Problem 77 - Chemistry The Science in Context

Question

Complete and balance the molecular equations for the precipitation reactions, if any, between the following pairs of reactants, and write the overall and net ionic equations. a. \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}(a q)+\mathrm{Na}_{2} \mathrm{SO}_{4}(a q) \rightarrow\) b. \(\mathrm{NiCl}_{2}(a q)+\mathrm{NH}_{4} \mathrm{NO}_{3}(a q) \rightarrow\) c. \(\operatorname{Fe} C l_{2}(a q)+N a_{2} S(a q) \rightarrow\) d. \(\operatorname{MgSO}_{4}(a q)+\mathrm{BaCl}_{2}(a q) \rightarrow\)

Step-by-Step Solution

Answer

For each given pair of reactants, we identified potential products using solubility rules, wrote and balanced the molecular equations, and wrote the overall and net ionic equations. The net ionic equations for each reaction are: a. \(\mathrm{Pb}^{2+}(a q)+\mathrm{SO}_{4}^{2-}(a q) \rightarrow \mathrm{PbSO}_{4}(s)\) b. No precipitation reaction occurred, as both potential products were soluble. c. \(\mathrm{Fe}^{2+}(a q)+\mathrm{S}^{2-}(a q) \rightarrow \mathrm{FeS}(s)\) d. \(\mathrm{Ba}^{2+}(a q)+\mathrm{SO}_{4}^{2-}(a q) \rightarrow \operatorname{BaSO}_{4}(s)\)

Step 1: a. Identify potential products using solubility rules

For the first reaction, the potential products are \(\mathrm{PbSO}_{4}\) and \(\mathrm{NaNO}_{3}\). Use solubility rules to determine if any precipitation occurs.

Step 2: a. Write the molecular equation

\(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}(a q)+\mathrm{Na}_{2}\mathrm{SO}_{4}(a q) \rightarrow \mathrm{PbSO}_{4}(s)+2 \mathrm{NaNO}_{3}(a q)\)

Step 3: a. Balance the molecular equation

The molecular equation is already balanced.

Step 4: a. Write the overall ionic equation

\(\mathrm{Pb}^{2+}(a q)+2 \mathrm{NO}_{3}^{-}(a q)+2 \mathrm{Na}^{+}(a q)+\mathrm{SO}_{4}^{2-}(a q) \rightarrow \mathrm{PbSO}_{4}(s)+2 \mathrm{Na}^{+}(a q)+2 \mathrm{NO}_{3}^{-}(a q)\)

Step 5: a. Write the net ionic equation

\(\mathrm{Pb}^{2+}(a q)+\mathrm{SO}_{4}^{2-}(a q) \rightarrow \mathrm{PbSO}_{4}(s)\) b, c, and d will follow the same steps as above.

Step 6: b. Identify potential products using solubility rules

For the second reaction, the potential products are \(\mathrm{NiNO}_{3}\) and \(\mathrm{NH}_{4}\mathrm{Cl}\). No precipitation occurs in this reaction, as both potential products are soluble. Therefore, there's no precipitation reaction and we can skip the rest of the steps for this reaction.

Step 7: c. Identify potential products using solubility rules

For the third reaction, the potential products are \(\mathrm{FeS}\) and \(\mathrm{NaCl}\).

Step 8: c. Write the molecular equation

\(\operatorname{Fe} C l_{2}(a q)+N a_{2} S(a q) \rightarrow \operatorname{Fe} S(s)+2 \mathrm{NaCl}(a q)\)

Step 9: c. Balance the molecular equation

The molecular equation is already balanced.

Step 10: c. Write the overall ionic equation

\(\mathrm{Fe}^{2+}(a q)+2 \mathrm{Cl}^{-}(a q)+2 \mathrm{Na}^{+}(a q)+\mathrm{S}^{2-}(a q) \rightarrow \mathrm{FeS}(s)+2 \mathrm{Na}^{+}(a q)+2 \mathrm{Cl}^{-}(a q)\)

Step 11: c. Write the net ionic equation

\(\mathrm{Fe}^{2+}(a q)+\mathrm{S}^{2-}(a q) \rightarrow \mathrm{FeS}(s)\)

Step 12: d. Identify potential products using solubility rules

For the fourth reaction, the potential products are \(\mathrm{MgCl}_{2}\) and \(\mathrm{BaSO}_{4}\).

