Question

Write two sets of equations (both molecular and total ionic) with different reactants that have the same net ionic equation as the following equation: $$ \mathrm{Ba}\left(\mathrm{NO}_{0}\right)_{2}(a q)+\mathrm{Na}_{2} \mathrm{CO}_{3}(a q) \longrightarrow \mathrm{BaCO}_{3}(s)+2 \mathrm{NaNO}_{3}(a q) $$

Short answer

First set: \( \mathrm{BaCl_2 + K_2CO_3 \longrightarrow BaCO_3 + 2 KCl} \). Second set: \( \mathrm{Ba (CH_3CO_2)_2 + (NH_4)_2CO_3 \longrightarrow BaCO_3 + 2 NH_4CH_3CO_2} \).

Step-by-step solution

- Identify the net ionic equation

First, identify the net ionic equation from the given reaction. The given reaction is: \( \mathrm{Ba} \left( \mathrm{NO}_3 \right)_2 (aq) + \mathrm{Na}_2 \mathrm{CO}_3 (aq) \longrightarrow \mathrm{BaCO}_3 (s) + 2 \mathrm{NaNO}_3 (aq) \) The net ionic equation removes spectator ions and shows only the species that participate directly in the reaction. In this case, the net ionic equation is: \[ \mathrm{Ba^{2+} (aq) + CO_3^{2-} (aq) \longrightarrow BaCO_3 (s)} \]

- Choose reactants for the first set of equations

Select different reactants that will yield the same net ionic equation. For instance, consider using \( \mathrm{BaCl_2} \) and \( \mathrm{K_2CO_3} \). The relevant molecular equation is: \( \mathrm{BaCl_2 (aq) + K_2CO_3 (aq) \longrightarrow BaCO_3 (s) + 2 KCl (aq)} \)

- Write the total ionic equation for the first set

Break down the molecular equation into its ionic components: \[ \mathrm{Ba^{2+} (aq) + 2 Cl^{-} (aq) + 2 K^{+} (aq) + CO_3^{2-} (aq) \longrightarrow BaCO_3 (s) + 2 K^{+} (aq) + 2 Cl^{-} (aq)} \] Remove the spectator ions to find the net ionic equation, which should match the net ionic equation identified in Step 1.

- Choose reactants for the second set of equations

Select another pair of reactants that will yield the same net ionic equation. For example, \( \mathrm{Ba (CH_3CO_2)_2} \) (barium acetate) and \( \mathrm{(NH_4)_2CO_3} \) (ammonium carbonate). The molecular equation is: \( \mathrm{Ba (CH_3CO_2)_2 (aq) + (NH_4)_2CO_3 (aq) \longrightarrow BaCO_3 (s) + 2 NH_4CH_3CO_2 (aq)} \)

- Write the total ionic equation for the second set

Break down the second molecular equation into its ionic components: \[ \mathrm{Ba^{2+} (aq) + 2 (CH_3CO_2)^{-}(aq) + 2 NH_4^{+}(aq) + CO_3^{2-} (aq) \longrightarrow BaCO_3 (s) + 2 NH_4^{+}(aq) + 2 (CH_3CO_2)^{-}(aq)} \] Remove the spectator ions to confirm that the net ionic equation is the same as the original one identified in Step 1.

Key concepts

Understanding Molecular Equations

Molecular equations give a full picture of the chemical reaction by showing all the reactants and products in their molecular forms. In the given exercise, the first molecular equation is:
$$ \text{Ba(NO}_3\text{)}_{2(aq)} + \text{Na}_{2}\text{CO}_3\text{(aq)} \rightarrow \text{BaCO}_3\text{(s)} + 2\text{NaNO}_3\text{(aq)} $$
Think of these equations as the big snapshot of what's happening. All compounds are represented before they dissociate in water. For the exercise, other valid sets of molecular equations would be:
$$ \text{BaCl}_2\text{(aq)} + \text{K}_2\text{CO}_3\text{(aq)} \rightarrow \text{BaCO}_3\text{(s)} + 2\text{KCl}\text{(aq)} $$
$$ \text{Ba(CH}_3\text{CO}_2\text{)}_{2(aq)} + \text{(NH}_4\text{)}_2\text{CO}_3\text{(aq)} \rightarrow \text{BaCO}_3\text{(s)} + 2\text{NH}_4\text{CH}_3\text{CO}_2\text{(aq)} $$
Notice that these equations fully display the reactants and products without specifying ionic forms.

