1. Provide the missing information in the following table

2. Provide the missing information in the following table

3. The transmittance of a solution is found to be 35.0%. What is the transmittance if the solution is diluted in half?4. The transmittance of a solution is found to be 85.0% when measured in a cell whose pathlength is 1.00 cm. What is the percent transmittance if the pathlength is increased to 10.00 cm?6. Chemical deviations to Beers law may occur when the concentration of an absorbing species is affected by the position of an equilibrium reaction. Consider a weak acid, HA, for which Ka is 2.105. Construct Beers law calibration curves of absorbance versus the total concentration of weak acid (Ctot = [HA] + [A]), using values for Ctot of 1.0105, 3.0105, 5.0105, 9.0105, 11105, and 13105 M for the following sets of conditions: (a) HA = A- = 2000, and solution is not buffered; (b) HA = 2000 and A- = 500, and solution is not buffered; and, (c) HA = 2000 and A- = 500, and solution is buffered to a pH of 4.50. Assume a constant pathlength of 1.00 cm for all samples. All values of have units of M1cm1.

10. One method for the analysis of Fe3+ that can be used with a variety of sample matrices, is to form the highly colored Fe3+thioglycolic acid complex. The complex absorbs strongly at 535 nm. Standardizing the method is accomplished using external standards. A 10.00 ppm Fe3+ working standard is prepared by transferring a 10-mL aliquot of a 100.0 ppm stock solution of Fe3+ to a 100-mL volumetric flask and diluting to volume. Calibration standards of 1.00, 2.00, 3.00, 4.00, and 5.00 ppm are then prepared by transferring appropriate amounts of the 100.0-ppm working solution into separate 50-mL volumetric flasks, each containing 5 mL of thioglycolic acid, 2 mL of 20% w/v ammonium citrate, and 5 mL of 0.22 M NH3. After diluting to volume and mixing, the absorbances of the external standards are measured against an appropriate blank. Samples are prepared for analysis by taking a portion known to contain approximately 0.1 g of Fe3+, dissolving in a minimum amount of HNO3, and diluting to volume in a 1-L volumetric flask. A 1.00-mL aliquot of this solution is transferred to a 50-mL volumetric flask, along with 5 mL of thioglycolic acid, 2 mL of 20% w/v ammonium citrate, and 5 mL of 0.22 M NH3 and diluted to volume. The absorbance of this solution is used to determine the concentration of Fe3+ in the sample.(a) What is an appropriate blank for this procedure? (b) Ammonium citrate is added to prevent the precipitation of Al3+. What effect would the presence of trace amount of Fe3+ in the ammonium citrate have on the reported concentration of iron in the sample? (c) Why does the procedure call for taking an amount of sample containing approximately 0.1 grams of Fe3+? (d) Unbeknownst to the analyst, the 100-mL volumetric flask used to prepare the 10.00 ppm working standard of Fe3+ has a volume that is significantly smaller than 100.0mL. What effect will this have on the reported concentration of iron in the sample?11. A spectrophotometric method for the analysis of iron has a linear calibration curve for standards of 0.00, 5.00, 10.00, 15.00, and 20.00 ppm. An iron ore sample with an expected iron content of 4060% w/w is to be analyzed by this method. An approximately 0.5-g sample is taken, dissolved in a minimum of concentrated HCl, and diluted to 1 L in a volumetric flask using distilled water. A 5.00-mL aliquot is removed with a pipet. To what volume (10, 25, 50, 100, 250, 500, or 1000 mL) should it be diluted to minimize the uncertainty in the analysis? Explain.12. In a recent paper, Lozano-Calero and colleagues describe a new method for the quantitative analysis of phosphorus in cola beverages.28 The method is based on the formation of an intensely blue-colored phosphomolybdate complex, (NH4)3[PO4(MoO3)12]. The complex is formed by adding (NH4)6Mo7O24 to the sample in the presence of a reducing agent, such as ascorbic acid. The concentration of the complex is determined spectrophotometrically at a wavelength of 830 nm, using a normal calibration curve as a method of standardization. In a typical analysis, a set of standard solutions containing known amounts of phosphorus was prepared by placing appropriate volumes of a 4.00-ppm solution of P2O5 in a 5-mL volumetric flask, adding 2 mL of the ascorbic acid reducing solution, and diluting to volume with distilled water. Cola beverages were prepared for analysis by pouring a sample into a beaker and allowing it to stand for 24 h to expel the dissolved gases (CO2). A 2.50-mL sample of the degassed sample was transferred to a 50-mL volumetric flask and diluted to volume. A 250-L aliquot of the diluted sample was then transferred to a 5-mL volumetric flask, treated with 2 mL of the ascorbic acid reducing solution, and diluted to volume with distilled water. The authors note that this method only can be applied to noncolored cola beverages, such as Crystal Pepsi. Clearly explain why this is the case. (b) How might you modify this method so that it could be applied to any cola beverage? (c) Why is it necessary to ensure that the dissolved gases have been removed? (d) How would you prepare a blank for this method? (e) The calibration curve reported by the authors yields a regression line of Abs = 0.02 + 0.72 (ppm P2O5)A sample of Crystal Pepsi, analyzed as described here, yields an absorbance of 0.565. What is the concentration of phosphorus, reported as parts per million of P, in the original sample of Crystal Pepsi?13. EDTA forms colored complexes with a variety of metal ions that may serve as the basis for a quantitative spectrophotometric method of analysis. The molar absorptivities of the EDTA complexes of Cu2+, Co2+, and Ni2+ at three wavelengths are summarized in the following table (all values of are in M1 cm1)

