# Chapter 14 Solution for Principles of Instrument Analysis

Topics: Titration, Elementary algebra, Chemistry Pages: 11 (1964 words) Published: September 30, 2012
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

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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.

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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

determined graphically or evaluated by performing least-squares on the linear portions and solving the two linear simultaneous equations. Least-squares analysis gives the following results. Points 1 to 4 b1 = slope = 1.626 × 10-1 a1 = intercept = 3.000× 10-4 y = a1 + b1x x = a 2 − a1 = 3.19 mL b1 − b2

Points 5 to 9
b2 = 1.300 × 10-4 a2 = 5.1928 × 10-1 y = a2 + b2x

3.19 mL × 2.44 × 10 -4

mmol Nitroso R 1 mmol Pd(II) × mL 2 mmol Nitroso R = 3.89 × 10–5 M 10.00 mL solution

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Fundamentals of Analytical Chemistry: 8th ed.
0.6 0.5 A 500 (corrected) 0.4 0.3 0.2 0.1 0 0 2 4 6 8 10 Volume of Nitroso R, mL

Chapter 14

14-7. From Problem 14-1, cx =

Ax csVs ( Ax + s − Ax )Vx
0.276 × 4.25 × 5.00 = 1.0912 ppm Co (0.491 − 0.276) × 25.00

Substituting numerical values gives cx = mg L

Percent Co = 500 mL × 1.0912

× 10−3

g L 100% × 10−3 × = 0.0149% mL 3.65 g mg

14-8. (a) A365 = 0.426 = 3529 × 1.00 × cCo + 3228 × 1.00 × cNi A700 = 0.026 = 428.9 × 1.00 × cCo + 10.2 × 1.00 × cNi

Rearranging the second equation gives
cCo = (0.026 – 10.2cNi)/428.9

= 6.062 × 10–5 – 2.378 × 10–2cNi Substituting into the first equation gives 0.426 = 3529(6.062 × 10–5 – 2.378 × 10–2cNi) + 3228cNi 0.426 = 0.2139 – 83.91cNi + 3228cNi cNi = 6.75 × 10–5 M

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Fundamentals of Analytical Chemistry: 8th ed. Substituting into the equation for cCo gives cCo= 6.062 × 10–5 – 2.378 × 10–2cNi = 6.062 × 10–5 – 2.378 × 10–2 × 6.75 × 10–5

Chapter 14

= 5.90 × 10–5 M (b) Proceeding in the same way, we obtain cCo = 1.88 × 10–4M

and

cNi = 3.99 × 10–5 M

14-9. At 475 nm, εA = 0.155/7.50 × 10–5 = 2067; εB = 0.702/4.25 × 10–5 = 16518 At 700 nm εA = 0.755/7.50 × 10–5 = 10067; εB = 0.091/4.25 × 10–5 = 2141 (a) 2067cA + 16518cB = 0.439/2.5 = 0.1756 B

10067cA + 2141cB = 1.025/2.5 = 0.410 Solving the two equations simultaneously gives, cA = 3.95 × 10–5 M and cB = 5.69 × 10–6 M

(b) In the same way,
cA = 2.98 × 10–5 M and cB = 1.23 × 10–6 M

14-10. (a)

At 485 nm,...