`K_(h)` of salt obtained from strong acid and weak base is `2 xx 10^(-5)`. The `K_(b)` of weak base is

Options

A. `2 xx 10^(-19)`
B. `5 xx 10^(-10)`
C. `2 xx 10^(-10)`
D. `5 xx 10^(-9)`

Explanation

To find the \( K_b \) of the weak base from the given \( K_h \) of the salt obtained from a strong acid and a weak base, we can use the relationship between \( K_h \), \( K_a \), and \( K_b \).

### Step-by-Step Solution:

1. **Understanding the relationship**:
The relationship between \( K_h \), \( K_a \), and \( K_b \) is given by the equation:
\[ K_h = \frac{K_w}{K_b} \]
where \( K_w \) is the ion product of water, which at 25°C is \( 1 \times 10^{-14} \).

2. **Given values**:
We are given:
\[ K_h = 2 \times 10^{-5} \]
and
\[ K_w = 1 \times 10^{-14} \]

3. **Rearranging the equation**:
We can rearrange the equation to solve for \( K_b \):
\[ K_b = \frac{K_w}{K_h} \]

4. **Substituting the values**:
Now, substituting the known values into the equation:
\[ K_b = \frac{1 \times 10^{-14}}{2 \times 10^{-5}} \]

5. **Calculating \( K_b \)**:
Performing the division:
\[ K_b = \frac{1}{2} \times 10^{-14 + 5} = \frac{1}{2} \times 10^{-9} = 0.5 \times 10^{-9} = 5 \times 10^{-10} \]

6. **Final answer**:
Thus, the \( K_b \) of the weak base is:
\[ K_b = 5 \times 10^{-10} \]