Question

A solution contains one or more of the following ions: \(\mathrm{Ag}^{+}\), \(\mathrm{Ca}^{2+},\) and \(\mathrm{Cu}^{2+}\). When you add sodium chloride to the solution, no precipitate forms. When you add sodium sulfate to the solution, a white precipitate forms. You filter off the precipitate and add sodium carbonate to the remaining solution, producing another precipitate. Which ions were present in the original solution? Write net ionic equations for the formation of each of the precipitates observed.

Short answer

The original solution contained Ca2+ and Cu2+ ions. The net ionic equations for the precipitate formations are: For calcium sulfate: Ca^{2+} (aq) + SO4^{2-} (aq) -> CaSO4(s) and For copper(II) carbonate: Cu^{2+} (aq) + CO3^{2-} (aq) -> CuCO3(s).

Step-by-step solution

Analyze the reaction with sodium chloride

Adding sodium chloride (NaCl) to the solution would lead to the potential formation of a precipitate with any ion that forms an insoluble chloride salt. Since no precipitate forms, we can infer that the ions do not form insoluble chlorides. Both AgCl and CaCl2 are soluble, but AgCl is actually not soluble in water, it is a common precipitate. Since no precipitate is formed, this suggests that Ag+ is not present in the solution.

Analyze the reaction with sodium sulfate

Adding sodium sulfate (Na2SO4) could lead to the formation of either Ag2SO4, CaSO4, or CuSO4. As there is a white precipitate when sodium sulfate is added, and we have already determined that Ag+ is not present, we focus on Ca2+ and Cu2+. The formation of a white precipitate indicates that CaSO4, which is slightly soluble, has formed, since the addition of Na2SO4 to a solution containing only soluble Cu2+ ions (CuSO4 is soluble) would not result in a precipitate. This suggests the presence of Ca2+ in the original solution. The net ionic equation for the precipitation reaction is Ca^{2+} (aq) + SO4^{2-} (aq) -> CaSO4(s).

Analyze the reaction with sodium carbonate

Adding sodium carbonate (Na2CO3) to the filtrate, which now contains no Ca2+, would form a precipitate only if an ion forms an insoluble carbonate. Of the remaining ions (since Ag+ has been ruled out and Ca2+ has been filtered), Cu2+ can form an insoluble carbonate, CuCO3. Thus Cu2+ is present in the original solution after Ca2+ has been filtered out. The net ionic equation for this reaction is Cu^{2+} (aq) + CO3^{2-} (aq) -> CuCO3(s).

Conclude which ions are present

Based on the results from the previous steps, we can conclude that the original solution contained Ca2+ and Cu2+ ions, but not Ag+ ions.

Key concepts

Qualitative Analysis of Ions

Qualitative analysis of ions is a cornerstone of analytical chemistry, helping scientists and students alike determine the presence or absence of particular ions in a sample. The exercise tackles this analysis by using selective precipitation reactions.

When sodium chloride (NaCl) is added to a mixture and no precipitate forms, it implies that soluble chlorides are in the mixture or, more importantly, specific ions that do not form precipitates with chloride are present. In the case of Ag⁺, which forms an insoluble chloride, AgCl, the absence of a precipitate upon NaCl addition hints at the absence of Ag⁺ ions.

To confirm the presence of calcium ions (Ca²⁺), sodium sulfate (Na₂SO₄) is used as it reacts with Ca²⁺ to form a white precipitate of calcium sulfate (CaSO₄), which is slightly soluble but still precipitates under certain conditions. The color and texture of the precipitate further guide the analysis, leading to the identification of Ca²⁺ in the original solution.

Finally, sodium carbonate (Na₂CO₃) is added to the filtrate to test for the presence of copper ions (Cu²⁺), as Cu²⁺ forms a blue-green precipitate with carbonate ions. The formation of a new precipitate upon the addition of sodium carbonate, after the removal of calcium ions, strongly indicates the presence of Cu²⁺ ions.

Net Ionic Equations

Net ionic equations are a simplified representation of chemical reactions that show only the species involved in the reaction. They exclude the spectator ions that do not participate in the reaction, clarifying the changes occurring at the molecular level.

In the exercise, the net ionic equation for the precipitation of calcium sulfate from calcium ions and sulfate ions is written as: \[ \text{Ca}^{2+} (aq) + \text{SO}_4^{2-} (aq) \rightarrow \text{CaSO}_4 (s) \].
This illustrates the direct interaction between calcium and sulfate ions leading to the formation of an insoluble salt. Similarly, the reaction between copper ions and carbonate ions is represented by another net ionic equation: \[ \text{Cu}^{2+} (aq) + \text{CO}_3^{2-} (aq) \rightarrow \text{CuCO}_3 (s) \].
Such equations are critical in helping students visualize and understand the process of precipitation and the resultant formation of a solid from aqueous solutions.

Solubility Product

The solubility product (K_sp) is a numerical value that represents the equilibrium between a solid ionic compound and its dissolved ions. This value is pivotal in predicting whether a precipitate will form under given conditions.

The notion of K_sp is tied to the solubility rules that state whether a compound is soluble or not. In the exercise, knowing that calcium sulfate has a low K_sp allows us to understand why it precipitates out of the solution when sodium sulfate is added to the solution containing calcium ions. On the other hand, copper sulfate has a higher K_sp, indicating that it is more soluble and will not precipitate under similar conditions.

A detailed understanding of the solubility product concept enables students to grasp why some ions remain in solution while others form a precipitate. In this scenario, it is the crux that explains the formation of precipitates in a qualitative analysis, further emphasised by the presence of Ca²⁺ and Cu²⁺ in the original solution, as deduced from the exercise.