Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 21: Problem 6

Four different octahedral chromium coordination compounds exist that all have the same oxidation state for chromium and have \(\mathrm{H}_{2} \mathrm{O}\) and \(\mathrm{Cl}^{-}\) as the ligands and counterions. When 1 mole of each of the four compounds is dissolved in water, how many moles of silver chloride will precipitate upon addition of excess \(\mathrm{AgNO}_{3} ?\)

Short Answer

Expert verified
Upon addition of excess \(\mathrm{AgNO}_{3}\) to the solution, 6 moles of silver chloride (\(\mathrm{AgCl}\)) will precipitate.

Step by step solution

01

Determine the possible structures of the coordination compounds

Octahedral chromium complexes can have 0, 1, 2, or 3 \(\mathrm{Cl}^-\) anions attached to the chromium atom. Due to the octahedral geometry, the remaining coordination sites will be occupied by water molecules. The four different octahedral complexes are: 1. \(\mathrm{[Cr(H_2O)_6]^{3+}}\) with 3 \(\mathrm{Cl}^-\) counterions. 2. \(\mathrm{[Cr(H_2O)_5Cl]^{2+}}\) with 2 \(\mathrm{Cl}^-\) counterions. 3. \(\mathrm{[Cr(H_2O)_4Cl_2]^+}\) with 1 \(\mathrm{Cl}^-\) counterion. 4. \(\mathrm{[Cr(H_2O)_3Cl_3]}\) with no counterions.
02

Find moles of Cl- ions released

When 1 mole of each complex is dissolved in water, the moles of \(\mathrm{Cl}^-\) ions released will be equal to the total moles of counterions present in the complexes. Therefore, the total moles of \(\mathrm{Cl}^-\) ions will be: Moles of \(\mathrm{Cl}^-\) ion = 3 (from complex 1) + 2 (from complex 2) + 1 (from complex 3) + 0 (from complex 4) = 6 moles of \(\mathrm{Cl}^-\) ions
03

Determine moles of AgCl precipitate

When excess \(\mathrm{AgNO}_3\) is added to the solution, each \(\mathrm{Cl}^-\) ion will react with \(\mathrm{Ag}^+\) to form insoluble \(\mathrm{AgCl}\) precipitate. The reaction is: \[ \mathrm{Ag^+ + Cl^- \rightarrow AgCl} \] Since there are 6 moles of \(\mathrm{Cl}^-\) ions in the solution and the stoichiometry of the reaction is 1:1, 6 moles of \(\mathrm{AgCl}\) will precipitate when excess \(\mathrm{AgNO}_3\) is added. So, upon addition of excess \(\mathrm{AgNO}_{3}\) to the solution, 6 moles of silver chloride (\(\mathrm{AgCl}\)) will precipitate.

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.

Start your free trial

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

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

Nickel can be purified by producing the volatile compound nickel tetracarbonyl. Nickel is the only metal that reacts with carbon monoxide at room temperature. Assuming this compound is overall neutral, what is the oxidation state of Ni in the compound? Deduce the formula of the compound.

The complex ion NiCl \(_{4}^{2-}\) has two unpaired electrons, whereas \(\mathrm{Ni}(\mathrm{CN})_{4}^{2-}\) is diamagnetic. Propose structures for these two complex ions.

Almost all metals in nature are found as ionic compounds in ores instead of being in the pure state. Why? What must be done to a sample of ore to obtain a metal substance that has desirable properties?

The equilibrium constant \(K_{\mathrm{a}}\) for the reaction $$\mathrm{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{3+}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \\\ \quad\quad\quad\quad\quad\quad\quad\quad\quad \mathrm{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)_{5}(\mathrm{OH})^{2+}(a q)+\mathrm{H}_{3} \mathrm{O}^{+}(a q)$$ is \(1.0 \times 10^{-5}\) a. Calculate the pH of a 0.10\(M\) solution of $\left[\mathrm{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right] \mathrm{Cl}_{3}$ b. Will a 1.0\(M\) solution of cobalt(Il) nitrate have a higher or lower pH than a 1.0\(M\) solution of cobalt (III) nitrate? Explain. c. \(\mathrm{Co}^{3+}\) complex ions are generally low-spin cases, whereas \(\mathrm{Co}^{2+}\) complex ions are generally high-spin cases. Explain. If this is the situation, how many unpaired electrons are present in \(\mathrm{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{3+}\) and \(\mathrm{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{2+} ?\)

\(\mathrm{CoCl}_{4}^{2-}\) forms a tetrahedral complex ion and \(\mathrm{Co}(\mathrm{CN})_{6}^{3-}\) forms an octahedral complex ion. What is wrong about the following statements concerning each complex ion and the \(d\) orbital splitting diagrams? a. \(\mathrm{CoCl}_{4}^{2-}\) is an example of a strong-field case having two unpaired electrons. b. Because \(\mathrm{CN}^{-}\) is a weak-field ligand, \(\mathrm{Co}(\mathrm{CN})_{6}^{3-}\) will be a low-spin case having four unpaired electrons.
See all solutions

Recommended explanations on Chemistry Textbooks

The Earths Atmosphere

Read Explanation

Chemistry Branches

Read Explanation

Physical Chemistry

Read Explanation

Organic Chemistry

Read Explanation

Making Measurements

Read Explanation

Chemical Reactions

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.

Sign-up for free