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Which of the following ions will always be a spectator ion in a precipitation reaction? (a) \(\mathrm{Cl}^{-},(\mathrm{b}) \mathrm{NO}_{3}^{-},(\mathrm{c}) \mathrm{NH}_{4}^{+},(\mathrm{d}) \mathrm{S}^{2-},\) (e) \(\mathrm{SO}_{4}^{2-}\). Explain briefly. [Section \(\left.4.2\right]\)

Short Answer

Expert verified
The ions that will always be spectator ions in a precipitation reaction are (b) NO₃⁻ and (c) NH₄⁺. This is because they almost always form soluble salts with other ions and do not participate in the formation of the precipitate.

Step by step solution

01

Define a Precipitation Reaction

A precipitation reaction is a type of chemical reaction in which two soluble salts in an aqueous solution form an insoluble solid called the precipitate. There are generally four types of ions present in the reaction: two cations (positive ions) and two anions (negative ions). One of the cation-anion pairs will form the precipitate, while the other pair will remain soluble in the solution as spectator ions.
02

Analyse each ion given in the exercise

We will individually check each given ion to see whether or not they will be a spectator ion in a precipitation reaction. (a) Cl⁻: Chloride ions (Cl⁻) can form soluble and insoluble salts. For example, silver chloride (AgCl) is insoluble, while sodium chloride (NaCl) is soluble. Therefore, Cl⁻ is not always a spectator ion. (b) NO₃⁻: Nitrate ions (NO₃⁻) form soluble salts with most cations. It is very rare for a nitrate salt to be insoluble. Therefore, NO₃⁻ is most likely to be a spectator ion in a precipitation reaction. (c) NH₄⁺: Ammonium ions (NH₄⁺) typically form soluble salts with most anions. It is very rare for an ammonium salt to be insoluble. Therefore, NH₄⁺ is most likely to be a spectator ion in a precipitation reaction. (d) S²⁻: Sulfide ions (S²⁻) can form both soluble and insoluble salts. For example, sodium sulfide (Na₂S) is soluble, while lead sulfide (PbS) is insoluble. Therefore, S²⁻ is not always a spectator ion. (e) SO₄²⁻: Sulfate ions (SO₄²⁻) can form soluble and insoluble salts depending on the cation it is paired with. For example, barium sulfate (BaSO₄) is insoluble, while sodium sulfate (Na₂SO₄) is soluble. Therefore, SO₄²⁻ is not always a spectator ion.
03

Determine the Spectator Ion

Based on our analysis in Step 2, we can conclude that ions (b) NO₃⁻ and (c) NH₄⁺ are the ions that will always be spectator ions in a precipitation reaction. This is because these ions almost always form soluble salts with other ions and do not participate in the formation of the precipitate.

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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 processes in chemistry where two solutions combine to form an insoluble solid, known as a precipitate. Imagine you're mixing two clear liquids, and suddenly, a solid forms and settles at the bottom. This solid is the precipitate. It can be fascinating to watch as the clear solutions turn cloudy, signaling the formation of a new substance. These reactions occur due to the interaction of cations and anions from the solutions. When the specific cation and anion pair to form an insoluble compound, other ions present, known as spectator ions, remain dissolved in the solution. Spectator ions simply "watch" the formation of the precipitate without participating in the reaction. Hence, their role is like that of an audience, observing the main event. Understanding precipitation reactions is essential in fields such as analytical chemistry and environmental science, where they might be used to remove pollutants from water.
Solubility Rules
Solubility rules are a set of guidelines that help predict whether a compound will dissolve in water. These rules are handy shortcuts for determining the likelihood of forming a precipitate in double displacement reactions. Some of the general rules include:
  • Most nitrate ( NO_{3}^{-} ) and acetate ( CH_{3}COO^{-} ) salts are soluble.
  • Salts containing alkali metal ions and ammonium ( NH_{4}^{+} ) are typically soluble.
  • Sulfates ( SO_{4}^{2-} ) are generally soluble, but there are exceptions like barium sulfate ( BaSO_{4} ).
  • Chlorides, bromides, and iodides are usually soluble, except when combined with silver ( Ag^{+} ), mercury ( Hg^{2+} ), and lead ( Pb^{2+} ).
Solubility rules act as a guide to foresee which ions will stay in solution and which will precipitate out. This knowledge is crucial in predicting the products of chemical reactions.
Nitrate Ions
Nitrate ions ( NO_{3}^{-} ) are some of the most stable and predictable ions in chemistry regarding solubility. They form soluble salts with nearly every cation. Think of nitrate ions as the free-spirited entities in the chemical world—they prefer to remain dissolved, happily mingling with other ions. In precipitation reactions, nitrate ions are typically spectator ions. Due to their high solubility, they will not form a precipitate; instead, they remain in the aqueous solution, not reacting further. This behavior makes nitrate ions reliable components in solutions when precipitating other compounds since they do not interfere by forming solid products themselves.
Ammonium Ions
Ammonium ions ( NH_{4}^{+} ) are another group of ions known for their solubility. Like nitrate ions, ammonium ions generally form soluble salts with various anions, emphasizing their non-reactive role in precipitation reactions. Due to their solubility in water, ammonium ions also often play the part of spectator ions. They typically don't participate in forming precipitates but instead remain in the solution. This property makes ammonium compounds useful in various chemical preparations where precipitation needs to be avoided. In biological and environmental contexts, ammonium ions are vital as they are a form of nitrogen plants can absorb. Thus, in chemistry classroom discussions, ammonium ions might not only appear in precipitation reactions but also in lessons involving biochemical cycles.

