Problem 134 - Chemistry The Molecular Science

Question

An unknown solution contains either calcium ions or strontium ions, but not both. Which one of these solutions could you use to tell whether the ions present are \(\mathrm{Ca}^{2+}\) or \(\mathrm{Sr}^{2+}\) ? Explain the reasoning behind your choice. $$\mathrm{NaOH}(\mathrm{aq}), \mathrm{H}_{2} \mathrm{SO}_{4}(\mathrm{aq}), \mathrm{H}_{2} \mathrm{~S}(\mathrm{aq})$$

Step-by-Step Solution

Answer

Use \(\mathrm{H}_{2} \mathrm{SO}_{4}(\mathrm{aq})\) to differentiate between \(\mathrm{Ca}^{2+}\) and \(\mathrm{Sr}^{2+}\) due to differing precipitate formation.

Step 1: Analyze the Options

The exercise provides three solutions: \(\mathrm{NaOH}(\mathrm{aq})\), \(\mathrm{H}_{2} \mathrm{SO}_{4}(\mathrm{aq})\), and \(\mathrm{H}_{2} \mathrm{~S}(\mathrm{aq})\). To determine which ions are present in the unknown solution, we assume each solution could cause a chemical reaction with either calcium or strontium ions.

Step 2: Reaction with NaOH

\(\mathrm{NaOH}(\mathrm{aq})\) could form precipitates with both \(\mathrm{Ca}^{2+}\) and \(\mathrm{Sr}^{2+}\). \(\mathrm{Ca(OH)}_2\) and \(\mathrm{Sr(OH)}_2\) both have low solubilities and could precipitate. Thus, \(\mathrm{NaOH}\) is not effective in distinguishing between \(\mathrm{Ca}^{2+}\) and \(\mathrm{Sr}^{2+}\).

Step 3: Reaction with H2S

\(\mathrm{H}_{2} \mathrm{S}(\mathrm{aq})\) will also not effectively differentiate between \(\mathrm{Ca}^{2+}\) and \(\mathrm{Sr}^{2+}\) since calcium sulfide and strontium sulfide are both sparingly soluble or soluble, and does not form a distinct visible precipitate in either case.

Step 4: Reaction with H2SO4

\(\mathrm{H}_{2} \mathrm{SO}_{4}(\mathrm{aq})\) reacts with \(\mathrm{Ca}^{2+}\) to form \(\mathrm{CaSO4}\), which is sparingly soluble and thus precipitates out of the solution. \(\mathrm{H}_{2} \mathrm{SO}_{4}(\mathrm{aq})\) reacts with \(\mathrm{Sr}^{2+}\) to form \(\mathrm{SrSO4}\), which is also sparingly soluble, but the solubility is lower than \(\mathrm{CaSO4}\), so it forms a more immediate and visible precipitate. Therefore, by examining the intensity and quickness of precipitation, we can differentiate between \(\mathrm{Ca}^{2+}\) and \(\mathrm{Sr}^{2+}\).

Key Concepts

Chemical Reactions

In chemistry, a chemical reaction is a process in which substances, known as reactants, are transformed into different substances, called products. This transformation occurs due to the breaking and forming of chemical bonds. A chemical reaction might result in the formation of a precipitate, the release or absorption of heat, or a color change. For the solution containing either calcium ions \(\mathrm{Ca}^{2+}\) or strontium ions \(\mathrm{Sr}^{2+}\), chemical reactions with other substances might help identify which ion is present. When different solutions are mixed, watching for the formation of a solid, known as a precipitate, helps determine if a chemical reaction has occurred. This is particularly useful in distinguishing different ions by their behavior in reactions.

Solubility Rules

Solubility rules are guidelines that help predict whether an ionic compound will dissolve in water. These rules are vital for understanding which compounds will form precipitates in a solution.

• Compounds like nitrates \((\mathrm{NO}_3^-)\) and most salts of sodium \((\mathrm{Na}^+)\), potassium \((\mathrm{K}^+)\), and ammonium \((\mathrm{NH}_4^+)\) are generally soluble.
• Sulfates \((\mathrm{SO}_4^{2-})\) are mostly soluble, except for those combined with barium \((\mathrm{Ba}^2+)\), calcium \((\mathrm{Ca}^2+)\), and lead \((\mathrm{Pb}^{2+})\).
• Hydroxides \((\mathrm{OH}^-)\) are typically insoluble, except for those of sodium, potassium, and occasionally calcium or strontium to a slight extent.

In our exercise, understanding that calcium sulfate \((\mathrm{CaSO}_4)\) and strontium sulfate \((\mathrm{SrSO}_4)\) are sparingly soluble allows us to predict that they form precipitates, distinguishing between the two ions.

Precipitation Reactions

Precipitation reactions occur when two solutions mix and form an insoluble solid, known as a precipitate. The solid separates from the liquid, often forming a visible cloudiness or sediment in the solution. These reactions are driven by the lower solubility of the formed compound compared to the reactants. In the provided problem, \(\mathrm{H}_{2} \mathrm{SO}_4(\mathrm{aq})\) is used to differentiate between \(\mathrm{Ca}^{2+}\) and \(\mathrm{Sr}^{2+}\) ions. This solution leads to the formation of sparingly soluble calcium sulfate and even less soluble strontium sulfate.

• Calcium sulfate forms a precipitate, but this reaction is slower and less distinct.
• Strontium sulfate forms a precipitate rapidly, making it more suitable for identification.

Overall, by observing the nature of precipitation, one can effectively identify the ions based on their ability to form a precipitate.