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Building a system of questions and exercises for some chapters in physical and chemical analysis - 6

- The earliest and most commonly used monochromatic radiation lamp in AAS measurement is the hollow cathode lamp, which only emits sensitive emission rays of the metal element that makes up the hollow cathode. The sensitive emission lines of an element are usually resonance lines. Therefore, the hollow cathode lamp is also called a resonance radiation source. It is the emission spectrum of elements in a poor gas environment.

- Structure of hollow cathode lamp:


Figure 2.5 Hollow cathode lamp

1. Glass bulb. 2. Anode 3. Cathode 4. Quartz window The hollow cathode lamp is a cylindrical glass bulb, 3 - 5cm in diameter, with

The window is made of glass or quartz. The anode is made of a metal bar. Both poles are placed in a glass bulb containing inert gas (argon or neon) with a small pressure (0.2 - 2MPa). The hollow cathode lamp is connected to a 300 - 500V current, which is stable and must have high stability. The discharge current of the lamp is usually a few milliamperes. When the lamp is working, the current density on the inside of the cathode is higher than the outside. Therefore, the opening of the cathode will glow. The cathode of the lamp is made of metal, a refractory alloy containing the element to be determined.


Question 22:What are the main components of a hollow cathode lamp? HDTL:

Hollow cathode lamp consists of 3 main parts: (see figure 2.5) Part 1 is the lamp body and window.

Part 2 is the cathode and anode electrodes.

Part 3 is the gas contained in the lamp. It is inert gas He, Ar or Nitrogen.

+ Body and shell: The lamp body consists of the lamp shell, window and base for the anode and cathode electrodes. The base is made of PVC plastic. The lamp body and shell are made of glass or quartz. The S-window of the lamp can be glass or quartz transparent in the UV or VIS range depending on the type of lamp of each element emitting the emission beam in which spectrum range.

+ Electrodes: The electrodes of the lamp are the cathode and the anode. The anode is made of inert and heat-resistant metals such as W or Pt. The cathode is made in the form of a cylinder or hollow tube with a diameter of 3 - 5mm, a length of 5 - 6mm and is made of the metal to be analyzed with high purity (at least 99.9%). The cathode wire is also W or Pt metal. The power source is a DC source with a voltage of 220 - 240V.

+ Gas in the lamp: The lamp must suck out all the air and replace it with an inert gas with a pressure of 5 - 15mmHg. The inert gas is argon, helium or nitrogen but must have a purity higher than 99.99%. The gas charged into the lamp must not emit a spectrum that affects the lamp's emission beam and when working under certain conditions, the ratio depends on each type of lamp and each metal element that makes up the hollow cathode. The working intensity of the hollow cathode lamp is usually from 3 - 50mA.

II.3. EXERCISES:

Form 1 : Determine the concentration of metal in the sample, when knowing the metal content in the standard solution and the absorbance.

Solution:

Applying the linear regression method according to the principle of least squares, we construct the standard line: A = aC + b.

Calculate a, b?


i

i

i

i

i

nnnnnnn

i

i

i

i

n AC A C

A C 2 A . C C

a =i 1 i 1 i 1

; b =

i 1 i 1 i 1 i 1

n n 2 n n 2

i

i

i

i

n C 2 C n C 2 C

i 1

i 1

i 1

i 1

Lesson 1:Nowadays, to determine the mercury contamination of aqueous solutions by atomic absorption method, people use the new flameless method of fogging. The apparatus consists of a mercury reduction flask connected to an absorption cuvette. Put 10ml of water sample in the mercury reduction flask and dilute to 100ml, then add 25ml of concentrated sulfuric acid and 10ml of 10% tin (II) sulfate solution, 0.25M sulfuric acid (this last solution is used as a reducing agent). The mercury is reduced to the elemental (atomic) state and transferred to the absorption cuvette by a stream of air, people let this stream of air pass through the solution in the flask to reduce the mercury. Finally, use a lamp

Using a hollow cathode as a source, the absorption of mercury atoms at a wavelength of 2537A0 was measured , the absorption reaching its maximum after nearly 3 minutes.

