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

Short Answer

Expert verified
In a precipitation reaction, NO鈧冣伝 (b) will always be a spectator ion since all nitrates are soluble and will not form any precipitates with any counter-ion.

Step by step solution

01

Analyzing the Solubility Rules

To find the spectator ion, let us analyze the solubility rules for each given ion: (a) Cl鈦: Most chloride salts are soluble, except for those containing Ag鈦, Pb虏鈦, and Hg鈧偮测伜 ions. (b) NO鈧冣伝: All nitrates (NO鈧冣伝) are soluble. (c) NH鈧勨伜: All ammonium salts (NH鈧勨伜) are soluble. (d) S虏鈦: Most sulfides are insoluble, except those of group 1 and 2 elements, and those of the ammonium ion (NH鈧勨伜). (e) SO鈧劼测伝: Most sulfates are soluble, except for those of Pb虏鈦, Sr虏鈦, Ba虏鈦, and Ca虏鈦.
02

Identifying the Always Soluble Ion

From the analysis of solubility rules, we can observe that: (a) Cl鈦 has a few exceptions in its solubility. (d) S虏鈦 is mostly insoluble, with a few exceptions. (e) SO鈧劼测伝 has some exceptions in its solubility. However, (b) NO鈧冣伝 and (c) NH鈧勨伜 are soluble with no exceptions. Thus, these two ions will always be spectator ions as they will not form any precipitate with any ion. Since the question is asking for just one spectator ion, we must choose the best fit from these two options.
03

Choosing the Best Fit

Between NO鈧冣伝(b) and NH鈧勨伜(c), we will choose NO鈧冣伝 as the always-soluble spectator ion because: - As a monovalent anion, it has the least probability of forming insoluble salts. Thus, the correct answer is (b) NO鈧冣伝.

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Key Concepts

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

Solubility Rules
Understanding the solubility rules is essential for predicting the outcomes of chemical reactions, especially precipitation reactions.

Soluble substances can dissolve in a solvent, such as water, whereas insoluble substances do not dissolve significantly. These rules are guidelines that help determine whether a particular salt (ionic compound) will dissolve in water. Certain patterns exist, such as most sodium, potassium, and ammonium salts being soluble, irrespective of the accompanying anion.

Chlorides, bromides, and iodides are also generally soluble, with notable exceptions like salts of silver, lead, and mercury. Importantly, nitrates (O3 ) and acetates are always soluble, making them common spectator ions in precipitation reactions since they do not contribute to the formation of precipitates.

Conversely, carbonates, phosphates, and sulfides are largely insoluble, except when paired with highly soluble cations like those of group 1 elements. Understanding these rules aids in predicting whether a salt will remain in solution or form a precipitate.
Precipitate Formation
Precipitate formation occurs when two aqueous solutions, containing different ions, react to form an insoluble product. This is a hallmark of precipitation reactions, a type of chemical reaction important in various fields, from environmental science to medicine.

When the ions of two soluble salts interact, and one of the possible products is an insoluble salt, a precipitate will form. The insolubility arises due to the low solubility product (Ksp) of the compound, a measure of the product of the ionic concentrations at saturation.

Predicting Precipitation

To predict whether a precipitation reaction will occur, one must consider the solubility rules and the potential outcomes of ion exchange when solutions are mixed. For instance, when solutions of silver nitrate and sodium chloride are mixed, silver chloride, an insoluble salt, precipitates out of the solution.

It's also crucial to note that temperature can influence the solubility of salts, and therefore, the conditions under which a reaction takes place are important for predicting precipitate formation.
Soluble Salts
Soluble salts are the ionic compounds that readily dissolve in a solvent, forming a homogeneous mixture called a solution. Salts are composed of cations and anions, and when these ions interact with the solvent molecules, specifically water in many cases, the salt dissociates into its component ions.

