Quantitative Chemical Analysis
Quantitative Chemical Analysis
9th Edition
ISBN: 9781464135385
Author: Daniel C. Harris
Publisher: W. H. Freeman
Question
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Chapter 15, Problem 15.BE
Interpretation Introduction

Interpretation:

The cell voltage at each volumes of added EDTA has to be calculated.

Concept Introduction:-

Cell Voltage:

Cell voltage is the difference between cathode potential and anode potential.

Ecell=E+-E-

The unreacted concentration in the solution is determined by

[X]=(fractionof[X]remaining)(initialconcentrationofX)(dilutionfactor)

Ion-selective electrode:

An ion-selective electrode (ISE) best known as a specific ion electrode (SIE), is a sensor (or transducer) that transforms the activity of a specific ion dissolved in a solution into an electrical potential.

Ion-selective electrode obeys following equation.

E=constant-0.05916log[CN-]

Expert Solution & Answer
Check Mark

Explanation of Solution

Given:

The given reactions are the small amount of HgY2- added to analyte equilibrates with a tiny amount of Hg2+

The titration reaction is a reduction reaction.

Hg2++Y4-HgY2-Kf=[HgY2-][Hg2+][Y4-]=1021.5...(1)

The Nernst equation for the cell is

E=E+-E-=(0.852-0.059162log(1[Hg2+]))-E-...(2)

Where [Hg2+] can be substituted with equation 1 that is [HgY2-]/Kf[Y4-]

E=E+-E-=(0.852-0.059162log([Y4-]Kf[HgY2-]))-E-=0.852-E--0.059162log(Kf[HgY2-])-0.059162log[Y4-]...(3)

To calculate the voltage we have to calculate the concentration of [HgY2-] and [Y4-] at every point.

The concentration of [HgY2-]=1.0×104-M .

If V=0 the dilution will be effected Kf(HgY2-)Kf(MgY2-)

[X]=(fractionof[X]remaining)(initialconcentrationofX)(dilutionfactor)

At 0mL: [Hg+][Hg2+][EDTA]Y4-Kf(forHgY4-)

[Y4-]Y4-[EDTA]=9.7×10-14M

Using equation 3,

E=0.852-0.241-0.059162log1021.51.0×10-4-0.059162log(9.7×10-14)=0.242V

E=0.242V

At 10.0mL:

Ve=25.0mL Thus 1025ofMg2+ is in the form MgY2-

And 1525ofMg

[Y4-]=[MgY2-][Mg2+]/Kf

[Y4-]=[10][15]/6.2×108=1.08×10-9M

[HgY2-]=(50.0mL60.0mL)(1.0×10-4M) = 8.33×10-5M

E=0.852-0.241-0.059162log1021.58.33×10-5-0.059162log(1.08×10-9)=0.120V

At 20.0mL:

[Y4-]=[MgY2-][Mg2+]/Kf=(205)/6.2×108=6.45×10-9M

[HgY2-]=[MgY2-][Mg2+]/Kf=(50.070.0)/1.0×10-4=7.14×10-5M

E=0.852-0.241-0.059162log1021.57.14×10-5-0.059162log(6.45×10-9)=0.095V

At 24.9mL:

[Y4-]=[MgY2-][Mg2+]/Kf=(24.90.1)/6.2×108=4.0×10-7M

[HgY2-]=[MgY2-][Mg2+]/Kf=(50.074.9)/1.0×10-4=6.48×10-5M

E=0.852-0.241-0.059162log1021.56.48×10-5-0.059162log(6.68×10-9)=0.041V

After 25mL , this is the equivalence point where [Mg2+]=[EDTA]

[Mg+][Mg2+][EDTA]Y4-Kf(forMgY4-)

(50.075.0)(0.0100)-xx2=1.85×108x=6.0×10-6M

[Y4-]=αY4-(6.0×106)=1.80×10-6M

[HgY2-]=(50.075.0)(1.0×104)=6.67×10-5M

E=0.852-0.241-0.059162log1021.56.67×10-5-0.059162log(1.0×10-6M)=0.021V

E=0.021V

At 26.0mL the excess EDTA in the solution has to be calculated.

[Y4-]Y4-(EDTA)=(0.30)[(1.0mL76.0mL)(0.0200M)]=7.89×10-6M

[HgY2-]=(50.076.0)(1.0×104)=6.58×10-5M

E=0.852-0.241-0.059162log1021.56.58×10-5-0.059162log(7.89×10-6M)=0.027V

E=-0.027V

Conclusion

The cell voltage at each volumes of added EDTA was calculated.

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