At `- 50^(@) C`, the self - ionization constant (ion product ) of `NH_(3)" is " K_(NH_(3)) = [NH_(4)^(+)] [NH_(2)^(-)] = 10^(-30) M^(2)`. How many amide ions are present per `mm^(3)` of pure liquid ammonia ?

Options

A. `600 "ions/mm"^(3)`
B. `6xx 1-^(6)" ions/mm"^(3)`
C. `6xx10^(4)"ions/mm"^(3)`
D. `60 "ions/mm"^(3)`

Explanation

To find the number of amide ions (NH2⁻) present per mm³ of pure liquid ammonia (NH₃) at -50°C, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Self-Ionization of Ammonia**:
The self-ionization of ammonia can be represented as:
\[ 2 \text{NH}_3 \rightleftharpoons \text{NH}_4^+ + \text{NH}_2^- \]

2. **Define the Ion Product Constant**:
The ion product constant \( K_{NH_3} \) is given as:
\[ K_{NH_3} = [\text{NH}_4^+][\text{NH}_2^-] = 10^{-30} \, \text{M}^2 \]

3. **Set Up the Equilibrium Concentrations**:
Assume the initial concentration of ammonia is 1 M (1 mole per liter). Let \( x \) be the concentration of NH₄⁺ and NH₂⁻ formed at equilibrium. Therefore, at equilibrium:
- Concentration of NH₃ = \( 1 - 2x \)
- Concentration of NH₄⁺ = \( x \)
- Concentration of NH₂⁻ = \( x \)

4. **Substitute into the Ion Product Expression**:
Substitute the equilibrium concentrations into the ion product expression:
\[ K_{NH_3} = x \cdot x = x^2 \]
Thus, we have:
\[ x^2 = 10^{-30} \]

5. **Solve for \( x \)**:
Taking the square root of both sides:
\[ x = \sqrt{10^{-30}} = 10^{-15} \, \text{M} \]
This means the concentration of NH₂⁻ ions is \( 10^{-15} \, \text{M} \).

6. **Convert Concentration to Number of Ions**:
To find the number of amide ions per mm³, we need to convert the concentration from moles per liter to ions per mm³. Using Avogadro's number (\( N_A = 6.022 \times 10^{23} \) ions/mol):
\[ \text{Number of ions per liter} = 10^{-15} \, \text{mol/L} \times 6.022 \times 10^{23} \, \text{ions/mol} \]
\[ = 6.022 \times 10^{8} \, \text{ions/L} \]

7. **Convert Liters to mm³**:
Since 1 L = \( 10^{6} \) mm³, we can convert the number of ions per liter to ions per mm³:
\[ \text{Number of ions per mm³} = \frac{6.022 \times 10^{8} \, \text{ions/L}}{10^{6}} = 602.2 \, \text{ions/mm³} \]

8. **Final Answer**:
Rounding to the nearest whole number, we find that the number of amide ions present per mm³ of pure liquid ammonia is approximately:
\[ \boxed{600} \, \text{ions/mm³} \]