Liquid ammonia (anhydrous NH_3(I) is often used as a solvent. Like water, ammoni

Question

Liquid ammonia (anhydrous NH_3(I) is often used as a solvent. Like water, ammonia undergoes auto ionization by the transfer of a proton: 2NH_3 NH_4^+ + NH_2^- K = 1 times 10^-33 at -50 degree C Since ammonia is better proton acceptor than water, the ionization of acids is relatively enhanced in liquid ammonia. For example, acetic acid is a strong acid in liquid ammonia. Estimate [NH_4^+] (mol/L) in pure (i.e. neutral) liquid ammonia at -50 degree C. mol/L Calculate [NH_4^+] (mol/L) in a 8.92 M solution of acetic acid in liquid ammonia at -50 degree C. mol/L Estimate [NH_2^-] (mol/L) in a 8.92 M solution of acetic acid in liquid ammonia at -50 degree C. mol/L

Explanation / Answer

2NH3(l) <--> NH4+ (aq) + NH2- (aq)

K = [NH4+] [NH2-] = S x S = S^2

10^ -33 = S^2

S = 3.16 x10^ -17 M = [NH4+]

2) we have 8.92 acetic acid , in liquid NH3 CH3COOH is strong acid and compltely dissociates to give H+

hence H+ given = 8.92 M

now NH3 (l) + H+ (aq) <---> NH4+ (aq)

hence [NH4+] = 8.92 M

3) we have [NH4+] [NH2-] = 10^ -33

8.92 x [NH2-] = 10^ -33                     ( In 8.92 acetic acid we got [NH4+] = 8.92 M)

[NH2-] = 1.12 x 10^ -34 M

( M is unit mol/L)

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