   Chapter 17.5, Problem 17.12CYU

Chapter
Section
Textbook Problem

Solid Pbl2 (Ksp = 9.8 × 10‒9) is placed in a beaker of water. After a period of time, the lead(II) concentration is measured and found to be 1.1 × 10−3 M. Has the system reached equilibrium? That is, is the solution saturated? If not, will more Pbl2 dissolve?

Interpretation Introduction

Interpretation:

Predict whether the reaction has reached the equilibrium or not. If not then predict will more PbI2 dissolve or not also calculate the additional amount of PbI2 that will dissolve to reach the equilibrium.

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 PbI2.

Given:

The value of solubility product constant,Ksp, for PbI2 is 9.8×109.

The concentration of Pb2+ ions present in the solution is s=1.1×103molL1.

PbI2 undergoes dissociation in water as follows,

PbI2(s)Pb2+(aq)+2I(aq)

One mole of PbI2 dissolves in water to give one mole of Pb2+ ions and two moles of I ions in the solution. Therefore the concentration of I ions in the solution will be twice the concentration of Pb2+ ions.

The concentration of I ions present in the solution is,

[I]=2s

Substitute 1.1×103molL1 for s.

[I]=2(1.1×103molL1)=2.2×103molL1

The expression of Q for Ksp is as follows,

Q=[Pb2+][I1]2

Substitute 1.1×103 for [Pb2+] and 2.2×103 for [I1].

Q=(1.1×103)(2.2×103)2=5.3×109

Calculate the value of solubility, s' for PbI2 that, is the concentration of [Pb2+] and [I1] present in the solution at equilibrium to make it saturated using the expression for Ksp.

The expression of Ksp for PbI2 is as follows,

Ksp=[Pb2+][I1]2

Substitute s' for [Pb2+], 2s' for [I1]

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