Using this information, determine (a) the concentration of Cu2+ in a solution that has an absorbance of 0.338 at a wavelength of 732.0 nm; (b) the concentrations of Cu2+ and Co2+ in a solution that has an absorbance of 0.453 at a wavelength of 732.0 nm and 0.107 at a wavelength of 462.9 nm; and (c) the concentrations of Cu2+, Co2+, and Ni2+ in a sample that has an absorbance of 0.423 at a wavelength of 732.0 nm, 0.184 at a wavelength of 462.9 nm, and 0.291 at a wavelength of 378.7 nm. The pathlength, b, for all measurements is 1.00 cm.14. The concentration of phenol in a water sample is determined by separating the phenol from nonvolatile impurities by steam distillation, followed by reacting with 4- aminoantipyrine and K3Fe(CN)6 at pH 7.9 to form a colored antipyrine dye. A phenol standard with a concentration of 4.00 ppm has an absorbance of 0.424 at a wavelength of 460 nm using a 1.00-cm cell. A water sample is steam-distilled, and a 50.00-mL aliquot of the distillate is placed in a 100-mL volumetric flask and diluted to volume with distilled water. The absorbance of this solution is found to be 0.394. What is the concentration of phenol (in parts per million) in the water sample?16. In the DPD colorimetric method for the free chlorine residual, which is reported as parts per million of Cl2, the oxidizing power of free chlorine converts the colorless amine N,N-diethyl-p-phenylenediamine to a colored dye that absorbs strongly over the wavelength range of 440580 nm. Analysis of a set of calibration standards gave the following results

A sample from a public water supply is analyzed to determine the free chlorine residual, giving an absorbance of 0.113.What is the free chlorine residual for the sample in parts per million Cl2?17. Brown and Lin reported a quantitative method for methanol based on its effect on the visible spectrum of methylene blue.30 In the absence of methanol, the visible spectrum for methylene blue shows two prominent absorption bands centered at approximately 610 nm and 660 nm, corresponding to the monomer and dimer, respectively. In the presence of methanol, the intensity of the dimers absorption band decreases, and that of the monomer increases. For concentrations of methanol between 0 and 30% v/v, the ratio of the absorbance at 663 nm, A663, to that at 610 nm, A610, is a linear function of the amount of methanol. Using the following standardization data, determine the %v/v methanol in a sample for which A610 is 0.75 and A663 is 1.07.

