Tube Number41Initial moles of Pb2+ (× 10-5)Initial moles of I (× 10-5)Equilibrium moles of I-(x 10-5)Precipitated moles of I-(x 10-5)Precipitated moles of Pb2+ (x 10-5)Equilibrium moles of Pb2+ (x 10-5)[Pb2*] at equilibriumK Pbl,spAverage K, Pbl,sp3. Tube #:234mL 0.0120 M Pb(NO,),5.005.005.005.002.003.004.005.00mL 0.0300 M KImL 0.20 M KNO,3.002.001.000.00Total vol. (mL)10.0010.0010.0010.00Absorbance of soln.0.3810.4200.4700.542[I] at equilibrium (M)

1) 5 mL of 0.012 M Pb(NO3)2 gives 0.012 *5mmol= 0.06 mmol of Pb2+.

Answered: oles of I-(x 10) noles of Pb2+ (x 10-5)…

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oles of I-(x 10) noles of Pb2+ (x 10-5) noles of Pb2+ (X 10-5)

Fundamentals Of Analytical Chemistry

9th Edition

ISBN:9781285640686

Author:Skoog

Publisher:Skoog

Chapter7: Statistical Data Treatment And Evaluation

Section: Chapter Questions

Problem 7.10QAP

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Concept explainers

Acid Base Titration

An acid-base titration is the method of quantitative analysis for determining the concentration of an acid and base by exactly neutralizing it with a standard solution of base or acid having known concentration.

Question

Tube Number
4
1
Initial moles of Pb2+ (× 10-5)
Initial moles of I (× 10-5)
Equilibrium moles of I-(x 10-5)
Precipitated moles of I-(x 10-5)
Precipitated moles of Pb2+ (x 10-5)
Equilibrium moles of Pb2+ (x 10-5)
[Pb2*] at equilibrium
K Pbl,
sp
Average K, Pbl,
sp
3.

Tube #:
2
3
4
mL 0.0120 M Pb(NO,),
5.00
5.00
5.00
5.00
2.00
3.00
4.00
5.00
mL 0.0300 M KI
mL 0.20 M KNO,
3.00
2.00
1.00
0.00
Total vol. (mL)
10.00
10.00
10.00
10.00
Absorbance of soln.
0.381
0.420
0.470
0.542
[I] at equilibrium (M)

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