Exploring Chemical Analysis
Exploring Chemical Analysis
5th Edition
ISBN: 9781429275033
Author: Daniel C. Harris
Publisher: Macmillan Higher Education
Question
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Chapter 12, Problem 12.20P
Interpretation Introduction

Interpretation:

Iterations have to be carried to find [Pb2+] and ionic strength of PbF2 by approximation.

Concept Introduction:

Expression to compute ionic strength is as follows:

  μ=12cizi2

Here

z denotes charge.

ci denotes concentration.

μ denotes ionic strength.

Expert Solution & Answer
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Explanation of Solution

Equilibrium for PbF2 written as follows:

  PbF2Pb2++2F

Assume [Pb2+] is same as [F] equal to x.

Corresponding Ksp expression in terms of activity coefficients is as written as follows:

  Ksp=[Pb2+][F]2        (1)

Substitute 3.6×108 for Ksp , x for [Pb2+] and 2x for [F] in equation (1).

  3.6×108=4x3x=2.08×103 M

Thus when ionic strength is assumed as zero, the value for [Pb2+] is 2.08×103 M.

Formula to compute ionic strength is as follows:

  μ=12(c1z12+c2z22)        (2)

Substitute +2 for z1, 2.08×103 M for c1, 1 for z2, 4.16×103 M for c2 in equation (2).

  μ=12((2.08×103 M)(+2)2+(4.16×103 M)(1)2)=6.24×103 M0.00624 M

Expression for linear interpolation is given as follows:

  unknown γ intervalΔγ=unknown μ intervalΔμ        (3)

Substitute 0.6650.742 for Δγ, 0.665γ for unknown γ interval, 0.010.005 for Δμ and 0.010.00624 for known μ interval equation (3).

  0.665γ0.6650.742=0.010.006240.010.005γ=0.723

Thus activity coefficient of Pb2+  ion is 0.723.

Substitute 0.9000.926 for Δγ, 0.900γ for unknown γ interval, 0.010.005 for Δμ and 0.010.00624 for known μ interval equation (3) to calculate γ for F ion.

  0.900γ0.9000.926=0.010.006240.010.005γ=0.919

Thus activity coefficient of F  ion is 0.919.

Corresponding Ksp expression in terms of activity coefficients is as written as follows:

  Ksp=[Pb2+](γPb2+)[F]2(γF)2        (4)

Substitute 0.723 for γPb2+, 0.919 for γF,  and 3.6×108 for Ksp and x for [Pb2+] and 2x for [F] in equation (4).

  3.6×108=4x3(0.723)(0.919)2x=2.45×103 M

Thus, value of [Pb2+] in this iteration is 2.45×103 M while value of [F] is 4.9×103 M.

For the next iteration,

Substitute +2 for z1, 2.45×103 M for c1, 1 for z2, 4.9×103 M for c2 in equation (2).

  μ=12((2.45×103 M)(+2)2+(4.9×103 M)(1)2)=7.35×103 M0.00735 M

For μ that includes 0.00735 M interval for γ is from 0.005 to 0.01.

Substitute 0.6650.742 for Δγ, 0.665γ for unknown γ interval, 0.010.005 for Δμ and 0.010.00735 for known μ interval equation (3).

  0.665γ0.6650.742=0.010.007350.010.005γ=0.705

Thus activity coefficient of Pb2+  ion is 0.705.

Substitute 0.9000.926 for Δγ, 0.900γ for unknown γ interval, 0.010.005 for Δμ and 0.010.00735 for known μ interval equation (3) to calculate γ for F ion.

  0.900γ0.9000.926=0.010.007350.010.005γ=0.914

Thus activity coefficient of F  ion is 0.914.

Substitute 0.705 for γPb2+, 0.914 for γF,  and 3.6×108 for Ksp and x for [Pb2+] and 2x for [F] in equation (4).

  3.6×108=4x3(0.705)(0.914)2x=2.48×103 M

Thus, value of [Pb2+] in this iteration is 2.48×103 M while value of [F] is 4.96×103 M.

For the next iteration,

Substitute +2 for z1, 2.48×103 M for c1, 1 for z2, 4.96×103 M for c2 in equation (2).

  μ=12((2.48×103 M)(+2)2+(4.96×103 M)(1)2)=7.44×103 M0.00744 M

For μ that includes 0.00744 M interval for γ is from 0.005 to 0.01.

Substitute 0.6650.742 for Δγ, 0.665γ for unknown γ interval, 0.010.005 for Δμ and 0.010.00744 for known μ interval equation (3).

  0.665γ0.6650.742=0.010.007440.010.005γ=0.704

Thus activity coefficient of Pb2+  ion is 0.704.

Substitute 0.9000.926 for Δγ, 0.900γ for unknown γ interval, 0.010.005 for Δμ and 0.010.00744 for known μ interval equation (3) to calculate γ for F ion.

  0.900γ0.9000.926=0.010.007440.010.005γ=0.913

Thus activity coefficient of F  ion is 0.913.

Substitute 0.704 for γCa2+, 0.913 for γF,  and 3.6×108 for Ksp and x for [Pb2+] and 2x for [F] in equation (4).

  3.6×108=4x3(0.704)(0.913)2x=2.48×103 M

At this stage no more iteration is required as concentration has not changed.

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