18. The concentration of the barbiturate barbital in a blood sample was determined by extracting 3.00 mL of the blood with 15 mL of CHCl3. The chloroform, which now contains the barbital, is then extracted with 10.0 mL of 0.045 M NaOH (pH 13). A 3.00-mL sample of the aqueous extract is placed in a 1.00-cm cell, and an absorbance of 0.115 is measured. The pH of the sample in the absorption cell is then adjusted to approximately 10 by adding 0.5 mL of 16% w/v NH4Cl, giving an absorbance of 0.023. When 3.00 mL of a standard barbital solution with a concentration of 3.0 mg/100 mL is taken through the same procedure, the absorbance at pH 13 is 0.295, whereas that at a pH of 10 is 0.002. Report the concentration of barbital, (milligrams per 100 mL), in the sample.19. Jones and Thatcher developed a spectrophotometric method for analyzing analgesic tablets containing aspirin, phenacetin, and caffeine. The sample is dissolved in CHCl3 and extracted with an aqueous solution of NaHCO3 to remove the aspirin. After the extraction is complete, the chloroform is then transferred to a 250-mL volumetric flask and diluted to volume with CHCl3. A 2.00-mL portion of this solution is diluted to volume in a 200-mL volumetric flask with CHCl3. The absorbance of the final solution is measured at wavelengths of 250 nm and 275 nm, at which the absorptivities, in ppm1 cm1, for caffeine and phenacetin are

Aspirin is determined by neutralizing the NaHCO3 in the aqueous solution and extracting the aspirin into CHCl3. The combined extracts are diluted to 500 mL in a volumetric flask. A 20.00-mL portion of the solution is placed in a 100-mL volumetric flask and diluted to volume with CHCl3. The absorbance of this solution is measured at 277 nm, where the absorptivity of aspirin is 0.00682 ppm1 cm1. An analgesic tablet treated by this procedure is found to have absorbances of 0.466 at 250 nm, 0.164 at 275 nm, and 0.600 at 277 nm when using a cell with a 1.00-cm pathlength. Report themilligrams of aspirin, caffeine, and phenacetin in the analgesic tablet.

20. The concentration of SO2 in a sample of air was determined by the p-rosaniline method. The SO2 was collected in a 10.00- mL solution of HgCl42, where it forms Hg(SO3)2 , by pulling the air through the solution for 75 min at a rate of 1.6 L/min.2

After adding p-rosaniline and formaldehyde, the colored solution was diluted to 25 mL in a volumetric flask. The absorbance was measured at 569 nm in a 1-cm cell, yielding a value of 0.485. A standard sample was prepared by substituting a 1.00-mL sample of a standard solution containing the equivalent of 15.00 ppm SO2 for the air sample. The absorbance of the standard was found to be 0.181. Report the concentration of SO2 in the air in parts per million. The density of air may be taken as 1.18 g/L.

23. The following table lists the molar absorptivities for the Arsenazo complexes of copper and barium at selected wavelengths. Determine the optimum wavelengths for the analysis of a mixture of copper and barium.

595 11900 7100600 15500 7200607 18300 7400611 19300 6900614 19300 7000620 17800 7100626 16300 8400635 10900 9900641 7500 10500645 5300 10000650 3500 8600655 2200 6600658 1900 6500665 1500 390067015002800

68018001500

26. The stoichiometry for a metalligand complex was determined by the method of continuous variations. A series of solutions was prepared in which the combined concentrations of the metal and ligand were held constant at 5.15 104 M. The absorbances of these solutions were measured at a wavelength at which only the metalligand complex absorbs. Using the following data, determine the formula of the complex.

Mole fraction M Mole fraction L Absorbance33. Hobbins reported the following calibration data for the flame atomic absorption analysis for phosphorus.

To determine the purity of a sample of Na2HPO4, a 2,469g sample is dissolved and diluted to volume in a 100-mL volumetric flask. Analysis of the resulting solution gives an absorbance of 0.135. What is the purity of the Na2HPO4?34. Bonert and Pohl reported results for the atomic absorption analysis of several metals in caustic suspensions produced during the manufacture of soda by the ammonia-soda process. The concentration of Cu was determined by acidifying a 200-mL sample of the caustic solution with 20 mL of concentrated HNO3, adding 1 mL of 27% w/v H2O2 and boiling for 30 min. The resulting solution was diluted to 500 mL, filtered, and analyzed by flame atomic absorption using matrix-matched standards. The results for a typical analysis are shown in the following table.

