Question

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

Answer 1

Hint: We know that, when two aqueous solutions having ions as solutes are combined, the ions might react with one another. These reactions are always double displacement reactions. The solvent molecules (water) do not react usually.

Complete step by step answer:
Let’s discuss the ionic equation. Chemists use ionic equations to show the details of reactions that involve ions in aqueous solutions. The difference between ionic equation and chemical equation is that substances that can form ions are written as ions in ionic equation.
Let’s understand the complete ionic equation. This is the ionic equation which shows all the particles of the solution as they exist.
Now, come to the question. The reactants are given as sodium bicarbonate and sulphuric acid. The chemical formula of sodium carbonate is ${\rm{NaHC}}{{\rm{O}}_{\rm{3}}}$ and sulphuric acid is ${{\rm{H}}_{\rm{2}}}{\rm{S}}{{\rm{O}}_{\rm{4}}}$
First we write the balanced chemical equation of reaction of sulphuric acid and sodium bicarbonate.
${\rm{2NaHC}}{{\rm{O}}_{\rm{3}}} + {{\rm{H}}_{\rm{2}}}{\rm{S}}{{\rm{O}}_{\rm{4}}} \to {\rm{N}}{{\rm{a}}_{\rm{2}}}{\rm{S}}{{\rm{O}}_{\rm{4}}} + 2{{\rm{H}}_{\rm{2}}}{\rm{O}} + 2{\rm{C}}{{\rm{O}}_2}$
Sodium bicarbonate, sulphuric acid and sodium sulphate are ionic compounds, so they exist as ions. Now, we have to write the complete ionic equation of the reaction.
${\rm{2N}}{{\rm{a}}^ + } + 2{\rm{HC}}{{\rm{O}}_{\rm{3}}}^ - + 2{{\rm{H}}^ + } + {\rm{S}}{{\rm{O}}_{\rm{4}}}^ - \to 2{\rm{N}}{{\rm{a}}^ + } + {\rm{S}}{{\rm{O}}_{\rm{4}}}^ - + 2{{\rm{H}}_{\rm{2}}}{\rm{O}} + 2{\rm{C}}{{\rm{O}}_2}$
Sodium ions and sulphate ions are spectator ions, so they will not participate in the reaction. So, the net ionic equation is,
${\rm{2HC}}{{\rm{O}}_{\rm{3}}}^ - + 2{{\rm{H}}^ + } \to 2{{\rm{H}}_{\rm{2}}}{\rm{O}} + 2{\rm{C}}{{\rm{O}}_2}$
${\rm{HC}}{{\rm{O}}_{\rm{3}}}^ - + {{\rm{H}}^ + } \to {{\rm{H}}_{\rm{2}}}{\rm{O}} + {\rm{C}}{{\rm{O}}_2}$
The above ionic equation is the balanced ionic equation.

So, the correct answer is “Option A”.

Note:
It is to be noted that the net ionic equation consists of only particles that participate in the reactions that means spectator ions are not present. In general, spectator ion is the ion that is present in both the reactant and product side of the reaction.