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##### Analytical Chemistry: Quantitative Analysis

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Quantitative chemistry question. Numbers 5, 6, and 8 on the attached document. Thanks!;Attachment Preview;Final Exam ss14.pdf Download Attachment;Chemistry 150;Analytical Chemistry: Quantitative Analysis;Final Exam (take home);200 pts. possible;Rules: You may use the following materials: (1) your textbook, (2) your class notes, and;(3) any downloaded items from the course website. Please do your own work, no;collaborating. Work the problems on the exam sheet and show all of your work. Please;hand in both your answers and the exam sheet. Good luck!;Due: No later than 5 pm Friday, September 12, 2014. Slide your exam under the door of;Prof. Burattos office (4148B Chemistry) or hand it to Prof. Buratto himself.;1. (15 pts.) A 3.00 L sample of urban air was bubbled through a solution containing 50.0;mL of 0.0116 M Ba(OH)2, which caused the CO2 in the sample to precipitate as;BaCO3. The excess base was back-titrated to a phenolphthalein endpoint with 23.6;mL of 0.0108 M HCl. Calculate the parts-per-million of CO2 in the air (i.e. mL;CO2/106 mL air), use 1.98 g/L as the density of CO2.;2. (15 pts.) Your first assignment at El Smello Perfume Co. is to determine the proper;setting for filling the bottles. The company wishes to keep profits high by using as;little perfume as possible, but a government regulation states that no more than 2.5%;of the bottles may contain less than the stated volume on the bottle. If the bottles you;are filling are stated to contain 100 mL, what is the minimum volume you should use;for filling if the standard deviation in the fill level is 1.6 mL. (Hint: assume an;infinite number of samples where the CI is and = 1.96 for 95% confidence;and = 2.58 for 99% confidence).;3. (20 pts.) Calculate the pZn of a solution prepared by mixing 25.0 mL of 0.0100 M;EDTA with 50.0 mL of 0.00500 M Zn2+. Assume that both the Zn2+ and EDTA;solutions are buffered with 0.100 M NH3 and 0.176 M NH4Cl.;4. (15 pts.) Calculate the pH resulting from mixing 15.00 mL of 0.800 M HIO3 (pKa =;0.77) with 45.00 mL of 0.0200 M NaOH.;5. (20 pts) Isotope dilution is a mass spectrometric technique in which a known amount;of an unusual isotope (called the spike) is added to an unknown as an internal;standard for quantitative analysis. The ratio of the isotopes is measured and from this;ratio the quantity of the element in the original unknown can be calculated. For;example, natural vanadium atom fractions 51V = 0.9975 and 50V = 0.0025. The;atomic fraction is defined as;atom fraction of 51V = atoms 51V/(atoms 51V + atoms 50V);A spike enriched in 50V has atom fractions 51V = 0.6391 and 50V = 0.3609.;a) Let A be 51V and B be 50V. Let Ax be the atom fraction of A in an unknown.;Let Bx be the atom fraction of B in the unknown. Let As and Bs be the;corresponding atom fractions in a spike. Let Cx be the total concentration of;all isotopes of vanadium (mol/g) in the unknown, and let Cs be the;concentration in the spike. After mixing mx grams of unknown with ms;grams of the spike, show that the ratio of isotopes in the mixture (denoted R);is given by;=;+;=;+;b) Solve the above equation for Cx to show that;=;c) A 0.40167 g sample of crude oil containing an unknown concentration of;natural vanadium was mixed with a 0.41946 g spike containing 2.2435;mol/g enriched with 50V (atom fractions: 51V = 0.6391 and 50V = 0.3609).;The measured isotope ratio by mass spectrometry was R = 51V /50V =;10.545. Determine the concentration of vanadium (mol/g) in the crude oil.;6. (15 pts.) Consider the electrochemical cell described below.;()|()|(,)|| |();The cell solution was made by mixing;25.0 mL of 4.00 mM;25.0 mL of 4.00 mM ();25.0 mL of 0.400 M acid,, with pKa = 9.50;25.0 mL of solution;The measured voltage was -0.440 V. Calculate the molarity of the KOH solution.;Assume that essentially all of the copper(I) is in the form (). For the right side;(the cathode), the half cell reaction is;() + () + 2 = 0.429;7. (20 pts.) A 3.67 g sample of insecticide was decomposed in acid, any As5+ was;reduced to As3+ and diluted to 250.0 mL in a volumetric flask. A 5.00 mL aliquot;was added to 125.0 mL of 0.0500 M KI buffered to pH 7. A coulometric titration was;carried out with electrically-generated I3-, which oxidized As3+ to As5+ according to;the reaction.;As3+ + I3-;3I- + As5+;The titration required 287 s at a constant current of 24.25 mA to reach the endpoint.;Calculate the percentage of As2O3 (197.84 g/mol) in the insecticide.;8. (15 pts.) An ion selective electrode used to measure;obeys the equation;= +;is also sensitive to and;57.1;log{[ ] + 0.002[ ] };2;When the electrode was immersed in 100.0 mL of unknown containing in;0.200, the reading was 194.6. When 1.00 of 1.07 10;(in 0.200) was added to the unknown, the reading increased to 200.7.;Find the concentration of in the original unknown.;9. (20 pts.) The sulfur in a 5.00 g sample of steel was evolved as H2S and then collected;in a solution of CdCl2 to produce the precipitate CdS. The CdS was then titrated with;excess I2 and the remaining I2 was then back titrated with 4.82 ml of 0.0510 M;sodium thiosulfate. If the concentration of the I2 is 0.0600 M and a total of 10.0 ml is;added, calculate the %S in the steel. The relevant reactions are;CdS + I2;I2 + 2S2O32-;S + Cd2+ + 2IS2O62- + 2I-;Titration (excess I2);Back Titration;10. (15 pts.) Inorganic phosphorous is spectrophotochemically measured in blood serum;by formation of a reduced heteropoly acid (heteropoly blue) and comparison of its;absorbance with a standard treated by the same procedure. If a 1.00 mL blood serum;sample gives an absorbance of 0.217, while a 2.00 mL aliquot of a standard;containing 91.2 mg KH2PO4 per liter gives an absorbance of 0.285, calculate the;milligram percent of P (milligrams per 100.0 mL) in the blood sample. Note: both;aliquots are diluted to the same volume prior to absorbance measurements.;11. (15 pts.) Butanoic acid has a partition coefficient of 3.0 (favoring benzene) when;distributed between water and benzene. Find the formal concentration of butanoic;acid in each phase when 100.0 mL of 0.10 M aqueous butanoic acid is extracted with;25.0 mL of benzene at a pH of 4.00.;12. (15 pts.) A 480 mg sample of cat food is digested with HNO3 and HClO4 and diluted;to 25.0 mL in a volumetric flask. Three aliquots of 2.00 mL each are drawn, to which;are added 0.1 mL, 0.2 mL and 0.3 mL spikes of 100.0 ppm Fe standard solution. The;absorbance of an unspiked 2.00 mL aliquot is measured by AAS, and the absorbance;of the spiked samples is measured as well. The data is presented in the Table below.;Correct the absorbance of the spiked samples for dilution and plot the absorbance;versus ppm Fe added. Extrapolate the graph to the point where absorbance is equal to;zero to determine the amount of Fe in the cat food sample in ppm.;Sample;original;+ 0.1 mL;+ 0.2 mL;+ 0.3 mL;A;0.186;0.295;0.391;0.478

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