CHEM 235TAKE HOME QUIZ

Prepared by:Janine V. SameloBSChem2

8-15. The following are the relative peak areas for chromatograms of standard solutions of methyl vinyl ketone (MVK).

Concentration MVK, mmol/L

Relative Peak Areas

0.500 3.761.50 9.162.50 15.033.50 20.424.50 25.335.50 31.97

a) Determine the coefficients of the best fit line using the least square method.

Ans: slope = 5.57intercept = 0.90

b) Construct an ANOVA table.

Concentration MVK, mmol/L Relative Peak Area xi

2 yi2 x1y1

0.5 3.76 0.25 14.1376 1.881.5 9.16 2.25 83.9056 13.742.5 15.03 6.25 225.9009 37.5753.5 20.42 12.25 416.9764 71.474.5 25.33 20.25 641.6089 113.9855.5 31.97 30.25 1022.0809 175.83518 105.67 71.5 2404.6103 414.485

a) Plot the least-squares line as well as the experimental points

d) A sample containing MVK yielded relative peak area of 10.3. Calculate the concentration of MVK in the solution. Ans: 1.68757 mmol/L

e) Assume that the result in (d) represents a single measurement as well as the mean of the four measurements. Calculate the respective absolute and relative standard deviations.

Ans:

single measurement:

Sc = 0.08139

cv = 4.82%

mean of four measurement:

Sc = 0.05193

cv = 3.07%

f) Repeat the calculations in (d) and (e) for a sample that gave a peak area of 22.8.

Ans: unknown = 3.93 mmol/L

sc = 0.08

CV = 2.03%

for four measurements,:sc = 0.05

CV = 1.26%

8-19. Water can be determined in solid samples by infrared spectroscopy. The water content of calcium sulfate hydrates is to be measured using calcium carbonate as an internal standard to compensate for some systematic errors in the procedure. A series of standard solutions containing calcium sulfate dihydrate and a constant known amount of the internal standard are prepared. The solution of the unknown water content is also prepared with the same amount of the internal standard. The absorbance of the dihydrate is measured at one wavelength ( ) along with that of the internal standard at another wavelength (. ) . The following results were obtained:

𝐴𝑠𝑎𝑚𝑝𝑙𝑒

Asample Astd % Water0.15 0.75 4.00.23 0.60 8.00.19 0.31 12.00.57 0.70 16.00.43 0.45 20.00.37 0.47 unknown

a)Plot the absorbance of the sample (Asample) vs. the % water and determine whether the plot is linear from the regression statistics

•Based from the graph we can see that the plot is not linear.

b) Plot the ratio / vs. % water and comment on whether use of the internal standard improves the linearity from that in part (a). If it improves the linearity, why?

• Upon the addition of internal standard, the graphs becomes linear. The ratio Asample/Astd replace the systematic errors that affects the sample and the internal standard

c) Calculate the percentage of water in the unknown using the internal standard data.(Ans:9.58%)

8-20. Potassium can be can be determined by flame emission spectrometry (flame photometry) using a lithium internal standard. The following data were obtained for standard solutions of KCl and an unknown containing a constant known amount of LiCl as the internal standard. All the intensities were corrected for a background by subtracting the intensity of the blank.

Concentration of K, ppm

Intensity ofK Emission

Intensity ofLi Emission

1.0 10.0 10.0

2.0 15.3 7.5

5.0 34.7 6.8

7.5 65.2 8.5

10.0 95.8 10.0

20.0 110.2 5.8

unknown 47.3 9.1

a) Plot the K emission intensity vs. the concentration of K and determine the linearity from the regression statistics.

b) Plot the ratio of K intensity to the Li intensity vs. the concentration of K and compare the resulting linearity to that in part (a). Why does the internal standard improve the linearity?

Ans: The graph with the ratio of K/Li intensity vs. concentration is very much linear compared to that without the internal standard because the ratio reduces the propagation of errors.

c) Calculate the concentration of k in the unknown.

Ans: 5.15 ppm

8-23. The following atomic absorption results were obtained for the determinations of Zn in a multivitamin tablet. All absorbance values were corrected for the appropriate reagent blank (cZn = 0.0 ng/mL). The mean value for the blank was 0.0000 with a standard deviation of 0.0047 absorbance units.

cZn ng/mL A

5.0 0.0519

5.0 0.0463

5.0 0.0485

10.0 0.0980

10.0 0.1033

10.0 0.0925

Tablet sample 0.0672

Tablet sample 0.0614

Tablet sample 0.0661

a) Find the mean absorbance values for the 5.0 and 10.0 ng/mL standards and for the tablet sample. Find the standard deviations of these values.

Mean 5.0= 0.0489 s=0.00280

Mean 10.0= 0097933 s= o.00540

Mean tablet samples= 0.0489 s=0.003081

b) Find the least-squares best line through the points at cZn = 0.0, 5.0 and 10.0 ng/ mL. Find the calibration sensitivity and the analytical sensitivity.

