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

Short Answer

Expert verified
The ions that will always be spectator ions in a precipitation reaction are (b) Nitrate ion (NO₃⁻) and (c) Ammonium ion (NH₄⁺), as they form soluble salts with every cation and anion, respectively, causing no precipitate to form in the reactions they are involved in.

Step by step solution

01

Ion (a): Chloride Ion (Cl-)

The solubility rule tells us that chloride ions generally form soluble salts except silver(I), mercury(I), and lead(II) chloride. Due to these few exceptions, it may sometimes participate in precipitation reactions and, thus, cannot be always considered a spectator ion.
02

Ion (b): Nitrate Ion (NO₃-)

The solubility rule states that nitrate ions form soluble salts with every cation. This means no precipitate is formed when nitrate is present, indicating that the nitrate ion will always be a spectator ion in precipitation reactions.
03

Ion (c): Ammonium Ion (NH₄⁺)

The solubility rule informs us that ammonium ions form soluble salts with every anion, which means that ammonium salts will always be dissolved in a solution, and no precipitate is formed. So, the ammonium ion can always be considered a spectator ion in precipitation reactions.
04

Ion (d): Sulfide Ion (S²-)

The solubility rule tells us that sulfide ions form insoluble salts with many cations such as calcium, barium, lead, and others. This means that the sulfide ion will often participate in precipitation reactions, so it cannot always be considered a spectator ion.
05

Ion (e): Sulfate Ion (SO₄²-)

The solubility rule states that sulfate ions form soluble salts with most cations, except calcium, barium, lead, mercury(I), and silver(I). Due to these exceptions, the sulfate ion often participates in precipitation reactions and cannot always be considered a spectator ion. Based on the analysis of each ion, we can conclude that: The ions that will always be spectator ions in a precipitation reaction are: (b) Nitrate ion (NO₃⁻) and (c) Ammonium ion (NH₄⁺).

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Key Concepts

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

Solubility Rules
Understanding solubility rules is essential when identifying spectator ions in precipitation reactions. These rules, which are based on empirical observations, provide guidelines to predict whether a compound will dissolve in water, thus forming a soluble salt, or form a precipitate, an insoluble substance that separates from the solution. For students grappling with chemistry homework, a solid handle on these rules can simplify the process of solving complex reaction problems.

Soluble salts are those which remain disassociated in ionic form in aqueous solutions. Some ions, like the nitrate ion (o₃⁻) and the ammonium ion (h₄⁺), have the distinctive property of forming soluble salts with virtually any other ion, thereby indicating their roles as spectator ions—ions that do not participate in the actual chemical change but exist in the same form on both the reactant and product sides of a chemical equation. In contrast, ions like the chloride ion (o⁻), while often forming soluble salts, have notable exceptions and can sometimes be involved in the formation of a precipitate. Thus, solubility rules are the key to unlocking the mysteries of which substances will form a new product and which will merely observe the reaction unfold.
Nitrate Ion
The nitrate ion (o₃⁻) is a common anion found in various inorganic compounds. According to solubility rules, it is known for its high solubility in water. The rules state that all nitrate salts are soluble, meaning they can dissolve in water without forming a precipitate. This distinctive chemical behavior makes the nitrate ion a universal spectator ion in precipitation reactions.

For students seeing the term 'spectator ion' in their textbook exercises, it's important to recognize that these ions do not change their oxidation state or form any part of the precipitate. They are essentially the uninvolved bystanders of the reaction. The reliability of the nitrate ion as a spectator ion simplifies the prediction process in stoichiometry and can be a relief during complex problem-solving. Given its role in reactions, if you encounter a nitrate ion in a precipitation reaction, you can confidently assume it will not contribute to the formation of the precipitate, hence its designation as a spectator ion.
Ammonium Ion
Like the nitrate ion, the ammonium ion (h₄⁺) also plays a predictable role in precipitation reactions due to solubility rules. These rules clearly endorse that all salts containing ammonium ions are soluble in water. For learners, this is another piece of chemical trivia that can greatly expedite solving chemistry equations. As the ammonium ion does not participate in the formation of a precipitate, it will always be present in the solution as a spectator ion.

When tackling exercises in chemistry, understanding that the ammonium ion is always a spectator ion lets you streamline your approach to balancing equations and predicting outcomes of reactions. This non-reactive nature is specifically beneficial when identifying the components in a mixture that will remain unchanged throughout the reaction process. The presence of ammonium ions indicates no precipitate formation with this ion, simplifying the analysis of the reaction's net ionic equation.

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Most popular questions from this chapter

The metal cadmium tends to form \(\mathrm{Cd}^{2+}\) ions. The following observations are made: (i) When a strip of zinc metal is placed in \(\mathrm{CdCl}_{2}(a q)\), cadmium metal is deposited on the strip. (ii) When a strip of cadmium metal is placed in \(\mathrm{Ni}\left(\mathrm{NO}_{3}\right)_{2}(a q)\), nickel metal is deposited on the strip. (a) Write net ionic equations to explain each of the preceding observations. (b) Which elements more closely define the position of cadmium in the activity series? (c) What experiments would you need to perform to locate more precisely the position of cadmium in the activity series?

Would you expect that anions would be physically closer to the oxygen or to the hydrogens of water molecules that surround it in solution?

The average concentration of gold in seawater is \(100 \mathrm{fM}\) (femtomolar). Given that the price of gold is \(\$ 1764.20\) per troy ounce ( 1 troy ounce \(=31.103 \mathrm{~g}\) ), how many liters of seawater would you need to process to collect \(\$ 5000\) worth of gold, assuming your processing technique captures only \(50 \%\) of the gold present in the samples?

Lead glass contains \(18-40 \%\) by mass of \(\mathrm{PbO}\) (instead of \(\mathrm{CaO}\) in regular glass). Lead glass is still used industrially, but "lead crystal" drinking goblets are no longer considered safe, as the lead may leach out and cause toxic responses in humans. A particular 286-g lead crystal goblet that holds \(450 \mathrm{~mL}\) of liquid is \(27 \%\) PbO by mass, and it leaches \(3.4\) micrograms of lead every time it is filled. How many grams of sodium sulfide would be required to decontaminate all the lead in the entire goblet?

Will precipitation occur when the following solutions are mixed? If so, write a balanced chemical equation for the reaction. (a) \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) and \(\mathrm{AgNO}_{3}\), (b) \(\mathrm{NaNO}_{3}\) and \(\mathrm{NiSO}_{4}\), (c) \(\mathrm{FeSO}_{4}\) and \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}\).

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