The following values ​​of absorbance were obtained for a series of standard solutions of mercury (II):

Hg content in standard solution, g

Absorbance

0.00

0.002

0.30

0.090

0.60

0.175

1.00

0.268

2.00

0.440

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Building a system of questions and exercises for some chapters in physical and chemical analysis - 6

The absorbance values ​​of the two water samples are 0.040 and 0.305 respectively. What is the content of mercury in each sample? What is the concentration ( g / ml ) of mercury in each sample? [28]

Prize:

Proceed to construct the standard line, linear regression according to the least squares principle A = aC + b:


STT

Hg content (C)

Absorbance (A)

AC

C 2

1

0.00

0.002

0.000

0.000

2

0.30

0.090

0.027

0.090

3

0.60

0.175

0.105

0.360

4

1.00

0.268

0.268

1,000

5

2.00

0.440

0.880

4,000

n=5

C = 3.9

A = 0.975

AC =1.28

C 2 = 5.45



nnn

n A i C i Ai C​​

a i 1 i 1 i 15.1, 28 0, 975.3, 9


= 0.2157

n n 2

5.5, 45 3.9 2

i

n C 2 C i

i 1

i 1


yes

i

A i C 2 A i C i C i

b i 1 i 1 i 1 i 1


0, 975.5, 45 1, 28.3, 9


= 0.02672

n n 2

5.5, 45 3.9 2

i

n C 2 C i

i 1

i 1


The resulting regression equation is: A = 0.2157C + 0.0267

* With A = 0.040 substituted into the equation to obtain the mercury content:

0.040 = 0.2157C + 0.02672

Therefore C = 0.0617 g corresponding to a concentration of 0.00617 g/ml.

* With A = 0.305 substituted into the equation to obtain the mercury content:

0.305 = 0.2157C + 0.02672

Therefore C = 1.2902 g corresponding to a concentration of 0.12902 g/ml.


Lesson 2:Atomic absorption spectrometry can be used to determine traces of heavy metals in fuel oil. To analyze a 5.000 g sample of used fuel oil, it is placed in a 25.00 ml volumetric flask, dissolved in 2-methyl-4-pentanol and brought to the mark with this solvent. The resulting solution is then nebulized in an air-acetylene flame. To determine copper and lead, hollow cathode lamps with emission lines of 324.7 and 283.3 nm are used, respectively. To obtain the calibration graphs, a series of standard solutions containing known amounts of copper and lead in the mixture corresponding to unused fuel oil and 2-methyl-4-pentanol are required. The regression equations for copper and lead in a 5.000 g sample of used fuel oil are constructed according to the following data:


Solution

Absorbance

at 283.3 nm (Pb)

at 324.7 nm (Cu)

a, Standard: 19.5 g / ml Pb 5.25 g / ml Cu

0.356

0.514

b, Standard: 4.00 g / ml Pb 4.00 g / ml Cu

0.073

0.392

c, Standard: 12.1 g / ml Pb 6.27 g / ml Cu

0.220

0.612

d, Standard: 8.50 g / ml Pb

0.155

0.101

đ, Standard: 15.2 g / ml Pb 2.40 g / ml Cu

0.277

0.232

e, Unknown

0.247

0.371

Prize:

Proceed to construct the standard line, linear regression according to the least squares principle A = aC + b:

Status

Lead content (C)

Absorbance (A)

AC

C 2

1

4.00

0.073

0.292

16,000

2

8.50

0.155

1,318

72,250

3

12.1

0.220

2,662

146,410

4

0.00

0.247

0.000

0.000

5

15.2

0.277

4,210

231,040

6

19.5

0.356

6,942

380,250


C = 59.3

A = 1.382

AC = 15.424

C 2 = 845,950

nnn

n A i C i A i C i

a =i 1 i 1 i 1=

6.15,424 1,328.59.3


= 0.1099

n n 2

6,845,950 59.3 2

i

i

n C 2 C

i 1

i 1


yes

A C 2 A . C C

iiii

i 1,328,845.95 15,424.59.3

b =i 1 i 1 i 1 i 1=

= 0.1339

n n 2

6,845,950 59.3 2

i

i

n C 2 C

i 1

i 1

The resulting regression equation is: A = 0.1099C + 0.1339


Status

Copper content (C)

