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Chapter 4: Problem 60

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

Douglasite is a mineral with the formula 2 \(\mathrm{KCl} \cdot \mathrm{FeCl}_{2}\) . 2 \(\mathrm{H}_{2} \mathrm{O}\) . Calculate the mass percent of douglasite in a 455.0 \(\mathrm{-mg}\) sample if it took 37.20 \(\mathrm{mL}\) of a \(0.1000-M \mathrm{AgNO}_{3}\) solution to precipitate all the \(\mathrm{Cl}^{-}\) as AgCl. Assume the douglasite is the only source of chloride ion.

Chlorisondamine chloride \(\left(\mathrm{C}_{14} \mathrm{H}_{20} \mathrm{Cl}_{6} \mathrm{N}_{2}\right)\) is a drug used in the treatment of hypertension. A 1.28-g sample of a medication containing the drug was treated to destroy the organic material and to release all the chlorine as chloride ion. When the filtered solution containing chloride ion was treated with an excess of silver nitrate, 0.104 g silver chloride was recovered. Calculate the mass percent of chlorisondamine chloride in the medication, assuming the drug is the only source of chloride.

Some of the substances commonly used in stomach antacids are \(\mathrm{MgO}, \mathrm{Mg}(\mathrm{OH})_{2},\) and \(\mathrm{Al}(\mathrm{OH})_{3.}\) a. Write a balanced equation for the neutralization of hydrochloric acid by each of these substances. b. Which of these substances will neutralize the greatest amount of 0.10 M HCl per gram?

A 6.50-g sample of a diprotic acid requires 137.5 mL of a 0.750 M NaOH solution for complete neutralization. Determine the molar mass of the acid.

Saccharin \(\left(\mathrm{C}_{7} \mathrm{H}_{5} \mathrm{NO}_{3} \mathrm{S}\right)\) is sometimes dispensed in tablet form. Ten tablets with a total mass of 0.5894 g were dissolved in water. The saccharin was oxidized to convert all the sulfur to sulfate ion, which was precipitated by adding an excess of barium chloride solution. The mass of BaSO\(_{4}\) obtained was 0.5032 g. What is the average mass of saccharin per tablet? What is the average mass percent of saccharin in the tablets?
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