Problem 24 - Chemistry The Central Science

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) \(\mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2}(a q)+\mathrm{K}_{2} \mathrm{SO}_{4}(a q) \longrightarrow\) (c) \(\mathrm{Fe}\left(\mathrm{NO}_{3}\right)_{2}(a q)+\mathrm{KOH}(a q) \longrightarrow\).

Step-by-Step Solution

Answer

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

Step 1: (a) Predict the products of the reaction

For the given reaction: \(\mathrm{Cr}_{2}\left(\mathrm{SO}_{4}\right)_{3}(a q)+\left(\mathrm{NH}_{4}\right)_{2} \mathrm{CO}_{3}(a q) \longrightarrow\) We can predict the products by exchanging the cations and anions: Chromium will combine with carbonate ion, and ammonium will combine with sulfate ion. So the products will be chromium(III) carbonate (Cr2(CO3)3) and ammonium sulfate ((NH4)2SO4).

Step 2: (a) Write the balanced molecular equation

The balanced molecular equation for this reaction is: $$ \mathrm{Cr}_{2}\left(\mathrm{SO}_{4}\right)_{3}(a q)+2\left(\mathrm{NH}_{4}\right)_{2} \mathrm{CO}_{3}(a q) \longrightarrow\mathrm{Cr}_{2}(\mathrm{CO}_{3})_{3}(s)+2(\mathrm{NH}_{4})_{2}\mathrm{SO}_{4}(a q) $$

Step 3: (a) Write the total ionic equation

The total ionic equation is obtained by breaking all the aqueous compounds into their respective ions: $$ 2\mathrm{Cr}^{3+}(a q)+3\mathrm{SO}_{4}^{2-}(a q)+4\mathrm{NH}_{4}^{+}(a q)+2\mathrm{CO}_{3}^{2-}(a q) \longrightarrow\mathrm{Cr}_{2}(\mathrm{CO}_{3})_{3}(s)+4\mathrm{NH}_{4}^{+}(a q)+2\mathrm{SO}_{4}^{2-}(a q) $$

Step 4: (a) Write the net ionic equation

The net ionic equation is obtained by canceling out the spectator ions found on both sides of the total ionic equation. In this case, the spectator ions are \(\mathrm{NH}_{4}^{+}(a q)\) and \(\mathrm{SO}_{4}^{2-}(a q)\). $$ 2\mathrm{Cr}^{3+}(a q)+2\mathrm{CO}_{3}^{2-}(a q) \longrightarrow\mathrm{Cr}_{2}(\mathrm{CO}_{3})_{3}(s) $$

Step 5: (b) Predict the products of the reaction

For the given reaction: \(\mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2}(a q)+\mathrm{K}_{2}\mathrm{SO}_{4}(a q) \longrightarrow\) We can predict the products by exchanging the cations and anions: Barium will combine with sulfate ion, and potassium will combine with nitrate ion. So the products will be barium sulfate (BaSO4) and potassium nitrate (KNO3).

Step 6: (b) Write the balanced molecular equation

The balanced molecular equation for this reaction is: $$ \mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2}(a q)+\mathrm{K}_{2}\mathrm{SO}_{4}(a q) \longrightarrow \mathrm{BaSO}_{4}(s)+2\mathrm{KNO}_{3}(a q) $$

Step 7: (b) Write the total ionic equation

The total ionic equation is obtained by breaking all the aqueous compounds into their respective ions: $$ \mathrm{Ba}^{2+}(a q)+2\mathrm{NO}_{3}^{-}(a q)+2\mathrm{K}^{+}(a q)+\mathrm{SO}_{4}^{2-}(a q) \longrightarrow\mathrm{BaSO}_{4}(s)+2\mathrm{K}^{+}(a q)+2\mathrm{NO}_{3}^{-}(a q) $$

Step 8: (b) Write the net ionic equation

The net ionic equation is obtained by canceling out the spectator ions found on both sides of the total ionic equation. In this case, the spectator ions are \(\mathrm{K}^{+}(a q)\) and \(\mathrm{NO}_{3}^{-}(a q)\). $$ \mathrm{Ba}^{2+}(a q)+\mathrm{SO}_{4}^{2-}(a q) \longrightarrow \mathrm{BaSO}_{4}(s) $$

Step 9: (c) Predict the products of the reaction

For the given reaction: \(\mathrm{Fe}\left(\mathrm{NO}_{3}\right)_{2}(a q)+\mathrm{KOH}(a q) \longrightarrow\) We can predict the products by exchanging the cations and anions: Iron will combine with hydroxide ion, and potassium will combine with nitrate ion. So the products will be iron(II) hydroxide (Fe(OH)2) and potassium nitrate (KNO3).

