W 460 nm dilution 100 mL c 0.538071 ppm 5 pts 1.2 ...

1.1 The concentration of phenol in a water sample is determined by separating the phenol from nonvolatile impurities by steam distillation, followed by reacting with 4-aminoantipyrine and K3Fe(CN)6 at pH 7.9 to form a colored antipyrine dye. A phenol standard with a concentration of 4.00 ppm has an absorbance of 0.424 at a wavelength of 460 nm using a 1.00-cm cell. A water sample is steam-distilled, and a 50.00-mL aliquot of the distillate is placed in a 100-mL volumetric flask and diluted to volume with distilled water. The absorbance of this solution is found to be 0.394. What is the concentration of phenol (in parts per million) in the water sample?

PhOH std A w b

4 ppm 0.424

460 nm 1 cm

PhOH samp 0.269036 ppm

aliquot

50 mL

dilution

100 mL

A

0.394

c

0.538071 ppm

5 pts

1.2 The determination of phosphate concentration in a washing powder requires first the hydrolysis of the tripolyphosphate components into the phosphate ion or its protonated forms. Then a quantitative colorimetric method based upon the absorption of the yellow complex of phosphate and ammonium vanadomolybdate can be used. A series of standard solutions was prepared. The complex was then developed in 10mL aliquots of these solutions by adding a 5.0mL aliquot of an ammonium vanadomolybdate solution. Measurements were made in a glass cell of 1.0 cm pathlength at 415 nm.

The following absorbance data were obtained.

Solutions C(mmol/L) A

S0

0

0

m

S1

0.1

0.15

b

S2

0.2

0.28

S3

0.3

0.4

S4

0.4

0.55

S5

0.5

0.7

1.377143 0.002381

a.

0.8

0.7

0.6

0.5

0.4

0.3

0.2

0.1

0 0

y = 1.3771x + 0.0024 R? = 0.9987

3pts(plot)

0.1

0.2

0.3

0.4

0.5

0.6

b. By the method of least squares, derive an equation relating absorbance to phosphate concentration.

A=1.3771c+0.0024

2pts(equation)

c. An unknown solution obtained by hydrolysis of 1 g of detergent treated in an identical way gave an A50 absorbance of 0.45. Determine the phosphate concentration of this solution.

1g

A

0.45

c

0.325035 mmol/L

3pts

2.1 To determine the concentrations (mol/L) of Co(NO3)2 (A) and Cr(NO3)3 (B) in an unknown sample, the following representative absorbance data were obtained

A (mol/L) B (mol/L) 510nm 575nm

0.15

0

0.714 0.097

0

0.06

0.298

0.757

Unknown Unknown 0.671

0.33

points given if solution is clearly presented

A

Abs

e

b

C

510 0.714

4.76

1

0.15

1pt

575 0.097 0.646667

1

0.15

1pt

B

510 0.298 4.966667

1

0.06

1pt

575 0.757 12.61667

1

0.06

1pt

2 equations, 2 unknown

at 510nm

0.671 =

4.76 bCA

+

4.966667 bCB

at 575nm

0.33 =

0.646667 bCA

+

12.61667 bCB

solve for CA by eliminating CB (also note b=1)

1.70452 =

12.09168 bCA

+

-

0.33 =

0.646667 bCA

+

1.37452 =

11.44501 CA

12.61667 bCB 12.61667 bCB

0

CA

0.120098 mol/L

1pt

substitute CA in one of the above equations to get CB

CB

0.02 mol/L

1pt

3.1

A100

63

96.4 ppm

a

b sample y

x

390

0.9

0.33

0.65 0.722222 0.366667

430

0.88

0.5

0.74 0.840909 0.568182

450

0.69

0.53

0.67 0.971014 0.768116

470

0.48

0.51

0.55 1.145833 1.0625

slope and intercept is clearly identified (or in equation form)

slope

0.611921

2pts

y-int

0.496934

2pts

points given if solution is clearly presented

Cb

58.98915 ppm

=E106*D99

1pt

Ca

31.30682 ppm

=E107*C99

1pt

3.2 Blanco and co-workers reported several examples of the application of multiwavelength linear regression analysis for the simultaneous determination of mixtures containing two components with overlapping spectra. For each of the following, determine the molar concentration of each analyte in the mixture.

a. A mixture of MnO4? and Cr2O72?, and standards of 1.0 x 10?4 M KMnO4 and 1.0 x 10?4 M K2Cr2O7 gave the following results

wl(nm) 266 288 320 350 360

M std C std

1.00E-04 1.00E-04

0.042

0.41

0.082 0.283

0.168 0.158

0.125 0.318

0.056 0.181

slope y-int

Cr2O72MnO4?

Mixture

y

x

0.766 18.2381 9.761905

0.571 6.963415 3.45122

0.422 2.511905 0.940476

0.672 5.376 2.544

0.366 6.535714 3.232143

slope and intercept is clearly identified (or in equation form)

1.78386

2pts

0.814653

2pts

points given if solution is clearly presented

1.78E-04 M

=E131*D124

1pt

8.15E-05 M

=E132*C124

1pt

b. Titanium and vanadium were determined by forming the H2O2 complexes. Results for a mixture of Ti(IV) and V(V) and standards of 63.1 ppm Ti(IV) and 96.4 ppm V(V) are listed in the following table Absorbances

Ti(std) V(std)

w(nm)

63.1

96.4 Mixture

y

x

390 0.895 0.326 0.651 0.727374 0.364246

430 0.884 0.497 0.743 0.840498 0.562217

450 0.694 0.528 0.665 0.958213 0.760807

470 0.481 0.512 0.547 1.137214 1.064449

510 0.173 0.374 0.314 1.815029 2.16185

slope and intercept is clearly identified (or in equation form)

slope

0.606898

2pts

y-int

0.499258

2pts

points given if solution is clearly presented

[V]

58.505 ppm

=E150*D143

1pt

[Ti]

31.5032 ppm

=E151*C143

1pt

c. Copper and zinc were determined by forming colored complexes with 2-pyridyl-azo-resorcinol (PAR). The absorbances for PAR, a mixture of Cu2+ and Zn2+, and standards of 1.0-ppm Cu2+ and 1.0-ppm Zn2+ are listed in the following table. Note that you must correct the absorbances for the metal for the absorbance due to the PAR.

Absorbances

w(nm) PAR

Cu std Zn std Mixture

480 0.211 0.698 0.971 0.656

496 0.137 0.732 1.018 0.668

510

0.1 0.732 0.891 0.627

526 0.072 0.602 0.672 0.498

540 0.056 0.387 0.306

0.29

corrected Absorbances

Cu std Zn std Mixture

w(nm)

1

1

y

x

480 0.487

0.76 0.445 0.913758 1.560575

496 0.595 0.881 0.531 0.892437 1.480672

510 0.632 0.791 0.527 0.833861 1.251582

526

0.53

0.6 0.426 0.803774 1.132075

540 0.331

0.25 0.234 0.706949 0.755287

slope and intercept is clearly identified (or in equation form)

slope

0.256297

2pts

y-int

0.513363

2pts

points given if solution is clearly presented

Zn

0.256297 ppm

=E178*D171

1pt

Cu

0.513363 ppm

=E179*C171

1pt

 ................
................

In order to avoid copyright disputes, this page is only a partial summary.

Google Online Preview   Download