A 3.00 L sample of urban air was bubbled through a solution

1. (15 pts.) A 3.00 L sample of urban air was bubbled through a solution containing 50.0mL of 0.0116 M Ba(OH)2 which caused the CO2 in the sample to precipitate asBaCO3. The excess base was back-titrated to a phenolphthalein endpoint with 23.6mL of 0.0108 M HCl. Calculate the parts-per-million of CO2 in the air (i.e. mLCO2/106 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 propersetting for filling the bottles. The company wishes to keep profits high by using aslittle 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 youare filling are stated to contain 100 mL what is the minimum volume you should usefor filling if the standard deviation in the fill level is 1.6 mL. (Hint: assume aninfinite number of samples where the CI isandfor 95% confidenceandfor 99% confidence). 3. (20 pts.) Calculate the pZn of a solution prepared by mixing 25.0 mL of 0.0100 MEDTA with 50.0 mL of 0.00500 M Zn2 . Assume that both the Zn2 and EDTAsolutions 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 amountof an unusual isotope (called the spike) is added to an unknown as an internalstandard for quantitative analysis. The ratio of the isotopes is measured and from thisratio the quantity of the element in the original unknown can be calculated. Forexample natural vanadium atom fractions 51V = 0.9975 and 50V = 0.0025. Theatomic 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 thecorresponding atom fractions in a spike. Let Cx be the total concentration ofall isotopes of vanadium ( mol/g) in the unknown and let Cs be theconcentration in the spike. After mixing mx grams of unknown with msgrams of the spike show that the ratio of isotopes in the mixture (denoted R)is given byb) Solve the above equation for Cx to show thatc) A 0.40167 g sample of crude oil containing an unknown concentration ofnatural vanadium was mixed with a 0.41946 g spike containing 2.2435mol/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. (20 pts.) Consider the electrochemical cell described below.The cell solution was made by mixing25.0 mL of 4.00 mM25.0 mL of 4.00 mM25.0 mL of 0.400 M acid 25.0 mL of with pKa = 9.50solutionThe 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: 7. (15 pts.) From the standard potentialTl e- = Tl(s)E = – 0.336 VDetermine the standard potential of.Tl2S(s) 2e- = 2Tl(s) S2Given that the Ksp for Tl2S is 1.2 x 10-22. 8. (15 pts.) An ion selective electrode used to measureobeys the equationis also sensitive toWhen the electrode was immersed in 100.0 mL of unknown containing the reading was. Whenof(in) was added to the unknown the reading increased toFind the concentration ofin the original unknown.andin. 9. (20 pts.) The sulfur in a 5.00 g sample of steel was evolved as H2S and then collectedin a solution of CdCl2 to produce the precipitate CdS. The CdS was then titrated withexcess I2 and the remaining I2 was then back titrated with 4.82 ml of 0.0510 Msodium thiosulfate. If the concentration of the I2 is 0.0600 M and a total of 10.0 ml isadded calculate the %S in the steel. The relevant reactions are:CdS I2I2 2S2O32-S Cd2 2IS2O62- 2I-Titration (excess I2)Back Titration 10. (15 pts.) Inorganic phosphorous is spectrophotochemically measured in blood serumby formation of a reduced heteropoly acid (heteropoly blue) and comparison of itsabsorbance with a standard treated by the same procedure. If a 1.00 mL blood serumsample gives an absorbance of 0.217 while a 2.00 mL aliquot of a standardcontaining 91.2 mg KH2PO4 per liter gives an absorbance of 0.285 calculate themilligram percent of P (milligrams per 100.0 mL) in the blood sample. Note: bothaliquots are diluted to the same volume prior to absorbance measurements. 11. (15 pts.) Butanoic acid has a partition coefficient of 3.0 (favoring benzene) whendistributed between water and benzene. Find the formal concentration of butanoicacid in each phase when 100.0 mL of 0.10 M aqueous butanoic acid is extracted with25.0 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 dilutedto 25.0 mL in a volumetric flask. Three aliquots of 2.00 mL each are drawn to whichare added 0.1 mL 0.2 mL and 0.3 mL spikes of 100.0 ppm Fe standard solution. Theabsorbance of an unspiked 2.00 mL aliquot is measured by AAS and the absorbanceof 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 absorbanceversus ppm Fe added. Extrapolate the graph to the point where absorbance is equal tozero to determine the amount of Fe in the cat food sample in ppm.Sampleoriginal 0.1 mL 0.2 mL 0.3 mLA0.1860.2950.3910.478