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A sample may contain any or all of the following ions: \(\mathrm{Hg}_{2}^{2+}\) \(\mathrm{Ba}^{2+}\) and \(\mathrm{Mn}^{2+}\) a. No precipitate formed when an aqueous solution of NaCl was added to the sample solution. b. No precipitate formed when an aqueous solution of \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) was added to the sample solution. c. A precipitate formed when the sample solution was made basic with NaOH. Which ion or ions are present in the sample solution?

Short Answer

Expert verified
Based on the reactions with NaCl, Na2SO4, and NaOH, the only ion present in the sample solution is \(Mn^{2+}\).

Step by step solution

01

Analyzing reaction with NaCl

When NaCl is added to the sample, it dissociates into Na+ and Cl- ions. If any of the cations (Hg2^2+, Ba^2+, or Mn^2+) form an insoluble chloride salt with Cl-, a precipitate should form. In this case, no precipitate is formed, which means that Hg2^2+ can't be present in the sample as Hg2Cl2 is insoluble in water.
02

Analyzing reaction with Na2SO4

When Na2SO4 is added to the sample, it dissociates into 2 Na+ and SO4^2- ions. If any of the remaining cations (Ba^2+ or Mn^2+) form an insoluble sulfate salt with SO4^2-, a precipitate should form. Since no precipitate is formed, this means that Ba^2+ can't be in the sample, as BaSO4 is insoluble in water.
03

Analyzing reaction with NaOH

When NaOH is added to the sample, it dissociates into Na+ and OH- ions. If any of the remaining cations (Mn^2+) form an insoluble hydroxide salt with OH-, a precipitate should form. A precipitate is formed in this case, which means that Mn^2+ must be present in the sample as Mn(OH)2 is insoluble in water.
04

Conclusion

Based on the reactions with NaCl, Na2SO4, and NaOH, the only ion present in the sample solution is Mn^2+.

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Key Concepts

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

Insolubility rules
Understanding insolubility rules is crucial in predicting whether a precipitate will form when different chemical ions are mixed. These rules are essentially guidelines that help predict the solubility of ionic compounds in water. Some general rules include:
  • Most nitrates (\( ext{NO}_3^- \)) and acetates are soluble.
  • Most salts containing alkali metal ions (such as Na\(^+\), K\(^+\)) and ammonium (\( ext{NH}_4^+ \)) are soluble.
  • Sulfates (\( ext{SO}_4^{2-} \)) are generally soluble, with exceptions like barium sulfate (\( ext{BaSO}_4 \)), which is insoluble.
  • Chlorides (\( ext{Cl}^- \)) are mostly soluble, except for those containing silver (\( ext{Ag}^+ \)), lead (\( ext{Pb}^{2+} \)), and mercury (\( ext{Hg}_2^{2+} \)).

These rules were applied in the exercise to identify the ions in the sample based on their reactions with NaCl, Na\(_2\)SO\(_4\), and NaOH. For example, when NaCl is added and no precipitate forms, it rules out \( ext{Hg}_2^{2+} \) because \( ext{Hg}_2 ext{Cl}_2 \) would typically precipitate.
Precipitation reactions
Precipitation reactions occur when two soluble salts in aqueous solutions are combined and form an insoluble solid, known as the precipitate. This process involves the exchange of ions between the salts. For a precipitation reaction to occur, the product of the concentrations of the ions involved must exceed the solubility product (\( K_{sp} \)) value for that precipitate.

The presence or absence of precipitates helps chemists identify specific ions. In the exercise, adding Na\(_2\)SO\(_4\) resulted in no precipitate, indicating that \( ext{Ba}^{2+} \) wasn't present because \( ext{BaSO}_4 \) would have precipitated. When NaOH was added, a precipitate formed, pointing to \( ext{Mn}^{2+} \) being in the sample since \( ext{Mn(OH)}_2 \) is insoluble.
Chemical ions
Chemical ions are atoms or molecules that have lost or gained electrons, resulting in a charge. Depending on their charge, ions can either be cations (\( ext{positive} \)), which have lost electrons, or anions (\( ext{negative} \)), which have gained electrons.

In the context of the textbook exercise, the ions involved were \( ext{Hg}_2^{2+} \), \( ext{Ba}^{2+} \), and \( ext{Mn}^{2+} \). Each of these ions reacts differently with other ions based on their chemical properties. Understanding these interactions can help predict which compounds will precipitate under specific conditions.
  • The reaction with NaCl targeted possible chloride formations.
  • Na\(_2\)SO\(_4\) helped identify sulfate interactions.
  • NaOH was tested for hydroxide interactions.
By analyzing these reactions, we learn about the presence and absence of specific ions in the solution, highlighting how chemical ion behavior is essential in qualitative analysis.

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Most popular questions from this chapter

Three students were asked to find the identity of the metal in a particular sulfate salt. They dissolved a 0.1472-g sample of the salt in water and treated it with excess barium chloride, resulting in the precipitation of barium sulfate. After the precipitate had been filtered and dried, it weighed 0.2327 g. Each student analyzed the data independently and came to different conclusions. Pat decided that the metal was titanium. Chris thought it was sodium. Randy reported that it was gallium. What formula did each student assign to the sulfate salt? Look for information on the sulfates of gallium, sodium, and titanium in this text and reference books such as the CRC Handbook of Chemistry and Physics. What further tests would you suggest to determine which student is most likely correct?

What mass of barium sulfate can be produced when 100.0 mL of a 0.100-M solution of barium chloride is mixed with 100.0 mL of a 0.100-M solution of iron(III) sulfate?

A student wants to prepare 1.00 L of a 1.00-M solution of NaOH (molar mass = 40.00 g/mol). If solid NaOH is available, how would the student prepare this solution? If 2.00 M NaOH is available, how would the student prepare the solution? To help ensure three significant figures in the NaOH molarity, to how many significant figures should the volumes and mass be determined?

What mass of iron(III) hydroxide precipitate can be produced by reacting 75.0 mL of 0.105 M iron(III) nitrate with 125 mL of 0.150 M sodium hydroxide?

Calculate the concentration of all ions present in each of the following solutions of strong electrolytes. a. 0.0200 mole of sodium phosphate in 10.0 mL of solution b. 0.300 mole of barium nitrate in 600.0 mL of solution c. 1.00 g of potassium chloride in 0.500 L of solution d. 132 g of ammonium sulfate in 1.50 L of solution

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