Determine the concentration of Cu in the caustic suspension. The determination of Cr was accomplished by acidifying a 200-mL sample of the caustic solution with 20 mL of concentrated HNO3, adding 0.2 g of Na2SO3 and boiling for 30 min. The Cr was isolated from the sample by adding 20 mL of NH3, producing a precipitate that includes the chromium as well as other oxides. The precipitate was isolated by filtration, washed, and transferred with wash water to a beaker. After acidifying with 10 mL of HNO3, the solution was evaporated to dryness. The residue was redissolved in a combination of HNO3 and HCl and evaporated to dryness. Finally, the residue was dissolved in 5 mL of HCl, filtered, diluted to volume in a 50-mL volumetric flask, and analyzedby atomic absorption using the method of standard additions. The atomic absorption results are summarized in the following table.

35. Quigley and Vernon report results for the determination of trace metals in sea water using a graphite furnace atomic absorption spectrophotometer. Calibration was achieved by the method of standard additions. The trace metals were first separated from their complex, high-salt matrix by coprecipitating with Fe3+. In a typical analysis a 5.00-mL portion of 2000-ppm Fe3+ was added to 1.00 L of sea water. The pH was adjusted to 9 using NH4OH, and the precipitate of Fe(OH)3 allowed to stand overnight. After isolating and rinsing the precipitate, the Fe(OH)3 and coprecipitated metals were dissolved in 2 mL of concentrated HNO3 and diluted to volume in a 50-mL volumetric flask. To analyze for Mn2+, a 1.00-mL sample of this solution was diluted to 100 mL in a volumetric flask. The following samples were injected into the graphite furnace and analyzed

Report the parts per billion of Mn2+ in the sample of sea water.36. The concentration of Na in plant materials may be determined by flame atomic emission. The material to be analyzed is prepared by grinding, homogenizing, and drying at 103 C. A sample of approximately 4 g is transferred to a quartz crucible and heated on a hot plate to char the organic material. The sample is heated in a muffle furnace at 550 C for several hours. After cooling to room temperature the residue is dissolved by adding 2 mL of 1:1 HNO3 and evaporated to dryness. The residue is redissolved in 10 mL of 1:9 HNO3, filtered, and diluted to 50 mL in a volumetric flask. The following data were obtained during a typical analysis for the concentration of Na in a 4.0264-g sample of oat bran

Determine the parts per million Na in the sample of oat bran.37. Gluodenis describes the use of ICP to analyze samples containing Pb and Ni in brass. The analysis for Pb uses external standards prepared from brass samples containing known amounts of lead. Results are shown in the following table.

What is the %w/w Pb in a sample of brass that gives an emission intensity of 9.25 . 104? The analysis for Ni uses an internal standard. Results for a typical calibration are shown in the following table.

What is the %w/w Ni in a sample for which the ratio of emission intensity is 1.10 * 103?Bi dch:1. Cung cp cc thng tin cn thiu trong bng di y:Bc sng(m)Tn s(s-1)S sng(cm-1)Nng lng(J/mol)

4,50.10-96,67.101622222224,42.10-17

2,26.10-71,33.1015442488,80.10-19

3,11.10-69,65.101332156,39.10-20

2,76.10-71,09.1015362327,20.10-19

2. Cung cp cc thng tin cn thiu trong bng di y:Nng cht phn tch(M)Mt quang truyn quaH s hp thu phn t(M-1cm-1)Pathlength(cm)

