(C) conjugate base is formed by the removal of a proton $(H^{+})$ from the acid.For $NH_3$,the removal of one $H^{+}$ ion results in the formation of

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(C) conjugate base is formed by the removal of a proton $(H^{+})$ from the acid.For $NH_3$,the removal of one $H^{+}$ ion results in the formation of the amide ion $(NH_2^{-})$.The reaction is: $NH_3 \rightleftharpoons NH_2^{-} + H^{+}$Therefore,the conjugate base of $NH_3$ is $NH_2^{-}$.