/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 16 When a solution of ammonium chlo... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

When a solution of ammonium chloride is added to a solution of lead(II) nitrate, \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}\), a white precipitate, lead(II) chloride, forms. Write a balanced net ionic equation for this reaction. Both ammonium chloride and lead(II) nitrate exist as dissociated ions in aqueous solution.

Short Answer

Expert verified
Net ionic equation: \(\text{Pb}^{2+} + 2\text{Cl}^- \rightarrow \text{PbCl}_2 (s)\).

Step by step solution

01

Write the Dissociation Equations

First, write the dissociation equations for ammonium chloride \((\text{NH}_4\text{Cl})\) and lead(II) nitrate \((\text{Pb(NO}_3\text{)}_2)\) in an aqueous solution.\[\text{NH}_4\text{Cl} \rightarrow \text{NH}_4^+ + \text{Cl}^-\] \[\text{Pb(NO}_3\text{)}_2 \rightarrow \text{Pb}^{2+} + 2\text{NO}_3^-\]This shows that both compounds fully dissociate in water.
02

Identify the Precipitate

From the problem, we know a white precipitate forms, which is lead(II) chloride \((\text{PbCl}_2)\). This compound is insoluble in water, causing it to precipitate out of solution.
03

Write the Formation of the Precipitate

Next, write the chemical equation showing the formation of the precipitate from the dissociated ions. The reaction of lead ions with chloride ions forms lead(II) chloride.\[\text{Pb}^{2+} + 2\text{Cl}^- \rightarrow \text{PbCl}_2 (s)\].
04

Write the Net Ionic Equation

The net ionic equation includes only the ions directly involved in forming the precipitate, excluding spectator ions. For this reaction, the net ionic equation is:\[\text{Pb}^{2+} + 2\text{Cl}^- \rightarrow \text{PbCl}_2(s)\]\(\text{NH}_4^+\) and \(\text{NO}_3^-\) are spectator ions and do not appear in the net ionic equation.

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 91Ó°ÊÓ!

Key Concepts

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

Precipitation Reactions
Precipitation reactions are a type of chemical reaction where two soluble substances react in solution to form an insoluble product, known as a precipitate. These reactions occur frequently in aqueous environments, where the precipitate can be observed as a solid forming from a clear solution. Precipitation reactions are important in various fields, such as chemistry, biology, and environmental science, due to their role in processes like water treatment and mineral formation.
In a typical precipitation reaction, two ionic compounds dissolve in water and exchange ions. When the combination of ions results in the formation of an insoluble solid, a precipitate forms. The solubility rules can help predict whether a compound will precipitate out of solution. For the reaction between ammonium chloride (\( ext{NH}_4 ext{Cl}\)) and lead(II) nitrate (\( ext{Pb(NO}_3 ext{)}_2\)), lead(II) chloride (\( ext{PbCl}_2\)) is the insoluble compound that precipitates.
  • A precipitation reaction involves mixing ions in solution to form an insoluble product.
  • The newly formed precipitate removes ions from the solution, leading to the formation of a solid.
  • Understanding precipitation reactions is key for predicting the outcomes of mixing different ionic solutions.
Ionic Dissociation
Ionic dissociation refers to the process where an ionic compound separates into its individual ions when dissolved in a solvent, typically water. This process is crucial in understanding how certain reactants interact to form products such as precipitates. In the example given, ammonium chloride and lead(II) nitrate dissociate into their respective ions when placed in water.
When ammonium chloride (\( ext{NH}_4 ext{Cl}\)) is dissolved, it separates into ammonium ions (\( ext{NH}_4^+\)) and chloride ions (\( ext{Cl}^-\)). Similarly, lead(II) nitrate (\( ext{Pb(NO}_3 ext{)}_2\)) dissociates into lead ions (\( ext{Pb}^{2+}\)) and nitrate ions (\( ext{NO}_3^-\)). This complete ion separation allows the different ions to freely float and interact within the solution, paving the way for possible reactions.
  • Ionic dissociation is essential for ionic compounds to interact in aqueous solutions.
  • The dissociation process ensures reactant ions are available for potential reactions.
  • Complete dissociation allows for mobility and reactivity of ions in solution.
Spectator Ions
Spectator ions are ions that exist in the same form on both the reactant and product sides of a chemical equation. They do not participate in the actual chemical reaction and thus remain unchanged throughout the process. Identifying spectator ions is crucial when determining the net ionic equation that reflects only the species directly involved in the reaction.
In the net ionic equation, these ions are left out, simplifying the equation to only include the ions that form the precipitate. In our exercise example, the ions (\( ext{NH}_4^+\)) and (\( ext{NO}_3^-\)) are spectator ions. They remain in solution without partaking in the formation of lead(II) chloride (\( ext{PbCl}_2\)) precipitate. As a result, they do not appear in the final net ionic equation.
  • Spectator ions are present in solution but do not engage in the reaction.
  • Ignoring spectator ions simplifies the chemical equation to show only the reactive entities.
  • By focusing on non-spectator ions, we highlight the actual chemical change taking place.

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

2,3,7,8 -Tetrachlorodibenzo-p-dioxin (TCDD) is a potent poison with the chemical formula \(\mathrm{C}_{12} \mathrm{H}_{4} \mathrm{Cl}_{4} \mathrm{O}_{2}\). The average lethal dose in humans is approximately \(2.9 \times 10^{-2}\) mg per kg of body weight. How many molecules of TCDD constitute a lethal dose for an 82 -kg individual?

Lead(II) nitrate and aluminum chloride react according to the following equation: \(3 \mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}+2 \mathrm{AlCl}_{3} \longrightarrow 3 \mathrm{PbCl}_{2}+2 \mathrm{Al}\left(\mathrm{NO}_{3}\right)_{3}\) In an experiment, 8.00 g of lead nitrate reacted with \(2.67 \mathrm{g}\) of aluminum chloride to give \(5.55 \mathrm{g}\) of lead chloride. (a) Which reactant was the limiting reagent? (b) What was the percent yield?

Ammonia is prepared industrially by the reaction of nitrogen and hydrogen according to the following equation: \(\mathrm{N}_{2}(\mathrm{g})+3 \mathrm{H}_{2}(\mathrm{g}) \longrightarrow 2 \mathrm{NH}_{3}(\mathrm{g})\) If \(29.7 \mathrm{kg}\) of \(\mathrm{N}_{2}\) is added to \(3.31 \mathrm{kg}\) of \(\mathrm{H}_{2}\) (a) Which reactant is the limiting reagent? (b) How many grams of the other reactant are left over? (c) How many grams of \(\mathrm{NH}_{3}\) are formed if the reaction goes to completion?

When solid carbon burns in a limited supply of oxygen gas, the gas carbon monoxide, \(\mathrm{CO}\) forms. This gas is deadly to humans because it combines with hemoglobin in the blood, making it impossible for the blood to transport oxygen. Write a balanced equation for the formation of carbon monoxide.

Aspirin is made by the reaction of salicylic acid with acetic anhydride. How many grams of aspirin are produced if \(85.0 \mathrm{g}\) of salicylic acid is treated with excess acetic anhydride?

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.