Question

Complete the right side of each of the following molecular equations. Then write the net ionic equations. Assume all salts formed are soluble. Acid salts are possible. (a) $$ 2 \mathrm{KOH}(a q)+\mathrm{H}_{3} \mathrm{PO}_{4}(a q) \longrightarrow $$ (b) $$ 3 \mathrm{H}_{2} \mathrm{SO}_{1}(a q)+2 \mathrm{Al}(\mathrm{OH})_{2}(s) \longrightarrow $$ (c) $$ 2 \mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}(a q)+\mathrm{Ca}(\mathrm{OH})_{2}(a q) \longrightarrow $$ (d) $$ \mathrm{H}_{2} \mathrm{SO}_{3}(a q)+\mathrm{NaOH}(a q) \longrightarrow $$

Short answer

The net ionic equations are: (a) \(\mathrm{H}^+(aq) + \mathrm{OH}^-(aq) \rightarrow \mathrm{H}_2\mathrm{O}(l)\); (b) No net ionic (s precipitate); (c) \(\mathrm{Ca}^{2+}(aq) + 2 \mathrm{C}_{2}\mathrm{H}_{3}\mathrm{O}_2^-(aq) \rightarrow \mathrm{Ca}(\mathrm{C}_{2}\mathrm{H}_{3}\mathrm{O}_2)_2(aq)\); (d) \(\mathrm{H}^+(aq) + \mathrm{OH}^-(aq) \rightarrow \mathrm{H}_2\mathrm{O}(l)\).

Step-by-step solution

Write Balanced Molecular Equations

\(2 \mathrm{KOH}(aq) + \mathrm{H}_{3}\mathrm{PO}_{4}(aq) \rightarrow \mathrm{K}_{3}\mathrm{PO}_{4}(aq) + 3\mathrm{H}_{2}\mathrm{O}(l)\)\(3 \mathrm{H}_{2}\mathrm{SO}_{4}(aq) + 2 \mathrm{Al}(\mathrm{OH})_{3}(s) \rightarrow \mathrm{Al}_{2}(\mathrm{SO}_{4})_{3}(aq) + 6\mathrm{H}_{2}\mathrm{O}(l)\)\(2 \mathrm{HC}_{2}\mathrm{H}_{3}\mathrm{O}_{2}(aq) + \mathrm{Ca}(\mathrm{OH})_{2}(aq) \rightarrow \mathrm{Ca}(\mathrm{C}_{2}\mathrm{H}_{3}\mathrm{O}_{2})_{2}(aq) + 2\mathrm{H}_{2}\mathrm{O}(l)\)\(\mathrm{H}_{2}\mathrm{SO}_{3}(aq) + \mathrm{NaOH}(aq) \rightarrow \mathrm{Na}_{2}\mathrm{SO}_{3}(aq) + 2\mathrm{H}_{2}\mathrm{O}(l)\)

Determine Ionic Forms in Aqueous Solution

For the soluble compounds, split into ions:- \(\mathrm{KOH}(aq) \rightarrow 2\mathrm{K}^+(aq) + 2\mathrm{OH}^-(aq)\)- \(\mathrm{H}_{3}\mathrm{PO}_{4}(aq) \rightarrow 3\mathrm{H}^+(aq) + \mathrm{PO}_{4}^{3-}(aq)\)- \(\mathrm{K}_{3}\mathrm{PO}_{4}(aq) \rightarrow 3\mathrm{K}^+(aq) + \mathrm{PO}_{4}^{3-}(aq)\)Similarly for other equations, dissociate each soluble compound.

Identify Spectator Ions

Spectator ions do not participate in the actual reaction:- For (a): \(\mathrm{K}^+(aq)\) and \(\mathrm{PO}_{4}^{3-}(aq)\) are spectators.- For (b), \(\mathrm{Al}^{3+}(aq)\) and \(\mathrm{SO}_{4}^{2-}(aq)\) are spectators.- For (c), \(\mathrm{Ca}^{2+}(aq)\) and \(\mathrm{C}_{2}\mathrm{H}_{3}\mathrm{O}_2^{-}(aq)\) are spectators.- For (d), \(\mathrm{Na}^+(aq)\) and \(\mathrm{SO}_{3}^{2-}(aq)\) are spectators.

