Estimate [NH4+](mol/L) in pure(ex neutral) liquid ammonia @-50°C
2 NH3 => NH4+ + NH2- ; K = 10^-33
17 0 0
-2x +x +x
(17 - 2x) (x) (x)
K = (x^2)/(17-2x) => (x^2) = (17-2x)K
=> (x^2) = 17K -2Kx
=> (x^2) + (2K)x - 17K = 0
using quadratic formula and substituting for K we get: x ~ 1.3 * 10^-16 M = [NH4+]
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