# Question 25981

May 16, 2017

I think you are on the right track...........

#### Explanation:

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 $N {H}_{4}^{+}$, whose conjugate base is $N {H}_{3}$. And we have the base, ""^(-)C-=N#, whose conjugate acid is $H C \equiv N$.

And for $b .$ we have the acid $H C l \left(a q\right)$, whose conjugate base is $C {l}^{-}$. And we have the base, $C {O}_{3}^{2 -}$, whose conjugate acid is $H C {O}_{3}^{-}$, $\text{bicarbonate ion}$.

And for $c .$ we have the acid $H C l$, 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, ${H}_{2} O$, the conjugate acid is ${H}_{3} {O}^{+}$, $\text{hydronium ion}$. Its conjugate base is $H {O}^{-}$, $\text{hydroxide ion}$. And the conjugate base of $H {O}^{-} \equiv {O}^{2 -}$.

And if we go to ammonia as a SOLVENT, we can invoke equivalent conjugate acid/base pairs for $N {H}_{4}^{+}$, and $N {H}_{2}^{-}$ (this amide base is TOO BASIC to exist in water).

Confused yet?