Question

Write balanced ionic and net ionic equations for these reactions. (a) \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{CO}_{3}(a q)+\mathrm{BaCl}_{2}(a q) \longrightarrow\) $$ \mathrm{NH}_{4} \mathrm{Cl}(a q)+\mathrm{BaCO}_{3}(s) $$ (b) \(\mathrm{CuCl}_{2}(a q)+\mathrm{NaOH}(a q) \longrightarrow\) $$ \mathrm{Cu}(\mathrm{OH})_{2}(s)+\mathrm{NaCl}(a q) $$ (c) \(\mathrm{FeSO}_{4}(a q)+\mathrm{Na}_{3} \mathrm{PO}_{4}(a q) \longrightarrow\) \(\mathrm{Fe}_{3}\left(\mathrm{PO}_{4}\right)_{2}(s)+\mathrm{Na}_{2} \mathrm{SO}_{4}(a q)\) (d) \(\mathrm{AgC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}(a q)+\mathrm{NiCl}_{2}(a q) \longrightarrow\) \(\mathrm{AgCl}(s)+\mathrm{Ni}\left(\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\right)_{2}(a q)\)

Short answer

The net ionic equations for the reactions are (a) \(CO_3^{2-}(aq) + Ba^{2+}(aq) \longrightarrow BaCO_3(s)\), (b) \(Cu^{2+}(aq) + 2OH^{-}(aq) \longrightarrow Cu(OH)_2(s)\), (c) \(3Fe^{2+}(aq) + 2PO_4^{3-}(aq) \longrightarrow Fe_3(PO_4)_2(s)\), and (d) \(Ag^{+}(aq) + Cl^{-}(aq) \longrightarrow AgCl(s)\).

Step-by-step solution

Write the balanced molecular equation for reaction (a)

First, write out the molecular equation with the correct formulas for each compound, and then balance it.\[(\mathrm{NH}_4)_2CO_3(aq) + BaCl_2(aq) \longrightarrow 2NH_4Cl(aq) + BaCO_3(s)\] The equation is balanced as written; each side has 2 ammonium ions, 1 carbonate ion, 1 barium ion, and 2 chloride ions.

Write the complete ionic equation for reaction (a)

Next, write out all the ionic species as they exist in aqueous solution.\[2NH_4^{+}(aq) + CO_3^{2-}(aq) + Ba^{2+}(aq) + 2Cl^{-}(aq) \longrightarrow 2NH_4^{+}(aq) + 2Cl^{-}(aq) + BaCO_3(s)\]

Write the net ionic equation for reaction (a)

To find the net ionic equation, cancel out ions that appear on both sides of the reaction, as they are spectator ions.\[CO_3^{2-}(aq) + Ba^{2+}(aq) \longrightarrow BaCO_3(s)\]

Repeat steps for reaction (b)

Begin with the balanced molecular equation. Repeat the process of writing the complete ionic and net ionic equations.\[CuCl_2(aq) + 2NaOH(aq) \longrightarrow Cu(OH)_2(s) + 2NaCl(aq)\] Complete ionic equation:\[Cu^{2+}(aq) + 2Cl^{-}(aq) + 2Na^{+}(aq) + 2OH^{-}(aq) \longrightarrow Cu(OH)_2(s) + 2Na^{+}(aq) + 2Cl^{-}(aq)\] Net ionic equation:\[Cu^{2+}(aq) + 2OH^{-}(aq) \longrightarrow Cu(OH)_2(s)\]

Repeat steps for reaction (c)

Follow the same procedure for reaction (c). Balanced molecular equation:\[FeSO_4(aq) + 2Na_3PO_4(aq) \longrightarrow Fe_3(PO_4)_2(s) + 3Na_2SO_4(aq)\] Complete ionic equation:\[Fe^{2+}(aq) + SO_4^{2-}(aq) + 6Na^{+}(aq) + 2PO_4^{3-}(aq) \longrightarrow Fe_3(PO_4)_2(s) + 6Na^{+}(aq) + 3SO_4^{2-}(aq)\] Net ionic equation:\[3Fe^{2+}(aq) + 2PO_4^{3-}(aq) \longrightarrow Fe_3(PO_4)_2(s)\]