1,4.10-40,15769,711201,00

7,5. 10-40,56327,47501,00

2,56.10-40,22559,64402,00

1,55. 10-30,16768,121,55,00

8,46.10-40,47833,35651,00

4,35. 10-30,67421,215500,10

1,20. 10-40,089981,37510,00

3. truyn qua ca mt dung dch c tm thy l 35%. truyn qua l g nu mt dung dch c pha long trong mt na?Bi gii: Ta c: I/Io=0,35 => A ban u = log(Io/I) = log(1/0,35)=0.456Khi dung dch c pha long mt na th nng gim i mt na, tc Cs=C/2 M A=.l.C => As=A/2= 0.456/2= 0,228 A=log(Io/I) => T= I/Io= 1/10A = 1/ 100.228= 59,2% Vy truyn qua khi pha long dung dch i mt na l 59,2%4. truyn qua ca mt dung dch c tm thy l 85% khi o trong mt m pathlength l 1cm. truyn qua l g nu pathlength c tng ln n 10 cm?Bi gii: Ta c: I/Io= 0,85 => A= log(Io/I) = log (1/0.85) = 0.0706 A= .l.C As= .ls.C => A/ As= l/ls=1/10 => As= 10. A= 10. 0,0706= 0,706=> I/Io=1/10A=1/100,706= 19,7%Vy truyn qua khi di cuvet l 10 cm l 19,7%6. lch ha ca nh lut Beer c th xy ra khi nng ca cht hp th b nh hng bi v tr ca cn bng phn ng. Xem xt mt axit yu , HA, cho Ka l 2.10-5. Xy dng ng nh chun nh lut Beer ca hp th so vi tng nng ca axit yu (Ctot = [HA] + [A]), s dng cc gi tr cho Ctot l 1.0105, 3.0105, 5.0105, 9.0105, 11105, and 13105 M i vi cc b sau y ca iu kin: (a) HA = A- = 2000, v dung dch ny khng c m; (b) HA = 2000 v A- = 500, v dung dch ny khng c m, (c) HA = 2000 v A- = 500v dung dch c m n pH l 4.50. Gi s mt pathlength khng i l 1.00 cm cho tt c cc mu. Tt c gi tr ca c n v of M1cm1.Bi gii: a) HA = A- = 2000, v dung dch ny khng c m, l= 1cmTheo : Ctot = [HA] + [A] Ta c A= AHA + AA_ = HA. CHA.l + A_. CA_.l A= 2000. CHA + 2000 . CA_ = 2000. (CHA + CA_) CHA + CA_ = 1.0105 ta c A1= 2000. 1. 10-5 = 0,02 CHA + CA_ = 3.0105 ta c A1= 2000. 3. 10-5 = 0,06 CHA + CA_ = 5.0105 ta c A1= 2000. 5. 10-5 = 0,10 CHA + CA_ = 9.0105 ta c A1= 2000. 9. 10-5 = 0,18 CHA + CA_ = 11.0105 ta c A1= 2000. 11. 10-5 = 0,22 CHA + CA_ = 13.0105 ta c A1= 2000. 13. 10-5 = 0,26b) HA = 2000 v A- = 500, v dung dch ny khng c m, l=1cmTa c A= AHA + AA_ = HA. CHA.l + A_. CA_.l A= 2000. CHA + 500 . CA_

B: CoCb: Co-a a aKa = a2 / (Co-a) = (CA_)2/ CHA = 2.10-5 => (CHA)2 = 2.10-5. CHAM theo ta c Ctot = [HA] + [A] = 1.10-5 => CHA = 10-5- CA_=> (CA_)2 = 2.10-5. (10-5- CA_) => CA_ = 7,32. 10-6 M CHA = 2,68.10-6M=> A= 2000. 2,68.10-6 +500. 7,32. 10-6 = 0,009 Ctot = [HA] + [A] = 3.10-5 => CHA = 3.10-5- CA_=> (CA_)2 = 2.10-5. (3.10-5- CA_) => CA_ = 1,65. 10-5 M CHA = 1,35.10-5 M=> A= 2000. 1,35.10-5 +500. 1,65. 10-5 = 0,035 Ctot = [HA] + [A] = 5.10-5 => CHA = 5.10-5- CA_=> (CA_)2 = 2.10-5. (5.10-5- CA_) => CA_ = 2,32. 10-5 M CHA = 2,68.10-5M=> A= 2000. 2,68.10-5 +500. 2,32. 10-5 = 0,065 Ctot = [HA] + [A] = 9.10-5 => CHA = 9.10-5- CA_=> (CA_)2 = 2.10-5. (9.10-5- CA_) => CA_ = 3,36. 10-5 M CHA = 5,64.10-5 M=> A= 2000. 5,64.10-5 +500. 3,36. 10-5 = 0,130 Ctot = [HA] + [A] = 11.10-5 => CHA = 11.10-5- CA_=> (CA_)2 = 2.10-5. (11.10-5- CA_) => CA_ = 3,8. 10-5 M CHA = 7,2.10-5 M=> A= 2000. 7,2.10-5 +500. 3,8. 10-5 = 0,163 Ctot = [HA] + [A] = 13.10-5 => CHA = 13.10-5- CA_=> (CA_)2 = 2.10-5. (13.10-5- CA_) => CA_ = 4,2. 10-5 M CHA = 8,8.10-5 M=> A= 2000. 8,8.10-5 +500. 8,8. 10-5 = 0,197(c) HA = 2000 v A- = 500v dung dch c m n pH l 4.50, l=1cmTa c A= AHA + AA_ = HA. CHA.l + A_. CA_.l A= 2000. CHA + 500 . CA_ Theo : dung dch m c pH= 4,5 => CH+= 10-4,5

