Calculate equilibrium constant for `NH_3(aq.)+H_2O hArr NH_4 +OH`
if equilibrium constant for `NH_4 OH+H hArr NH_4 +H_2O k_1=1.8 xx 10^9`
Also `2H_2O hArr H_3O+OH K_2 =1xx 10^(-14) "at" 25^@C`

Options

A. `1.8 xx 10^(-10)`
B. `1.8 xx 10^(-5)`
C. `1.8 xx 10^(-9)`
D. `1.8 xx 10^(-14)`

Explanation

To calculate the equilibrium constant for the reaction:

\[ NH_3(aq) + H_2O \rightleftharpoons NH_4^+ + OH^- \]

we will use the given equilibrium constants for the following reactions:

1. \( NH_4OH + H^+ \rightleftharpoons NH_4^+ + H_2O \) with \( K_1 = 1.8 \times 10^9 \)
2. \( 2H_2O \rightleftharpoons H_3O^+ + OH^- \) with \( K_2 = 1 \times 10^{-14} \)

### Step-by-Step Solution:

**Step 1: Write the first reaction in terms of \( NH_4OH \)**

We can express \( NH_4OH \) in terms of \( NH_3 \) and \( H_2O \):

\[ NH_4OH \rightleftharpoons NH_3 + H_2O \]

This reaction can be considered as the reverse of the first reaction provided. Therefore, the equilibrium constant for this reaction will be the inverse of \( K_1 \):

\[ K_{NH_4OH} = \frac{1}{K_1} = \frac{1}{1.8 \times 10^9} \]

**Step 2: Write the second reaction**

The second reaction is:

\[ 2H_2O \rightleftharpoons H_3O^+ + OH^- \]

This reaction has an equilibrium constant \( K_2 = 1 \times 10^{-14} \).

**Step 3: Combine the reactions**

Now, we can combine the two reactions. When we add the modified first reaction and the second reaction, we get:

1. \( NH_4OH \rightleftharpoons NH_3 + H_2O \) (reverse of reaction 1)
2. \( 2H_2O \rightleftharpoons H_3O^+ + OH^- \)

Adding these gives:

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

The \( H_2O \) cancels out on both sides, leading to:

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

**Step 4: Calculate the equilibrium constant for the combined reaction**

The equilibrium constant for the combined reaction is the product of the equilibrium constants of the individual reactions:

\[ K = K_{NH_4OH} \times K_2 = \left( \frac{1}{1.8 \times 10^9} \right) \times (1 \times 10^{-14}) \]

Calculating this gives:

\[ K = \frac{1 \times 10^{-14}}{1.8 \times 10^9} = \frac{1}{1.8} \times 10^{-23} \]

\[ K \approx 5.56 \times 10^{-24} \]

### Final Answer:

The equilibrium constant \( K \) for the reaction \( NH_3(aq) + H_2O \rightleftharpoons NH_4^+ + OH^- \) is approximately:

\[ K \approx 5.56 \times 10^{-24} \]