Liquid ammonia ionizes to slight extent. At -50°C, its self-ionization constant, K = [NH4+][NH2-] = 10^-30 M^2. How many amide ions are present per ml of pure liquid ammonia? (Na = 6 × 10^23)

Question

Prashant Kotian · Accepted Answer

Correct answer: C. In pure liquid ammonia, the self-ionization reaction is: 2 ⇌ NH4+ + NH2-. Since the solution is neutral, we have: [NH4+] = [NH2-].
Given: K = [NH4+][NH2-] = 10^-30 M^2.
Therefore, [NH2-]^2 = 10^-30 => [NH2-] = 10^-15 M.
In 1 ml of pure liquid ammonia (which is 10^-3 L), the moles of amide ions is: Moles = 10^-15 mol/L × 10^-3 L = 10^-18 mol.
The number of amide ions = Moles × Na = 10^-18 × 6 × 10^23 = 6 × 10^5 amide ions.

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