Question

On the basis of the general solubility rules given in Table \(7.1,\) write a balanced molecular equation for the precipitation reactions that take place when the following aqueous solutions are mixed. Underline the formula of the precipitate (solid) that forms. If no precipitation reaction is likely for the reactants given, explain why. a. ammonium chloride, \(\mathrm{NH}_{4} \mathrm{Cl}\), and sulfuric acid, \(\mathrm{H}_{2} \mathrm{SO}_{4}\) b. potassium carbonate, \(\mathrm{K}_{2} \mathrm{CO}_{3},\) and tin(IV) chloride, \(\mathrm{SnCl}_{4}\) c. ammonium chloride, \(\mathrm{NH}_{4} \mathrm{Cl}\), and lead(II) nitrate, \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}\) d. copper(II) sulfate, \(\mathrm{CuSO}_{4}\), and potassium hydroxide, KOH e. sodium phosphate, \(\mathrm{Na}_{3} \mathrm{PO}_{4}\), and chromium(III) chloride, \(\mathrm{Cr} \mathrm{Cl}_{3}\) f. ammonium sulfide, \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{S},\) and iron(III) chloride, \(\mathrm{FeCl}_{3}\)

Short answer

a. No precipitation reaction will occur, as no insoluble products are formed. b. \(K_{2}CO_{3} + SnCl_{4} \rightarrow \underline{SnCO_{3}}(s) + 2KCl(aq)\) c. \(NH_{4}Cl + Pb(NO_{3})_{2} \rightarrow \underline{PbCl_{2}}(s) + 2NH_{4}NO_{3}(aq)\) d. \(CuSO_{4} + 2KOH \rightarrow \underline{Cu(OH)_{2}}(s) + K_{2}SO_{4}(aq)\) e. \(2CrCl_{3} + 3Na_{3}PO_{4} \rightarrow \underline{Cr_{2}(PO_{4})_{3}}(s) + 9NaCl(aq)\) f. \(2(NH_{4})_{2}S + 3FeCl_{3} \rightarrow \underline{Fe_{2}S_{3}}(s) + 6NH_{4}Cl(aq)\)

Step-by-step solution

a. Reaction between ammonium chloride (NH4Cl) and sulfuric acid (H2SO4)

Ammonium chloride and sulfuric acid are both soluble in water, but since they don't create any insoluble products, no precipitation reaction will occur.

b. Reaction between potassium carbonate (K2CO3) and tin(IV) chloride (SnCl4)

When we mix these two solutions, the following reaction occurs: \(K_{2}CO_{3} + SnCl_{4} → SnCO_{3} + 2KCl\) However, SnCO3 is not soluble and will precipitate: \(K_{2}CO_{3} + SnCl_{4} \rightarrow \underline{SnCO_{3}}(s) + 2KCl(aq)\)

c. Reaction between ammonium chloride (NH4Cl) and lead(II) nitrate (Pb(NO3)2)

Mixing these two solutions, we get: \(NH_{4}Cl + Pb(NO_{3})_{2} \rightarrow PbCl_{2} + 2NH_{4}NO_{3}\) The product PbCl2 is insoluble, so it will precipitate: \(NH_{4}Cl + Pb(NO_{3})_{2} \rightarrow \underline{PbCl_{2}}(s) + 2NH_{4}NO_{3}(aq)\)

d. Reaction between copper(II) sulfate (CuSO4) and potassium hydroxide (KOH)

Mixing these solutions results in the following reaction: \(CuSO_{4} + 2KOH \rightarrow Cu(OH)_{2} + K_{2}SO_{4}\) Copper(II) hydroxide (Cu(OH)2) is insoluble, so it will precipitate: \(CuSO_{4} + 2KOH \rightarrow \underline{Cu(OH)_{2}}(s) + K_{2}SO_{4}(aq)\)

e. Reaction between sodium phosphate (Na3PO4) and chromium(III) chloride (CrCl3)

Mixing these solutions results in the following reaction: \(2CrCl_{3} + 3Na_{3}PO_{4} \rightarrow Cr_{2}(PO_{4})_{3} + 9NaCl\) Chromium(III) phosphate (Cr2(PO4)3) is insoluble, so it will precipitate: \(2CrCl_{3} + 3Na_{3}PO_{4} \rightarrow \underline{Cr_{2}(PO_{4})_{3}}(s) + 9NaCl(aq)\)

f. Reaction between ammonium sulfide ((NH4)2S) and iron(III) chloride (FeCl3)