Decoding Total Ionic Equations

Total ionic equations break down all aqueous substances into their respective ions. This provides a more granular look at the reaction. Based on the exercise:
The total ionic equation for the reaction:
$$ \text{Ba(NO}_3\text{)}_{2(aq)} + \text{Na}_2\text{CO}_3\text{(aq)} \rightarrow \text{BaCO}_3\text{(s)} + 2\text{NaNO}_3\text{(aq)} $$
becomes:
$$ \text{Ba}^{2+}\text{(aq)} + 2\text{NO}_3{}\text{(aq)} + 2\text{Na}^{+}\text{(aq)} + \text{CO}_3{}^{2-}\text{(aq)} \rightarrow \text{BaCO}_3\text{(s)} + 2\text{Na}^{+}\text{(aq)} + 2\text{NO}_3{}\text{(aq)} $$
In this step, we express all reactants and products that are strong electrolytes in their dissociated forms. Similarly, breaking down other molecular equations:
For: $$ \text{BaCl}_2\text{(aq)} + \text{K}_2\text{CO}_3\text{(aq)} \rightarrow \text{BaCO}_3\text{(s)} + 2\text{KCl}\text{(aq)} $$
we get:
$$ \text{Ba}^{2+}\text{(aq)} + 2\text{Cl}^{-}\text{(aq)} + 2\text{K}^{+}\text{(aq)} + \text{CO}_3{}^{2-}\text{(aq)} \rightarrow \text{BaCO}_3\text{(s)} + 2\text{K}^{+}\text{(aq)} + 2\text{Cl}^{-}\text{(aq)} $$
Remove spectator ions to reveal the net ionic equation next.

Understanding Chemical Reactions

A chemical reaction involves the rearrangement of atoms to form new substances. The molecular equations mentioned above show this at a macroscopic level. The total ionic and net ionic equations zoom in to depict the exact particles involved.
In a reaction like:
$$ \text{BaCl}_2\text{(aq)} + \text{K}_2\text{CO}_3{(aq)} \rightarrow \text{BaCO}_3\text{(s)} + 2\text{KCl}\text{(aq)} $$
Barium ions (\text{Ba}^{2+}) react with carbonate ions (\text{CO}_3{}^{2-}) to form the solid barium carbonate (\text{BaCO}_3), indicating a precipitation reaction. This shows how ions in solution form a solid.
Chemical reactions can be varied, such as acid-base reactions, redox reactions, and precipitation reactions like our example here. They form the core of much of chemistry learning and applications.

What are Spectator Ions?

Spectator ions are ions in the solution that do not participate in the formation of the precipitate or the reaction, essentially 'watching' the other ions react. For example, in the total ionic equation:
$$ \text{Ba}^{2+}\text{(aq)} + 2\text{Cl}^{-}\text{(aq)} + 2\text{K}^{+}\text{(aq)} + \text{CO}_3{}^{2-}\text{(aq)} \rightarrow \text{BaCO}_3\text{(s)} + 2\text{K}^{+}\text{(aq)} + 2\text{Cl}^{-}\text{(aq)} $$
$$ 2\text{Cl}^{-} $$ and $$ 2\text{K}^{+} $$ are spectator ions because they appear unchanged on both sides of the equation. They do not affect the formation of the precipitate (the solid product) and can be omitted to simplify the net ionic equation. Removing these spectator ions highlights the actual reacting species and the real transformation taking place during the reaction.
Identifying and removing spectator ions simplifies the understanding of chemical reactions, focusing on the core chemical change.