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Separate samples of a solution of an unknown soluble ionic compound are treated with \(\mathrm{KCl}, \mathrm{Na}_{2} \mathrm{SO}_{4},\) and \(\mathrm{NaOH}\). A precipitate forms only when \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) is added. Which cations could be present in the unknown soluble ionic compound?

Short Answer

Expert verified
The cations that could be present in the unknown soluble ionic compound are \(\mathrm{Ba}^{2+}\) and \(\mathrm{Pb}^{2+}\), as they both form insoluble sulfate salts (\(\mathrm{Ba}\mathrm{SO}_{4}\) and \(\mathrm{Pb}\mathrm{SO}_{4}\)) when reacting with \(\mathrm{Na}_{2}\mathrm{SO}_{4}\) and do not form precipitates with \(\mathrm{KCl}\) or \(\mathrm{NaOH}\).

Step by step solution

01

Identify precipitates with \(\mathrm{Na}_{2} \mathrm{SO}_{4}\)

Begin by analyzing the reagent, \(\mathrm{Na}_{2}\mathrm{SO}_{4}\), and its solubility with different cations. Based on solubility rules, we know that some cations will form a precipitate with sulfate ions (\(\mathrm{SO}_{4}^{2-}\)). Some common insoluble sulfate salts are: 1. \(\mathrm{Ag}_{2}\mathrm{SO}_{4}\) 2. \(\mathrm{Ba}\mathrm{SO}_{4}\) 3. \(\mathrm{Ca}\mathrm{SO}_{4}\) 4. \(\mathrm{Pb}\mathrm{SO}_{4}\)
02

Check the conditions in the problem

We are given that the unknown ionic compound forms precipitates when treated with \(\mathrm{Na}_{2}\mathrm{SO}_{4}\) but not with \(\mathrm{KCl}\) and \(\mathrm{NaOH}\). This information helps us eliminate some possibilities. For example, \(\mathrm{Ag}^{+}\) would react with both \(\mathrm{KCl}\) and \(\mathrm{Na}_{2}\mathrm{SO}_{4}\), forming \(\mathrm{AgCl}\) and \(\mathrm{Ag}_{2}\mathrm{SO}_{4}\) respectively. Since no precipitate is formed with \(\mathrm{KCl}\), we can eliminate \(\mathrm{Ag}^{+}\) from the possibilities. Similarly, \(\mathrm{Ca}^{2+}\) would react with both \(\mathrm{Na}_{2}\mathrm{SO}_{4}\) and \(\mathrm{NaOH}\), forming \(\mathrm{Ca}\mathrm{SO}_{4}\) and \(\mathrm{Ca(OH)}_{2}\) respectively. Since no precipitate is formed with \(\mathrm{NaOH}\), we can eliminate \(\mathrm{Ca}^{2+}\) from the possibilities.
03

Identify possible cations in the unknown compound

Based on our analysis, we are left with two possible cations from our original list, which only form precipitates with \(\mathrm{Na}_{2}\mathrm{SO}_{4}\): 1. \(\mathrm{Ba}^{2+}\), forming \(\mathrm{Ba}\mathrm{SO}_{4}\) 2. \(\mathrm{Pb}^{2+}\), forming \(\mathrm{Pb}\mathrm{SO}_{4}\) Therefore, the cations that could be present in the unknown soluble ionic compound are \(\mathrm{Ba}^{2+}\) and \(\mathrm{Pb}^{2+}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Precipitation Reactions
Precipitation reactions are fascinating chemical reactions where two aqueous solutions mix and form an insoluble solid, known as a precipitate. This type of reaction is commonly used in laboratory settings to identify the presence of certain ions in a solution. A precipitate is usually formed when the product of the ion concentrations in the solution exceeds the solubility product constant (Ksp) for the solid compound.

Imagine mixing two clear solutions, and suddenly a cloudy substance appears. That's a precipitate forming! For example, if you mix sodium sulfate ( Na_2SO_4 ) with a solution containing barium ions ( Ba^{2+} ), you'll see barium sulfate ( BaSO_4 ) form as a precipitate. Why? Because BaSO_4 is insoluble in water.

This reaction is not only visually intriguing but is also key for detecting ions in various unknown solutions. By adding specific reagents, such as KCl, Na_2SO_4, and NaOH, to an unknown ionic compound, one can observe which combinations result in precipitate formation, helping to deduce the possible ions present.
Sulfate Compounds
Sulfate compounds are a major class of ionic compounds, where the sulfate ion ( SO_4^{2-} ) combines with various cations. The solubility of these compounds in water varies significantly, depending on the cation they are paired with.

