Question

Write balanced net ionic equations for the reactions that occur in each of the following cases. Identify the spectator ion or ions in each reaction. (a) \(\mathrm{Cr}_{2}\left(\mathrm{SO}_{4}\right)_{3}(a q)+\left(\mathrm{NH}_{4}\right)_{2} \mathrm{CO}_{3}(a q) \longrightarrow\) (b) \(\operatorname{Ba}\left(\mathrm{NO}_{3}\right)_{2}(a q)+\mathrm{K}_{2} \mathrm{SO}_{4}(a q) \longrightarrow\) (c) \(\mathrm{Fe}\left(\mathrm{NO}_{3}\right)_{2}(a q)+\mathrm{KOH}(a q) \longrightarrow\)

Short answer

(a) The balanced net ionic equation is: \(2\mathrm{Cr}^{3+}(aq)+3\mathrm{CO}_{3}^{2-}(aq) \longrightarrow \mathrm{Cr}_{2}(\mathrm{CO}_3)_3(s)\). The spectator ions are \(\mathrm{NH}_{4}^{+}\) and \(\mathrm{SO}_{4}^{2-}\). (b) The balanced net ionic equation is: \(\mathrm{Ba}^{2+}(aq)+\mathrm{SO}_{4}^{2-}(aq) \longrightarrow \mathrm{BaSO}_4(s)\). The spectator ions are \(\mathrm{K}^{+}\) and \(\mathrm{NO}_{3}^{-}\). (c) The balanced net ionic equation is: \(\mathrm{Fe}^{2+}(aq)+2\mathrm{OH}^{-}(aq) \longrightarrow \mathrm{Fe(OH)}_2(s)\). The spectator ions are \(\mathrm{K}^{+}\) and \(\mathrm{NO}_{3}^{-}\).

Step-by-step solution

Predict the products

In this reaction, chromium sulfate (Cr2(SO4)3) reacts with ammonium carbonate ((NH4)2CO3). The sulfate ions (\(\mathrm{SO}_{4}^{2-}\)) will react with the ammonium ions (\(\mathrm{NH}_{4}^{+}\)), and the chromium ions (\(\mathrm{Cr}^{3+}\)) will react with the carbonate ions (\(\mathrm{CO}_{3}^{2-}\)). The resulting products will be ammonium sulfate and chromium carbonate.

Write the balanced molecular equation

Now we write the balanced molecular equation for this reaction: \(\mathrm{Cr}_{2}\left(\mathrm{SO}_{4}\right)_{3}(eq)+3\left(\mathrm{NH}_{4}\right)_{2} \mathrm{CO}_{3}(eq) \longrightarrow \mathrm{Cr}_{2}(\mathrm{CO}_3)_3(s)+6\mathrm{NH}_4\mathrm{SO}_4(eq)\)

Write the total ionic equation

We'll now write the total ionic equation, showing each ion in the aqueous solutions: \(2\mathrm{Cr}^{3+}(eq)+6\mathrm{SO}_{4}^{2-}(eq)+6\mathrm{NH}_{4}^{+}(eq)+3\mathrm{CO}_{3}^{2-}(eq) \longrightarrow \mathrm{Cr}_{2}(\mathrm{CO}_3)_3(s)+6\mathrm{NH}_4^{+}(eq)+6\mathrm{SO}_4^{2-}(eq)\)

Identify the spectator ions and write the balanced net ionic equation

The spectator ions are the ones that do not participate in the reaction; they are the same on both sides of the equation. In this case, the spectator ions are \(\mathrm{NH}_{4}^{+}\) and \(\mathrm{SO}_{4}^{2-}\). So, the balanced net ionic equation is: \(2\mathrm{Cr}^{3+}(eq)+3\mathrm{CO}_{3}^{2-}(eq) \longrightarrow \mathrm{Cr}_{2}(\mathrm{CO}_3)_3(s)\) (b) $\operatorname{Ba}\left(\mathrm{NO}_{3}\right)_{2}(aq)+\mathrm{K}_{2} \mathrm{SO}_{4}(aq) \longrightarrow$

Predict the products

In this reaction, barium nitrate (Ba(NO3)2) reacts with potassium sulfate (K2SO4). The barium ions (\(\mathrm{Ba}^{2+}\)) will react with the sulfate ions (\(\mathrm{SO}_{4}^{2-}\)), and the potassium ions (\(\mathrm{K}^{+}\)) will react with the nitrate ions (\(\mathrm{NO}_{3}^{-}\)). The resulting products will be barium sulfate and potassium nitrate.

