Equilibrium constant for reaction `NH_(4)OH(aq)+H^(+)(aq)hArr NH_(4)^(+)(aq)+H_(2)O(l)`
`1.8xx19^(9)`.
Hence equilibrium constant for ionization `NH_(3)+H_(2)OhArr NH_(4)^(+)(aq)+OH^(-)(aq)` is `x xx 10^(-6)`. The value of 'x' is

Answer

To solve the problem, we need to find the equilibrium constant for the ionization of ammonia (NH₃) in water, given the equilibrium constant for the dissociation of ammonium hydroxide (NH₄OH) and hydrogen ions (H⁺).

### Step-by-Step Solution:

1. **Write the given reaction and its equilibrium constant**:
The reaction provided is:
\[ \text{NH}_4\text{OH}(aq) + \text{H}^+(aq) \rightleftharpoons \text{NH}_4^+(aq) + \text{H}_2\text{O}(l) \]
The equilibrium constant (K) for this reaction is given as:
\[ K = 1.8 \times 10^9 \]

2. **Write the reverse reaction**:
The reverse reaction can be written as:
\[ \text{NH}_4^+(aq) + \text{H}_2\text{O}(l) \rightleftharpoons \text{NH}_4\text{OH}(aq) + \text{H}^+(aq) \]
The equilibrium constant for this reverse reaction (K') is the reciprocal of K:
\[ K' = \frac{1}{K} = \frac{1}{1.8 \times 10^9} \]

3. **Write the second reaction for ionization of ammonia**:
The ionization of ammonia in water can be represented as:
\[ \text{NH}_3(aq) + \text{H}_2\text{O}(l) \rightleftharpoons \text{NH}_4^+(aq) + \text{OH}^-(aq) \]
Let’s denote the equilibrium constant for this reaction as K''.

4. **Relate K'' to K'**:
The relationship between K' and K'' can be established by noting that NH₄OH can dissociate into NH₄⁺ and OH⁻. Therefore, we can express K'' in terms of K' and the ion product of water (K_w):
\[ K'' = K' \cdot K_w \]
where \( K_w \) (the ion product of water) is approximately \( 1 \times 10^{-14} \).

5. **Calculate K''**:
Substitute K' and K_w into the equation:
\[ K'' = \left(\frac{1}{1.8 \times 10^9}\right) \cdot (1 \times 10^{-14}) \]
Simplifying this gives:
\[ K'' = \frac{1 \times 10^{-14}}{1.8 \times 10^9} = \frac{1}{1.8} \times 10^{-14 - 9} = \frac{1}{1.8} \times 10^{-23} \]
\[ K'' \approx 0.555 \times 10^{-23} \approx 5.55 \times 10^{-24} \]

6. **Convert K'' to the required form**:
We need to express K'' in the form \( x \times 10^{-6} \):
\[ 5.55 \times 10^{-24} = x \times 10^{-6} \]
To convert \( 10^{-24} \) to \( 10^{-6} \), we multiply by \( 10^{-18} \):
\[ x = 5.55 \times 10^{-24 + 18} = 5.55 \times 10^{-6} \]

7. **Final value of x**:
Since we need to express this as \( x \times 10^{-6} \), we find:
\[ x = 5.55 \]

### Final Answer:
The value of \( x \) is approximately **5.55**.