Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 16: Problem 87

In a group 1 analysis, a student obtained a precipitate containing both \(\mathrm{AgCl}\) and \(\mathrm{PbCl}_{2}\). Suggest one reagent that would enable her to separate \(\operatorname{AgCl}(s)\) from \(\mathrm{PbCl}_{2}(s)\)

Short Answer

Expert verified
A suitable reagent that can be used to separate AgCl from PbCl2 is aqueous ammonia (NH3). This compound will dissolve the AgCl while leaving the PbCl2 untouched.

Step by step solution

01

Understanding the Concept

The solubility rules of chemical compounds states that all chlorides (Cl-) are soluble, except for silver (Ag+), lead (Pb2+) and mercury (Hg2 2+). However, AgCl and PbCl2 are more soluble in concentrated solutions of certain anions than in water. So, it is possible to choose a reagent which forms a complex with either Ag+ or Pb2+ ion to make it soluble.
02

Choosing the Reagent

Silver chloride (AgCl) can be solubilised in aqueous ammonia (NH3), as it forms a soluble complex ion with the silver ion, [Ag(NH3)2]+. The reaction is as follows: \n\n AgCl(s) + 2NH3(aq) -> [Ag(NH3)2]+(aq) + Cl−(aq) \n\nLead chloride (PbCl2), on the other hand, does not form a similar soluble complex in aqueous ammonia. Thus, adding aqueous ammonia to the mixture of AgCl and PbCl2 will selectively dissolve AgCl, thus effectively separating the two compounds.
03

Seperation Process

The seperation process can be carried out by simply adding ammonia solution to the mixture and stir it well. Then, allow the solution to stand for some time to let the undissolved PbCl2 to settle down at the bottom, this process is known as precipiation. The resulted clear solution above the precipitate can be decanted or can be separated by filtration, which will contain the dissolved AgCl. Thus, separation of AgCl and PbCl2 can be achieved.

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

A student carried out an acid-base titration by adding \(\mathrm{NaOH}\) solution from a buret to an Erlenmeyer flask containing HCl solution and using phenolphthalein as indicator. At the equivalence point, she observed a faint reddish-pink color. However, after a few minutes, the solution gradually turned colorless. What do you suppose happened?

Equal volumes of \(0.12 M \mathrm{AgNO}_{3}\) and \(0.14 \mathrm{M} \mathrm{ZnCl}_{2}\) solution are mixed. Calculate the equilibrium concentrations of \(\mathrm{Ag}^{+}, \mathrm{Cl}^{-}, \mathrm{Zn}^{2+},\) and \(\mathrm{NO}_{3}^{-}\)

A 0.2688 -g sample of a monoprotic acid neutralizes \(16.4 \mathrm{~mL}\) of \(0.08133 \mathrm{M}\) KOH solution. Calculate the molar mass of the acid.

A \(0.054 M \mathrm{HNO}_{2}\) solution is titrated with a \(\mathrm{KOH}\) solution. What is \(\left[\mathrm{H}^{+}\right]\) at half way to the equivalence point?

Write the solubility product expression for the ionic compound \(\mathrm{A}_{x} \mathrm{~B}_{y}\)
See all solutions

Recommended explanations on Chemistry Textbooks

Chemistry Branches

Read Explanation

Nuclear Chemistry

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Physical Chemistry

Read Explanation

Inorganic Chemistry

Read Explanation

Organic Chemistry

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