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

Question

Liquid ammonia (anhydrous NH3(i) is often used as a solvent. Like water, ammonia undergoes auto ionization by the transfer of a proton: 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 (NH4^+J (mol/L) in pure (i.e. neutral) liquid ammonia at -50degree C. Calculate [NH4^+] (mol/L) in a 1.49 times 10 7 M solution of acetic acid in liquid ammonia at -50 degree C. Estimate [NH2] (mol/L) in a 1.49 times 10^-7 M solution of acetic acid in liquid ammonia at -50 degree C.

Explanation / Answer

In neutral liquid ammonia:

Equilibrium constant expression for NH3 is K = [NH4+][NH2-]

In neutral liquid ammonia, [NH4+] = [NH2-] = a M (say)

1x10-33 = a2 . So, a = sqroot(1x10-33 ) = 3.17x10-17

So, [NH4+] = 3.17x10-17 mol/L

In 1.49x10-7 M acetic acid solution:

The reaction is CH3CO2H + NH3 ---------> NH4+ + CH3CO2-

Since acetic acid in liquid ammonia is strong acid, [NH4+] = [ CH3CO2H]

So, [NH4+] = 1.49x10-7 mol/L

In 1.49x10-7 M acetic acid solution:

[NH4+] = 1.49x10-7 mol/L

Since [NH4+][NH2-] = 1x10-33 ,

[NH2-] = (1x10-33 )/[NH4+] = (1x10-33 )/(1.49x10-7 ) = 6.71x10-27

So, [NH2-] = 6.71x10-27 mol/L

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