
How to write an equation to show ionization?
Answer
550.5k+ views
Hint As we know that ionization reaction is the reaction in which an atom basically acquires a positive or negative charge by gaining or losing electrons. And this results in the electrically charged ion atom called an ion.
Complete Step by step solution:
- As we know that the main difference between chemical reaction and ionization reaction is that the substances that can form ions are basically written as ions in the ionic equation.
- Let us take an example in order to show how to write an equation to show ionization:
- We can write the balanced chemical equation of the reaction of sulphuric acid and sodium bicarbonate as:
$2NaHC{{O}_{3}}+{{H}_{2}}S{{O}_{4}}\to N{{a}_{2}}S{{O}_{4}}+2{{H}_{2}}O+2C{{O}_{2}}$
- As we know that sodium sulphate, sulphuric acid and sodium bicarbonate are ionic compounds, hence they exist as ions.
- Now, we can write the ionic equation of the reaction as:
$2Na+2HC{{O}_{3}}^{-}+2{{H}^{+}}+SO_{4}^{-}\to 2N{{a}^{+}}+SO_{4}^{-}+2{{H}_{2}}O+2C{{O}_{2}}$
- Here, we can see that sulphate ions and sodium ions are spectator ions, hence they do not participate in the reaction. Hence, we can write the net ionic equation as:
$\begin{align}
& 2HC{{O}_{3}}^{-}+2{{H}^{+}}\to 2{{H}_{2}}O+2C{{O}_{2}} \\
& HC{{O}_{3}}^{-}+{{H}^{+}}\to {{H}_{2}}O+C{{O}_{2}} \\
\end{align}$ - Hence, in this way we can write an equation to show ionization.
Note:
- We should note here that the net ionic equation has only particles that are found to participate in the reactions that means spectator ions are not present. Basically the spectator ion is the ion which is present in both the reactant as well as in the product side of the reaction.
Complete Step by step solution:
- As we know that the main difference between chemical reaction and ionization reaction is that the substances that can form ions are basically written as ions in the ionic equation.
- Let us take an example in order to show how to write an equation to show ionization:
- We can write the balanced chemical equation of the reaction of sulphuric acid and sodium bicarbonate as:
$2NaHC{{O}_{3}}+{{H}_{2}}S{{O}_{4}}\to N{{a}_{2}}S{{O}_{4}}+2{{H}_{2}}O+2C{{O}_{2}}$
- As we know that sodium sulphate, sulphuric acid and sodium bicarbonate are ionic compounds, hence they exist as ions.
- Now, we can write the ionic equation of the reaction as:
$2Na+2HC{{O}_{3}}^{-}+2{{H}^{+}}+SO_{4}^{-}\to 2N{{a}^{+}}+SO_{4}^{-}+2{{H}_{2}}O+2C{{O}_{2}}$
- Here, we can see that sulphate ions and sodium ions are spectator ions, hence they do not participate in the reaction. Hence, we can write the net ionic equation as:
$\begin{align}
& 2HC{{O}_{3}}^{-}+2{{H}^{+}}\to 2{{H}_{2}}O+2C{{O}_{2}} \\
& HC{{O}_{3}}^{-}+{{H}^{+}}\to {{H}_{2}}O+C{{O}_{2}} \\
\end{align}$ - Hence, in this way we can write an equation to show ionization.
Note:
- We should note here that the net ionic equation has only particles that are found to participate in the reactions that means spectator ions are not present. Basically the spectator ion is the ion which is present in both the reactant as well as in the product side of the reaction.
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