B: Co 10-4,5Cb: Co-a a + 10-4,5 aKa = a.( a+ 10-4,5) / (Co-a) = (CA_).( CA_ + 10-4,5)/ CHA = 2.10-5 => (CA_).( CA_ + 10-4,5) = 2.10-5 CHAM theo ta c Ctot = [HA] + [A] = 1.10-5 => CHA = 10-5- CA_=> (CA_).( CA_ + 10-4,5) = 2.10-5. (10-5- CA_) => CA_ = 3,62.10-6 M=> CHA= 6,38.10-6 M=> A= 2000. 6,38.10-6 +500. 3,62. 10-6 = 0,014 Ctot = [HA] + [A] = 3.10-5 => CHA = 3.10-5- CA_=> (CA_).( CA_ + 10-4,5) = 2.10-5. (3.10-5- CA_) => CA_ = 9,77.10-6 M=> CHA= 2,02.10-5 M=> A= 2000. 2,02.10-5 +500. 9,77. 10-6 = 0,045 Ctot = [HA] + [A] = 5.10-5 => CHA = 5.10-5- CA_=> (CA_).( CA_ + 10-4,5) = 2.10-5. (5.10-5- CA_) => CA_ = 1,5.10-5 M=> CHA= 3,5.10-5 M=> A= 2000. 3,5.10-5 +500. 1,5. 10-5 = 0,078 Ctot = [HA] + [A] = 9.10-5 => CHA = 9.10-5- CA_=> (CA_).( CA_ + 10-4,5) = 2.10-5. (9.10-5- CA_) => CA_ = 2,38.10-5 M=> CHA= 6,62.10-5 M=> A= 2000. 6,62.10-5 +500. 2,38. 10-5 = 0,144 Ctot = [HA] + [A] = 11.10-5 => CHA = 11.10-5- CA_=> (CA_).( CA_ + 10-4,5) = 2.10-5. (11.10-5- CA_) => CA_ = 2,72.10-5 M=> CHA= 8,28.10-5 M=> A= 2000. 8,28.10-5 +500. 2,72. 10-5 = 0,179 Ctot = [HA] + [A] = 13.10-5 => CHA = 13.10-5- CA_=> (CA_).( CA_ + 10-4,5) = 2.10-5. (13.10-5- CA_) => CA_ = 3,13.10-5 M=> CHA= 9,87.10-5 M=> A= 2000. 9,87.10-5 +500. 3,13. 10-5 = 0,21310. Mt phng php phn tch Fe3+ c th c s dng mt lot cc mu nn, to thnh phc mu mnh Fe3+thioglycolic. Cc phc hp th mnh ti 535 nm. Phng php chun ha ny c thc hin bng cch s dng cc tiu chun ngoi. C 10ml dung dch gc Fe3+ 100ppm cho vo bnh nh mc 100 ml v pha long n vch th thu c 10 ppm Fe3+ tiu chun lm vic. Tiu chun hiu chun 1.00, 2.00, 3.00, 4.00, v 5.00 ppm c chun b bng cch chuyn 10-mL lng cht ca 100.0 ppm dung dch gc Fe3+ vo bnh nh mc ring bit 50 mL, mi bnh c cha 5 ml acid thioglycolic, 2 ml 20% w / v ammonium citrate, v 5 mL NH30,22 M.Sau khi pha long n vch v trn, mt quang ca cc tiu chun bn ngoi c xc nh i vi mt dung dch trng thch hp.Cc mu c chun b phn tch bng cch ly mt phn bit c cha khong 0,1 g Fe3+, ha tan trong mt lng ti thiu ca HNO3, v pha long n vch trong bnh nh mc 1-L. 1,00 ml lng cht ca dung dch ny c chuyn vo mt bnh nh mc 50-mL, cng vi 5 ml acid thioglycolic, 2 ml 20% w / v ammonium citrate, v 5 mL NH30,22 M v pha long n vch. Mt quang ca gii php ny c s dng xc nh nng ca Fe3+trong mu.