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Chapter 4: Problem 93

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

Expert verified
The precipitates formed at each step of the experiment are: 1. AgCl after adding HCl: \( \mathrm{Ag}^{+} + \mathrm{Cl}^{-} \rightarrow \mathrm{AgCl鈫搣 \) 2. SrSO鈧� after adding H鈧係O鈧�: \( \mathrm{Sr}^{2+} + \mathrm{SO}_{4}^{2-} \rightarrow \mathrm{SrSO}_{4}鈫� \) Mn虏鈦� must be absent in the original solution, as no precipitate is observed upon adding NaOH.

Step by step solution

01

Determine the precipitate after HCl is added

In this step, consider the precipitation reactions that may occur when HCl is added to the solution. Recall that chlorides are generally soluble, except for silver chloride (AgCl), lead(II) chloride (PbCl鈧�), and mercury(I) chloride (Hg鈧侰l鈧�). Comparing this information with the possible cations present in the solution, we can conclude that the precipitate formed after HCl addition is AgCl. The reaction that occurs is: \[ \mathrm{Ag}^{+} + \mathrm{Cl}^{-} \rightarrow \mathrm{AgCl鈫搣 \] The remaining ions in the solution are Ni虏鈦�, Sr虏鈦�, and possibly Mn虏鈦�.
02

Determine the precipitate after H鈧係O鈧� is added

Next, consider the precipitation reactions that may occur when H鈧係O鈧� is added to the solution. It is important to note that most sulfates are soluble, except for lead(II) sulfate (PbSO鈧�), barium sulfate (BaSO鈧�), calcium sulfate (CaSO鈧�), and strontium sulfate (SrSO鈧�). As the solution contains Sr虏鈦�, it will form a precipitate with the sulfate anion. The reaction that occurs is: \[ \mathrm{Sr}^{2+} + \mathrm{SO}_{4}^{2-} \rightarrow \mathrm{SrSO}_{4}鈫� \] The remaining ions in the solution are Ni虏鈦� and possibly Mn虏鈦�.
03

Determine if there's a precipitate after NaOH is added

Lastly, consider the precipitation reactions that may occur when NaOH is added to the solution. Hydroxides are generally insoluble, except for those of the Group 1 elements and the ammonium ion (NH鈧勨伜). Transition metal hydroxides, like Ni(OH)鈧� and Mn(OH)鈧�, are insoluble. The fact that no precipitate is observed when NaOH is added indicates that Mn虏鈦� must be absent in the original solution. Since the reaction with NaOH does not yield any precipitate, we can deduce that Ni(OH)鈧� must also be absent from the original solution.
04

Conclusion

After analyzing the reactions occurring at each step of the experiment, the precipitates formed are determined to be: 1. AgCl after adding HCl 2. SrSO鈧� after adding H鈧係O鈧� Mn虏鈦� must be absent in the original solution, as no precipitate is observed upon adding NaOH.

