   Chapter 17, Problem 68PS

Chapter
Section
Textbook Problem

You have 95 mL of a solution that has a lead(II) concentration of 0.0012 M. Will PbCl2 precipitate when 1.20 g of solid NaCl is added?

Interpretation Introduction

Interpretation:

Whether the precipitation of PbCl2 occurs or not on addition of 1.20 g NaCl to 95ml  water containing 0.0012 M 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 volume of solution is 95.0 ml

Given mass of added NaCl is 1.20 g .

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+][Cl]2 (1)

NaCl is strong electrolyte and when dissolved in water dissociates completely into its constituent ions. Therefore the concentration of the sodium ions and chloride ions coming from NaCl is equal to the initial concentration of NaCl.

NaCl(s)Na+(aq)+Cl(aq)

The concentration of sodium chloride (NaCl) is calculated as follows,

[NaCl]=wMV  (2)

Here,

• w is the given mass of sodium chloride.
• M is the gram molecular mass of the sodium chloride.
• is the volume of the solution in liter.

The volume of the solution is,

V =(95.0 ml)(1.00 L1000 ml)=95×103L

Gram molecular mass of sodium chloride is 40.0 gmol

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