Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 21: Problem 77

When aqueous KI is added gradually to mercury(II) nitrate, an orange precipitate forms. Continued addition of KI causes the precipitate to dissolve. Write balanced equations to explain these observations. (Hint: \(\mathrm{Hg}^{2+}\) reacts with \(\mathrm{I}^{-}\) to form \(\mathrm{HgI}_{4}{ }^{2-}\).) Would you expect \(\mathrm{HgL}_{4}{ }^{2-}\) to form colored solutions? Explain.

Short Answer

Expert verified
The two balanced chemical equations are: 1) \(2KI(aq) + Hg(NO_{3})_{2}(aq) \rightarrow HgI_{2}(s) + 2KNO_{3}(aq)\) 2) \(HgI_{2}(s) + 2KI(aq) \rightarrow HgI_{4}^{2-}(aq) + 2K^{+}(aq)\) HgL₄⁠^{2⁠−} may form colored solutions if the halide (L) has similar properties to iodide ions when attached to Hg²⁺. More information is needed to confidently predict the color formation of HgL₄⁠^{2⁠−} complexes.

Step by step solution

01

Write the first balanced chemical equation

Initially, aqueous KI is added to mercury(II) nitrate, and an orange precipitate of mercury(II) iodide forms. The balanced chemical equation for this reaction is: \[2KI(aq) + Hg(NO_{3})_{2}(aq) \rightarrow HgI_{2}(s) + 2KNO_{3}(aq)\]
02

Write the second balanced chemical equation

Upon further addition of KI, the orange precipitate dissolves because Hg⁠^{2+} ions combine with I⁠^− ions to form the HgI₄⁠^{2⁠−} complex ion. The balanced chemical equation for this reaction is: \[HgI_{2}(s) + 2KI(aq) \rightarrow HgI_{4}^{2-}(aq) + 2K^{+}(aq)\]
03

Determine if Hg⁠L₄⁠^{2⁠−} would form colored solutions

The HgI₄⁠^{2⁠−} complex ion forms colored solutions because it absorbs certain wavelengths of light, resulting in the observed orange color. If the halides (L) in the HgL₄⁠^{2⁠−} complex ions have similar properties to iodide ions when attached to Hg⁠²⁺, then they would also form colored solutions. Without more information about the specific ligand (L), we cannot confidently predict if Hg⁠L₄⁠^{2⁠−} would form colored solutions, although it is likely due to the similar behavior with iodide 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.

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

Sketch and explain the most likely pattem for the crystal field diagram for the complex ion trans-diamminetetracyanonickelate(II), where \(\mathrm{CN}^{-}\) produces a much stronger crystal field than \(\mathrm{NH}_{3}\). Explain completely and label the \(d\) orbitals in your diagram. Assume the \(\mathrm{NH}_{3}\) ligands lie on the \(z\) axis.

Give formulas for the following. a. Hexakis(pyridine)cobalt(III) chloride b. Pentaammineiodochromium(III) iodide c. Tris(ethylenediamine)nickel(II) bromide d. Potassium tetracyanonickelate(II) e. Tetraamminedichloroplatinum(IV) tetrachloroplatinate(II)

Ammonia and potassium iodide solutions are added to an aqueous solution of \(\mathrm{Cr}\left(\mathrm{NO}_{3}\right)_{3} .\) A solid is isolated (compound \(\mathrm{A}\) ), and the following data are collected: i. When \(0.105 \mathrm{~g}\) of compound \(\mathrm{A}\) was strongly heated in excess \(\mathrm{O}_{2}, 0.0203 \mathrm{~g} \mathrm{CrO}_{3}\) was formed. ii. In a second experiment it took \(32.93 \mathrm{~mL}\) of \(0.100 \mathrm{M} \mathrm{HCl}\) to titrate completely the \(\mathrm{NH}_{3}\) present in \(0.341 \mathrm{~g}\) compound \(\mathrm{A}\). iii. Compound A was found to contain \(73.53 \%\) iodine by mass. iv. The freezing point of water was lowered by \(0.64^{\circ} \mathrm{C}\) when \(0.601\) g compound \(A\) was dissolved in \(10.00 \mathrm{~g} \mathrm{H}_{2} \mathrm{O}\) \(\left(K_{\mathrm{f}}=1.86^{\circ} \mathrm{C} \cdot \mathrm{kg} / \mathrm{mol}\right)\) What is the formula of the compound? What is the structure of the complex ion present? (Hints: \(\mathrm{Cr}^{3+}\) is expected to be sixcoordinate, with \(\mathrm{NH}_{3}\) and possibly \(\mathrm{I}^{-}\) as ligands. The \(\mathrm{I}^{-}\) ions will be the counterions if needed.)

Draw structures of each of the following. a. cis-dichloroethylenediamineplatinum(II) b. trans-dichlorobis(ethylenediamine) cobalt(II) c. cis-tetraamminechloronitrocobalt(III) ion d. trans-tetraamminechloronitritocobalt(III) ion e. trans-diaquabis(ethylenediamine) copper(II) ion

A certain first-row transition metal ion forms many different colored solutions. When four coordination compounds of this metal, each having the same coordination number, are dissolved in wa ter, the colors of the solutions are red, yellow, green, and blue Further experiments reveal that two of the complex ions are para magnetic with four unpaired electrons and the other two are dia magnetic. What can be deduced from this information about th four coordination compounds?
See all solutions

Recommended explanations on Chemistry Textbooks

Inorganic Chemistry

Read Explanation

The Earths Atmosphere

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Physical Chemistry

Read Explanation

Nuclear Chemistry

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