Question

Suppose you have a solution that might contain any or. all of the following cations: \(\mathrm{Ni}^{2+}, \mathrm{Ag}^{+}, \mathrm{Sr}^{2+}\), and \(\mathrm{Mn}^{2+}\). Addition of \(\mathrm{HCl}\) solution causes a precipitate to form. After filtering off the precipitate, \(\mathrm{H}_{2} \mathrm{SO}_{4}\) solution is added to the resultant solution and another precipitate forms. This is filtered off, and a solution of \(\mathrm{NaOH}\) is added to the resulting solution. No precipitate is observed. Which ions are present in each of the precipitates? Which of the four ions listed above must be absent from the original solution?

Short answer

In summary, after adding \(\mathrm{HCl}\), the precipitate contains \(\mathrm{Ag}^{+}\) ions as \(\mathrm{AgCl}\). When adding \(\mathrm{H}_{2} \mathrm{SO}_{4}\), the precipitate contains \(\mathrm{Sr}^{2+}\) ions as \(\mathrm{SrSO}_{4}\). No precipitate was observed after adding \(\mathrm{NaOH}\). Therefore, \(\mathrm{Ni}^{2+}\) ions must be absent from the original solution, and the final filtrate contains \(\mathrm{Mn}^{2+}\) ions.

Step-by-step solution

Analyzing the reaction with \(\mathrm{HCl}\)

The first reagent we add to the solution is \(\mathrm{HCl}\). The formation of a precipitate after adding \(\mathrm{HCl}\) suggests that chloride ions react with one or more of the cations. By analyzing the solubility rules, we find that \(\mathrm{Ag}^{+}\) will form a precipitate when mixed with chloride ions: \[\mathrm{Ag}^{+} + \mathrm{Cl}^- \rightarrow \mathrm{AgCl}\] No other cations listed will form precipitates under these conditions.

Analyzing the reaction with \(\mathrm{H}_{2} \mathrm{SO}_{4}\)

Next, the filtrate (the remaining solution after filtering out the \(\mathrm{AgCl}\) precipitate) is treated with \(\mathrm{H}_{2} \mathrm{SO}_{4}\), and another precipitate forms. This suggests that sulfate ions react with one or more remaining cations. From the solubility rules, we find that \(\mathrm{Sr}^{2+}\) will form a precipitate with sulfate ions: \[\mathrm{Sr}^{2+} + \mathrm{SO}_{4}^{2-} \rightarrow \mathrm{SrSO}_{4}\] None of the other two remaining cations will form a precipitate with the sulfate ions.

Analyzing the reaction with \(\mathrm{NaOH}\)

Finally, the filtrate after removing the \(\mathrm{SrSO}_{4}\) precipitate is treated with \(\mathrm{NaOH}\), but no precipitate is observed. This means that the remaining cations in the solution do not form insoluble hydroxide compounds. If \(\mathrm{Ni}^{2+}\) ions were present, they'd form a precipitate with hydroxide ions: \[\mathrm{Ni}^{2+} + 2\mathrm{OH}^{-} \rightarrow \mathrm{Ni(OH)}_{2}\] Thus, we can conclude that \(\mathrm{Ni}^{2+}\) ions must be absent from the original solution.

Determining the remaining cation

As we've already identified the absence of \(\mathrm{Ni}^{2+}\) ions and the presence of \(\mathrm{Ag}^{+}\) and \(\mathrm{Sr}^{2+}\) ions in the previous steps, the only remaining cation is \(\mathrm{Mn}^{2+}\). Since no precipitate formed after adding \(\mathrm{NaOH}\), we can conclude that \(\mathrm{Mn}^{2+}\) ions are present in the final filtrate. To summarize the results: - The precipitate formed after adding \(\mathrm{HCl}\) contains \(\mathrm{Ag}^{+}\) ions as \(\mathrm{AgCl}\). - The precipitate formed after adding \(\mathrm{H}_{2} \mathrm{SO}_{4}\) contains \(\mathrm{Sr}^{2+}\) ions as \(\mathrm{SrSO}_{4}\). - The \(\mathrm{Ni}^{2+}\) ions are absent from the original solution. - The final filtrate contains \(\mathrm{Mn}^{2+}\) ions.

Key concepts

Solubility Rules

Understanding solubility rules is key in predicting whether two substances will form a precipitate when mixed. Solubility rules are simplified guidelines that help identify which ionic compounds will dissolve in water and which will not. For example:

• Most nitrate (\( ext{NO}_3^-\)) compounds are soluble.
• \( ext{Cl}^-, ext{Br}^-,\) and \( ext{I}^-\) compounds are generally soluble, except when combined with ions like \( ext{Ag}^+, ext{Pb}^{2+},\) and \( ext{Hg}_2^{2+}\).
• Sulfates (\( ext{SO}_4^{2-}\)) are usually soluble, but not with \( ext{Sr}^{2+}, ext{Ba}^{2+}, ext{Pb}^{2+},\) or \( ext{Ca}^{2+}\).

In our exercise, these rules allowed us to predict the formation of \( ext{AgCl}\) and \( ext{SrSO}_4\) as precipitates. Both are insoluble in water according to the typical solubility rules, leading to their precipitation from the solution.

Qualitative Analysis

Qualitative analysis is a method used in chemistry to identify the chemical components of a substance. This often involves forming and examining a precipitate to determine the presence of specific ions.
During our cation precipitation test, the qualitative analysis involved sequentially adding chemicals that would precipitate specific ions out of a mixture in a known order. This method allowed us to determine which ions were present in the original solution.
Here's how it worked:

• Adding \( ext{HCl}\) allowed for the precipitation of \( ext{Ag}^+\) ions as \( ext{AgCl}\).
• Then, \( ext{H}_2 ext{SO}_4\) was used to remove \( ext{Sr}^{2+}\) ions as \( ext{SrSO}_4\).
• Finally, the absence of precipitation with \( ext{NaOH}\) confirmed the absence of \( ext{Ni}^{2+}\) ions, and consequently confirmed the presence of \( ext{Mn}^{2+}\) remaining in solution.

Qualitative analysis is precise in determining the nature of substances by making them react under different controlled conditions.

Chemical Reactions

Chemical reactions are processes in which substances, called reactants, are transformed into different substances, or products. These reactions involve the rearrangement of atoms and can often be identified by changes such as the formation of a precipitate.
In our cation precipitation testing exercise, several chemical reactions occurred to form different precipitates. Each reaction provides clues about the composition of the original mixture. Consider the reactions observed:

• When \( ext{HCl}\) was added to the solution, it reacted with \( ext{Ag}^+\) ions to form the solid \( ext{AgCl}\) precipitate. This reaction is an example of a double displacement reaction.
• Subsequent addition of \( ext{H}_2 ext{SO}_4\) led to a reaction with \( ext{Sr}^{2+}\) ions forming \( ext{SrSO}_4\), another solid precipitate, highlighting another such double-replacement reaction.
• Lastly, \( ext{NaOH}\) was added and no additional precipitate was noted, meaning no reactions occurred involving the remaining ions and hydroxide ions.

Through these reactions, we identified the components in the initial solution and deduced information about their solubility and presence.