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Chapter 4: Problem 53

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 barium (Ba²⁺) and strontium (Sr²⁺), as they form a precipitate with SO₄²⁻ but not with Cl⁻ or OH⁻.

Step by step solution

01

Identify possible precipitation reactions

When KCl, Na2SO4, and NaOH are added to the solution, the ions may react with the anions/cations to form precipitates. The precipitation reactions can be defined as: 1. KCl added: K⁺ + unknown anion → precipitate 2. Na2SO4 added: 2Na⁺ + unknown cation → precipitate 3. NaOH added: Na⁺ + unknown anion → precipitate
02

Analyze possible cations in the given scenarios

In scenario 1 and 3, no precipitate forms, meaning the unknown anion doesn't react with K⁺ or OH⁻ ions. In scenario 2, the formation of a precipitate indicates that the unknown cation reacts with SO₄²⁻ ions. The common cations that react with SO₄²⁻ ions to form precipitates are Ba²⁺, Sr²⁺, Pb²⁺, and Ca²⁺. Thus, the unknown soluble ionic compound must have a cation that forms a precipitate when SO₄²⁻ is added but not when Cl⁻ or OH⁻ is added. Let's analyze the possible precipitation reactions for each cation.
03

Check the solubility rules for the possible cations

Ba²⁺ - BaCl₂: soluble (Ba²⁺ has no precipitate with Cl⁻) - BaSO₄: insoluble (formation of precipitate) - Ba(OH)₂: soluble (Ba²⁺ has no precipitate with OH⁻) Sr²⁺ - SrCl₂: soluble (Sr²⁺ has no precipitate with Cl⁻) - SrSO₄: insoluble (formation of precipitate) - Sr(OH)₂: soluble (Sr²⁺ has no precipitate with OH⁻) Pb²⁺ - PbCl₂: insoluble (Pb²⁺ forms a precipitate with Cl⁻) - PbSO₄: insoluble (formation of precipitate) - Pb(OH)₂: insoluble (Pb²⁺ forms a precipitate with OH⁻) Ca²⁺ - CaCl₂: soluble (Ca²⁺ has no precipitate with Cl⁻) - CaSO₄: slightly insoluble (formation of precipitate) - Ca(OH)₂: insoluble (Ca²⁺ forms a precipitate with OH⁻)
04

Match the cations with the observed reactions

Observations: 1. No precipitate with KCl (no precipitate with Cl⁻) 2. Precipitate with Na2SO4 (precipitate with SO₄²⁻) 3. No precipitate with NaOH (no precipitate with OH⁻) From the solubility rules analyzed in step 3, the cations that match the observations are Ba²⁺ and Sr²⁺. Ba²⁺ 1. No precipitate with Cl⁻ 2. Precipitate with SO₄²⁻ 3. No precipitate with OH⁻ Sr²⁺ 1. No precipitate with Cl⁻ 2. Precipitate with SO₄²⁻ 3. No precipitate with OH⁻
05

Conclusion

The cations that could be present in the unknown soluble ionic compound are barium (Ba²⁺) and strontium (Sr²⁺).

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Most popular questions from this chapter

Consider the reaction between oxygen \(\left(\mathrm{O}_{2}\right)\) gas and magnesium metal to form magnesium oxide. Using oxidation states, how many electrons would each oxygen atom gain, and how many electrons would each magnesium atom lose? How many magnesium atoms are needed to react with one oxygen molecule? Write a balanced equation for this reaction.

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A \(10.00-\mathrm{g}\) sample consisting of a mixture of sodium chloride and potassium sulfate is dissolved in water. This aqueous mixture then reacts with excess aqueous lead(II) nitrate to form \(21.75 \mathrm{~g}\) of solid. Determine the mass percent of sodium chloride in the original mixture.
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