0 2 4 6 8 10 120

0.02

0.04

0.06

0.08

0.1

0.12

f(x) = 0.00980000000000002 x − 0.000900000000000033R² = 0.998988944966152

Absorbance vs. Concentration of Zn

Series1Linear (Series1)

Concentration of Zn, ng/mL

Ab

so

rba

nce

A = 0.0098cZn – 0.00002Calibration sensitivity = 0.0098 mL/ngAnalytical sensitivity at 5.0 ng/mL = 3.47 mL/ng

c)Find the detection limit for a k value of 3. To what level of confidence does this corresponds?

Ans: DL = 1.44 ng/mL

d)Find the concentration of Zn in the tablet sample and the standard deviation in the concentration.

Ans:cZn = 6.63 ng/mL, s = 0.005

21-24. A 4.00-mL aliquot of HNO2 0.05000 M is diluted to 75.00-mL and titrated with 0.0800 M Ce+4 . The pH of the solution is maintained at 1.00 throughout the titration; the formal potential of the cerium system is 1.44 V.

a) Calculate the potential of the indicator electrode with respect to a saturated calomel reference electrode after the addition of 5.00, 10.00, 15.00, 25.00, 40.00, 49.00, 49.50, 49.60, 49.70, 49.80, 49.90,49.95, 49.99, 50.00, 50.01, 50.05, 50.10, 50.20, 50.30,50.40, 50.50, 51.00, 60.00, 75.00, and 90.00 mL of cerium(IV)

Vol Ce(IV), mL

E vs. SCE, V Vol Ce(IV), mL

E vs. SCE, V

5.00 0.58 50.00 0.80

10.00 0.59 50.01 0.98

15.00 0.60 50.05 1.02

25.00 0.61 50.10 1.04

40.00 0.63 50.20 1.05

49.00 0.66 50.30 1.06

49.50 0.67 50.40 1.07

49.60 0.67 50.50 1.08

49.70 0.67 51.00 1.10

49.80 0.68 60.00 1.15

49.90 0.69 75.00 1.18

49.95 0.70 90.00 1.19

49.99 0.72

b) Draw the titration curve for these data.

0 10 20 30 40 50 60 70 80 90 1000

0.2

0.4

0.6

0.8

1

1.2

1.4

Titration Curve

Series1

Vol Ce(IV), mL

E v

s.

SC

E,

V

c) Generate a first and second derivative curve of these data. Does the volume at which the second- derivative curve crosses zero correspond to the theoretical equivalence point? Why or why not?

0 10 20 30 40 50 60 70 80 900

2

4

6

8

10

12

14

16

18

20

First-Derivative Plot

Series1

Vol Ce(IV), ml

∆E/∆

V,

V/m

L

0 10 20 30 40 50 60 70 80 90

-2000

-1500

-1000

-500

0

500

1000

1500

Second-Derivative Plot

Series1

Vol Ce(IV), ml

∆E2/∆

V2,

V2/m

L2

21-28. A lithium ion-selective electrode gave the potential given in the table for the following standard solutions of LiCl and the three samples of unknown concentration.

Solution a (Li+)Potential vs. SCE,

mV

0.100 +1.0

0.050 -30.0

0.010 -60.0

0.001 -138.0

unknown 1 -48.5

unknown 2 -75.3

a) Draw the calibration curve of electrode potential versus log a (Li+) and determine if the electrode follows the Nernst equation.

b) Use the linear least-squares procedure to determine the concentrations of the two unknowns.Ans:Unknown 1 = 0.0199 MUnknown 2 = 0.00788 M

26-19. The indicator HIn has an acid dissociation constant of at ordinary temperatures. The accompanying absorbance of the data are for solutions of the indicator measured in 1.00 cm cells in strongly acidic and strongly alkaline media.

λ nmAbsorbance

pH 1.00 pH 13.00420 0.355 0.050445 0.567 0.068450 0.658 0.076455 0.656 0.085470 0.614 0.116510 0.353 0.223550 0.119 0.324570 0.068 0.352585 0.044 0.360595 0.032 0.361610 0.019 0.355650 0.014 0.284

Estimate the wavelength at which absorption by the indicator becomes independent of pH ( the so-called isosbestic point).

26-24. Solutions of P and Q individually obey Beer’s Law over a large concentration range. Spectral data for these species in 1.00-cm cells are

λ nmAbsorbance

P Q400 0.078 0.500420 0.087 0.592440 0.096 0.599460 0.102 0.590480 0.106 0.564500 0.110 0.515520 0.113 0.433540 0.116 0.343580 0.170 0.170600 0.264 0.100620 0.326 0.055640 0.359 0.030660 0.373 0.030680 0.370 0.035700 0.346 0.063

a) Plot an absorption spectrum for a solution that is in P and in Q.

10000150002000025000300000

0.2

0.4

0.6

Absorption spectrum

wavelength

Ab

sorb

an

ce

b) Calculate the absorbance (1.00 cm cells) at 440 nm of a solution that is in P and in Q.