Write the balanced molecular equation

Now we write the balanced molecular equation for this reaction: \(\mathrm{Ba}(\mathrm{NO}_3)_2(eq)+\mathrm{K}_2\mathrm{SO}_4(eq) \longrightarrow \mathrm{BaSO}_4(s)+2\mathrm{KNO}_3(eq)\)

Write the total ionic equation

We'll now write the total ionic equation, showing each ion in the aqueous solutions: \(\mathrm{Ba}^{2+}(eq)+2\mathrm{NO}_{3}^{-}(eq)+2\mathrm{K}^{+}(eq)+\mathrm{SO}_{4}^{2-}(eq) \longrightarrow \mathrm{BaSO}_4(s)+2\mathrm{K}^{+}(eq)+2\mathrm{NO}_3^-(eq)\)

Identify the spectator ions and write the balanced net ionic equation

The spectator ions are the ones that do not participate in the reaction; they are the same on both sides of the equation. In this case, the spectator ions are \(\mathrm{K}^{+}\) and \(\mathrm{NO}_{3}^{-}\). So, the balanced net ionic equation is: \(\mathrm{Ba}^{2+}(eq)+\mathrm{SO}_{4}^{2-}(eq) \longrightarrow \mathrm{BaSO}_4(s)\) (c) $\mathrm{Fe}\left(\mathrm{NO}_{3}\right)_{2}(aq)+\mathrm{KOH}(aq) \longrightarrow$

Predict the products

In this reaction, iron(II) nitrate (Fe(NO3)2) reacts with potassium hydroxide (KOH). The iron(II) ions (\(\mathrm{Fe}^{2+}\)) will react with the hydroxide ions (\(\mathrm{OH}^{-}\)), and the potassium ions (\(\mathrm{K}^{+}\)) will react with the nitrate ions (\(\mathrm{NO}_{3}^{-}\)). The resulting products will be iron(II) hydroxide and potassium nitrate.

Write the balanced molecular equation

Now we write the balanced molecular equation for this reaction: \(\mathrm{Fe}(\mathrm{NO}_3)_2(eq)+2\mathrm{KOH}(eq) \longrightarrow \mathrm{Fe(OH)}_2(s)+2\mathrm{KNO}_3(eq)\)

Write the total ionic equation

We'll now write the total ionic equation, showing each ion in the aqueous solutions: \(\mathrm{Fe}^{2+}(eq)+2\mathrm{NO}_{3}^{-}(eq)+2\mathrm{K}^{+}(eq)+2\mathrm{OH}^{-}(eq) \longrightarrow \mathrm{Fe(OH)}_2(s)+2\mathrm{K}^{+}(eq)+2\mathrm{NO}_3^-(eq)\)

Identify the spectator ions and write the balanced net ionic equation

The spectator ions are the ones that do not participate in the reaction; they are the same on both sides of the equation. In this case, the spectator ions are \(\mathrm{K}^{+}\) and \(\mathrm{NO}_{3}^{-}\). So, the balanced net ionic equation is: \(\mathrm{Fe}^{2+}(eq)+2\mathrm{OH}^{-}(eq) \longrightarrow \mathrm{Fe(OH)}_2(s)\)

Key concepts

Chemical Reactions

In understanding chemical reactions, we enter the fascinating world of how substances interact to form new products. A chemical reaction is a process that involves the rearrangement of the molecular or ionic structure of a substance, as opposed to a change in physical form or a nuclear reaction.

Every chemical reaction involves two key components: reactants and products. Reactants are the starting materials that react to produce new substances, while products are the end result of the reaction. In the given exercise, we observe that compounds like chromium sulfate and ammonium carbonate react to form new compounds such as ammonium sulfate and chromium carbonate.

Writing a balanced chemical equation is critical in depicting what is happening in the reaction. The equation must follow the law of conservation of mass, meaning the number of atoms of each element remains the same before and after the reaction. This ensures that the equation is balanced for both reactants and products. The step-by-step solutions for writing balanced equations for the given problems effectively demonstrate this practice. To aid understanding, think of a chemical equation as a recipe for a reaction, with specific amounts of reactants combining to yield precise quantities of products.

Spectator Ions

A deep dive into spectator ions reveals why not all participants in a chemical reaction undergo a transformation. Spectator ions are ions that exist in the same form on both the reactant and product sides of a chemical equation; they do not partake in the actual chemical change. Instead, they 'watch' the reaction occur without being changed or used up at all.

The significance of identifying spectator ions comes into play when writing net ionic equations. These equations show only the species that genuinely partake in the reaction, providing a clearer picture of the actual chemical changes occurring. In each of the reactions given in the exercise, certain ions remain unchanged and are thus labeled as spectators, like \(\mathrm{NH}_{4}^{+}\) and \(\mathrm{SO}_{4}^{2-}\) in the reaction between chromium sulfate and ammonium carbonate. An improved understanding of spectator ions can be achieved by always looking for ions that appear unchanged on both sides of the equation, as they are the ones simply 'observing' the action.

Balanced Equations

Mastering the concept of balanced equations is essential in chemistry to ensure that what is consumed as reactants has an equivalent expression in the products. To balance a chemical equation, we follow the principle that matter cannot be created or destroyed; it can only be transformed. This is reflected in the practice of balancing the number of atoms of each element involved in the reaction.

Let's review the procedure with the example: for reaction (a), the balanced molecular equation is meticulously composed to ensure that the number of every atom is the same on both sides. Then, to condense this information, we write the net ionic equation, omitting the spectator ions, to underscore the actual chemical change taking place. It's beneficial to approach balancing equations like a puzzle or mathematics problem, rearranging coefficients (never altering subscripts) until each atom is accounted for symmetrically on either side of the reaction arrow. This foundation in chemical problem-solving is an invaluable skill set for students as it underpins much of what's done in the study of chemistry.