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Chemistry

9th Edition
Steven S. Zumdahl
ISBN: 9781133611097

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Chemistry

9th Edition
Steven S. Zumdahl
ISBN: 9781133611097
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,

Initialconcentration=Volumeofcompound(mL)×Givenconcentration[Volumeoforiginalsolution(mL)+(Volumeofcompoundadded(mL))] (1)

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

InitialconcentrationofPb2+=Volumeofcompound(mL)×Givenconcentration[Volumeoforiginalsolution(mL)+(Volumeofcompoundadded(mL))]=(50mL)(0.10M)50mL+50mL=0.050M_

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

InitialconcentrationofCl=Volumeofcompound(mL)×Givenconcentration[Volumeoforiginalsolution(L)+(Volumeofcompoundadded(L))]=50mL(1.0)M50mL+50mL=0.50M_

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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