Repeat steps for reaction (d)

Apply the same method for the final reaction (d). Molecular equation:\[AgC_2H_3O_2(aq) + NiCl_2(aq) \longrightarrow 2AgCl(s) + Ni(C_2H_3O_2)_2(aq)\] Complete ionic equation:\[Ag^{+}(aq) + C_2H_3O_2^{-}(aq) + Ni^{2+}(aq) + 2Cl^{-}(aq) \longrightarrow 2AgCl(s) + Ni^{2+}(aq) + 2C_2H_3O_2^{-}(aq)\] Net ionic equation:\[Ag^{+}(aq) + Cl^{-}(aq) \longrightarrow AgCl(s)\]

Key concepts

Balanced Ionic Equations

Balanced ionic equations are a cornerstone of understanding chemical reactions. These equations showcase the conservation of mass and charge by ensuring that the number of atoms of each element, and the total charge, are the same on both sides of the reaction equation. This balance is crucial for accurately depicting what occurs during a chemical reaction.

When writing a balanced ionic equation, we separate the compounds into their constituent ions if they are soluble in water (aqueous). This helps in visualizing the actual species that are interacting in the solution. It's important to recognize that some compounds, typically solids, liquids, gases, or weak electrolytes, do not dissociate into ions and are written in their molecular forms. For example, in the precipitation reaction (a), \(BaCO_3(s)\) does not split into ions because it is insoluble and forms a precipitate.

Using subscripts and coefficients, the equation is balanced to ensure that the number of atoms for each element is the same on both the reactant and product sides. Coefficients adjust the number of molecules, and subscripts represent the number of atoms within a molecule or a polyatomic ion. In essence, each atom and charge is accounted for from the reactants (start of reaction) to the products (end of reaction).

Chemical Reaction Stoichiometry

Chemical reaction stoichiometry refers to the quantitative relationship between the substances involved in a chemical reaction. It is based on the law of conservation of mass, which states that in a chemical reaction, the total mass of the reactants equals the total mass of the products. Stoichiometry allows chemists to predict the amount of product that will form in a reaction from a given amount of reactants, as well as to determine the necessary amounts of reactants to create a desired quantity of product.

In the context of the balanced equations provided in the exercise, stoichiometry involves using the coefficients of each substance to understand their molar ratios. For instance, in reaction (c), \(2Na_3PO_4\) react with \(FeSO_4\) to produce \(Fe_3(PO_4)_2\) and \(3Na_2SO_4\), indicating a stoichiometric ratio of 2:1:1:3. These ratios are crucial when determining how much of each reactant is needed and can be used to calculate theoretical yields—how much product should form under ideal conditions.

Precipitation Reactions

Precipitation reactions are a type of chemical reaction where two solutions are mixed, resulting in the formation of an insoluble solid called a precipitate. A classic sign of a precipitation reaction is the sudden appearance of a cloudy or solid substance in what was previously a clear solution. This occurs when the product of the reactants' interaction is not soluble in the solvent, typically water.

In our example reaction (a), \(BaCO_3\) forms as a precipitate when solutions of \((NH_4)_2CO_3\) and \(BaCl_2\) are mixed. The net ionic equation provides insight into the driving force of the reaction—the combination of \(Ba^{2+}\) ions and \(CO_3^{2-}\) ions to form the insoluble \(BaCO_3\). Students can identify the precipitate in a reaction by referring to solubility rules, which indicate whether common ionic compounds are soluble or insoluble in water. Understanding these rules is vital for predicting which products will precipitate in a reaction.

Precipitation reactions are not only important in academic exercises but also have real-world applications such as in the treatment of wastewater, where they are used to remove harmful ions from the water.