Problem 81 - Chemistry The Molecular Nature of Matter

Question

Complete and balance the following molecular equations, being careful to apply the solubility rules. Write balanced ionic and net ionic equations for the reactions. (a) \(\mathrm{FeSO}_{4}(a q)+\mathrm{K}_{3} \mathrm{PO}_{4}(a q) \longrightarrow\) (b) \(\mathrm{AgC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}(a q)+\mathrm{AlCl}_{3}(a q) \longrightarrow\) (c) chromium(III) chloride \(+\) barium hydroxide

Step-by-Step Solution

Answer

Balanced molecular equations: (a) 3FeSO4 + 2K3PO4 -> Fe3(PO4)2 + 3K2SO4, (b) 3AgC2H3O2 + AlCl3 -> 3AgCl + Al(C2H3O2)3, (c) CrCl3 + 3Ba(OH)2 -> 2BaCl2 + 3Cr(OH)3. Net ionic equations: (a) 3Fe^2+ + 2PO4^3- -> Fe3(PO4)2(s), (b) 3Ag+ + 3Cl- -> 3AgCl(s), (c) Cr^3+ + 3OH- -> Cr(OH)3(s).

Step 1: Write the Products of the Reaction

Predict the products of the reaction using the solubility rules. For reaction (a), FeSO4 + K3PO4 will produce K2SO4 and Fe3(PO4)2. For reaction (b), AgC2H3O2 + AlCl3 will produce AgCl and Al(C2H3O2)3. For reaction (c), CrCl3 + Ba(OH)2 will form Cr(OH)3 and BaCl2.

Step 2: Balance the Molecular Equations

Balance each equation by adjusting coefficients. Balanced molecular equations: (a) 3FeSO4 + 2K3PO4 -> Fe3(PO4)2 + 3K2SO4, (b) 3AgC2H3O2 + AlCl3 -> 3AgCl + Al(C2H3O2)3, (c) CrCl3 + 3Ba(OH)2 -> 2BaCl2 + 3Cr(OH)3.

Step 3: Write the Total Ionic Equations

Dissociate all soluble compounds into their ions: (a) 3Fe^2+ + 3SO4^2- + 6K+ + 2PO4^3- -> Fe3(PO4)2(s) + 6K+ + 3SO4^2-, (b) 3Ag+ + 3C2H3O2- + Al^3+ + 3Cl- -> 3AgCl(s) + Al^3+ + 3C2H3O2-, (c) Cr^3+ + 3Cl- + 3Ba^2+ + 6OH- -> 3Ba^2+ + 6Cl- + 3Cr(OH)3(s).

Step 4: Write the Net Ionic Equations

Cancel out the spectator ions to write the net ionic equations. Net ionic equations: (a) 3Fe^2+ + 2PO4^3- -> Fe3(PO4)2(s), (b) 3Ag+ + 3Cl- -> 3AgCl(s), (c) Cr^3+ + 3OH- -> Cr(OH)3(s).

Key Concepts

Solubility Rules

Understanding solubility rules is crucial when predicting whether a substance will dissolve in water, which directly affects the outcomes of chemical reactions. These rules guide us to determine whether a compound will remain intact or dissociate into ions when mixed with a solvent.

For example, a general rule is that most salts containing alkali metal ions (like sodium and potassium) and ammonium (h4^+) are soluble. Conversely, many silver, lead, and mercury salts are typically insoluble. Sulfates (so4^2-) are largely soluble, except for those of calcium, strontium, and barium. Similarly, carbonates (cO3^2-) and phosphates (po4^3-) are usually insoluble, with exceptions for salts containing alkali metal ions and ammonium.

Applying these rules, in the given exercise, K3PO4 is soluble and dissociates into ions while Fe3(PO4)2 is insoluble and remains a solid precipitate. Recognizing which compounds dissolve and which ones don't helps in writing balanced ionic equations and predicting reaction products.

Ionic Equations

Ionic equations enhance our understanding of what undergoes change during a chemical reaction by showing the dissociated forms of soluble ionic compounds. They differentiate between the free-floating ions present in a solution and the compounds that remain intact. Whereas a molecular equation may represent all reactants and products as compounds, an ionic equation explicitly illustrates which species exist as ions in the solution.

For instance, the equation 3Fe^2+ + 3SO4^2- + 6K+ + 2PO4^3- -> Fe3(PO4)2(s) + 6K+ + 3SO4^2- from step 3 of the solution lists soluble ionic compounds like FeSO4 as dissociated into Fe^2+ and SO4^2- ions, and K3PO4 into K+ and PO4^3- ions. Writing the total ionic equation is an intermediate step that leads us closer to finding the net ionic equation, which reveals the essence of the reaction.

Net Ionic Equations

Net ionic equations simplify chemical reactions down to the most essential components, which participate directly in the chemical change. They are obtained by eliminating spectator ions—ions that appear on both sides of the ionic equation and do not participate in the reaction.

When writing a net ionic equation, you effectively strip away the chaff, leaving only the grains. Spectator ions, like the potassium ions (K+) and sulfate ions (SO4^2-) in our example reaction (a), do not change during the course of the reaction and are therefore not included in the net ionic equation. The result, 3Fe^2+ + 2PO4^3- -> Fe3(PO4)2(s), conveys the formation of the iron(III) phosphate precipitate, the key process occurring in the reaction mixture.

Understanding net ionic equations helps students to focus on the substances that are influential in the chemical reaction, offering a clearer picture of the reaction's stoichiometry and the principles governing the chemical process.