Question

Prediet whether reactions should occur between aqueous solutions of the following compounds. If so, write balanced molecular and net ionic equations for the reactions. Include physical states. (a) \(\mathrm{Sr}\left(\mathrm{NO}_{3}\right)_{2}\) and \(\mathrm{H}_{2} \mathrm{SO}_{4}\) (b) \(\mathrm{Zn}\left(\mathrm{NO}_{1}\right)_{2}\) and \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) (c) \(\mathrm{CuSO}_{4}\) and \(\mathrm{BaS}\) (d) \(\mathrm{NaHCO}_{5}\) and \(\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}\)

Short answer

Reactions occur for combinations (a), (c) and (d). For combinations (a) and (c), the reactions are double displacement, while for (d), the reaction is neutralization. Combination (b) doesn't undergo any reactions.

Step-by-step solution

Decide the Types of the Reaction

Let's check each pairs individually: (a) \(\mathrm{Sr(NO_3)_2}\) and \(\mathrm{H_2SO_4}\): These salts will break into their respective ions in the water. Sr(NO3)2 breaks into Sr^2+ and NO3^- ions. H2SO4 breaks into 2H+ and SO4^2- ions. We'll have a double displacement reaction here. Using rules of solubility, SrSO4 is not soluble in water. So, a reaction will occur. (b) \(\mathrm{Zn(NO_3)_2}\) and \(\mathrm{Na_2SO_4}\): Similarly, this will also involve a double displacement reaction. ZnSO4 produced as a possible product is soluble, hence no reaction. (c) \(\mathrm{CuSO_4}\) and \(\mathrm{BaS}\): Here, we'll have double displacement reaction. Possible product CuS isn't soluble hence a reaction will take place. (d) \(\mathrm{NaHCO_3}\) and \(\mathrm{CH_{3}CO_2H}\): This will be a neutralization reaction since it involves an acid and a base. Reaction will take place.

Write Balanced Molecular and Net Ionic Equations

(a) Molecular equation: \(\mathrm{Sr(NO_3)_2} + \mathrm{H_2SO_4} \rightarrow \mathrm{SrSO_4} + 2\mathrm{HNO_3}\) Ionic equation: \(\mathrm{Sr^{2+} + SO_4^{2-}} \rightarrow \mathrm{SrSO_4(s)}\) (b) No reaction. (c) Molecular equation: \(\mathrm{CuSO_4 + BaS \rightarrow CuS + BaSO_4}\) Ionic equation: \(\mathrm{Cu^{2+} + S^{2-} \rightarrow CuS(s)}\) (d) Molecular equation: \(\mathrm{NaHCO_3 + CH_3CO_2H \rightarrow CH_3CO_2Na + H_2O + CO_2}\) Ionic equation: \(\mathrm{HCO_3^- + H^+ \rightarrow H_2O + CO_2}\)

Key concepts

Molecular Equations

Molecular equations provide a broad overview of a chemical reaction, showing the complete form of the reactants and products. They’re essential when you first learn about a reaction because they summarize the process without focusing on individual ions or changes that occur in the solution.
For instance, when strontium nitrate (\(\text{Sr(NO}_3 ext{)}_2 \)) reacts with sulfuric acid (\(\text{H}_2\text{SO}_4 \)), the molecular equation is: \[\text{Sr(NO}_3 ext{)}_2 + \text{H}_2\text{SO}_4 \rightarrow \text{SrSO}_4 + 2\text{HNO}_3\]
This equation outlines how molecules combine and what new compounds they form. It helps in predicting the outcome of reactions based on the initial materials involved.
Molecular equations are a clear communication tool - showing which compounds participate in a reaction and resulting products.

Net Ionic Equations

Net ionic equations strip away the spectator ions and focus solely on the ions participating in the reaction. This provides a fuller understanding of the actual chemical changes occurring.
Start by breaking down the molecular equation into its ionic components, then remove any ions that don’t change from reactant to product:

• Ions that do not participate in the reaction are known as spectator ions.

For example, in our \(\text{Sr(NO}_3 ext{)}_2\) and \(\text{H}_2\text{SO}_4\) reaction, the net ionic equation becomes: \[ \text{Sr}^{2+} + \text{SO}_4^{2-} \rightarrow \text{SrSO}_4(s) \]
Net ionic equations highlight the driving force behind a reaction. They are beneficial in understanding why reactions occur, showing only the production and consumption of ions that form a precipitate, gas, or water.

Double Displacement Reactions

Double displacement reactions are a type of chemical reaction where two compounds exchange ions to form two new compounds. These reactions are fundamental in predicting the outcome of mixing different solutions.
Typically, these reactions occur in aqueous solutions and lead to the formation of a precipitate, gas, or non-electrolyte. They follow a general form: \[ AB + CD \rightarrow AD + CB \]
The reaction between \(\text{CuSO}_4 \) and \(\text{BaS} \) is an excellent example, where \(\text{Cu}^{2+}\) and \(\text{S}^{2-}\) form insoluble \(\text{CuS} \).
These reactions are prevalent in everyday life and various industries, from wastewater treatment to the manufacturing of ceramics, and even in the biology of living organisms.
Essentially, understanding double displacement can facilitate learning more complex reaction types.

Solubility Rules

Solubility rules are essential to predicting whether a precipitation reaction will occur during mixing solutions. They serve as guidelines for identifying if certain compounds are soluble or insoluble in water.
For instance, any compound containing alkali metal ions or ammonium (like \(\text{NH}_4^+\)) is usually soluble. Additionally, nitrates and acetates are generally soluble.

• Key exceptions can include salts like \(\text{BaSO}_4\), \(\text{CaSO}_4\), and \(\text{PbSO}_4\), which tend to be insoluble despite containing sulfate ions.

In the reaction of \(\text{Sr(NO}_3 ext{)}_2 \) and \(\text{H}_2\text{SO}_4 \), the formation of \(\text{SrSO}_4 \) as a precipitate is predicted using these rules, because \(\text{SrSO}_4 \) is insoluble.
Applying solubility rules allows chemists to predict and balance reactions more effectively.
They help in identifying potential outcomes and are crucial in laboratory settings where predicting the results of new reactions is necessary.