/*! 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 4 You add an aqueous solution of l... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 4: Problem 4

You add an aqueous solution of lead nitrate to an aqueous solution of potassium iodide. Draw highly magnified views of each solution individually, and the mixed solution, including any product that forms. Write the balanced equation for the reaction.

Short Answer

Expert verified
In the individual aqueous solutions, water molecules surround the ions of lead nitrate (Pb(NO鈧�)鈧�) and potassium iodide (KI). When these solutions are mixed, Pb虏鈦� reacts with I鈦� to form solid lead iodide (PbI鈧�) while the remaining K鈦� and NO鈧冣伝 ions stay dissolved in water. The balanced equation for this reaction is: \(Pb(NO_{3})_{2}(aq) + 2KI(aq) \rightarrow PbI_{2}(s) + 2KNO_{3}(aq)\)

Step by step solution

01

Identify the Chemical Formulae of the Aqueous Solutions

First, let's establish the chemical formula for each aqueous solution. Lead nitrate has the formula Pb(NO鈧�)鈧�, and potassium iodide has the formula KI.
02

Draw the Highly Magnified Views of Individual Solutions

For an aqueous solution of lead nitrate, we will have polar water molecules surrounding the lead (Pb虏鈦�) and nitrate (NO鈧冣伝) ions. Similarly, for potassium iodide, the water molecules will surround the potassium (K鈦�) and iodide (I鈦�) ions. Pb(NO鈧�)鈧� solution: Water molecules are interacting with the Pb虏鈦� and NO鈧冣伝 ions. KI solution: Water molecules are interacting with K鈦� and I鈦� ions.
03

Combine the Solutions and Show Ion Interaction with Water Molecules

When lead nitrate and potassium iodide solutions are combined, a chemical reaction occurs between the Pb虏鈦� ions and the I鈦� ions. They form a solid product called lead iodide (PbI鈧�). The remaining ions in the solution are K鈦� and NO鈧冣伝, which remain dissolved in the water. Mixed solution: Water molecules are surrounding the K鈦� and NO鈧冣伝 ions, while Pb虏鈦� and I鈦� ions have combined to form a solid product - lead iodide (PbI鈧�).
04

Write the Balanced Equation for the Reaction

To write the balanced equation for the reaction, recognize that lead nitrate and potassium iodide react to form lead iodide and potassium nitrate: \(Pb(NO_{3})_{2}(aq) + 2KI(aq) \rightarrow PbI_{2}(s) + 2KNO_{3}(aq)\) Here, the solid lead iodide (PbI鈧�) forms a precipitate, while the potassium nitrate (KNO鈧�) remains in a dissolved state in the aqueous solution.

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 91影视!

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

A 25.00-mL sample of hydrochloric acid solution requires 24.16 mL of 0.106 M sodium hydroxide for complete neutralization. What is the concentration of the original hydrochloric acid solution?

A 450.0 -mL sample of a \(0.257-M\) solution of silver nitrate is mixed with 400.0 \(\mathrm{mL}\) of 0.200\(M\) calcium chloride. What is the concentration of \(\mathrm{Cl}^{-}\) in solution after the reaction is complete?

Balance the following oxidation鈥搑eduction reactions that occur in basic solution. a. \(\mathrm{Al}(s)+\mathrm{MnO}_{4}^{-}(a q) \rightarrow \mathrm{MnO}_{2}(s)+\mathrm{Al}(\mathrm{OH})_{4}-(a q)\) b. \(\mathrm{Cl}_{2}(g) \rightarrow \mathrm{Cl}^{-}(a q)+\mathrm{OCl}^{-}(a q)\) c. \(\mathrm{NO}_{2}^{-}(a q)+\mathrm{Al}(s) \rightarrow \mathrm{NH}_{3}(g)+\mathrm{AlO}_{2}^{-}(a q)\)

A student mixes four reagents together, thinking that the solutions will neutralize each other. The solutions mixed together are 50.0 mL of 0.100 M hydrochloric acid, 100.0 mL of 0.200 M of nitric acid, 500.0 mL of 0.0100 M calcium hydroxide, and 200.0 mL of 0.100 M rubidium hydroxide. Did the acids and bases exactly neutralize each other? If not, calculate the concentration of excess \(\mathrm{H}^{+}\) or \(\mathrm{OH}^{-}\) ions left in solution.

A solution of permanganate is standardized by titration with oxalic acid \(\left(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\right) .\) It required 28.97 \(\mathrm{mL}\) of the permanganate solution to react completely with 0.1058 g of oxalic acid. The unbalanced equation for the reaction is $$\mathrm{MnO}_{4}^{-}(a q)+\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}(a q) \stackrel{\mathrm{Acidic}}{\longrightarrow} \mathrm{Mn}^{2+}(a q)+\mathrm{CO}_{2}(g)$$ What is the molarity of the permanganate solution?
See all solutions

Recommended explanations on Chemistry Textbooks

Inorganic Chemistry

Read Explanation

Chemical Analysis

Read Explanation

Kinetics

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Chemical Reactions

Read Explanation

The Earths Atmosphere

Read Explanation
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.