For `NH_(3)`, `K_(b)=1.8xx10^(-5)`. `K_(a)` for `NH_(4)^(+)` would be

Options

A. `1.8xx10^(5)`
B. `5.56xx10^(5)`
C. `1.8xx10^(10)`
D. `5.56xx10^(-10)`

Explanation

To find the acid dissociation constant \( K_a \) for \( NH_4^+ \) (the conjugate acid of ammonia \( NH_3 \)), we can use the relationship between the base dissociation constant \( K_b \) and the acid dissociation constant \( K_a \) through the ion product of water \( K_w \).

### Step-by-Step Solution:

1. **Identify the Constants**:
- We are given \( K_b \) for ammonia \( NH_3 \):
\[ K_b = 1.8 \times 10^{-5} \]
- The ion product of water \( K_w \) at 25°C is:
\[ K_w = 1.0 \times 10^{-14} \]

2. **Use the Relationship**:
- The relationship between \( K_a \), \( K_b \), and \( K_w \) is given by:
\[ K_a \times K_b = K_w \]
- Rearranging this equation to solve for \( K_a \):
\[ K_a = \frac{K_w}{K_b} \]

3. **Substitute the Values**:
- Now substitute the known values into the equation:
\[ K_a = \frac{1.0 \times 10^{-14}}{1.8 \times 10^{-5}} \]

4. **Perform the Calculation**:
- First, calculate the division:
\[ K_a = \frac{1.0}{1.8} \times 10^{-14 + 5} = \frac{1.0}{1.8} \times 10^{-9} \]
- Simplifying \( \frac{1.0}{1.8} \) gives approximately \( 0.555 \):
\[ K_a \approx 0.555 \times 10^{-9} \]
- Therefore, we can express this as:
\[ K_a \approx 5.56 \times 10^{-10} \]

5. **Final Result**:
- The \( K_a \) for \( NH_4^+ \) is:
\[ K_a \approx 5.56 \times 10^{-10} \]