Mixing these solutions results in the following reaction: \(2(NH_{4})_{2}S + 3FeCl_{3} \rightarrow Fe_{2}S_{3} + 6NH_{4}Cl\) Iron(III) sulfide (Fe2S3) is insoluble and will precipitate: \(2(NH_{4})_{2}S + 3FeCl_{3} \rightarrow \underline{Fe_{2}S_{3}}(s) + 6NH_{4}Cl(aq)\)

Key concepts

Molecular Equations

In the study of chemistry, molecular equations provide a macroscopic view of chemical reactions. They present the compounds involved in a reaction as molecules or formula units, instead of ions. When writing a molecular equation, it's essential to consider the reactants and products as whole entities.
For example, when ammonium chloride \((\text{NH}_4\text{Cl})\) and lead(II) nitrate \((\text{Pb(NO}_3)_2)\) react, the molecular equation would be written as:

• \(\text{NH}_4\text{Cl} + \text{Pb(NO}_3)_2 \rightarrow \text{PbCl}_2 + 2\text{NH}_4\text{NO}_3\)

The reactants and products are expressed in their entirety, allowing us to see the reaction taking place at the molecular level.
Understanding molecular equations is crucial as they provide a clear picture of which substances are present and how they transform during the chemical reaction, aiding in predicting the products and the potential formation of a precipitate.

Solubility Rules

Solubility rules are a set of guidelines that help predict whether a compound will dissolve in water. These rules are vital for determining which products will form precipitates in a chemical reaction. Some general rules include:

• Most nitrate \((\text{NO}_3^-\)) salts are soluble.
• Salts containing alkali metal cations \((\text{e.g., Na}^+, \text{K}^+)\) and the ammonium ion \((\text{NH}_4^+)\) are usually soluble.
• Most chloride, bromide, and iodide salts are soluble, except those of silver, lead \((\text{Pb}^{2+})\), and mercury \((\text{Hg}_2^{2+})\).
• Carbonates \((\text{CO}_3^{2-})\) and phosphates \((\text{PO}_4^{3-})\) are generally insoluble, except when paired with alkali metals or ammonium.

These rules help predict that when solutions like potassium carbonate \((\text{K}_2\text{CO}_3)\) and tin(IV) chloride \((\text{SnCl}_4)\) are mixed, tin(IV) carbonate \((\text{SnCO}_3)\) will form as an insoluble precipitate.

Chemical Equations

Chemical equations are symbolic representations of chemical reactions. They display the reactants on the left, the products on the right, and sometimes the states of matter using symbols \((\text{(s)}, \text{(l)}, \text{(g)}, \text{(aq)})\). Writing balanced chemical equations is crucial because it ensures the conservation of mass – where the number of atoms for each element is the same on both sides of the equation.
A typical equation might look like this:

• \(\text{CuSO}_4 + 2\text{KOH} \rightarrow \text{Cu(OH)}_2(\text{s}) + \text{K}_2\text{SO}_4(\text{aq})\)

This shows that copper(II) sulfate reacts with potassium hydroxide to produce copper(II) hydroxide and potassium sulfate. The copper(II) hydroxide forms a precipitate, indicating an interaction guided by solubility principles.
Chemical equations also help scientists and students understand the quantitative relationships in chemical reactions, allowing them to predict the amounts of reactants needed and products formed.

Insoluble Compounds

Insoluble compounds are substances that do not dissolve significantly in a solvent, usually water. In the context of aqueous solutions, these compounds form solid precipitates, which can be identified and separated by their cloudy appearance.
In the reaction of sodium phosphate \((\text{Na}_3\text{PO}_4)\) and chromium(III) chloride \((\text{CrCl}_3)\), an insoluble compound – chromium(III) phosphate \((\text{Cr}_2(\text{PO}_4)_3)\) – forms:

• \(2\text{CrCl}_3 + 3\text{Na}_3\text{PO}_4 \rightarrow \underline{\text{Cr}_2(\text{PO}_4)_3}(\text{s}) + 9\text{NaCl}(\text{aq})\)

This precipitate formation is a key example of how solubility rules apply to predict insoluble products.
Understanding the concept of insoluble compounds aids in predicting reaction outcomes and designing strategies to recover or utilize these solids in laboratory and industrial settings.