Question

The chemical reactions that are used in the anion spot tests in this experiment are for the most part simple precipitation or acid-basc reactions. Given the information in cach test procedure, try to write the net ionic equation for the key reaction in each test. A. \(\mathrm{CO}_{3}^{2-}\) B. \(\mathrm{SO}_{4}^{2-}\) C. \(\mathrm{PO}_{4}^{3-}\) (Reactants are \(\mathrm{HPO}_{4}^{2-}, \mathrm{NH}_{4}^{+}, \mathrm{MoO}_{4}^{2-},\) and \(\mathrm{H}^{+} ;\) products are \(\left(\mathrm{NH}_{4}\right)_{3} \mathrm{PO}_{4} \cdot 12 \mathrm{MoO}_{3}\) and \(\mathrm{H}_{2} \mathrm{O}\) no oxidation or reduction occurs.) D. \(\mathrm{SO}_{3}^{2-}\) E. \(\mathrm{SCN}^{-}\) F. \(\mathrm{Cl}^{-}\) G. \(\mathrm{NO}_{3}^{-}\) (Take as reactants \(\mathrm{Al}\) and \(\mathrm{NO}_{3}\); as products \(\mathrm{NH}_{3}\) and \(\mathrm{AlO}_{2}\); final cquation also contains \(\mathrm{OH}^{-}\) and \(\mathrm{H}_{2} \mathrm{O}\) as reactants.)

Short answer

To summarize, here are the net ionic equations for each anion test: A. Carbonate Ion: \(Ca^{2+}(aq) + CO_{3}^{2-}(aq) \rightarrow CaCO_{3}(s)\) B. Sulfate Ion: \(Ba^{2+}(aq) + SO_{4}^{2-}(aq) \rightarrow BaSO_{4}(s)\) C. Phosphate Ion: \(3NH_{4}^{+}(aq) + HPO_{4}^{2-}(aq) + 12MoO_{4}^{2-}(aq) \rightarrow (NH_{4})_{3}PO_{4}\cdot12MoO_{3}(s) + H_{2}O(l)\) D. Sulfite Ion: \(Ba^{2+}(aq) + SO_{3}^{2-}(aq) \rightarrow BaSO_{3}(s)\) E. Thiocyanate Ion: \(3SCN^{-}(aq) + Fe^{3+}(aq) \rightarrow Fe(SCN)_{3}(aq)\) F. Chloride Ion: \(Ag^{+}(aq) + Cl^{-}(aq) \rightarrow AgCl(s)\) G. Nitrate Ion: \(Al(s) + 3NO_{3}^{-}(aq) + 6OH^{-}(aq) \rightarrow NH_{3}(aq) + 3AlO_{2}^{-}(aq) + 3H_{2}O(l)\)

Step-by-step solution

A. Carbonate Ion, CO3²⁻

We have to write the net ionic equation for the reaction involving the carbonate ion, CO3²⁻. Suppose the carbonate ion reacts with calcium ion (Ca²⁺) to form a precipitate of calcium carbonate. The balanced molecular equation is: \(Ca^{2+}(aq) + CO_{3}^{2-}(aq) \rightarrow CaCO_{3}(s)\) Now, we'll write the complete ionic equation: \(Ca^{2+}(aq) + CO_{3}^{2-}(aq) \rightarrow CaCO_{3}(s)\) There are no spectator ions, so the net ionic equation is: \(Ca^{2+}(aq) + CO_{3}^{2-}(aq) \rightarrow CaCO_{3}(s)\)

B. Sulfate Ion, SO4²⁻

Similarly, let's consider the reaction of sulfate ion (SO4²⁻) with barium ion (Ba²⁺) to form a precipitate of barium sulfate. The balanced molecular equation is: \(Ba^{2+}(aq) + SO_{4}^{2-}(aq) \rightarrow BaSO_{4}(s)\) Now, we'll write the complete ionic equation: \(Ba^{2+}(aq) + SO_{4}^{2-}(aq) \rightarrow BaSO_{4}(s)\) There are no spectator ions, so the net ionic equation is: \(Ba^{2+}(aq) + SO_{4}^{2-}(aq) \rightarrow BaSO_{4}(s)\)

C. Phosphate Ion, PO4³⁻

For the reaction involving phosphate ion (PO4³⁻) and ammonium ion (NH4⁺), we're given the reactants and products. The balanced molecular equation is: \(3NH_{4}^{+}(aq) + HPO_{4}^{2-}(aq) + 12MoO_{4}^{2-}(aq) \rightarrow (NH_{4})_{3}PO_{4}\cdot12MoO_{3}(s) + H_{2}O(l)\) Now, we'll write the complete ionic equation: \(3NH_{4}^{+}(aq) + HPO_{4}^{2-}(aq) + 12MoO_{4}^{2-}(aq) \rightarrow (NH_{4})_{3}PO_{4}\cdot12MoO_{3}(s) + H_{2}O(l)\) There are no spectator ions, so the net ionic equation is: \(3NH_{4}^{+}(aq) + HPO_{4}^{2-}(aq) + 12MoO_{4}^{2-}(aq) \rightarrow (NH_{4})_{3}PO_{4}\cdot12MoO_{3}(s) + H_{2}O(l)\)

