Question

Self-ionisation of liquid ammonia occurs as $ 2N{{H}_{3}}\rightleftharpoons NH_{4}^{+}+NH_{2}^{-}; $ $ K={{10}^{-10}} $ In this solvent, an acid might be:

• A. $ NH_{4}^{+} $
• B. $ N{{H}_{3}} $
• C. any species that will form $ NH_{4}^{+} $
• D. all the above

Answer 1

Answer: (A)

$ N{{H}_{3}}+N{{H}_{3}}\,\,\,\,\rightleftharpoons \underset{\begin{smallmatrix} \text{proton}\,\text{donar} \\ \,\,\,\,\,\,\,\,\,\,\text{(acid)} \end{smallmatrix}}{\mathop{NH_{4}^{+}}}\,+NH_{2}^{-} $ Hence, $ NH_{4}^{+} $ can donate proton and is an acid.