Absorbance (A)

AC

C 2

1

0.00

0.101

0.000

0.000

2

2.40

0.232

0.557

5,760

3

0.00

0.371

0.000

0.000

4

4.00

0.392

1,568

16,000

5

5.25

0.514

2,699

27,563

6

6.27

0.612

3,837

39,313


C = 17.92

A = 2.222

AC = 8.661

C 2 = 88.635

nnn

n A i C i A i C i

a =i 1 i 1 i 1=

6.8,661 2,222.17.92


= 0.0557

n n 2

6.88,635 17.92 2

i

i

n C 2 C

i 1

i 1

yes

A C 2 A . C C

iiii

i 2,222.88,635 8,661.17.92

b =i 1 i 1 i 1 i 1=

= 0.1981

n n 2

6.88,635 17.92 2

i

i

n C 2 C

i 1

i 1

The resulting regression equation is: A = 0.0557C + 0.1981

Lesson 3 : To determine the copper content in the FFDT sample (dry free fat sample obtained from the joint extract), 11.23 mg of FFDT was mixed in a 5.00 ml flask with 0.75 M HNO3 solution to the mark. Atomic absorption spectroscopy was performed using an air-acetylene flame to obtain an absorbance of 0.023 at a wavelength of 324.8 nm. Standard solutions were measured under similar conditions, and the results were as follows:

ppm Cu

0.000

0.100

0.200

0.300

0.400

0.500

0.600

0.700

1,000

Absorbance

0.000

0.006

0.013

0.020

0.026

0.033

0.039

0.046

0.066

What is the concentration of copper (  g/gram FFDT)?

Solution: Construct the standard line and linear regression according to the least squares principle A = aC + b:

Status

Copper content (C)

Absorbance (A)

AC

C 2

1

0.000

0.000

0.000

0.000

2

0.100

0.006

0.001

0.010

3

0.200

0.013

0.003

0.040

4

0.300

0.020

0.006

0.090

5

0.400

0.026

0.010

0.160

6

0.500

0.033

0.017

0.250

7

0.600

0.039

0.023

0.360

8

0.700

0.046

0.032

0.490

9

1,000

0.066

0.066

1,000

n = 9

C = 3,800

A = 0.249

AC = 0.158

C 2 =

2,400

nnn

n A i C i A i C i

a =i 1 i 1 i 1=

9.0,158 0,249.3,800


= 0.0665

n n 2

9.2,400 3,800 2

i

i

n C 2 C

i 1

i 1


yes

A C 2 A . C C

iiii

i 0.249.2.400 0.158.3.800

b =i 1 i 1 i 1 i 1=

= - 0.00039

n n 2

9.2,400 3,800 2

i

i

n C 2 C

i 1

i 1

The regression equation obtained is: A = 0.0665C - 0.00039 (C is ppm Cu concentration) replace A = 0.023

C = 0.352 ppm = 0.352g/ml


Cu content:

0.352 g / ml .5 ml0.01123 grams

= 157 gCu/gram FFDT.

Lesson 4:An analyst has given the results of data for constructing a standard curve for phosphorus analysis using flame atomic absorption spectroscopy:

ppm P

2130

4260

6400

8530

Absorbance

0.048

0.110

0.173

0.230


To determine the purity of the Na 2 HPO 4 sample , dissolve 2.469 grams of the sample in water, transfer it to a 100.00 ml volumetric flask and make up to the mark. The atomic absorption of the resulting color is 0.135. Determine the purity of the Na 2 HPO 4 sample .


Prize:

Proceed to construct the standard line, linear regression according to the least squares principle A = aC + b:

Status

Content

phosphorus (C)

Absorbance

(A)

AC

C 2

1

2130

0.048

102,2400

4536900

2

4260

0.110

468,6000

18147600

6400

0.173

1107,2000

40960000

4

8530

0.230

1961,9000

72760900

n =4

C = 21320

A = 0.561

AC = 3639.940

C 2 = 136405400

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