values:
\[ K_a = \frac{1 \times 10^{-14}}{1 \times 10^{-5}} = 1 \times 10^{-9} \]

3. **Calculate \( pK_a \)**:
The \( pK_a \) is calculated using the formula:
\[ pK_a = -\log K_a \]
Substituting the value of \( K_a \):
\[ pK_a = -\log(1 \times 10^{-9}) \]
Using the logarithmic property:
\[ pK_a = -(-9) = 9 \]

4. **Final Result**:
Thus, the \( pK_a \) of the base is \( 9 \).

### Summary:
- \( K_b = 1 \times 10^{-5} \)
- \( K_a = 1 \times 10^{-9} \)
- \( pK_a = 9 \)