The dissolved ions are uniformly distributed throughout the solution, which can conduct electricity due to the presence of these mobile charged particles. Solubility is not the same for all salts; it varies widely depending on the specific ions involved. Some salts, such as table salt (sodium chloride), dissolve easily in water because of their strong interactions with water molecules. Others, like calcium carbonate, are not very soluble due to their larger lattice energies, which require more energy to overcome than what water can provide.

The solubility of a salt is crucial in many fields, for example, in water treatment, where certain salts must be dissolved to treat hardness, or in medical applications where electrolyte solutions are needed for patient hydration. Moreover, the concept of soluble salts also plays a prominent role in understanding the ion-exchange processes, such as those used in water softeners and other filtration systems.

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Most popular questions from this chapter

Write the balanced molecular and net ionic equations for each of the following neutralization reactions: \begin{equation} \begin{array}{l}{\text { (a) Aqueous acetic acid is neutralized by aqueous barium }} \\ {\text { hydroxide. }} \\ {\text { (b) Solid chromium(II) hydroxide reacts with nitrous acid. }} \\ {\text { (c) Aqueous nitric acid and aqueous ammonia react. }}\end{array} \end{equation}

Specify what ions are present in solution upon dissolving each of the following substances in water: (a) \(\mathrm{FeCl}_{2}\) ,\((\mathbf{b}) \mathrm{HNO}_{3,}(\mathbf{c})\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4},(\mathbf{d}) \mathrm{Ca}(\mathrm{OH})_{2.}\)

You know that an unlabeled bottle contains an aqueous solution of one of the following: \(A g N O_{3},\) CaCl \(_{2},\) or \(A l_{2}\left(S O_{4}\right)_{3}\) . A friend suggests that you test a portion of the solution with \(\mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2}\) and then with NaCl solutions. According to your friend's logic, which of these chemical reactions could occur, thus helping you identify the solution in the bottle? (a) Barium sulfate could precipitate. (b) Silver chloride could precipitate. (c) silver sulfate could precipitate. (d) More than one, but not all, of the reactions described in answers a-c could occur. (e) All three reactions described in answers a-c could occur.

Glycerol, \(\mathrm{C}_{3} \mathrm{H}_{8} \mathrm{O}_{3},\) is a substance used extensively in the manufacture of cosmetic s, foodstuffs, antifreeze, and plastics. Glycerol is a water-soluble liquid with a density of 1.2656 \(\mathrm{g} / \mathrm{mL}\) at \(15^{\circ} \mathrm{C}\) . Calculate the molarity of a solution of glycerol made by dissolving 50.000 \(\mathrm{mL}\) glycerol at \(15^{\circ} \mathrm{C}\) in enough water to make 250.00 \(\mathrm{mL}\) of solution.

Which of the following are redox reactions? For those that are, indicate which element is oxidized and which is reduced. For those that are not, indicate whether they are precipitation or neutralization reactions.\begin{equation} \begin{array}{c}{\text { (a) } \mathrm{P}_{4}(s)+10 \mathrm{HClO}(a q)+6 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow} \\\ \quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad{\quad 4 \mathrm{H}_{3} \mathrm{PO}_{4}(a q)+10 \mathrm{HCl}(a q)}\end{array} \end{equation} \begin{equation} \begin{array}{l}\quad\quad\quad\quad\quad\quad{\text { (b) } \mathrm{Br}_{2}(l)+2 \mathrm{K}(s) \longrightarrow 2 \mathrm{KBr}(s)} \\\\\quad\quad\quad\quad\quad\quad {\text { (c) } \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}(l)+3 \mathrm{O}_{2}(g) \longrightarrow 3 \mathrm{H}_{2} \mathrm{O}(l)+2 \mathrm{CO}_{2}(g)}\end{array} \end{equation} \begin{equation} \begin{array}{r}\quad\quad\quad\quad{\text { (d) } \mathrm{ZnCl}_{2}(a q)+2 \mathrm{NaOH}(a q) \longrightarrow \mathrm{Zn}(\mathrm{OH})_{2}(s)+} \\ {2 \mathrm{NaCl}(a q)}\end{array} \end{equation}

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