EBK EXPLORING CHEMICAL ANALYSIS
EBK EXPLORING CHEMICAL ANALYSIS
5th Edition
ISBN: 9781319416942
Author: Harris
Publisher: VST
Question
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Chapter 12, Problem 12.21P

(a)

Interpretation Introduction

Interpretation:

Iterations have to be carried to find values of [Ca2+] and [SO42] until concentration does not changes.

Concept Introduction:

Expression to compute ionic strength is as follows:

  μ=12cizi2

Here

z denotes charge.

ci denotes concentration.

μ denotes ionic strength.

(a)

Expert Solution
Check Mark

Explanation of Solution

Equilibrium for CaSO4 written as follows:

  CaSO4Ca2++SO42

Assume [Ca2+] is same as [SO42] equal to x.

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

  Ksp=[Ca2+][SO42]        (1)

Substitute 2.4×105 for Ksp and x for both [Ca2+] and [SO42] in equation (1).

  2.4×105=x2x=4.8989×103 M

Formula to compute ionic strength is as follows:

  μ=12(c1z12+c2z22)        (2)

Substitute +2 for z1, 4.8989×103 M for c1, 2 for z2, 4.8989×103 M for c2 in equation (2).

  μ=12((4.8989×103 M)(+2)2+(4.8989×103 M)(2)2)=19.6×103 M0.0196 M

For Ca2+ ion, interval for μ that includes 0.0196 M is 0.485 to 0.675 and corresponding interval for γ is from 0.05 to 0.01.

Expression for linear interpolation is given as follows:

  unknown γ intervalΔγ=unknown μ intervalΔμ        (3)

Substitute 0.4850.675 for Δγ, 0.485γ for unknown γ interval, 0.050.01 for Δμ and 0.050.0196 for known μ interval equation (3).

  0.485γ0.4850.675=0.050.01960.050.01γ=0.629

Thus activity coefficient of Ca2+  ion is 0.629.

Substitute 0.4450.660 for Δγ, 0.445γ for unknown γ interval, 0.050.01 for Δμ and 0.050.0196 for known μ interval equation (3) to calculate γ for SO42 ion.

  0.445γ0.4450.660=0.050.01960.050.01γ=0.6084

Thus activity coefficient of SO42  ion is 0.6084.

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

  Ksp=[Ca2+](γCa2+)[SO42](γSO42)        (4)

Substitute 0.629 for γCa2+, 0.608 for γSO42,  and 2.4×105 for Ksp and x for both [Ca2+] and [SO42] in equation (4).

  2.4×105=x2(0.629)(0.608)x=7.92×103 M

For the next iteration,

Substitute +2 for z1, 7.92×103 M for c1, 2 for z2, 7.92×103 M for c2 in equation (2).

  μ=12((7.92×103 M)(+2)2+(7.92×103 M)(2)2)=31.7×103 M0.0317 M

For Ca2+ ion, interval for μ that includes 0.0317 M is 0.485 to 0.675 and corresponding interval for γ is from 0.05 to 0.01.

Substitute 0.4850.675 for Δγ, 0.485γ for unknown γ interval, 0.050.01 for Δμ and 0.050.0317 for known μ interval equation (3).

  0.485γ0.4850.675=0.050.03170.050.01γ=0.572

Thus activity coefficient of Ca2+  ion is 0.572.

Substitute 0.4450.660 for Δγ, 0.445γ for unknown γ interval, 0.050.01 for Δμ and 0.050.0317 for known μ interval equation (3) to calculate γ for SO42 ion.

  0.445γ0.4450.660=0.050.03170.050.01γ=0.543

Thus activity coefficient of SO42  ion is 0.543.

Substitute 0.572 for γCa2+, 0.543 for γSO42,  and 2.4×105 for Ksp and x for both [Ca2+] and [SO42] in equation (4).

  2.4×105=x2(0.572)(0.573)x=8.79×103 M

For the next iteration,

Substitute +2 for z1, 8.79×103 M for c1, 2 for z2, 8.79×103 M for c2 in equation (2).

  μ=12((8.79×103 M)(+2)2+(8.79×103 M)(2)2)=35.16×103 M0.03516 M

For Ca2+ ion, interval for μ that includes 0.03516 M is 0.485 to 0.675 and corresponding interval for γ is from 0.05 to 0.01.

Substitute 0.4850.675 for Δγ, 0.485γ for unknown γ interval, 0.050.01 for Δμ and 0.050.03516 for known μ interval equation (3).

  0.485γ0.4850.675=0.050.03170.050.01γ=0.555

Thus activity coefficient of Ca2+  ion is 0.572.

Substitute 0.4450.660 for Δγ, 0.445γ for unknown γ interval, 0.050.01 for Δμ and 0.050.03516 for known μ interval equation (3) to calculate γ for SO42 ion.

  0.445γ0.4450.660=0.050.03170.050.01γ=0.525

Thus activity coefficient of SO42  ion is 0.525.

Substitute 0.555 for γCa2+, 0.525 for γSO42,  and 2.4×105 for Ksp and x for both [Ca2+] and [SO42] in equation (4).

  2.4×105=x2(0.555)(0.525)x=9.08×103 M

For the next iteration,

Substitute +2 for z1, 9.08×103 M for c1, 2 for z2, 9.08×103 M for c2 in equation (2).

  μ=12((9.08×103 M)(+2)2+(9.08×103 M)(2)2)=36.32×103 M0.0363 M

For Ca2+ ion, interval for μ that includes 0.0363 M is 0.485 to 0.675 and corresponding interval for γ is from 0.05 to 0.01.

Substitute 0.4850.675 for Δγ, 0.485γ for unknown γ interval, 0.050.01 for Δμ and 0.050.0363 for known μ interval equation (3).

  0.485γ0.4850.675=0.050.03630.050.01γ=0.550

Thus activity coefficient of Ca2+  ion is 0.550.

Substitute 0.4450.660 for Δγ, 0.445γ for unknown γ interval, 0.050.01 for Δμ and 0.050.0363 for known μ interval equation (3) to calculate γ for SO42 ion.

  0.445γ0.4450.660=0.050.03630.050.01γ=0.519

Thus activity coefficient of SO42  ion is 0.519.

Substitute 0.550 for γCa2+, 0.519 for γSO42,  and 2.4×105 for Ksp and x for both [Ca2+] and [SO42] in equation (4).

  2.4×105=x2(0.550)(0.519)x=9.16×103 M9.2 mM

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

(b)

Interpretation Introduction

Interpretation:

Iterations have to be carried to find values of [Ca2+] and [SO42] until concentration does not changes.

Concept Introduction:

Refer to part (a).

(b)

Expert Solution
Check Mark

Explanation of Solution

Equilibrium for CaSO4 is written as follows:

  CaSO4Ca2++SO42

Concentration of [Ca2+] and [SO42] are 9.2 mM and rest of the millimoles concentration can be attributed to formation of ion pair CaSO4(aq) as CaSO4 is only sparingly soluble salt.

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