Principles of Instrumental Analysis
7th Edition
ISBN: 9781305577213
Author: Douglas A. Skoog, F. James Holler, Stanley R. Crouch
Publisher: Cengage Learning
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Question
Chapter 9, Problem 9.20QAP
Interpretation Introduction
Interpretation:
Precipitate protein is obtained by treating 5.00 mL sample of blood with trichloroacetic acid. The final solution has pH of 3 and this solution is extracted with two portions of 5 ml of methyl isobutyl
Concept introduction:
Beer Lambert law gives the relationship between absorbance and concentration of a sample. For the calculation of concentration, the following the equation of straight line as follows:
Where,
y = absorbance
x = concentration of sample
m is slope of graph plotted between absorbance and concentration of the sample.
And,
C is intercept of the graph.
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3. A total of 5 mL of blood samples were treated with trichloroacetic acid to precipitate
proteins. After centrifugation, the resulting solution is adjusted
to pH 3 and extracted with 5 mL of methyl isobutyl ketone containing the
ammonium pyrolydine dithiocarbamate (APDC) complexing compound. The extract was
then measured its absorbance using an atomic absorption spectrometer (AAS) at 7.283.3
nm with an absorbance of 0.502. A total of 5 mL of standard Pb solution with
concentrations of 0.4 and 0.6 ppm was also given the same treatment and resulted in
absorbance of 0.396 and 0.599 ppm. Determine the pb concentration in the sample in ppm
units!
A 5.00-mL sample of blood was treated with trichloroacetic acid to precipitate proteins. After centrifugation, the resulting solution was brought to pH 3 and extracted with two 5-mL portions of methyl isobutyl ketone containing the lead-complexing agent APCD. The extract was aspirated directly into an air/acetylene flame and yielded an absorbance of 0.527 at 283.3 nm. Five-milliliter aliquots of standard solutions containing 0.400 and 0.600 ppm of lead were treated in the same way and yielded absorbances of 0.396 and 0.599. Find the concentration of lead in the sample in ppm assuming that Beer’s law is followed.
A 10.00mL sample of alcoholic ethyl acetate was diluted to 100.00 mL. 20.00 mL was aliquoted and mixed with 40.00 mL of 0.04672 M KOH. The resulting mixture was heated for 2 hours.
CH3COOC2H5 + OH- → CH3COO- + C2H5OH
After cooling, the excess OH was back titrated with 3.41 mL of 0.05042 M H2SO4.
Answer the following:
Calculate the number of moles of OH- that reacted with ethyl acetate.
Calculate the number of moles of ethyl acetate in the 20.00 mL solution.
What is the mass of ethyl acetate (FW=88.11 g/mol) in the original 10.00 mL sample?
Chapter 9 Solutions
Principles of Instrumental Analysis
Ch. 9 - Prob. 9.1QAPCh. 9 - Prob. 9.2QAPCh. 9 - Why is an electrothermal atomizer more sensitive...Ch. 9 - Prob. 9.4QAPCh. 9 - Prob. 9.5QAPCh. 9 - Prob. 9.6QAPCh. 9 - Prob. 9.7QAPCh. 9 - Prob. 9.8QAPCh. 9 - Prob. 9.9QAPCh. 9 - Prob. 9.10QAP
Ch. 9 - Prob. 9.11QAPCh. 9 - Prob. 9.12QAPCh. 9 - Prob. 9.13QAPCh. 9 - Prob. 9.14QAPCh. 9 - Prob. 9.15QAPCh. 9 - Prob. 9.16QAPCh. 9 - Prob. 9.17QAPCh. 9 - In the concentration range of 1 to 100 g/mL P,...Ch. 9 - Prob. 9.19QAPCh. 9 - Prob. 9.20QAPCh. 9 - Prob. 9.21QAPCh. 9 - The chromium in an aqueous sample was determined...Ch. 9 - Prob. 9.23QAP
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