Analytical Chemistry Final Exam

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Since CE separates analytes by their size to charge ratio we need to look at

the charge of the analytes first. Since they are all of different charges we can

use that to identify the order in which they will migrate. The order in which they

will pass the detector will be:

1

– benzylammonium

2

– benzyl alcohol

3

– benzoate

The reason for this that the negative electrode is at the outlet of the capillary

and the positive electrode is at the inlet. As such cations are attracted to the

outlet, and anions are slightly repelled by it. The bulk flow of solution pulls the

all species (including neutral molecules) towards the outlet at the same rate.

+

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Electroosmotic flow

Detector

Name : RedID # :


 

Show your work, answers alone will not receive partial credit. Where indicated, please enter your answer the box beneath the question. There are blank pages near the end of the exam to use as scratch paper. Answers without the units indicated (when necessary) will not be graded.

The last page of the exam contains a selection of data tables and equations for your use – feel free to remove the page from the exam.

 

Question

Value

Score

1

4

 

2

6

 

3

5

 

4

6

 

5

8

 

6

6

 

7

5

 

8

6

 

9

6

 

10

8

 

Total

60

 

By signing below I acknowledge that I will abide by the SDSU Academic Code of Honor as it pertains to academic integrity. For this exam I will not make use of any unauthorized materials, nor will I engage in any practices that otherwise fall within the broader definition of “cheating”.

_____________________________________________________________________________________

Signature Date

If this pledge is not signed the instructor reserves the right to not grade the exam and award a score of 0 for this exam in the calculation of the final grade.


  1. (4 points) Following a series of measurements and calculations you have determined that the concentration of calcium in a water source is 26.812316 ppm ± 0.0046641%. Express the result with the absolute uncertainty, and the correct number of significant figures.


 

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  1. (6 points) The concentration of CO2 in the atmosphere is currently 412 ppm. The density of the atmosphere is 0.0012 g/mL. What is the molar concentration of CO2 in our atmosphere?


 

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  1. (5 points) A stock solution of sodium sulfite has a concentration of 12.049 ±0.004 mM. Using a pipette with a volume of 10.0023 ±0.0008 mL, you transfer an aliquot of the solution to a 250.0 mL ±0.04% volumetric flask and dilute the solution to volume with water. What is the concentration (including the absolute uncertainty) of the diluted sodium sulfite solution?


 

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  1. (6 points) What is the pH of a solution prepared by mixing 125 mL of 45.02 mM disodium phosphate, with 45.00 mL of 60.31 mM trilithium phosphate?


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  1. word image 701 (8 points) You need to prepare 300.0 mL of a buffer with a concentration of 37.90 mM and a pH of 9.34. At your disposal are the following chemicals:

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

  1. Accounting for the activity of the ions in the solution, what is the concentration of iron free in solution for a solution containing 12.06 mM KNO3 (an inert salt) and saturated with Fe3(PO4)2 (Ksp = 1×10–36)? Note: the iron and phosphate ions do not contribute to the overall ionic strength of the solution.

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7)

(5

points)

A 25.00 mL solution containing 18.00 mM

arsenic

acid is titrated with 47.35

mL of a 15.00 mM solution of potassium hydroxide. What is the pH of the solution

once the titrant is added?

 

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  1. A solution of bromide in base can be oxidized by reaction with rhodium(VI) – see half reactions below. A 20.00 mL solution containing 12.00 mM Br– and buffered to pH 9.85, is titrated with 15.00 mL of a 8.00 mM Rh6+ solution. What potential will be measured by a potentiometer with a platinum wire is as the cathode, and a S.C.E. (0.241 V) as the anode?


BrO

+ H

2

O + 2e

 

Br

+ 2OH

E° = 0.766 V

Rh

6+

+ 3e

 

Rh

3+

E° = 1.48 V

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Replicate

Chemist

A

Chemist

B

1

15.87

15.76

2

15.79

15.80

3

15.91

15.73

4

15.83

15.74

5

15.80

15.72

6

15.8

15.78

AVG

15.83

15.76

STDEV

0.05

0.03

  1. word image 2280 Two analytical chemists are measuring the Soil PCB Content in ppb amount of PCB contamination in the soil of an old industrial facility. They each use a different analytical approach to measure fractions of the same soil sample. Tabulated to the right are the results that they obtained. Are their results equivalent at the 95% confidence level? Support your answer

mathematically. Note: the data has already been Grubb’s tested and all values are valid.

 

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10)(8 points) As part of a 2008 research study wastewater going to a treatment plant was sampled for illicit drugs during a music festival in Spain. The water was found to contain large amounts of MDMA (ecstasy), benzoylecgonine (the main metabolite of cocaine). The analysis was done with a liquid chromatography (LC) instrument.


 

Compound

Concentration (µg/L)

Signal

MDMA

8.46

15,610.0

Benzoylecgonine

9.94

110,051

Mesityl oxide

23.13

36,205

word image 702 In order to quantify the drugs and metabolites LC Calibration Data the LC was tested with a known mixture of MDMA, benzoylecgonine, and an internal standard mesityl oxide. Results for the testing of the instrument are tabulated to the right.


