Both HCO{3}− and HS− are amphoteric. Write an equation to show how each substance can...

$HC{O}_3^{-}(ag)+HS(ag)\rightleftharpoons C{O}_3^{2-}(ag)+H_{2}S(aq)$
Let ${\displaystyle HC{O}_{\left\{3\right\}}^{-}}$ be a $Br\mathrm{\varnothing }$ nsted-Lowry acid and let $H{S}^{-}$ be a $Br\mathrm{\varnothing }$ nsted-Lowry base. The conjugate base of the acid is then determined by removing an H atom and
a unit of positive charge. The conjugate acid of the base is then determined by adding an H atom anda unit of positive charge.
$HC{O}_3^{-}(ag)+H{S}^{-}(ag)\rightleftharpoons H_{2}C{O}_3(ag)+S^{-2}(aq)$
Let ${\displaystyle HC{O}_{\left\{3\right\}}^{-}}$ be a ​​​​​​​$Br\mathrm{\varnothing }$ nsted-Lowry base and let $H{S}^{-}$
be a $Br\mathrm{\varnothing }$ nsted-Lowry acid. The conjugate acid of the
base is then determined by adding an H atom anda unit of positive charge. The conjugate base of the acid is then determined by removing an H atom and a unit of positive charge.

Explanation:
According to Bronsted and Lowry an acid is a proton donor and base is a proton acceptor.
Therefore, the equation for the reaction will be as follows.
$HC{O}_3^{-}+H{S}^{-}=H_{2}C{O}_3+S^{2}t$
Here, ${\displaystyle HC{O}_{\left\{3\right\}}^{-}}$ acts as a base as it is accepting a proton and ${\displaystyle H{S}^{-}}$ acts as an acid because it is donating a proton.