Question

Write the balanced ionic equation for the reaction of sodium bicarbonate with sulphuric acid.

Answer 1

The molecular equation for the chemical change is:
$NaHC{O}_3+H_{2}S{O}_4\to N{a}_{2}S{O}_4+H_{2}O+C{O}_2$
$NaHC{O}_3,H_{2}S{O}_4$ and $N{a}_{2}S{O}_4$ are ionic compounds; so these are written in ionic forms.
$N{a}^{+}+HC{O}_3^{-}+2H^{+}+S{O}_4^{2-}\to 2N{a}^{+}+S{O}_4^{2-}+H_{2}O+C{O}_2$
$N{a}^{+}$ and $S{O}_4^{2-}$ ions are spectator ions; hence these shall not appear in the final equation.
$HC{O}_3^{-}+2H^{+}\to H_{2}O+C{O}_2$
To make equal charges on both sides, $HC{O}_3^{-}$ should have a coefficient $2$.
$2HC{O}_3^{-}+2H^{+}\to H_{2}O+C{O}_2$
In order to balance the hydrogen and carbon on both sides, the molecules of $H_{2}O$ and $C{O}_2$ should have a coefficient $2$ respectively.
$2HC{O}_3^{-}+2H^{+}=2H_{2}O+2C{O}_2$
or $HC{O}_3^{-}+H^{+}=H_{2}O+C{O}_2$
This is the balanced ionic equation.