The conjugate base of an acid is simply the original acid LESS a proton, #H^+#. And likewise the conjugate acid of a base is the original base PLUS a proton. As with any chemical process, both mass and charge are conserved.
So for #a.# we have the acid #NH_4^+#, whose conjugate base is #NH_3#. And we have the base, #""^(-)C-=N#, whose conjugate acid is #HC-=N#.
And for #b.# we have the acid #HCl(aq)#, whose conjugate base is #Cl^-#. And we have the base, #CO_3^(2-)#, whose conjugate acid is #HCO_3^-#, #"bicarbonate ion"#.
And for #c.# we have the acid #HCl#, whose conjugate base we have already identified.
Note that all I have done here is to add (conjugate acid) or subtract (conjugate base) a proton, and conserved charge.
For water, #H_2O#, the conjugate acid is #H_3O^+#, #"hydronium ion"#. Its conjugate base is #HO^-#, #"hydroxide ion"#. And the conjugate base of #HO^(-)-=O^(2-)#.
And if we go to ammonia as a SOLVENT, we can invoke equivalent conjugate acid/base pairs for #NH_4^+#, and #NH_2^(-)# (this amide base is TOO BASIC to exist in water).
Confused yet?
And see here and links.