Ans:

For 3.86 x 10-5 M in P:A = 0.04334

For 5.37 x 10-4 M in Q:A = 1.357

c) Calculate the absorbance (1.00-cm cells) at 620 nm of a solution that is 1.89 x 10-4 M in P and 6.84 x 10-4 M in Q.

Ans:

For 1.89 x 10-4 M in P:A = 0.7206For 6.84 x 10-4 M in Q:A = 0.1587

26-26. A standard solution was put through appropriate dilutions to give the concentrations of iron shown below. The iron(II)-1,10-phenanthroline complex was then formed in 25.0-mL aliquots of these solutions, following which each was diluted to 50.0 mL. The following absorbances (1.00-cm cells) were recorded at 510 nm

Fe(II) concentration in

original solutions, ppm4.00 0.16010.0 0.39016.0 0.63024.0 0.95032.0 1.26040.0 1.580

a) Plot a calibration curve from these data.

b) Use the method of least squares to find an equation relating absorbance and concentration of iron (II).

Ans: A = 0.03949cFe – 0.001008

c) Calculate the standard deviation of the slope and intercept.

Ans: sm = 1.1 x 10-4

sb = 2.7 x 10-3

26-32. Mercury(II) forms a 1:1 complex with triphenyltetrazolium chloride (TTC) that exhibits an absorption maximum at 255 nm. The mercury (II) in a soil sample was extracted into an organic solvent containing an excess of TTC, and the resulting solution was diluted to 100.0 mL in a volumetric flask. Five-mL aliquots of the analyte solution were then transferred to six 25-mL volumetric flasks. A standard solution was then prepared that was in Hg(II). Volumes of the standard solution shown in the table were then pipetted into the volumetric flasks, and each solution was then diluted to 25.00 mL. The absorbance of each solution was measured at 255 nm in 1.00-cm quartz cells.

volume standard solution, mL

0.00 0.5822.00 0.6894.00 0.7676.00 0.8698.00 1.009

10.00 1.127

a) Enter the data into a spreadsheet, and construct a standard-additions plot of the data.

0 2 4 6 8 10 120

0.5

1

1.5

f(x) = 0.0541 x + 0.57R² = 0.992314380023588

Plot

Volume

Ab

sorb

an

ce

b)Determine the slope and intercept of the line.Ans:

slope = 0.05406intercept = 0.57036

c) Determine the standard deviation of the slope and the intercept

Ans: standard deviation of the slope:

sm = 0.002381 standard deviation of the intercept:

sb = 0.01442

d) Calculate the concentration of Hg(II) in the analyte solution.Ans:

cx = 4.22 x 10-5 M

23-44) Consider the peaks for pentafluorobenzene and benzene in the chromatogram shown above. The elution time for unretained solute is 1.06 min. The open tubular column is 30.0 m in length and 0.530 mm in diameter, with a layer of stationary phase 3.0 m thick on the inner wall.

a)Find the adjusted retention times and capacity factors for both compounds.

Ans:

t’r = 11.92 min

t’r = 12.14 min

k= 11.25

k= 11.45

b) Find the relative retention.

Ans: 1.01c ) Find the separation factor.

Ans: 1.01

d) Measuring W½ on the chromatogram, find the number of plates, N1 and N2, and the plate height for these two compounds.

W½ C6HF5 = 0.45 cm W½ C6H6 = 0.5 cmNC6HF5 = 4609 NC6H6 = 3861

e) Measuring the width, w, at the baseline on the chromatogram, find the number of plates for these two compounds.

W C6HF5 = 0.765 cm W C6H6 = 0.85 cmN = 4606 N = 3859

f) Use your answer in part (e) to find the resolution between the two peaks.Ans:

N=4216 Rs=0.27

g) Using the number of plates N= √N1N2, with the values from part (e) and the observed separation factor, calculate what you resolution should be compare your answer with the measured resolution in part (f).

23-45. A layer with negligible thickness containing 10.0nmol of methanol (D = 1.6x10-9m2/s) was placed in a tube of water 5.00cm in diameter and allowed to spread by diffusion . Using Equation 23-25, prepare a graph showing the Gaussian concentration profile of the methanol zone after 1.00, 10.0 and 100 min. Prepare a second graph showing the same experiment with the enzyme ribonuclease (D = 0.12x10-9m2/s).

23-46. A 0.25mm-diameter open tubular gas chromatography column is located with stationary phase that is 0.25 µm thick. The diffusion coefficient for a compound with a capacity factor k’ = 10 is Dm = 1.0 x 10-5m2/s in the gas phase and Ds = 1.0x10-9m2/s in the stationary phase. Consider longitudal diffusion and finite equilibration time in the mobile and stationary phases as sources of broadening . Prepare a graph showing the plate height from each of these three sources and the total plate height as a function of linear flow rate (2cm/s to 1m/s). Then change the stationary phase thickness to 2µm and repeat the calculations. Explain the difference in the two results.

23-47. Consider two Gaussian peaks with relative areas 4:1. Construct a set of graphs to show the overlapping peaks if the resolution is 0.5, 1 or 2.