Chapter 14 Solution for Principles of Instrument Analysis
Skoog/Holler/Crouch Principles of Instrumental Analysis, 6th ed.
Chapter 14 Instructor’s Manual
CHAPTER 14
14-1. Letting the subscript x stand for the unknown solution, x + s stand for the unknown plus standard, and Vt the total volume of solution, we can write
Ax = ε bcxVx / Vt Ax + s = ε b(cxVx + csVs ) / Vt
Dividing the first equation by the second and rearranging gives cx = Ax csVs 0.656 × 25.7 ×10.0 = = 21.1 ppm ( Ax + s − Ax )Vx (0.976 − 0.656) × 25.0
14-2. Using the equation developed in problem 14-1, we can write cCu 2+ = 0.723 × 2.75 × 1.00 = 2.0497 ppm (0.917 − 0.723) × 5.00
For dilute solutions, 1 ppm = 1 mg/L, so
Percent Cu = 200 mL × 2.0497 14-3.
mg L
× 10−3
g L 100% × 10−3 × = 0.0684% mL 0.599 g mg
End point
1
Fundamentals of Analytical Chemistry: 8th ed.
Chapter 14
There should be little or no absorbance until the end point after which the absorbance should increase approximately linearly. A green filter should be used because the red permanganate solution absorbs green light. 14-4.
End Point
Absorbance
A green filter is used because the red Fe(SCN)2+ absorbs green light.
Volume SCN-
14-5. The absorbance should decrease approximately linearly with
Absorbance
titrant volume until the end point. After the end point the absorbance becomes independent of titrant volume.
End Point
Volume EDTA
14-6. The data must be corrected for dilution so
Acorr = A500 ×
10.00 mL + V 10.00 mL
For 1.00 mL
Acorr = 0.147 ×
10.00 mL + 1.00 mL = 0.162 10.00 mL
Acorr is calculated for each volume in the same way and the following results are obtained.
2
Fundamentals of Analytical Chemistry: 8th ed.
Chapter 14
Vol, mL A500 Acorr Vol, mL A500 Acorr 0 0 0 5.00 0.347 0.521 1.00 0.147 0.162 6.00 0.325 0.520 2.00 0.271 0.325 7.00 0.306 0.520 3.00 0.375 0.488 8.00 0.289 0.520 4.00 0.371 0.519 These data are plotted below. The point of intersection of the linear portion of the plot can be
Chapter 14 Instructor’s Manual
CHAPTER 14
14-1. Letting the subscript x stand for the unknown solution, x + s stand for the unknown plus standard, and Vt the total volume of solution, we can write
Ax = ε bcxVx / Vt Ax + s = ε b(cxVx + csVs ) / Vt
Dividing the first equation by the second and rearranging gives cx = Ax csVs 0.656 × 25.7 ×10.0 = = 21.1 ppm ( Ax + s − Ax )Vx (0.976 − 0.656) × 25.0
14-2. Using the equation developed in problem 14-1, we can write cCu 2+ = 0.723 × 2.75 × 1.00 = 2.0497 ppm (0.917 − 0.723) × 5.00
For dilute solutions, 1 ppm = 1 mg/L, so
Percent Cu = 200 mL × 2.0497 14-3.
mg L
× 10−3
g L 100% × 10−3 × = 0.0684% mL 0.599 g mg
End point
1
Fundamentals of Analytical Chemistry: 8th ed.
Chapter 14
There should be little or no absorbance until the end point after which the absorbance should increase approximately linearly. A green filter should be used because the red permanganate solution absorbs green light. 14-4.
End Point
Absorbance
A green filter is used because the red Fe(SCN)2+ absorbs green light.
Volume SCN-
14-5. The absorbance should decrease approximately linearly with
Absorbance
titrant volume until the end point. After the end point the absorbance becomes independent of titrant volume.
End Point
Volume EDTA
14-6. The data must be corrected for dilution so
Acorr = A500 ×
10.00 mL + V 10.00 mL
For 1.00 mL
Acorr = 0.147 ×
10.00 mL + 1.00 mL = 0.162 10.00 mL
Acorr is calculated for each volume in the same way and the following results are obtained.
2
Fundamentals of Analytical Chemistry: 8th ed.
Chapter 14
Vol, mL A500 Acorr Vol, mL A500 Acorr 0 0 0 5.00 0.347 0.521 1.00 0.147 0.162 6.00 0.325 0.520 2.00 0.271 0.325 7.00 0.306 0.520 3.00 0.375 0.488 8.00 0.289 0.520 4.00 0.371 0.519 These data are plotted below. The point of intersection of the linear portion of the plot can be