 

When the wastewater was tested, a 10.00 mL

 

Wastewater Analysis Data

 

 

 

 

2

µ= x ± ts sd = ∑(di d) ey = ex21 +ex22 texp = µx n n−1s

n

x

i

x

(

)

2

n

1

s = texp = K = e−ΔG0RT ΔG H TΔS

d

n

sd

G = xout x F = sA2 t = xA xB × nAnB t = xA xB

exp exp 2 expexp 2 2 s sB spool nA +nB sA + sB

nA nB

%s2 s2 (2 A + B *

n

A

1

(

)

s

A

2

+

n

B

1

(

)

s

B

2

n

A

+

n

B

2

&nA nB )2 2 v = 2 2 −2 spool = %ey = %ex1 +%ex2

+ sA2 . + sB2 .

– 0 – 0 ,nA / ,nB / +

nA +1 nB +1

E =K+ 0.05916 log([A]+KA,I [I]zzAI ) KA,I = [Az]zAI “!H+#$= zA

K

a

1

K

a

2

F

+

K

a

1

K

w

K

a

1

+

F

x

=

b

±

b

2

4

a

c

[I]

SA Sspike 2a

=

word image 2282CA CA %’VA (*+Cspike %’Vspike (*

&VT ) & VT )

SA Sspike

=

CA %’VO (* CA %’VO (*+Cspike %’Vspike (*

F(0.05, nnum, ndenom) for a Two-Tailed F-Test

.onuodenom “m 1 2 3 4 5 6

1

2

3

4

5

6

647.8 799.5

864.2

899.6

921.8

937.1

38.51 39.00

39.17

39.25

39.30

39.33

17.44 16.04

15.44

15.10

14.88

14.73

12.22 10.65

9.979

9.605

9.364

9.197

10.01 8.434

7.764

7.388

7.146

6.978

8.813 7.260 6.599 6.227 5.988 5.820

word image 2283 &VF ) &VF ) & VF )

SA kACA K×CIS K = CIS × SA

= =

SIS kISCIS CA CA SIS

Page

Table 6.2 Effective Diameters (a) for Selected Ions

Ion Effective Diameter (nm)

H3O+

0.9

Li+

0.6

Na+, IO3, HSO3, HCO3, H2PO4

0.45

OH, F, SCN, HS, ClO3, ClO4, MnO4

0.35

K+, Cl, Br, I, CN, NO2, NO3

0.3

Cs+, Tl+, Ag+, NH4+

0.25

Mg2+, Be2+

0.8

Ca2+, Cu2+, Zn2+, Sn2+, Mn2+, Fe2+, Ni2+, Co2+

0.6

Sr2+, Ba2+, Cd2+, Hg2+, S2–

0.5

Pb2+, CO32–, SO32–

0.45

Hg22+, SO42–, S2O32–, CrO42–, HPO42–

0.40

Al3+, Fe3+, Cr3+

0.9

PO43–, Fe(CN)63–

0.4

Zr4+, Ce4+, Sn4+

1.1

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Fe(CN)64– 0.5

Source: Kielland, J. J. Am. Chem. Soc. 1937, 59, 1675–1678.

t-Test Values

Values of t for…

…a confidence interval of:

 

90%

95%

…an α value of:

Degrees of Freedom

 

0.10

0.05

−0.51(zA)2 µ logγA =

1 6.314 12.706 1+3.3×αA× µ

2 2.920

4.303

3 2.353

3.182

4 2.132

2.776

5 2.015

2.571

6 1.943

2.447

word image 703S2 =S2 +S2 total samp. meth.

57

La

138.9

58

Ce

140.1

59

Pr

140.9

60

Nd

144.2

61

Pm

[145]

62

Sm

150.4

63

Eu

152.0

64

Gd

157.3

65

Tb

158.9

66

Dy

162.5

67

Ho

164.9

68

Er

167.3

69

Tm

168.9

70

Yb

173.0

89

Ac

[227]

90

Th

232.0

91

Pa

231.0

92

U

238.0

93

Np

[237]

94

Pu

[244]

95

Am

[243]

96

Cm

[247]

97

Bk

[247]

98

Cf

[251]

99

Es

[252]

100

Fm

[257]

101

Md

[258]

102

No

[259]

*

Lanthanoids

**

Actinoids

 

Name

Structure*

pKa†

Acetic acid

(ethanoic acid)

Ammonia

CH3CO2NHH3

NH4%

4.756

9.245

Page

2.31

As

O

H

O

Arsenic acid HOOH 7.05

(hydrogen arsenate)11.9

CyclohexylamineNH%3 10.567

Hypobromous acid (hydrogen hypobromite)

HOBr

8.63

Hypochlorous acid

(hydrogen hypochlorite)

HOCl

7.53

Hypoiodous acid

(hydrogen hypoiodite)

HOI

10.64

2.148

P

O

Phosphoric acid* HOOH 7.198

(hydrogen phosphate)12.375

OH

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