Quantitative Chemical Analysis
Quantitative Chemical Analysis
9th Edition
ISBN: 9781464135385
Author: Daniel C. Harris
Publisher: W. H. Freeman
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Chapter 12, Problem 12.6P

(a)

Interpretation Introduction

Interpretation:

The equivalence volume of EDTA titration should be calculated.

Concept introduction:

Volumetric principle:

The volume and concentration of unknown solution is determined by it is titrate with known volume and concentration solution.

The volume and concentration of unknown solution is required equivalent volume and concentration of known solution in the volumetric titration.

V1M1=V2M2

Where,

V1 is volume of known solution

N1 is concentration of known solution

V1 is volume of unknown solution

V1 is concentration of unknown solution

(a)

Expert Solution
Check Mark

Answer to Problem 12.6P

The volume of solution at equivalent point is 100.0mL

Explanation of Solution

To determine the volume of solution at equivalent point in EDTA titration.

Given,

Volume of known solution is 100.0 mL

Concentration of known solution is 0.0500 M

Concentration of EDTA solution 0.0500 M

pH=9.0

According to the volumetric principle,

Ve×0.0500M=100.0 mL×0.0500M=100.0 mL×0.0500M0.500M=100.0mL

The given values are plugged in above equation to give a volume of solution at equivalent point in EDTA titration.

Conclusion

The equivalence volume of EDTA titration was calculated.

(b)

Interpretation Introduction

Interpretation:

The concentration of Mn+ given solution should be calculated.

Concept introduction:

Molarity:

The number of moles of solute present in liter of volume of solution is given by molarity.

Molarity=molesVolume(L)

(b)

Expert Solution
Check Mark

Answer to Problem 12.6P

The concentration of Mn+ is given solution is 0.0167M

Explanation of Solution

To determine the concentration of Mn+ is given solution.

Given,

Equivalent point in EDTA titration is 100.0 mL

Volume of solution is V=12Ve

Where,

12 is a fraction remaining

Total volume of solution is,

V=12Ve=1002=50=100+50=150

The concentration of Mn+ ion is,

[Mnn+]=(12)×(0.0500M)×(100150)=0.0167M

The fraction remaining and volume of solution are plugged in above equation to give the concentration of Mn+ ion.

Conclusion

The concentration of Mn+ was calculated.

(c)

Interpretation Introduction

Interpretation:

At 9pH , the degree of dissociation αY4- of EDTA should be given.

Concept Information:

Degree of dissociation:

The ratio of mole of reactant that underwent to dissociating to mole of initial reactant is known as degree of dissociation.

In EDTA the degree of dissociation is,

αY4-=[Y4-][H6Y2+]+[H5Y+]+[H4Y]+[H3Y-]+[H2Y2-]+[H3Y3-]+[Y4-]=[Y4-][EDTA]

If pH is fixed, the degree of dissociation αY4- is constant and Kf will be Kf'

The formation constantKf'=αY4-Kf

Where,

Kf' is the conditional formation constant

αY4- is degree of dissociation

(c)

Expert Solution
Check Mark

Explanation of Solution

To give the degree of dissociation αY4- at 9pH

Given,

From the standard data table the αY4- of EDTA at 9pH is 0.041

Conclusion

At 9pH , the degree of dissociation αY4- of EDTA was given.

(d)

Interpretation Introduction

Interpretation:

The conditional formation constant should be calculated.

Concept Information:

Degree of dissociation:

The ratio of mole of reactant that underwent to dissociating to mole of initial reactant is known as degree of dissociation.

In EDTA the degree of dissociation is,

αY4-=[Y4-][H6Y2+]+[H5Y+]+[H4Y]+[H3Y-]+[H2Y2-]+[H3Y3-]+[Y4-]=[Y4-][EDTA]

If pH is fixed, the degree of dissociation αY4- is constant and Kf will be Kf'

The formation constantKf'=αY4-Kf

Where,

Kf' is the conditional formation constant

αY4- is degree of dissociation

(d)

Expert Solution
Check Mark

Explanation of Solution

To calculate the conditional formation constant

Given,

The formation constant Kf is 1012.00

From the standard data table the αY4- of EDTA at 9pH is 0.041

The formation constantKf'=αY4-Kf=1012.00×0.041=4.1×1010

The degree of dissociation and formation constant are plugged ion above equation to give conditional formation constant.

Conclusion

The conditional formation constant was calculated.

(e)

Interpretation Introduction

Interpretation:

At V=Ve , the concentration of Mn+ ion should be calculated.

Concept Information:

Conditional Formation Constant:

In the reaction of metal with ligand, the equilibrium constant is called as formation constant or the stability constant.

The formation constant for above complex reaction is,

Kf=[MYn-4][Mn+][Y4-]

Where,

Kf is formation constant

[M+] is concentration of metal ion

[Y4-] is concentration of ligand

[MYn4-] is concentration of the acid

If pH is constant, the degree of dissociation αY4- is constant and Kf will be Kf'

The formation constantKf'=αY4-Kf

Where,

Kf' is the conditional formation constant

αY4- is degree of dissociation

(e)

Expert Solution
Check Mark

Answer to Problem 12.6P

At V=Ve , the concentration of Mn+ ion is 7.8×10-7

Explanation of Solution

To calculate the concentration of Mn+ ion at V=Ve

Given,

Total volume of solution is 200mL

Concentration of titrants is 0.0500M

Kf=4.1×1010

The concentration of MYn-4 is,

[MYn-4]=0.0500M(100200)=0.0250M

We know,

Kf=[MYn-4][Mn+][Y4-]4.1×1010=0.0250-xx2x=7.8×10-7

The calculated concentration of MYn-4 and Kf are plugged in above equation to give concentration of Mn+ ion at V=Ve .

Conclusion

At V=Ve , the concentration of Mn+ ion was calculated.

(f)

Interpretation Introduction

Interpretation:

At V = 1.100 Ve , the concentration of Mn+ ion should be calculated.

Concept Information:

Conditional Formation Constant:

In the reaction of metal with ligand, the equilibrium constant is called as formation constant or the stability constant.

The formation constant for above complex reaction is,

Kf=[MYn-4][Mn+][Y4-]

Where,

Kf is formation constant

[M+] is concentration of metal ion

[Y4-] is concentration of ligand

[MYn4-] is concentration of the acid

If pH is constant, the degree of dissociation αY4- is constant and Kf will be Kf'

The formation constantKf'=αY4-Kf

Where,

Kf' is the conditional formation constant

αY4- is degree of dissociation

(f)

Expert Solution
Check Mark

Answer to Problem 12.6P

At V = 1.100 Ve , the concentration of Mn+ ion is 2.4×10-10

Explanation of Solution

To calculate the concentration of Mn+ ion at V = 1.100 Ve

Given,

Total volume of EDTA is 100mL

Concentration of EDTA is 0.0500M

Kf=4.1×1010

The concentration of EDTA is,

[EDTA]=0.0500M(10.0210)=2.38×103M

The concentration of [MYn-4] is,

[MYn-4]=0.0500M(100210)=3.38×102M

We know,

Kf=[MYn-4][Mn+][Y4-]4.1×1010=[2.38×10-2][mn+][2.38×10-3][mn+]=2.4×10-10

The calculated concentrations of MYn-4 , [Y4-] and Kf are plugged in above equation to give concentration of Mn+ ion at V = 1.100 Ve .

Conclusion

At V = 1.100 Ve , the concentration of Mn+ ion was calculated.

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