2nh3(l) → nh4+(am) + nh2−(am) where (am) represents solvation by ammonia. at the boiling point (−33°c), kam = 5.1 × 10−27. calculate [nh4+] at this temperature.

Answer is: concentration of ammonium ions are 7,14·10⁻¹⁴ M.
Chemical reaction: 2NH₃(l) → NH₄⁺(am) + NH₂⁻(am).
Kam = 5,1·10⁻²⁷.
[NH₄⁺] · [NH₂⁻] = x; equilibrium concentration of cations and anions.
Kam = [NH₄⁺] · [NH₂⁻].
Kam = x².
x = [NH₄⁺] = √5,1·10⁻²⁷.
[NH₄⁺] = 7,14·10⁻¹⁴ M.

Sorry i dont understand the question can you rephase it?