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Liquid ammonia (anhydrous NH3(l)) is often used as a solvent. Like water, ammonia undergoes autoionization 

2NH3  NH4+ + NH2-        K = 1×10-33 at -50°C

Estimate [NH4+] (mol/L) in pure (i.e. neutral) liquid ammonia at -50°C.

Feb 15th, 2015

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+]


Feb 15th, 2015

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