   Chapter 17, Problem 65PS

Chapter
Section
Textbook Problem

You have a solution that has a lead(II) ion concentration of 0.0012 M. If enough soluble chloride-containing salt is added so that the Cl− concentration is 0.010 M, will PbCl2 precipitate?

Interpretation Introduction

Interpretation:

Whether the precipitation of PbCl2 occurs or not on addition of 0.010M chloride ions to the solution of lead ions has to be predicted.

Concept introduction:

Solubility product constant,Ksp is equilibrium constant and is defined as the product of the equilibrium concentration of the ions of the salt raised to the power of their coefficients in the balanced chemical equation.

The expression for Ksp of a salt is given as,

AxBy(s)xAy+(aq)+yBx(aq)Ksp=[Ay+]x[Bx]y

Here,

• [Ay+] and [Bx] are equilibrium concentration.

Reaction quotient, Q, for a reaction is defined as the product of the concentration of the ions at any time of the reaction (other than equilibrium time ) of the salt raised to the power of their coefficients in the balanced chemical equation.

The expression for Q of a salt is given as,

AxBy(s)xAy+(aq)+yBx(aq)Ksp=[Ay+]x[Bx]y

Here,

• [Ay+] and [Bx] are the concentration at any time except equilibrium.
1. 1. If Q=Ksp, this implies that the solution is saturated solution and the concentration of the ions have reached their maximum limit.
2. 2. If Q<Ksp, this implies that the solution is not saturated and more salt can be added to the solution or the salt present in the solution already will dissolve more until the precipitation starts.
3. 3. If Q>Ksp, this implies that the solution is oversaturated and precipitation of salt will occur.
Explanation

Value of reaction quotient,Q, is calculated and compared with the value of Ksp for PbCl2.

Given:

Refer to the Appendix J in the textbook for the value of  Ksp.

The value of solubility product constant,Ksp, for PbCl2 is 1.7×105.

The concentration of Pb2+ ions present in the solution is 0.0012molL1.

The concentration of Cl ions added to the solution is 0.010molL1.

When PbCl2 is dissolved in water it undergoes dissociation in as follows,

PbCl2(s)Pb2+(aq)+2Cl(aq)

The Q expression for PbCl2 is,

Q=[Pb2+][Cl1]2

Substitute 0

Still sussing out bartleby?

Check out a sample textbook solution.

See a sample solution

The Solution to Your Study Problems

Bartleby provides explanations to thousands of textbook problems written by our experts, many with advanced degrees!

Get Started 