Question

Write balanced complete ionic and net ionic equations for each reaction. \(\mathrm{a} . \mathrm{K}_{2} \mathrm{SO}_{4}(a q)+\mathrm{Cal}_{2}(a q) \longrightarrow \mathrm{CaSO}_{4}(s)+\mathrm{KI}(a q)\) b. \(\mathrm{NH}_{4} \mathrm{Cl}(a q)+\mathrm{NaOH}(a q) \longrightarrow\) \(\mathrm{H}_{2} \mathrm{O}(l)+\mathrm{NH}_{3}(g)+\mathrm{NaCl}(a q)\) c. \(\operatorname{AgNO}_{3}(a q)+\mathrm{NaCl}(a q) \longrightarrow \mathrm{AgCl(s} )+\mathrm{NaNO}_{3}(a q)\) \(\begin{aligned} \text { d. } \mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}(a q)+& \mathrm{K}_{2} \mathrm{CO}_{3}(a q) \\ & \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{CO}_{2}(g)+\mathrm{KC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}(a q) \end{aligned}\)

Short answer

a. Net Ionic: \((\mathrm{Ca}^{2+}(aq) + \mathrm{SO}_{4}^{2-}(aq)) \longrightarrow (\mathrm{CaSO}_{4}(s))\); b. Net Ionic: \((\mathrm{NH}_{4}^{+}(aq) + \mathrm{OH}^{-}(aq)) \longrightarrow (\mathrm{NH}_{3}(g) + \mathrm{H}_{2}\mathrm{O}(l))\); c. Net Ionic: \((\mathrm{Ag}^{+}(aq) + \mathrm{Cl}^{-}(aq)) \longrightarrow (\mathrm{AgCl}(s))\); d. Net Ionic: Same as the complete ionic equation.

Step-by-step solution

Reaction a - Write the Complete Ionic Equation

Dissociate all strong electrolytes (compounds that dissolve to form ions in solution). For the reaction \(\mathrm{K}_{2}\mathrm{SO}_{4}(aq)+\mathrm{CaI}_{2}(aq) \longrightarrow \mathrm{CaSO}_{4}(s)+\mathrm{KI}(aq)\), the complete ionic equation is \((2\mathrm{K}^{+}(aq) + \mathrm{SO}_{4}^{2-}(aq) + \mathrm{Ca}^{2+}(aq) + 2\mathrm{I}^{-}(aq)) \longrightarrow (\mathrm{CaSO}_{4}(s) + 2\mathrm{K}^{+}(aq) + 2\mathrm{I}^{-}(aq))\)

Reaction a - Write the Net Ionic Equation

Cancel out the spectator ions (ions that appear on both sides of the equation in the same form) to find the net ionic equation. The spectator ions are \(\mathrm{K}^{+}\) and \(\mathrm{I}^{-}\). The net ionic equation is \((\mathrm{Ca}^{2+}(aq) + \mathrm{SO}_{4}^{2-}(aq)) \longrightarrow (\mathrm{CaSO}_{4}(s))\)

Reaction b - Write the Complete Ionic Equation

For the reaction \(\mathrm{NH}_{4}\mathrm{Cl}(aq)+\mathrm{NaOH}(aq) \longrightarrow \mathrm{H}_{2}\mathrm{O}(l)+\mathrm{NH}_{3}(g)+\mathrm{NaCl}(aq)\), the complete ionic equation is \((\mathrm{NH}_{4}^{+}(aq) + \mathrm{Cl}^{-}(aq) + \mathrm{Na}^{+}(aq) + \mathrm{OH}^{-}(aq)) \longrightarrow (\mathrm{H}_{2}\mathrm{O}(l) + \mathrm{NH}_{3}(g) + \mathrm{Na}^{+}(aq) + \mathrm{Cl}^{-}(aq))\)

Reaction b - Write the Net Ionic Equation

Cancel out the spectator ions to find the net ionic equation. The spectator ions are \(\mathrm{Na}^{+}\) and \(\mathrm{Cl}^{-}\). The net ionic equation is \(\mathrm{NH}_{4}^{+}(aq) + \mathrm{OH}^{-}(aq) \longrightarrow \mathrm{NH}_{3}(g) + \mathrm{H}_{2}\mathrm{O}(l)\)

Reaction c - Write the Complete Ionic Equation

For the reaction \(\mathrm{AgNO}_{3}(aq)+\mathrm{NaCl}(aq) \longrightarrow\mathrm{AgCl}(s)+\mathrm{NaNO}_{3}(aq)\), the complete ionic equation is \((\mathrm{Ag}^{+}(aq) + \mathrm{NO}_{3}^{-}(aq) + \mathrm{Na}^{+}(aq) + \mathrm{Cl}^{-}(aq)) \longrightarrow (\mathrm{AgCl}(s) + \mathrm{Na}^{+}(aq) + \mathrm{NO}_{3}^{-}(aq))\)

