Chapter 9 Flashcards

1
Q

Define releasing agent

A r a i a s w r w a i s t r i i w t a i a AA e.

A

A r a i a s w r w a i s t r i i w t a i a AA e.

A releasing agent is a substance which reacts with an interfering substance to remove its interference with the analyte in an AA experiment.

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2
Q

Define protective agent

A p a b t t a t p a v s w a i t a o t a.

A

A p a b t t a t p a v s w a i t a o t a.

A protective agent binds to the analyte to produce a volatile species which aids in the atomization of the analyte.

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3
Q

Define ionization suppressor

A i s i a s a t t A s w p i o t a.

A

A i s i a s a t t A s w p i o t a.

An ionization suppressor is a species added to the AA sample which prevents ionization of the analyte.

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4
Q

Define atomization

A i t p b w a s i v a d i a u b h.

A

A i t p b w a s i v a d i a, u b h.

Atomization is the process by which a sample is vaporized and decomposed into atoms, usually by heat.

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5
Q

Define pressure broadening

P b r t t b o a l d t c w o s.

A

P b r t t b o a l d t c w o s.

Pressure broadening refers to the broadening of atomic lines due to collisions with
other species.

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6
Q

Define hollow-cathode lamp

H-c l a h v l w a t a a a m c m f t a o i i a l p o N o A.

A

H-c l a h v l w a t a a a m c m f t a o i i a l p o N o A.

Hollow-cathode lamps are high voltage lamps with a tungsten anode and a metal cathode made from the analyte of interest in an low pressure of Ne or Ar.

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7
Q

Define sputtering

S i t p b w i g i a h c l c w a r a f t c. T a a r f t r e b t h c l.

A

S i t p b w i g i a h c l c w a r a f t c. T a a r f t r e b t h c l.

Sputtering is the process by which ionized gas in a hollow cathodes lamp collides with and releases atoms from the cathode. These atoms are responsible for the radiation emitted by the hollow cathode lamp.

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8
Q

Define self-absorption

S-a i t a o r b t s w t r c f a e b t s a n f t l.

A

S-a i t a o r b t s w t r c f a e b t s a n f t l.

Self-absorption is the absorption of radiation by the sample where the radiation comes from an emission by the sample and not from the lamp.

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9
Q

Define spectral interference

S i i d t a/e o a i s t l w t r o t m o t a o i.

A

S i i d t a/e o a i s t l w t r o t m o t a o i.

Spectral interference is due to absorption/emission of an interfering species that lies within the resolution of the monochromator of the analyte of interest.

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10
Q

Define chemical interference

C i i d t c p d a t e t a c o t a.

A

C i i d t c p d a t e t a c o t a.

Chemical interference is due to chemical processes during atomization that effect the absorption characteristics of the analyte.

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11
Q

Define radiation buffer

A r b i a s a i e t b s a s w b t e i t s.

A

A r b i a s a i e t b s a s w b t e i t s.

A radiation buffer is a substance added in excess to both sample and standards which buffers the effect in the sample.

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12
Q

Define Doppler broadening

D b i b o a a o e l d t t m o t a t o a f t d.

A

D b i b o a a o e l d t t m o t a t o a f t d.

Doppler broadening is broadening of atomic absorption or emission lines due to the movement of the analyte towards or away from the detector.

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13
Q

Why is an electrothermal atomizer more sensitive than a flame atomizer?

A

u

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14
Q

Use Equation 7-13 for the resolving power of a grating monochromator to estimate the theoretical minimum size of a diffractIon grating that would provide a profile of an atomic absorption line at 500 nm haying a line width of 0.002 nm. Assume that the grating is to be used in the first order and that It has been ruled at 2400 grooves/mm.

A

b

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15
Q

For the flame shown in Figure 9-3, calculate the relative intensity of the 766.5-nm emission line for potassium at the following heights above the flame (assume no ionization):

(a) 2.0 cm
(b) 3.0 cm
(c) 4,0 cm
(d) 50 cm

A

l

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16
Q

Assume that the absorption signal shown in Figure 9-7 were obtained for 2-uL aliquots of standards and sample.

Calculate the concentration in parts per million
of lead in the sample of canned orange juice

A

g

17
Q

A 5.00-mL sample of blood was treated with trichloroacetic acid to precipitate proteins. After centrifugation, the resulting solution was brought to a pH of 3 and was extracted with two 5-mL portions of methyl isobutyl ketone containing the organic lead complexing agent APCD. The extract was aspirated directly into an air-acetylene flame yielding an absorbance of 0.444 at 283.3 nm. Five-milliliter
aliquots of standard solutions containing 0.250 and 0.450 ppm Pb were treated in the same way and yielded absorbances of 0.396 and 0.599.

Calculate the concentration Pb (ppm) in the sample assuming that Beer’s law is followed

A

Setting up two equations in two unknowns

  1. 599 = 0.450m+ b
  2. 396 = 0.250m+ b

Subtracting the second equation from the first gives
0.203 = 0.200 m
or the slope m= 0.203/0.200 = 1.015

Then using this value in the first equation gives
b= 0.599 – 0.450 ×1.015 = 0.14225

The unknown Pb concentration is then x= (0.444 – 0.14225)/1.015 = 0.297 ppm Pb