Sulfate compounds like sodium sulfate ( Na_2SO_4 ) are highly soluble in water, meaning they dissolve easily without forming a precipitate. However, other sulfate compounds, such as barium sulfate ( BaSO_4 ) and lead sulfate ( PbSO_4 ), are very insoluble and readily form a precipitate when mixed in solution.

Knowing which sulfate compounds are insoluble in water is essential for conducting precipitation reactions and accurately identifying the cations present in an unknown compound. When a sulfate compound precipitates, it often signals the presence of one of these less soluble cations, like Ba^{2+} or Pb^{2+} , which forms a distinct solid in the solution.
Ionic Compounds
Ionic compounds are made up of positively charged ions (cations) and negatively charged ions (anions) that are held together by strong electrostatic forces. They form a lattice structure and typically exist in solid form at room temperature.

These compounds dissolve in water by breaking apart into their individual ions. This property is crucial because it enables the ions to participate in chemical reactions, such as precipitation reactions and neutralization reactions.
  • Cations: They carry a positive charge and can include metal ions like Na^+ , Ba^{2+} , and Pb^{2+} .
  • Anions: They carry a negative charge, with common examples including chloride ( Cl^- ) and sulfate ( SO_4^{2-} ) ions.
Understanding the interactions between different ions helps in predicting whether a compound will remain in solution or form a precipitate.

In the context of our exercise, knowing that only certain combinations of anions and cations will form a solid is crucial. It enables one to deduce which ions are most likely present in an unknown ionic compound based on observed precipitation outcomes.

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Most popular questions from this chapter

What volume of \(0.0521 M B a(O H)_{2}\) is required to neutralize exactly \(14.20 \mathrm{mL}\) of \(0.141 M \mathrm{H}_{3} \mathrm{PO}_{4} ?\) Phosphoric acid contains three acidic hydrogens.

The unknown acid \(\mathrm{H}_{2} \mathrm{X}\) can be neutralized completely by \(\mathrm{OH}^{-}\) according to the following (unbalanced) equation: $$ \mathrm{H}_{2} \mathrm{X}(a q)+\mathrm{OH}^{-}(a q) \longrightarrow \mathrm{X}^{2-}(a q)+\mathrm{H}_{2} \mathrm{O}(l) $$ The ion formed as a product, \(\mathrm{X}^{2-},\) was shown to have 36 total electrons. What is element X? Propose a name for \(\mathrm{H}_{2} \mathrm{X}\). To completely neutralize a sample of \(\mathrm{H}_{2} \mathrm{X}, 35.6 \mathrm{~mL}\) of \(0.175 \mathrm{M}\) \(\mathrm{OH}^{-}\) solution was required. What was the mass of the \(\mathrm{H}_{2} \mathrm{X}\) sample used?

A \(230 .\) mL sample of a \(0.275-M \mathrm{CaCl}_{2}\) solution is left on a hot plate overnight; the following morning, the solution is \(1.10 \mathrm{M}\). What volume of water evaporated from the 0.275 \(M\) \(\mathrm{CaCl}_{2}\) solution?

Specify which of the following equations represent oxidationreduction reactions, and indicate the oxidizing agent, the reducing agent, the species being oxidized, and the species being reduced. a. \(\mathrm{CH}_{4}(g)+\mathrm{H}_{2} \mathrm{O}(g) \rightarrow \mathrm{CO}(g)+3 \mathrm{H}_{2}(g)\) b. \(2 \mathrm{AgNO}_{3}(a q)+\mathrm{Cu}(s) \rightarrow \mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}(a q)+2 \mathrm{Ag}(s)\) c. \(\mathrm{Zn}(s)+2 \mathrm{HCl}(a q) \rightarrow \mathrm{ZnCl}_{2}(a q)+\mathrm{H}_{2}(g)\) d. \(2 \mathrm{H}^{+}(a q)+2 \mathrm{CrO}_{4}^{2-}(a q) \rightarrow \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q)+\mathrm{H}_{2} \mathrm{O}(i)\)

A 0.500 -L sample of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) solution was analyzed by taking a 100.0-mL aliquot and adding \(50.0 \mathrm{mL}\) of \(0.213 \mathrm{M}\) NaOH. After the reaction occurred, an excess of \(\mathrm{OH}^{-}\) ions remained in the solution. The excess base required \(13.21 \mathrm{mL}\) of \(0.103 M\) HCI for neutralization. Calculate the molarity of the original sample of \(\mathrm{H}_{2} \mathrm{SO}_{4}\). Sulfuric acid has two acidic hydrogens.

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