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 value of 'x' in the equilibrium constant expression for the ionization of ammonia, given the equilibrium constant for the reaction involving ammonium hydroxide.

### Step-by-Step Solution:

1. **Understand the Given Reaction**:
The reaction provided is:
\[ NH_4OH(aq) + H^+(aq) \rightleftharpoons NH_4^+(aq) + H_2O(l) \]
The equilibrium constant \( K \) for this reaction is given as \( 1.8 \times 10^9 \).

2. **Write the Expression for the Equilibrium Constant**:
The equilibrium constant \( K \) for the above reaction can be expressed as:
\[ K = \frac{[NH_4^+][H_2O]}{[NH_4OH][H^+]} \]
Since water is a pure liquid, its concentration is constant and can be omitted from the expression:
\[ K = \frac{[NH_4^+]}{[NH_4OH][H^+]} \]

3. **Identify the Ionization Reaction**:
We are interested in the ionization of ammonia:
\[ NH_3 + H_2O \rightleftharpoons NH_4^+ + OH^- \]
The equilibrium constant for this reaction is denoted as \( K' \).

4. **Write the Expression for \( K' \)**:
The equilibrium constant \( K' \) for the ionization of ammonia can be expressed as:
\[ K' = \frac{[NH_4^+][OH^-]}{[NH_3][H_2O]} \]
Again, since water is a pure liquid, we can simplify this to:
\[ K' = \frac{[NH_4^+][OH^-]}{[NH_3]} \]

5. **Relate \( K' \) and \( K \)**:
From the first reaction, we can relate the two equilibrium constants using the ion-product of water \( K_w \):
\[ K' = \frac{K_w}{K} \]
where \( K_w = [H^+][OH^-] = 1.0 \times 10^{-14} \).

6. **Substitute the Values**:
Substitute \( K \) and \( K_w \) into the equation:
\[ K' = \frac{1.0 \times 10^{-14}}{1.8 \times 10^9} \]

7. **Calculate \( K' \)**:
\[ K' = 1.0 \times 10^{-14} \times \frac{1}{1.8 \times 10^9} = \frac{1.0}{1.8} \times 10^{-14 + 9} = \frac{1.0}{1.8} \times 10^{-5} \]
\[ K' \approx 0.555 \times 10^{-5} = 5.55 \times 10^{-6} \]

8. **Express \( K' \) in terms of \( x \)**:
We know that \( K' = x \times 10^{-6} \). Therefore, we can equate:
\[ 5.55 \times 10^{-6} = x \times 10^{-6} \]
Thus, \( x \approx 5.55 \).

9. **Final Value**:
Rounding to the nearest whole number, we find:
\[ x \approx 6 \]

### Final Answer:
The value of \( x \) is approximately **6**.