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Most popular questions from this chapter

A sample of \(1.50 \mathrm{~g}\) of lead(II) nitrate is mixed with \(125 \mathrm{~mL}\) of \(0.100 \mathrm{M}\) sodium sulfate solution. (a) Write the chemical equation for the reaction that occurs. (b) Which is the limiting reactant in the reaction? (c) What are the concentrations of all ions that remain in solution after the reaction is complete?

Three solutions are mixed together to form a single solution. One contains \(0.2 \mathrm{~mol} \mathrm{~Pb}\left(\mathrm{CH}_{3} \mathrm{COO}\right)_{2},\) the second contains \(0.1 \mathrm{~mol} \mathrm{Na}_{2} \mathrm{~S},\) and the third contains \(0.1 \mathrm{~mol} \mathrm{CaCl}_{2}\). (a) Write the net ionic equations for the precipitation reaction or reactions that occur. (b) What are the spectator ions in the solution?

Write balanced molecular and net ionic equations for the following reactions, and identify the gas formed in each: (a) solid cadmium sulfide reacts with an aqueous solution of sulfuric acid; (b) solid magnesium carbonate reacts with an aqueous solution of perchloric acid.

Lanthanum metal forms cations with a charge of \(3+.\) Consider the following observations about the chemistry of lanthanum: When lanthanum metal is exposed to air, a white solid (compound \(\mathrm{A}\) ) is formed that contains lanthanum and one other element. When lanthanum metal is added to water, gas bubbles are observed and a different white solid (compound \(\mathrm{B}\) ) is formed. Both \(\mathrm{A}\) and \(\mathrm{B}\) dissolve in hydrochloric acid to give a clear solution. When either of these solutions is evaporated, a soluble white solid (compound C) remains. If compound \(\mathrm{C}\) is dissolved in water and sulfuric acid is added, a white precipitate (compound D) forms. (a) Propose identities for the substances \(\mathrm{A}, \mathrm{B}, \mathrm{C},\) and \(\mathrm{D} .(\mathbf{b})\) Write net ionic equations for all the reactions described. (c) Based on the preceding observations, what can be said about the position of lanthanum in the activity series (Table 4.5\() ?\)

In each of the following pairs, indicate which has the higher concentration of I \(^{-}\) ion: (a) \(0.10 \mathrm{M} \mathrm{BaI}_{2}\) or \(0.25 \mathrm{M}\) KI solution, (b) \(100 \mathrm{~mL}\) of \(0.10 \mathrm{M}\) KI solution or \(200 \mathrm{~mL}\) of \(0.040 \mathrm{M} \mathrm{ZnI}_{2}\) solution, (c) \(3.2 \mathrm{M}\) HI solution or a solution made by dissolving \(145 \mathrm{~g}\) of NaI in water to make \(150 \mathrm{~mL}\) of solution.

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