The ionisation constant for `NH_(4)^(+)` in water is `5.0xx10^(-10)" at " 25^(@)C`. The rate constant for the reaction of `NH_(4)^(+) and OH^(-)` to form `NH_(3) and H_(2)O" at " 25^(@)C` is `3.0xx10^(10)" Lmol"^(-1)s^(-1)`. The rate constant for proton transfer from water to `NH_(3)` is `x xx 10^(5)`. The value of 'x' is

Answer

To solve the problem, we need to find the value of 'x' in the rate constant for proton transfer from water to NH3, which is given as \( x \times 10^5 \). Let's break this down step by step.

### Step 1: Write the ionization reaction of \( NH_4^+ \)
The ionization of \( NH_4^+ \) in water can be represented as:
\[ NH_4^+ + H_2O \rightleftharpoons NH_3 + H_3O^+ \]
The ionization constant \( K_a \) for this reaction is given as:
\[ K_a = 5.0 \times 10^{-10} \]

### Step 2: Write the equilibrium expression for \( K_a \)
The equilibrium expression for \( K_a \) is:
\[ K_a = \frac{[NH_3][H_3O^+]}{[NH_4^+]} \]

### Step 3: Write the reaction for \( NH_3 \) and water
The reaction of \( NH_3 \) with water can be represented as:
\[ NH_3 + H_2O \rightleftharpoons NH_4^+ + OH^- \]
The rate constant for this reaction is denoted as \( K_b \).

### Step 4: Relate \( K_a \) and \( K_b \)
Using the relationship between \( K_w \), \( K_a \), and \( K_b \):
\[ K_w = K_a \times K_b \]
Where \( K_w = 1.0 \times 10^{-14} \) at 25°C. Rearranging gives:
\[ K_b = \frac{K_w}{K_a} = \frac{1.0 \times 10^{-14}}{5.0 \times 10^{-10}} = 2.0 \times 10^{-5} \]

### Step 5: Use the given rate constant for the backward reaction
The rate constant for the reaction of \( NH_4^+ \) and \( OH^- \) to form \( NH_3 \) and \( H_2O \) is given as:
\[ K_b = 3.0 \times 10^{10} \, L \, mol^{-1} \, s^{-1} \]

### Step 6: Relate \( K_b \) and \( K_f \)
We know that:
\[ K_b = \frac{K_f}{K_b} \]
Rearranging gives:
\[ K_f = K_b \times K_b \]

### Step 7: Substitute the values
Substituting the values we have:
\[ K_f = (2.0 \times 10^{-5}) \times (3.0 \times 10^{10}) = 6.0 \times 10^{5} \]

### Step 8: Find the value of 'x'
From the problem, we know that:
\[ K_f = x \times 10^5 \]
Thus, equating gives:
\[ 6.0 \times 10^5 = x \times 10^5 \]
This implies:
\[ x = 6 \]

### Final Answer
The value of \( x \) is:
\[ \boxed{6} \]