The `K_(b)` of weak base is `10^(-4)`. The `["salt"]` to `["base"]` ratio to be maintained to keep the `P^(H)` of buffer solution as `9` is .

Options

A. `1 : 4`
B. `4 : 2`
C. `1 : 10`
D. `10 : 1`

Explanation

To solve the problem, we need to determine the ratio of the concentration of salt to the concentration of the base in a buffer solution that maintains a pH of 9, given that the \( K_b \) of the weak base is \( 10^{-4} \).

### Step-by-Step Solution:

1. **Determine \( pK_b \)**:
The \( K_b \) of the weak base is given as \( 10^{-4} \). To find \( pK_b \), we use the formula:
\[ pK_b = -\log(K_b) \]
Substituting the value of \( K_b \):
\[ pK_b = -\log(10^{-4}) = 4 \]

2. **Calculate \( pOH \)**:
We know that \( pH + pOH = 14 \). Given that \( pH = 9 \):
\[ pOH = 14 - pH = 14 - 9 = 5 \]

3. **Use the Henderson-Hasselbalch equation for basic buffers**:
The Henderson-Hasselbalch equation for a basic buffer is given by:
\[ pH = pK_b + \log\left(\frac{[\text{salt}]}{[\text{base}]}\right) \]
However, since we have \( pOH \), we can rearrange this to:
\[ pOH = pK_b + \log\left(\frac{[\text{base}]}{[\text{salt}]}\right) \]
We can express it in terms of \( pK_b \):
\[ 5 = 4 + \log\left(\frac{[\text{base}]}{[\text{salt}]}\right) \]

4. **Rearranging the equation**:
Subtract \( 4 \) from both sides:
\[ 1 = \log\left(\frac{[\text{base}]}{[\text{salt}]}\right) \]

5. **Convert from logarithmic to exponential form**:
To eliminate the logarithm, we convert it to exponential form:
\[ \frac{[\text{base}]}{[\text{salt}]} = 10^1 = 10 \]

6. **Finding the ratio of salt to base**:
The ratio of salt to base can be found by taking the reciprocal:
\[ \frac{[\text{salt}]}{[\text{base}]} = \frac{1}{10} \]
Therefore, the ratio of salt to base is \( 1:10 \).

### Final Answer:
The ratio of salt to base concentration to maintain a pH of 9 is \( 1:10 \).