Question

To precipitate \(\mathrm{Zn}^{2+}\) from \(\mathrm{Zn}\left(\mathrm{NO}_{3}\right)_{2}(\mathrm{aq}),\) add (a) \(\mathrm{NH}_{4} \mathrm{Cl} ;\) (b) \(\mathrm{MgBr}_{2} ;\) (c) \(\mathrm{K}_{2} \mathrm{CO}_{3} ;\) (d) \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}\).

Short answer

Only option (c) \(\mathrm{K}_{2} \mathrm{CO}_{3}\) will precipitate \(\mathrm{Zn}^{2+}\) from \(\mathrm{Zn}\left(\mathrm{NO}_{3}\right)_{2}(\mathrm{aq})\). The rest of the options will not form a precipitate with it.

Step-by-step solution

Analyze Option (a)

Option (a) involves adding \(\mathrm{NH}_{4} \mathrm{Cl}\) to the solution. Since \(\mathrm{NH}_{4}^{+}\) and \(\mathrm{Cl}^{-}\) are both common ions that form soluble compounds, they will not produce a precipitate with \(\mathrm{Zn}^{2+}\). Therefore, option (a) will not precipitate \(\mathrm{Zn}^{2+}\).

Analyze Option (b)

Option (b) involves adding \(\mathrm{MgBr}_{2}\) to the solution. Here, \(\mathrm{Mg}^{2+}\) is a common cation that forms soluble compounds, while \(\mathrm{Br}^{-}\) is a common anion, also forming soluble compounds. Therefore, adding \(\mathrm{MgBr}_{2}\) will not produce a precipitate with \(\mathrm{Zn}^{2+}\). Hence, option (b) will not precipitate \(\mathrm{Zn}^{2+}\).

Analyze Option (c)

Option (c) involves adding \(\mathrm{K}_{2} \mathrm{CO}_{3}\) to the solution. Here, \(\mathrm{CO}_{3}^{2-}\) tends to form insoluble compounds according to the solubility rules. Therefore, adding \(\mathrm{K}_{2} \mathrm{CO}_{3}\) will produce a precipitate with \(\mathrm{Zn}^{2+}\). Hence, option (c) will precipitate \(\mathrm{Zn}^{2+}\). It is noteworthy that \(\mathrm{K}^{+}\) forms soluble compounds, but it does not influence the precipitation of \(\mathrm{Zn}^{2+}\).

Analyze Option (d)

Option (d) involves adding \((\mathrm{NH}_{4})_{2} \mathrm{SO}_{4}\) to the solution. Both \(\mathrm{NH}_{4}^{+}\) and \(\mathrm{SO}_{4}^{-}\) form soluble compounds and will not produce a precipitate with \(\mathrm{Zn}^{2+}\). Therefore, option (d) will not precipitate \(\mathrm{Zn}^{2+}\).

Key concepts

Zinc Ion Precipitation

In chemistry, a precipitate is formed when a solid is created from a solution during a chemical reaction. The goal of a zinc ion precipitation reaction is to cause zinc ions (\( \mathrm{Zn}^{2+} \)) to form a solid, insoluble compound that will settle out of the solution. The exercise provided involves testing which chemical reagents can lead to this result.To precipitate zinc ions, one needs to pair them with an appropriate anion that forms an insoluble compound with zinc. In the original exercise, zinc ions in zinc nitrate react with other ions that either produce soluble or insoluble compounds. Among options like \( \mathrm{NH}_{4} \mathrm{Cl} \), \( \mathrm{MgBr}_{2} \), and \( \mathrm{K}_{2} \mathrm{CO}_{3} \), the potassium carbonate is effective because carbonate ions \( \mathrm{CO}_{3}^{2-} \) can react with zinc ions to form \( \mathrm{ZnCO}_{3} \), which is insoluble and thus precipitates out of solution.

Solubility Rules

When predicting whether a compound will dissolve in water or precipitate out of it, solubility rules become fundamental tools in the chemist's kit. These rules summarize common patterns regarding which ionic compounds are soluble (dissolve) and which are insoluble (form precipitates).

• Most nitrate \( \mathrm{NO}_{3}^{-} \), acetate \( \mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2}^{-} \), and alkali metal compounds are soluble in water.
• Chlorides \( \mathrm{Cl}^{-} \), bromides \( \mathrm{Br}^{-} \), and iodides \( \mathrm{I}^{-} \) are generally soluble, except when paired with silver, lead, or mercury ions.
• Sulfates \( \mathrm{SO}_{4}^{2-} \) are often soluble except for those of barium, strontium, lead, calcium, and mercury.
• Carbonates \( \mathrm{CO}_{3}^{2-} \), phosphates \( \mathrm{PO}_{4}^{3-} \), and sulfides \( \mathrm{S}^{2-} \) are typically insoluble except when combined with alkali metals or ammonium.

These rules help predict the outcomes of mixing various ions in an aqueous solution, such as in the determination of zinc ion precipitation.

Insoluble Compounds

Let's discuss insoluble compounds, which are essential in forming precipitates. When a compound does not dissolve in a solvent like water, it is regarded as insoluble and forms solid particles in the solution. Precipitation reactions often rely on these insoluble compounds.A classic example is zinc carbonate \( \mathrm{ZnCO}_{3} \), formed when zinc ions \( \mathrm{Zn}^{2+} \) and carbonate ions \( \mathrm{CO}_{3}^{2-} \) meet in solution. Zinc carbonate is insoluble in water, meaning it does not dissolve and instead forms a visible solid—a precipitate.In chemical equations, this process can be represented as follows:\[\mathrm{Zn}^{2+}_{(aq)} + \mathrm{CO}_{3}^{2-}_{(aq)} \rightarrow \mathrm{ZnCO}_{3}_{(s)}\]Understanding which compounds are insoluble is crucial when predicting whether certain reactants will lead to precipitation in a solution. This knowledge allows chemists to manipulate reactions to obtain desired products, purify mixtures, or conduct analyses.