Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Identify the spectator ions in the complete ionic equation. $$ \begin{aligned} 2 \mathrm{~K}^{+}(a q)+\mathrm{S}^{2-}(a q)+\mathrm{Pb}^{2+}(a q)+2 \mathrm{NO}_{3}^{-}(a q) \longrightarrow & \longrightarrow \\ & \mathrm{PbS}(s)+2 \mathrm{~K}^{+}(a q)+2 \mathrm{NO}_{3}^{-}(a q) \end{aligned} $$

Short Answer

Expert verified
The spectator ions in the given equation are \( \mathrm{K}^{+} \) and \( \mathrm{NO}_{3}^{-} \).

Step by step solution

01

Understand Spectator Ions

Spectator ions are ions that exist in the same form on both the reactant and product side of a chemical equation. They do not participate in the chemical reaction and therefore do not influence the reaction outcome.
02

Identify the Ions on Both Sides of the Equation

To find spectator ions, look at the ions that appear on both sides of the equation without changing their oxidation states. Write down all the ions present in the reactants and the products.
03

Compare the Ions on Both Sides

Go through the list of ions and compare each ion on the reactant side to the product side. The ions that are present on both sides and do not change their state are the spectator ions.
04

Identify the Spectator Ions

By comparing, we can see that the ions \( \mathrm{K}^{+} \) and \( \mathrm{NO}_{3}^{-} \) are present on both sides of the equation with the same oxidation state and do not participate in the formation of products. These are the spectator ions.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Complete Ionic Equation
When we delve into the realm of chemistry, the complete ionic equation serves as a detailed representation of a chemical reaction, divulging the individual ions involved in the process. Unlike the traditional molecular equation, which presents compounds in their simplest form, the complete ionic equation breaks down soluble ionic compounds into the ions they form in aqueous solution.

Imagine a dance where each participant is shown individually instead of being part of a group routine. That's what the complete ionic equation does; it showcases each ion 'dancing' solo before they partner up to form the final products.

For instance, when a soluble salt dissolves in water, it separates into anions and cations, which are explicitly displayed in the complete ionic equation. Such meticulous detailing is crucial for identifying spectator ions—the ions that witness the reaction without taking an active part. This comprehensive approach allows students not only to visualize the reaction more clearly but also to understand which components play a central role and which simply 'watch from the sidelines'.
Chemical Reaction Representation
The art of portraying a chemical reaction hinges on the ability to accurately represent the transformation of reactants into products. This skill is pivotal in communicating the essence of the reaction to those learning chemistry. Through various representation methods, such as molecular equations, complete ionic equations, and net ionic equations, students uncover the dynamics of how substances interact on a molecular level.

Picture a storyboard for a movie, showcasing key frames that tell the story. Similarly, each representation method tells a fragment of the reaction's story, from the reactants' initial encounter to the final scene where products are formed. This multi-angled narrative equips students with a diverse toolkit for understanding and predicting the outcomes of chemical reactions. Some species alter their state, some change partners—such as in double displacement reactions—while the spectator ions remain unchanged, bearing witness but not acting. This intricate depiction is instrumental in highlighting the nuanced nature of chemical processes.
Oxidation States in Chemistry
Grasping the concept of oxidation states is akin to understanding the balance of power within a molecule or ion. Oxidation states are numerical values that describe the degree of oxidation or reduction an atom has undergone. They provide insight into the electron distribution among atoms in a compound and are pivotal when balancing chemical equations.

Think of it as keeping score in a sports game, with electrons representing points that can be lost or gained. The oxidation state tells us who is ahead, who is behind, and by how much. In ionic compounds, identifying oxidation states is relatively straightforward as they often correspond to the charge of the ion. For instance, in the ion \(\mathrm{K}^{+}\), potassium has an oxidation state of +1, indicating it has lost one electron.

In chemical reactions, keeping track of oxidation states helps determine which elements are being oxidized or reduced. For a clear understanding of electron flow in a reaction, which can help predict reactivity, identifying changes in oxidation states is essential. Moreover, this understanding is vital for students as they differentiate between ions participating in a reaction and spectator ions, the latter maintaining their oxidation state throughout the course of the reaction.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Write a balanced chemical equation for each chemical reaction. (a) Solid lead(II) sulfide reacts with aqueous hydrochloric acid to form solid lead(II) chloride and dihydrogen sulfide gas. (b) Gaseous carbon monoxide reacts with hydrogen gas to form gaseous methane \(\left(\mathrm{CH}_{4}\right)\) and liquid water. (c) Solid iron(III) oxide reacts with hydrogen gas to form solid iron and liquid water. (d) Gaseous ammonia \(\left(\mathrm{NH}_{3}\right)\) reacts with gaseous oxygen to form gaseous nitrogen monoxide and gaseous water.

NO is a pollutant emitted by motor vehicles. It is formed by the reaction: (a) \(\mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{NO}(g)\) Once in the atmosphere, NO (through a series of reactions) adds one oxygen atom to form \(\mathrm{NO}_{2}\). \(\mathrm{NO}_{2}\) then interacts with UV light according to the reaction: (b) \(\mathrm{NO}_{2}(g) \underset{\mathrm{UV} \text { light }}{\mathrm{NO}(g)}+\mathrm{O}(g)\) These freshly formed oxygen atoms then react with \(\mathrm{O}_{2}\) in the air to form ozone \(\left(\mathrm{O}_{3}\right)\), a main component of smog: (c) \(\mathrm{O}(g)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{O}_{3}(g)\) Classify each of the preceding reactions \((a, b, c)\) as a synthesis, decomposition, single-displacement, or doubledisplacement reaction.

Predict the products of each reaction and write balanced complete ionic and net ionic equations for each. If no reaction occurs, write NO REACTION. (a) \(\mathrm{NaI}(a q)+\mathrm{Hg}_{2}\left(\mathrm{NO}_{3}\right)_{2}(a q) \longrightarrow\) (b) \(\mathrm{HClO}_{4}(a q)+\mathrm{Ba}(\mathrm{OH})_{2}(a q) \longrightarrow\) (c) \(\mathrm{Li}_{2} \mathrm{CO}_{3}(a q)+\mathrm{NaCl}(a q) \longrightarrow\) (d) \(\mathrm{HCl}(a q)+\mathrm{Li}_{2} \mathrm{CO}_{3}(a q)\)

Complete and balance each combustion reaction. (a) \(\mathrm{S}(s)+\mathrm{O}_{2}(g) \longrightarrow\) (b) \(\mathrm{C}_{7} \mathrm{H}_{16}(l)+\mathrm{O}_{2}(g)\) (c) \(\mathrm{C}_{4} \mathrm{H}_{10} \mathrm{O}(l)+\mathrm{O}_{2}(g) \longrightarrow\) (d) \(\mathrm{CS}_{2}(l)+\mathrm{O}_{2}(g)\)

Pair each cation on the left with an anion on the right that will form an insoluble compound with it and write a formula for the insoluble compound. Use each anion only once. $$ \begin{array}{ll} \mathrm{Ag}^{+} & \mathrm{SO}_{4}{ }^{2-} \\ \mathrm{Ba}^{2+} & \mathrm{Cl}^{-} \\ \mathrm{Cu}^{2+} & \mathrm{CO}_{3}{ }^{2-} \\ \mathrm{Fe}^{3+} & \mathrm{S}^{2-} \end{array} $$

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free