Question

A solution contains one or more of the following ions: \(\mathrm{Hg}_{2}^{2+}\), \(\mathrm{Ba}^{2+},\) and \(\mathrm{Fe}^{2+} .\) When you add potassium chloride to the solution, a precipitate forms. The precipitate is filtered off, and you add potassium sulfate to the remaining solution, producing no precipitate. When you add potassium carbonate to the remaining solution, a precipitate forms. Which ions were present in the original solution? Write net ionic equations for the formation of each of the precipitates observed.

Short answer

The ions present in the original solution were Hg2^2+ and Fe^2+. The net ionic equations for the precipitates are: Hg2^2+ + 2Cl- → Hg2Cl2(s) and Fe^2+ + CO3^2- → FeCO3(s).

Step-by-step solution

- Identify the Relevant Solubility Rules

First, understand the solubility rules which indicate that compounds containing chloride (Cl-), sulfate (SO4^2-), and carbonate (CO3^2-) ions have different solubility properties depending on the cation they are paired with. Barium sulfate and carbonates of Ba^2+, Hg2^2+, and Fe^2+ are generally insoluble, while chlorides are soluble except for Hg2^2+.

- Analyze the Precipitate Formed by Potassium Chloride

Knowing that chlorides are mostly soluble, the precipitate formed when potassium chloride is added indicates that the cation forms an insoluble chloride. Given the options, the precipitate must be mercury(I) chloride (Hg2Cl2), as it is one of the few insoluble chlorides. This suggests the presence of Hg2^2+ in the original solution.

- Analyze the Absence of a Precipitate with Potassium Sulfate

Since potassium sulfate did not form a precipitate, the cation that could form an insoluble sulfate (BaSO4) was not present in the solution. This means that the Ba^2+ ion was not present in the original solution.

- Analyze the Precipitate Formed by Potassium Carbonate

The formation of a precipitate upon the addition of potassium carbonate suggests the presence of a cation that forms an insoluble carbonate. Since we have already identified Hg2^2+ and ruled out Ba^2+, the remaining ion Fe^2+ must form the precipitate, i.e., iron(II) carbonate (FeCO3). This implies the presence of Fe^2+ in the original solution.

- Write the Net Ionic Equations for the Precipitate Formation

For the first precipitate with potassium chloride, the net ionic equation is: \( \mathrm{Hg}_{2}^{2+} + 2Cl^{-} \rightarrow \mathrm{Hg}_{2}\mathrm{Cl}_{2}(s) \) For the precipitate with potassium carbonate, the net ionic equation is: \( \mathrm{Fe}^{2+} + \mathrm{CO}_{3}^{2-} \rightarrow \mathrm{FeCO}_{3}(s) \)

Key concepts

Solubility Rules

Understanding solubility rules is crucial in the study of solution chemistry. Solubility rules help predict whether a substance will dissolve in a particular solvent, such as water. These are a set of guidelines that determine the ability of ionic compounds to form aqueous solutions.

For example, an essential rule to remember is that salts containing nitrate (o3^-), acetate (o3^-), and most of the alkali metal ions are generally soluble in water. Contrarily, carbonates (o3^2-), phosphates (o3^3-), sulfides, and hydroxides are often insoluble, with exceptions for compounds involving alkali metals and ammonium (o3^+).

In our exercise, the solubility rules dictate that while chlorides are largely soluble, some exceptions include those formed with lead (o3^{2+}), silver (o3^+), and mercury(I) (o3_{2}^{2+}). Barium sulfate is generally insoluble, helping us conclude that o3^{2+} was not present in the original solution since adding potassium sulfate caused no precipitate to form.

Always consult a comprehensive solubility chart when dealing with chemical reactions in solutions to make informed predictions about the formation of precipitates or clear solutions.

Net Ionic Equations

Net ionic equations provide a more straightforward representation of chemical reactions in ionic solutions by only showcasing the species that undergo the chemical change. To write a net ionic equation, follow these steps:

• Write the balanced molecular equation for the reaction.
• Break down all the soluble ionic compounds into their respective ions to form the complete ionic equation.
• Exclude the spectator ions, which do not take part in the actual reaction, to see the net effect.

For the problem at hand, the net ionic equation for the reaction between mercury(I) ions and chloride ions is simply the combination of o3_{2}^{2+} and 2o3^- ions to form solid mercury(I) chloride (o3_{2}o3_{2}(s)). This clarity ensures we understand that the solid substance observed is a result of this specific reaction without any unnecessary components involved.

Precipitate Formation

Precipitate formation is a common observation in chemical reactions involving two solutions that contain ions. A precipitate is an insoluble solid that forms when two ionic compounds in solution react and produce an insoluble product.

In many reactions, ions come together to form a compound that does not dissolve in the solution, thus creating a precipitate. Determining the presence of specific ions in a solution, in the exercise we are discussing, can be done by adding compounds that would form precipitates with those ions.

When potassium chloride was added to the solution, mercury(I) chloride precipitated, while potassium sulfate added afterward caused no such effect, which implied the absence of barium ions. The subsequent addition of potassium carbonate and the resulting precipitate confirmed the presence of iron(II) ions, as the insoluble iron(II) carbonate formed. Recognizing these signs of precipitate formation is vital in analytical chemistry, as it is often employed in qualitative analysis to identify the constituents of a solution.