Write the Net Ionic Equations

Eliminate the spectator ions to find the net ionic equations:(a) \(\mathrm{H}^+(aq) + \mathrm{OH}^-(aq) \rightarrow \mathrm{H}_2\mathrm{O}(l)\)(b) \(3\mathrm{H}^+(aq) + 2\mathrm{Al}(\mathrm{OH})_{3}(s) \rightarrow \mathrm{Al}_2(\mathrm{SO}_4)_{3}(aq) + 6\mathrm{H}_2\mathrm{O}(l)\)(c) \(2\mathrm{C}_{2}\mathrm{H}_{3}\mathrm{O}_2^-(aq) + \mathrm{Ca}^{2+}(aq) \rightarrow \mathrm{Ca}(\mathrm{C}_{2}\mathrm{H}_{3}\mathrm{O}_2)_{2}(aq)\)(d) \(\mathrm{H}^+(aq) + \mathrm{OH}^-(aq) \rightarrow \mathrm{H}_2\mathrm{O}(l)\)

Key concepts

Balanced Molecular Equations

In chemical reactions, a balanced molecular equation represents the complete chemical formula of all reactants and products involved. Every chemical equation must be balanced, meaning that the number of atoms of each element must be the same on both sides of the equation. This ensures that mass is conserved, following the law of conservation of mass. Balancing involves adjusting coefficients to achieve equal numbers of each type of atom. For instance, in the equation \(2 \mathrm{KOH}(aq) + \mathrm{H}_3\mathrm{PO}_4(aq) \rightarrow \mathrm{K}_{3}\mathrm{PO}_{4}(aq) + 3\mathrm{H}_{2}\mathrm{O}(l)\), the coefficients are adjusted so that there are equal numbers of each type of atom, including \(K\), \(O\), \(H\), and \(P\), on both sides.

Spectator Ions

Spectator ions are the ions in a chemical reaction that do not undergo any chemical change and remain in the solution unchanged. They are present in the same form on both the reactant and product sides of a balanced equation. Identifying spectator ions is a crucial step in simplifying net ionic equations. For example, in the reaction between \(\mathrm{KOH}\) and \(\mathrm{H}_3\mathrm{PO}_4\), the ions \(\mathrm{K}^+\) and \(\mathrm{PO}_4^{3-}\) do not participate in the reaction as they are present in the same form in reactants and products. Keeping track of spectator ions allows chemists to focus on the species that actually participate in the chemical change.

Ionic Forms

When a substance dissolves in water and forms an aqueous solution, the compound often dissociates into its ionic forms. In writing ionic forms, we express the dissolved ions separately. This step is important because it helps in identifying the actual entities that participate in the chemical reaction. For instance, in the ionization of \(\mathrm{KOH}\), it dissociates into \(\mathrm{K}^+(aq)\) and \(\mathrm{OH}^-(aq)\), and \(\mathrm{H}_3\mathrm{PO}_4\) dissociates to form \(3\mathrm{H}^{+}(aq)\) and \(\mathrm{PO}_4^{3-}(aq)\). By representing each compound in its ionic form, we can easily determine which ions are involved in the actual chemical transformation.

Chemical Reactions

Chemical reactions entail the transformation of reactants into products through the breaking and forming of chemical bonds. Each reaction can be categorized by the type of change occurring, such as acid-base, redox, or precipitation reactions, among others. Understanding the nature of the reaction helps to predict the products and the conditions necessary for the reaction to occur. For example, when \(\mathrm{H}_2\mathrm{SO}_3\) reacts with \(\mathrm{NaOH}\), a neutralization reaction occurs, resulting in the formation of \(\mathrm{H}_2\mathrm{O}\) and \(\mathrm{Na}_2\mathrm{SO}_3\). Recognizing the reaction type helps in understanding the process at a molecular level, leading to better control in practical applications.