According to the equations below, what is the conjugate base of HCO_3^- ?

${H}_{2} C {O}_{3} \left(a q\right) + {H}_{2} O \left(l\right) r i g h t \le f t h a r p \infty n s H C {O}_{3}^{-} \left(a q\right) + {H}_{3} {O}^{+} \left(a q\right)$ $H C {O}_{3}^{-} \left(a q\right) + {H}_{2} O \left(l\right) r i g h t \le f t h a r p \infty n s C {O}_{3}^{2 -} \left(a q\right) + {H}_{3} {O}^{+} \left(a q\right)$

Dec 9, 2016

The conjugate base of $H C {O}_{3}^{-}$ is $C {O}_{3}^{2 -}$

Explanation:

To find the conjugate base of ANY acid, all you have to do is remove a proton, and conserve mass and charge; simply take ${H}^{+}$ away.

e.g. ${H}_{2} S {O}_{4} \rightarrow H S {O}_{4}^{-}$

$H N {O}_{3} \rightarrow N {O}_{3}^{-}$

${H}_{3} P {O}_{4} \rightarrow {H}_{2} P {O}_{4}^{-}$

$N {H}_{3} \rightarrow N {H}_{2}^{-}$

What is the conjugate acid of $C {O}_{3}^{2 -}$ ion?

It seems I have asnwered this before.