D. Sulfite Ion, SO3²⁻

Let's consider the reaction of sulfite ion (SO3²⁻) with barium ion (Ba²⁺) to form a precipitate of barium sulfite. The balanced molecular equation is: \(Ba^{2+}(aq) + SO_{3}^{2-}(aq) \rightarrow BaSO_{3}(s)\) Now, we'll write the complete ionic equation: \(Ba^{2+}(aq) + SO_{3}^{2-}(aq) \rightarrow BaSO_{3}(s)\) There are no spectator ions, so the net ionic equation is: \(Ba^{2+}(aq) + SO_{3}^{2-}(aq) \rightarrow BaSO_{3}(s)\)

E. Thiocyanate Ion, SCN⁻

Let's consider the reaction of thiocyanate ion (SCN⁻) with iron(III) ion (Fe³⁺) to form the complex ion iron(III) thiocyanate. The balanced molecular equation is: \(3SCN^{-}(aq) + Fe^{3+}(aq) \rightarrow Fe(SCN)_{3}(aq)\) Now, we'll write the complete ionic equation: \(3SCN^{-}(aq) + Fe^{3+}(aq) \rightarrow Fe(SCN)_{3}(aq)\) There are no spectator ions, so the net ionic equation is: \(3SCN^{-}(aq) + Fe^{3+}(aq) \rightarrow Fe(SCN)_{3}(aq)\)

F. Chloride Ion, Cl⁻

Let's consider the reaction of chloride ion (Cl⁻) with silver ion (Ag⁺) to form a precipitate of silver chloride. The balanced molecular equation is: \(Ag^{+}(aq) + Cl^{-}(aq) \rightarrow AgCl(s)\) Now, we'll write the complete ionic equation: \(Ag^{+}(aq) + Cl^{-}(aq) \rightarrow AgCl(s)\) There are no spectator ions, so the net ionic equation is: \(Ag^{+}(aq) + Cl^{-}(aq) \rightarrow AgCl(s)\)

G. Nitrate Ion, NO3⁻

For the reaction involving nitrate ion (NO3⁻) and aluminum (Al), we're given the reactants and products. The balanced molecular equation is: \(Al(s) + 3NO_{3}^{-}(aq) + 6OH^{-}(aq) \rightarrow NH_{3}(aq) + 3AlO_{2}^{-}(aq) + 3H_{2}O(l)\) Now, we'll write the complete ionic equation: \(Al(s) + 3NO_{3}^{-}(aq) + 6OH^{-}(aq) \rightarrow NH_{3}(aq) + 3AlO_{2}^{-}(aq) + 3H_{2}O(l)\) There are no spectator ions, so the net ionic equation is: \(Al(s) + 3NO_{3}^{-}(aq) + 6OH^{-}(aq) \rightarrow NH_{3}(aq) + 3AlO_{2}^{-}(aq) + 3H_{2}O(l)\)

Key concepts

Precipitation Reactions

When we mix two solutions and observe the formation of a solid, it's often the result of a precipitation reaction. These reactions are paramount in chemistry because they allow for the separation of different ions from a solution. Precipitates are insoluble ionic compounds that form when two solutions containing soluble salts are mixed, and one of the product salts has a very low solubility in water.

For example, combining a solution of calcium ions with one containing carbonate ions results in the formation of calcium carbonate as a precipitate. The net ionic equation, which only includes the species that undergo a chemical change, confirms this: \(Ca^{2+}(aq) + CO_{3}^{2-}(aq) \rightarrow CaCO_{3}(s)\). Here, the aqueous ions come together to form a solid carbonate that settles out of solution. Understanding these equations helps students discern between the reactants and the products, which is crucial for analyzing and predicting chemical reactions.

Acid-Base Reactions

Acid-base reactions, also known as neutralization reactions, involve an acid and a base that react to form water and a salt. This classic chemistry event usually results in the formation of a molecule of water (H₂O) and an ionic compound known as a salt. These reactions are fundamental to various processes, from industrial manufacturing to biological systems.

For instance, when hydrochloric acid (\(HCl\)) is neutralized by sodium hydroxide (\(NaOH\)), the result is sodium chloride (\(NaCl\)) and water. For students, mastering the net ionic equations of these reactions, such as \(H^{+}(aq) + OH^{-}(aq) \rightarrow H_{2}O(l)\), is essential for understanding how acids and bases interact at the ionic level to create neutral compounds.

Chemical Equilibrium

Chemical equilibrium occurs when a chemical reaction's forward and reverse reactions proceed at the same rate, resulting in no net change in the concentration of reactants and products over time. It's a dynamic state, not a static one, with continuous movement of molecules forming reactants and products, but in such a way that the macroscopic properties remain constant.

For students learning about chemical equilibrium, it's important to recognize that reactions can reach an equilibrium position that favors either the reactants or the products. This position is expressed by the equilibrium constant, K, which reflects the ratio of the concentration of products to the concentration of reactants. The ability to calculate and interpret the equilibrium constant provides insight into the extent of a reaction and helps predict how changes in conditions, such as concentration or temperature, can shift the equilibrium according to Le Chatelier's principle.