   Chapter 16, Problem 56E

Chapter
Section
Textbook Problem

# A solution is prepared by mixing 50.0 mL of 0.10 M Pb(NO3)2 with 50.0 mL of 1.0 M KCl. Calculate the concentrations of Pb2+ and Cl− at equilibrium. [Ksp for PbCl2(s) is 1.6 × 10−5.]

Interpretation Introduction

Interpretation: The concentration and volume of Pb(NO3)2,KCl and solubility product of PbCl2 is given. The equilibrium concentration of Pb2+ and Cl is to be calculated.

Concept introduction: The formation of solid in a solution is known as precipitation.

Solubility product, Ksp is defined as the concentration of ions in a saturated solution where each ion is raised to the power of their coefficients. Ion product, Q is defined as the initial concentration of ions in any solution where each ion is raised to the power of their coefficients.

The relation between ion product, Q and solubility product, Ksp is as follows.

• If Ksp>Q , then no precipitate will form.
• If Ksp<Q , then precipitate will form.
• If Ksp=Q , then the solution will be in just saturation.
Explanation

Explanation

To determine: The equilibrium concentration of Pb2+(aq) and Cl(aq) in a solution prepared by mixing 50mL of 0.10M Pb(NO3)2 to 50mL of 1.0M KCl .

The initial concentration of Pb2+ is 0.050M_ and initial concentration of Cl(aq) is 0.50M_ .

Given

Concentration of Pb(NO3)2 is 0.10M .

Concentration of KCl is 1.0M .

Solubility product of PbCl2 is 1.6×105 .

Volume of Pb(NO3)2 is 50mL .

Volume of KCl is 50mL .

The PbCl2 will only precipitate out during mixing of Pb(NO3)2 and KCl if,

Ksp<Q

It is assumed that the concentration of Pb2+ and Cl(aq) is same as the concentration of Pb(NO3)2 and KCl . Hence, concentration of Pb2+ and Cl(aq) is,

[Pb2+]=0.10M[Cl]=1.0M

Formula

The initial concentration of compound is calculated using the formula,

Substitute the values of volume, given concentration of Pb2+ and volume of original solution in the above equation.

Substitute the values of volume, given concentration of Cl and volume of original solution in equation (1).

The ion product of PbCl2 is 0.0125_ .

Formula

The ion product of PbCl2 is calculated as,

Q=[Pb2+]0[Cl]02

Where,

• Q is ion product.
• [Pb2+]0 is initial concentration of Pb2+ .
• [Cl]0 is initial concentration of Cl .

Substitute the value of [Pb2+]0 and [Cl]0 in the above expression.

Q=[Pb2+]0[Cl]02=(0

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