$\begin{array}{l}{\mathrm{H}}_2{\mathrm{CO}}_3\left(aq\right)+{\mathrm{H}}_2\mathrm{O}\left(l\right)⟶{\mathrm{HCO}}_3^{-}\left(aq\right)+{\mathrm{H}}_3{\mathrm{O}}^{+}\left(aq\right)\\ {\mathrm{HCO}}_3^{-}\left(aq\right)+{\mathrm{H}}_2\mathrm{O}\left(l\right)⟶{\mathrm{CO}}_3^{2-}\left(aq\right)+{\mathrm{H}}_3{\mathrm{O}}^{+}\left(aq\right)\end{array}$

According to the equations above, what is the conjugate base of ${\mathrm{HCO}}_3^{-}$?

Given equations are,

$\begin{array}{l}{\mathrm{H}}_2{\mathrm{CO}}_3\left(aq\right)+{\mathrm{H}}_2\mathrm{O}\left(l\right)⟶{\mathrm{HCO}}_3^{-}\left(aq\right)+{\mathrm{H}}_3{\mathrm{O}}^{+}\left(aq\right)\\ {\mathrm{HCO}}_3^{-}\left(aq\right)+{\mathrm{H}}_2\mathrm{O}\left(l\right)⟶{\mathrm{CO}}_3^{2-}\left(aq\right)+{\mathrm{H}}_3{\mathrm{O}}^{+}\left(aq\right)\end{array}$

Here, hydrogen carbonate ion (HCO3–) can be obtained by the diprotic acid and is amphiprotic. 

In this,

Conjugate acid =${\mathrm{H}}_2{\mathrm{CO}}_3$, 

Conjugate base = ${{\mathrm{CO}}^{2–}}_3$