To find the number of amide ions present per mm³ of pure liquid ammonia at -50°C, we can follow these steps:
### Step 1: Understand the Self-Ionization of Ammonia
The self-ionization of ammonia can be represented as:
\[
2 \text{NH}_3 \rightleftharpoons \text{NH}_4^+ + \text{NH}_2^-
\]
Here, one molecule of ammonia ionizes to form one ammonium ion (\(\text{NH}_4^+\)) and one amide ion (\(\text{NH}_2^-\)).
### Step 2: Set Up the Expression for the Ion Product Constant
The ion product constant for ammonia at -50°C is given as:
\[
K_{NH_3} = [\text{NH}_4^+][\text{NH}_2^-] = 10^{-30} \, \text{M}^2
\]
### Step 3: Define the Concentrations
Let \(x\) be the concentration of both \(\text{NH}_4^+\) and \(\text{NH}_2^-\) at equilibrium. Since both ions are produced in equal amounts:
\[
[\text{NH}_4^+] = x \quad \text{and} \quad [\text{NH}_2^-] = x
\]
### Step 4: Substitute into the Ion Product Expression
Substituting the concentrations into the ion product expression gives:
\[
K_{NH_3} = x \cdot x = x^2
\]
Thus,
\[
x^2 = 10^{-30}
\]
### Step 5: Solve for \(x\)
To find \(x\), take the square root of both sides:
\[
x = \sqrt{10^{-30}} = 10^{-15} \, \text{M}
\]
This means the concentration of amide ions (\([\text{NH}_2^-]\)) is \(10^{-15} \, \text{M}\).
### Step 6: Convert Concentration to Number of Ions per mm³
To find the number of amide ions per mm³, we need to convert the concentration from moles per liter to ions per mm³:
1. **Convert Molarity to Number of Ions**:
- 1 mole contains \(6.022 \times 10^{23}\) ions.
- Therefore, the number of ions in \(10^{-15}\) moles is:
\[
\text{Number of ions} = 10^{-15} \, \text{moles} \times 6.022 \times 10^{23} \, \text{ions/mole}
\]
\[
= 6.022 \times 10^{8} \, \text{ions}
\]
2. **Convert Liters to mm³**:
- Since \(1 \, \text{L} = 10^6 \, \text{mm}^3\), the concentration in mm³ is:
\[
\text{Concentration in ions/mm}^3 = \frac{6.022 \times 10^{8} \, \text{ions}}{10^6 \, \text{mm}^3} = 6.022 \times 10^{2} \, \text{ions/mm}^3
\]
\[
= 602.2 \, \text{ions/mm}^3
\]
### Final Answer
Thus, the number of amide ions present per mm³ of pure liquid ammonia is approximately:
\[
\boxed{602} \, \text{ions/mm}^3
\]