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Key Concepts

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

Chemical Precipitation
Chemical precipitation is a useful method in qualitative chemical analysis. It involves adding a reagent to a solution causing one or more substances to convert to a solid form, known as a precipitate. This process is used to separate and identify different ions in a mixture.
In the context of the given exercise, precipitation reactions help identify which cations are present in the solution.
For instance, we added
  • HCl: to the solution and observed the formation of a precipitate, indicating the presence of the cation (\( \text{Ag}^+ \)) as (\( \text{AgCl} \)). Chlorides like (\( \text{AgCl} \)) are known to be less soluble.
  • \( \text{H}_2\text{SO}_4 \): led to the formation of a precipitate of (\( \text{SrSO}_4 \)) indicating that strontium ions (\( \text{Sr}^{2+} \)) are present.
Solubility Rules
Understanding solubility rules is vital in predicting whether a substance will precipitate from a solution. These rules give us a general guide on whether common ionic compounds are soluble or insoluble in water.
Specific ions form precipitates when certain reagents are added due to low solubility.
  • Chlorides (\(\text{Cl}^-\)): are generally soluble except when paired with (\( \text{Ag}^+, \text{Pb}^{2+}, \text{and Hg}^{2+} \)). Therefore, the formation of (\( \text{AgCl} \)) is expected.
  • Sulfates (\(\text{SO}_4^{2-}\)): most sulfates are soluble; however, those of (\( \text{Ba}^{2+}, \text{Sr}^{2+}, \text{and Pb}^{2+} \)) tend not to dissolve in water.
  • Hydroxides (\(\text{OH}^-\)): are generally insoluble, except for those containing alkaline metal cations like sodium and potassium.
Cation Identification
To identify cations in a mixture, one can use selective precipitation by adding reagents that precipitate specific ions.
The sequence and choice of reagents are crucial in the separation process.
  • The formation of (\( \text{AgCl} \)) upon adding (\( \text{HCl} \)) confirms the presence of (\( \text{Ag}^+ \)).
  • The appearance of (\( \text{SrSO}_4 \)) after introducing (\( \text{H}_2\text{SO}_4 \)) identifies (\( \text{Sr}^{2+} \)).
  • The absence of any precipitate upon adding (\( \text{NaOH} \)) suggests that (\( \text{Mn}^{2+} \)) and (\( \text{Ni}^{2+} \)) are unlikely to be in the solution.
It鈥檚 a step-by-step process with careful observation to determine the presence or absence of specific ions.
Precipitation Reactions
Precipitation reactions occur when two solutions are mixed, and an insoluble solid forms. This is due to the low solubility of particular combinations of cations and anions.
Let's detail each reaction's role as applied in this exercise:
  • \( \text{HCl} \) addition: resulted in the precipitation of (\( \text{AgCl} \)) due to its insolubility in water.
  • \( \text{H}_2\text{SO}_4 \) addition: led to (\( \text{SrSO}_4 \)) precipitating, indicating the low solubility of strontium sulfate.
  • \( \text{NaOH} \) addition: no precipitate formed, meaning the hydrolysis didn't cause precipitation of (\( \text{Ni(OH)}_2 \)) or (\( \text{Mn(OH)}_2 \)), due to the possible absence of these ions.
Each step is a test for special conditions that highlight the presence of certain ions by exclusion or inclusion.

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Most popular questions from this chapter

What mass of \(\mathrm{KCl}\) is needed to precipitate the silver ions from \(15.0 \mathrm{~mL}\) of \(0.200 \mathrm{M} \mathrm{AgNO}_{3}\) solution?

Can oxidation occur without accompanying reduction? Explain.

(a) What volume of \(0.115 \mathrm{M} \mathrm{HClO}_{4}\) solution is needed to neutralize \(50.00 \mathrm{~mL}\) of \(0.0875 \mathrm{M} \mathrm{NaOH}\) ? (b) What volume of \(0.128 \mathrm{M} \mathrm{HCl}\) is needed to neutralize \(2.87 \mathrm{~g}\) of \(\mathrm{Mg}(\mathrm{OH})_{2} ?\) (c) If \(25.8 \mathrm{~mL}\) of \(\mathrm{AgNO}_{3}\) is needed to precipitate all the \(\mathrm{Cl}^{-}\) ions in a \(785-\mathrm{mg}\) sample of \(\mathrm{KCl}\) (forming \(\mathrm{AgCl}\), what is the molarity of the \(\mathrm{AgNO}_{3}\) solution? (d) If \(45.3 \mathrm{~mL}\) of \(0.108 \mathrm{M} \mathrm{HCl}\) solution is needed to neutralize a solution of \(\mathrm{KOH}\), how many grams of \(\mathrm{KOH}\) must be present in the solution?

Formic acid, HCOOH, is a weak electrolyte. What solute particles are present in an aqueous solution of this compound? Write the chemical equation for the ionization of \(\mathrm{HCOOH}\).

You know that an unlabeled bottle contains a solution of one of the following: \(\mathrm{AgNO}_{3}, \mathrm{CaCl}_{2}\), or \(\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}\) A friend suggests that you test a portion of the solution with \(\mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2}\) and then with \(\mathrm{NaCl}\) solutions. Explain how these two tests together would be sufficient to determine which salt is present in the solution.
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