Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 16: Problem 84

Explain, with balanced ionic equations, why (a) \(\mathrm{CuI}_{2}\) dissolves in ammonia solution, (b) AgBr dissolves in \(\mathrm{NaCN}\) solution, (c) \(\mathrm{HgCl}_{2}\) dissolves in KCl solution.

Short Answer

Expert verified
The balanced ionic equations for the dissolution are as follows: (a) \(\mathrm{CuI}_2 + 4NH_3 ⟶ \mathrm{[Cu(NH_3)_4]}^{2+} + 2I^−\)(b) \(\mathrm{AgBr} + 2CN^- ⟶ \mathrm{[Ag(CN)_2]}^− + Br^-\)(c) \(\mathrm{HgCl}_{2} + Cl^- ⟶ \mathrm{HgCl}_{3}^-\)

Step by step solution

01

Dissolution of \(\mathrm{CuI}_2\) in ammonia solution

The first reaction involves the complex formation of Copper(II) ion with four molecules of Ammonia. It can be expressed in a reaction as follows: \[\mathrm{CuI}_2 + 4NH_3 ⟶ \mathrm{[Cu(NH_3)_4]}^{2+} + 2I^−\]where \(\mathrm{CuI}_2\) breaks down into Copper(II) ions and iodide ions. These Copper(II) ions then react with ammonia molecules to form a complex ion.
02

Dissolution of AgBr in \(\mathrm{NaCN}\) solution

In the second part of the exercise, AgBr reacts with \(\mathrm{NaCN}\) to form soluble complex ion. The actual ionic equation is:\[\mathrm{AgBr} + 2CN^- ⟶ \mathrm{[Ag(CN)_2]}^− + Br^-\]The \(\mathrm{NaCN}\) breaks up into sodium ions and cyanide ions. The cyanide ions then react with the silver ion part of the silver bromide to form a soluble complex ion.
03

Dissolution of \(\mathrm{HgCl}_{2}\) in KCl solution

In the final sub-problem, \(\mathrm{HgCl}_{2}\) reacts with KCl and forms a soluble ionic complex. This reaction is represented by the following ionic equation:\[\mathrm{HgCl}_{2} + Cl^- ⟶ \mathrm{HgCl}_{3}^-\]The \(\mathrm{KCl}\) breaks up into potassium ions and chloride ions. The chloride ions then react with the mercuric ions to form a soluble complex ion.

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

Which of the following solutions can act as a buffer. (a) \(\mathrm{KCl} / \mathrm{HCl}\), (b) \(\mathrm{KHSO}_{4} / \mathrm{H}_{2} \mathrm{SO}_{4}\) (c) \(\mathrm{Na}_{2} \mathrm{HPO}_{4} /\) \(\mathrm{NaH}_{2} \mathrm{PO}_{4}\) (d) \(\mathrm{KNO}_{2} / \mathrm{HNO}_{2} ?\)

How can we predict whether a precipitate will form when two solutions are mixed?

The \(\mathrm{p} K_{2} \mathrm{~s}\) of two monoprotic acids \(\mathrm{HA}\) and \(\mathrm{HB}\) are 5.9 and \(8.1,\) respectively. Which of the two is the stronger acid?

Compare the molar solubility of \(\mathrm{Mg}(\mathrm{OH})_{2}\) in water and in a solution buffered at a pH of 9.0 .

The iodide impurity in a \(4.50-\mathrm{g}\) sample of a metal nitrate is precipitated as silver iodide. If \(5.54 \mathrm{~mL}\) of \(0.186 M\) AgNO \(_{3}\) solution is needed for the precipitation, calculate the mass percent of iodide in the sample.
See all solutions

Recommended explanations on Chemistry Textbooks

Inorganic Chemistry

Read Explanation

Organic Chemistry

Read Explanation

Kinetics

Read Explanation

The Earths Atmosphere

Read Explanation

Making Measurements

Read Explanation

Chemical Analysis

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