Reaction c - Write the Net Ionic Equation

Cancel out the spectator ions to find the net ionic equation. The spectator ions are \(\mathrm{Na}^{+}\) and \(\mathrm{NO}_{3}^{-}\). The net ionic equation is \((\mathrm{Ag}^{+}(aq) + \mathrm{Cl}^{-}(aq)) \longrightarrow (\mathrm{AgCl}(s))\)

Reaction d - Write the Complete Ionic Equation

For the reaction \(\mathrm{HC}_{2}\mathrm{H}_{3}\mathrm{O}_{2}(aq)+\mathrm{K}_{2}\mathrm{CO}_{3}(aq) \longrightarrow \mathrm{H}_{2}\mathrm{O}(l)+\mathrm{CO}_{2}(g)+\mathrm{KC}_{2}\mathrm{H}_{3}\mathrm{O}_{2}(aq)\), note that acetic acid \(\mathrm{HC}_{2}\mathrm{H}_{3}\mathrm{O}_{2}\) is a weak acid, so it remains undissociated. The complete ionic equation is \((2\mathrm{K}^{+}(aq) + \mathrm{CO}_{3}^{2-}(aq) + \mathrm{HC}_{2}\mathrm{H}_{3}\mathrm{O}_{2}(aq)) \longrightarrow (\mathrm{H}_{2}\mathrm{O}(l) + \mathrm{CO}_{2}(g) + 2\mathrm{K}^{+}(aq) + 2\mathrm{C}_{2}\mathrm{H}_{3}\mathrm{O}_{2}^{-}(aq))\)

Reaction d - Write the Net Ionic Equation

No ions are canceled in this equation since acetic acid does not dissociate completely. The net ionic equation remains the same as the complete ionic equation.

Key concepts

Chemical Reaction

When studying chemistry, one of the foundational concepts is understanding a chemical reaction. This is a process in which one or more substances, known as reactants, are converted into one or more different substances, called products. The transformation involves breaking and forming chemical bonds, leading to changes in the composition and properties of the materials involved.

For example, in a textbook exercise, you might be asked to write balanced equations for reactions such as a mixture of potassium sulfate and calcium iodide yielding calcium sulfate and potassium iodide. The balanced chemical equation reflects the law of conservation of mass, ensuring that the number of atoms for each element is the same on both sides of the equation. It is the starting point for writing ionic and net ionic equations, highlighting the role of different ions in the reaction.

Ionic Equation

An ionic equation offers a more detailed perspective of a chemical reaction that involves ionic compounds. Ionic compounds dissociate into their constituent ions when dissolved in water. The ionic equation explicitly shows these ions as they exist in solution, separately representing the cations and anions.

An example would be the dissolution of calcium iodide in water, where it is represented as \(\mathrm{Ca}^{2+}(aq) + 2\mathrm{I}^{-}(aq)\). The aqueous designation (aq) indicates that the ions are solvated by water. For teaching purposes, it's important to clarify that each ionic species is written as it exists in the solution to fully understand the reactions taking place, both in individual states and when combined into a new product.

Spectator Ions

Within these reactions, not all ions undergo a chemical change. Those ions that do not participate in the actual reaction are known as spectator ions. They are present in the solution but stay unchanged before and after the reaction. Because they don't directly impact the outcome of the reaction, they can be omitted when writing a net ionic equation.

Using bullet points to illustrate, consider the following steps for identifying spectator ions:

• Write out the complete ionic equation for the reaction.
• Distinguish between the ions that form products from those that do not change.
• Omit the ions that are identical on both sides of the equation – these are your spectator ions.

Understanding spectator ions aids in simplifying equations and focusing on the actual chemical changes occurring, essential for predicting reaction results and engineering specific outcomes in practical applications.

Solubility Rules

The concept of solubility rules is crucial when predicting whether a compound will dissolve in water to form ions, or if it will result in a solid precipitate. These rules help determine the outcome of ionic reactions and are essential when writing complete and net ionic equations.

Some general rules include:

• Most nitrate \(\mathrm{NO}_{3}^{-}\) salts are soluble.
• Alkali metal ions and ammonium \(\mathrm{NH}_{4}^{+}\) salts are typically soluble.
• Chlorides \(\mathrm{Cl}^{-}\), bromides \(\mathrm{Br}^{-}\), and iodides \(\mathrm{I}^{-}\) are soluble, except when paired with silver, lead, or mercury.
• Carbonates \(\mathrm{CO}_{3}^{2-}\) and phosphates \(\mathrm{PO}_{4}^{3-}\) are largely insoluble, with the exception of those involving alkali metals and ammonium.

Correct application of these solubility rules is evident in our textbook exercises, where they guide us to predict the formation of insoluble compounds, like calcium sulfate or silver chloride, leading to the creation of a precipitate in a reaction.