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 analyze the given data and apply the relevant equations for ionization constants and rate constants.

### Step 1: Understand the Ionization 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: Understand the Reaction with \( OH^- \)

The reaction of \( NH_4^+ \) with \( OH^- \) to form \( NH_3 \) and \( H_2O \) can be represented as:

\[ NH_4^+ + OH^- \rightleftharpoons NH_3 + H_2O \]

The rate constant for this reaction is given as:

\[ k_f = 3.0 \times 10^{10} \, L \, mol^{-1} \, s^{-1} \]

### Step 3: Relate Ionization Constants and Rate Constants

The relationship between the ionization constant of a base (\( K_b \)) and the ionization constant of an acid (\( K_a \)) is given by:

\[ K_w = K_a \times K_b \]

Where \( K_w \) is the ionization constant of water at 25°C, which is:

\[ K_w = 1.0 \times 10^{-14} \]

### Step 4: Calculate \( K_b \) for \( NH_3 \)

From the relationship above, we can express \( K_b \) as:

\[ K_b = \frac{K_w}{K_a} \]

Substituting the known values:

\[ K_b = \frac{1.0 \times 10^{-14}}{5.0 \times 10^{-10}} = 2.0 \times 10^{-5} \]

### Step 5: Use the Rate Constants to Find \( k_b \)

The ionization constant for the base can also be expressed in terms of the rate constants for the forward and backward reactions:

\[ K_b = \frac{k_f}{k_b} \]

Where \( k_b \) is the rate constant for the backward reaction. Rearranging gives:

\[ k_b = \frac{k_f}{K_b} \]

Substituting the known values:

\[ k_b = \frac{3.0 \times 10^{10}}{2.0 \times 10^{-5}} = 1.5 \times 10^{15} \, L \, mol^{-1} \, s^{-1} \]

### Step 6: Relate \( k_b \) to the Proton Transfer Constant

The rate constant for proton transfer from water to \( NH_3 \) is given as \( x \times 10^5 \). Since we have calculated \( k_b \) as \( 1.5 \times 10^{15} \), we can express this as:

\[ k_b = x \times 10^5 \]

### Step 7: Solve for \( x \)

To find \( x \), we can rearrange the equation:

\[ x = \frac{1.5 \times 10^{15}}{10^5} = 1.5 \times 10^{10} \]

Thus, the value of \( x \) is:

\[ x = 1.5 \times 10^{10} \]

### Final Answer

The value of \( x \) is **1.5**.