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

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Solution

The molecular equation for the chemical change is:
$N a H C O_{3} + H_{2} S O_{4} \to N a_{2} S O_{4} + H_{2} O + C O_{2}$
$N a H C O_{3}, H_{2} S O_{4}$ and $N a_{2} S O_{4}$ are ionic compounds; so these are written in ionic forms.
$N a^{+} + H C O_{3}^{-} + 2 H^{+} + S O_{4}^{2 -} \to 2 N a^{+} + S O_{4}^{2 -} + H_{2} O + C O_{2}$
$N a^{+}$ and $S O_{4}^{2 -}$ ions are spectator ions; hence these shall not appear in the final equation.
$H C O_{3}^{-} + 2 H^{+} \to H_{2} O + C O_{2}$
To make equal charges on both sides, $H C O_{3}^{-}$ should have a coefficient $2$ .
$2 H C O_{3}^{-} + 2 H^{+} \to H_{2} O + C O_{2}$
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.
$2 H C O_{3}^{-} + 2 H^{+} = 2 H_{2} O + 2 C O_{2}$
or $H C O_{3}^{-} + H^{+} = H_{2} O + C O_{2}$
This is the balanced ionic equation.

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