Question #25981

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Question #25981

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Answer 1 · 2017-05-16T16:39:03

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

Explanation:

The conjugate base of an acid is simply the original acid LESS a proton, $H^{+}$ $H^+$ . 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 $a .$ $a.$ we have the acid $N H_{4}^{+}$ $NH_4^+$ , whose conjugate base is $N H_{3}$ $NH_3$ . And we have the base, $^{-} C \equiv N$ $""^(-)C-=N$ , whose conjugate acid is $H C \equiv N$ $HC-=N$ .

And for $b .$ $b.$ we have the acid $H C l (a q)$ $HCl(aq)$ , whose conjugate base is $C l^{-}$ $Cl^-$ . And we have the base, $C O_{3}^{2 -}$ $CO_3^(2-)$ , whose conjugate acid is $H C O_{3}^{-}$ $HCO_3^-$ , $bicarbonate ion$ $"bicarbonate ion"$ .

And for $c .$ $c.$ we have the acid $H C l$ $HCl$ , 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, $H_{2} O$ $H_2O$ , the conjugate acid is $H_{3} O^{+}$ $H_3O^+$ , $hydronium ion$ $"hydronium ion"$ . Its conjugate base is $H O^{-}$ $HO^-$ , $hydroxide ion$ $"hydroxide ion"$ . And the conjugate base of $H O^{-} \equiv O^{2 -}$ $HO^(-)-=O^(2-)$ .

And if we go to ammonia as a SOLVENT, we can invoke equivalent conjugate acid/base pairs for $N H_{4}^{+}$ $NH_4^+$ , and $N H_{2}^{-}$ $NH_2^(-)$ (this amide base is TOO BASIC to exist in water).

Confused yet?

[And see here and links.](https://socratic.org/questions/what-is-the-conjugate-acid-base-relationship-of-h2po4-and-hpo4)

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Question #25981

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