Step 10: (c) Write the balanced molecular equation

The balanced molecular equation for this reaction is: $$ \mathrm{Fe}\left(\mathrm{NO}_{3}\right)_{2}(a q)+2\mathrm{KOH}(a q) \longrightarrow\mathrm{Fe}(\mathrm{OH})_{2}(s)+2\mathrm{KNO}_{3}(a q) $$

Step 11: (c) Write the total ionic equation

The total ionic equation is obtained by breaking all the aqueous compounds into their respective ions: $$ \mathrm{Fe}^{2+}(a q)+2\mathrm{NO}_{3}^{-}(a q)+2\mathrm{K}^{+}(a q)+2\mathrm{OH}^{-}(a q) \longrightarrow\mathrm{Fe}(\mathrm{OH})_{2}(s)+2\mathrm{K}^{+}(a q)+2\mathrm{NO}_{3}^{-}(a q) $$

Step 12: (c) Write the net ionic equation

The net ionic equation is obtained by canceling out the spectator ions found on both sides of the total ionic equation. In this case, the spectator ions are \(\mathrm{K}^{+}(a q)\) and \(\mathrm{NO}_{3}^{-}(a q)\). $$ \mathrm{Fe}^{2+}(a q)+2\mathrm{OH}^{-}(a q) \longrightarrow\mathrm{Fe}(\mathrm{OH})_{2}(s) $$

Key Concepts

Spectator Ions

In any given chemical reaction, spectator ions are ions that do not participate in the formation of the precipitate and are present unchanged on both sides of the reaction equation. Identifying them is crucial for simplifying a reaction to its net ionic form. Unlike other ions, spectator ions don't directly affect the reaction or the outcome, serving as bystanders in the process.
For instance, in reaction (a)

• The ions \( \mathrm{NH}_{4}^{+} \) and \( \mathrm{SO}_{4}^{2-} \)are identified as spectator ions because they appear on both sides of the total ionic equation.

This means they do not participate in forming the solid product of \( \mathrm{Cr}_{2}(\mathrm{CO}_{3})_{3}(s) \).
Such ions are essential to consider when writing the net ionic equation as they can be "cancelled out" or omitted. Doing this helps in understanding the actual chemical transformation taking place. Maintaining the role of spectator ions helps illustrate the flow and dissociation of ions in solution.

Precipitation Reactions

Precipitation reactions occur when two aqueous solutions are mixed, and an insoluble solid, known as a precipitate, forms. This is a critical concept in chemistry, as it demonstrates how insoluble materials emerge in solutions from the reaction of known soluble reactants.
An example from the exercise is reaction (b):

• Barium ions \( \mathrm{Ba}^{2+} \)and sulfate ions \( \mathrm{SO}_{4}^{2-} \)combine to form the precipitate barium sulfate \( \mathrm{BaSO}_{4}(s) \).

This reaction illustrates how the ionic compounds in the solution can transform into a solid.
Precipitation reactions are typically governed by solubility rules because only specific ions form precipitates. It's essential to understand these principles to determine when a reaction will produce a precipitate. Knowing which compounds will form a precipitate helps predict the outcome of mixing solutions.

Chemical Equations

Chemical equations are written representations of chemical reactions, showing the reactants and products. They must be balanced to reflect the law of conservation of mass, which dictates that matter is neither created nor destroyed. The balanced molecular equation paints the complete picture, showing all reactants and products.
Taking reaction (c) as an example:

• The balanced molecular equation\( \mathrm{Fe}\left(\mathrm{NO}_{3}\right)_{2}(aq)+2\mathrm{KOH}(aq) \longrightarrow \mathrm{Fe}(\mathrm{OH})_{2}(s)+2\mathrm{KNO}_{3}(aq) \)

Clearly outlines the transformation and interaction of ions.
Once balanced, these equations can be broken down into total and net ionic equations. Total ionic equations split aqueous compounds into their respective ions. The net ionic equation, on the other hand, strips away the spectator ions, highlighting the essence of the reaction. Practicing these steps with chemical equations enhances understanding of how reactants combine and what products they form.