Step 13: d. Write the molecular equation

\(\operatorname{MgSO}_{4}(a q)+\mathrm{BaCl}_{2}(a q) \rightarrow \mathrm{MgCl}_{2}(a q)+\operatorname{BaSO}_{4}(s)\)

Step 14: d. Balance the molecular equation

The molecular equation is already balanced.

Step 15: d. Write the overall ionic equation

\(\mathrm{Mg}^{2+}(a q)+\mathrm{SO}_{4}^{2-}(a q)+\mathrm{Ba}^{2+}(a q)+2 \mathrm{Cl}^{-}(a q) \rightarrow \mathrm{Mg}^{2+}(a q)+2 \mathrm{Cl}^{-}(a q)+\operatorname{BaSO}_{4}(s)\)

Step 16: d. Write the net ionic equation

\(\mathrm{Ba}^{2+}(a q)+\mathrm{SO}_{4}^{2-}(a q) \rightarrow \operatorname{BaSO}_{4}(s)\)

Key Concepts

Solubility Rules

Understanding solubility rules is essential for predicting the outcome of precipitation reactions in chemistry. Solubility rules provide guidelines to help us determine whether a compound will dissolve in water (soluble) or form a solid (insoluble, often forming a precipitate).

For example, most nitrates (\texttt{NO}\(_3^-\)) are soluble, while most sulfates (\texttt{SO}\(_4^{2-}\)) are soluble except for those of barium, strontium, and lead. Knowing these rules can help us predict that lead sulfate (\texttt{PbSO}\(_4\)), for example, will likely form a precipitate. This is exactly what we see in our first reaction, where lead nitrate reacts with sodium sulfate resulting in the formation of solid lead sulfate. We can refer to these guidelines to decisively conclude if a precipitation will happen when certain ions are mixed.

Molecular Equations

A molecular equation shows the complete chemical formulas of the reactants and the products without indicating their ionic character. This type of equation provides an overall picture of the chemical reaction involved.

When writing a molecular equation, it is important to ensure the reactants and products are correctly written based on the principle of conservation of mass—meaning the number of atoms of each element must be the same on both sides of the equation. In our exercise's example, we see the molecular equation for the reaction of lead(II) nitrate with sodium sulfate, forming lead sulfate and sodium nitrate.

Ionic Equations

Moving beyond the molecular approach, ionic equations give us a clearer picture of how the reactants are ionized in solution. Soluble substances are represented by their constituent ions, while insoluble substances are written in their molecular form since they do not dissociate.

An ionic equation can highlight the species that are actually involved in the reaction, helping to illuminate the precise changes that occur during the chemical process. For instance, lead nitrate and sodium sulfate are presented as ions in the ionic equation because they are soluble, but lead sulfate, being insoluble, remains in its molecular form. This approach further aids in tracing the movement of individual ions during the course of the reaction.

Net Ionic Equations

The net ionic equation simplifies the ionic equation even further by removing the spectator ions—those that do not participate in the actual precipitation reaction. These equations highlight the essence of the chemical change occurring in a reaction.

In the net ionic equation, only the ions that form the precipitate are displayed. This method clearly shows the transformation without excess information, leading to a straightforward representation of the reaction. For example, in the reaction between lead(II) nitrate and sodium sulfate, the net ionic equation only includes the lead ion and the sulfate ion coming together to form lead sulfate, omitting the sodium and nitrate ions as they remain in solution and are not involved in the formation of the precipitate.

Chemical Reaction Balancing

The principle of the conservation of mass dictates that chemical reactions must be balanced—meaning the number of atoms for each element in the reactants side must equal the number in the products side. While our textbook examples show reactions that are already balanced, it is crucial to understand how to achieve this.

Start by writing down the unbalanced equation. Count the number of atoms of each element on both sides. Use coefficients to balance the elements one at a time, checking and adjusting coefficients where necessary to achieve balance. It's important not to change the subscripts of the compounds, as this would change their identity. Instead, only adjust the coefficients, which indicate the number of molecules or moles involved in the reaction. Balancing chemical equations ensures that the law of conservation of mass is followed, a fundamental concept in chemistry.