Auto-ionisation of liquid `NH_(3)` is
`2NH_(3) hArr NH_(4)^(o+) +NH_(2)^(Theta)`
with `K_(NH_(3)) = [NH_(4)^(o+)] [NH_(2)^(Theta)] = 10^(-30) at -50^(@)C` Number of amide ions `(NH_(2)^(Theta))`, present per `mm^(3)` of pure liquid `NH_(3)` is

Options

A. (a) `602`
B. (b) `301`
C. (c) `200`
D. (d) `100`

Explanation

To solve the problem regarding the auto-ionisation of liquid ammonia (NH₃), we will follow these steps:

### Step 1: Write the equilibrium expression
The auto-ionisation reaction of ammonia is given by:
\[ 2 \text{NH}_3 \rightleftharpoons \text{NH}_4^+ + \text{NH}_2^- \]

The equilibrium constant \( K_{NH_3} \) for this reaction is expressed as:
\[ K_{NH_3} = [\text{NH}_4^+][\text{NH}_2^-] \]

### Step 2: Set up the equilibrium concentrations
Let the concentration of both ions (\(\text{NH}_4^+\) and \(\text{NH}_2^-\)) at equilibrium be \( x \). Since the stoichiometric coefficients for both ions are equal, we can express:
\[ K_{NH_3} = x \cdot x = x^2 \]

### Step 3: Substitute the value of \( K_{NH_3} \)
We are given that:
\[ K_{NH_3} = 10^{-30} \]
Thus, we can write:
\[ x^2 = 10^{-30} \]

### Step 4: Solve for \( x \)
Taking the square root of both sides gives:
\[ x = \sqrt{10^{-30}} = 10^{-15} \]

This means that the concentration of both \(\text{NH}_4^+\) and \(\text{NH}_2^-\) ions in the liquid ammonia is \( 10^{-15} \, \text{mol/L} \).

### Step 5: Convert concentration to number of ions per mm³
To find the number of amide ions (\(\text{NH}_2^-\)) per mm³, we need to convert the concentration from mol/L to mol/mm³.

1 L = \( 10^3 \) mL = \( 10^6 \) mm³, so:
\[ 10^{-15} \, \text{mol/L} = 10^{-15} \, \text{mol} / 10^6 \, \text{mm}^3 = 10^{-21} \, \text{mol/mm}^3 \]

### Step 6: Calculate the number of ions using Avogadro's number
The number of ions in 1 mm³ can be calculated using Avogadro's number (\( N_A = 6.022 \times 10^{23} \, \text{ions/mol} \)):
\[ \text{Number of ions} = \text{Concentration} \times N_A \]
\[ \text{Number of ions} = 10^{-21} \, \text{mol/mm}^3 \times 6.022 \times 10^{23} \, \text{ions/mol} \]

### Step 7: Perform the multiplication
Calculating this gives:
\[ \text{Number of ions} = 10^{-21} \times 6.022 \times 10^{23} = 6.022 \times 10^{2} \approx 602 \]

Thus, the number of amide ions (\(\text{NH}_2^-\)) present per mm³ of pure liquid NH₃ is approximately **602**.

### Final Answer:
The number of amide ions (\(\text{NH}_2^-\)) present per mm³ of pure liquid NH₃ is **602**.

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