(A) Mt dung dch trng thch hp cho quy trnh ny l g?(B) Ammonium citrate c thm vo ngn chn s kt ta ca Al3+.Cc hiu ng vo nng bo co ca st trong mu nu c mt du vt ca tp cht Fe 3+trong citrate amoni l g?(C) Ti sao cc th tc xc nh rng mu c cha khong 0,1 g Fe3+?(D) Khng nh cc nh phn tch, cc bnh nh mc 100 ml dng chun b 10,00 ppm tiu chun lm vic ca Fe3+c mt khi lng nh hn so vi 100,0 mL.Hiu ng g s c trong nng bo co ca st trong mu?11. Mt phng php quang ph phn tch cc cht st c mt ng cong hiu chun tuyn tnh cho cc tiu chun l 0.00, 5.00, 10.00, 15.00, 20.00 vi n v mg Fe / L (ppm).Mt mu qung st 40-60% w / w c phn tch bng phng php ny. Ly khong 0,5g mu, ha tan trong mt lng ti thiu HCl m c v pha long trong bnh nh mc1 lt bng nc ct. 5.00 mL c ly ra bng Pipet. Nn chn nhng th tch no sau y 10, 25, 50, 100, 250, 500, hoc 1000 mL pha long gim thiu s bt nh trong phn tch?Gii thch.12. Lozano-Calero v ng nghip m t mt phng php phn tch nh lng pht pho trong ung cola da trn s hnh thnh ca cc phc phosphomolybdate c mu xanh m, (NH4)3[PO4(Moo3)12]. Cc phc c hnh thnh bng cch thm (NH4)6Mo7O24 vo mu trong s c mt ca mt cht kh acid ascorbic. Nng ca phc c xc nh quang ph bc sng 830 nm, s dng mt ng cong hiu chun bnh thng nh mt phng php tiu chun ha.Trong phn tch in hnh, mt b cc dung dch chun c cha mt lng bit ca pht pho c chun b bng cch t th tch thch hp ca 4,00 ppm dung dch ca P2O5trong bnh nh mc 5 mL, thm 2 ml dung dch acid ascorbic dung dch gim, v pha long n vch bng nc ct. ung Cola c chun b phn tch bng cch mt mu vo mt cc thy tinh v cho php n ng yn trong 24 gi nh ui tan CO2. Ly 2.50 ml mu kh kh cho vo mt bnh nh mc 50-mL v pha long n vch. Ly 250 L ca mu c pha long sau c chuyn vo mt bnh nh mc 5 mL, nghin cu vi 2 ml acid ascorbic dung dch gim, v pha long n vch bng nc ct.(A) Cc tc gi lu rng phng php ny c th c p dng vi ung cola khng mu.Gii thch l do ti sao iu ny l ng.(B) Lm th no bn c th thay i phng php ny n c th c p dng cho bt k ung cola?(C) Ti sao cn thit loi b cc kh ho tan?(D) ngh mt dung dch trng thch hp cho phng php ny?(E) Bo co ca tc gi mt ng cong hiu chunA= -0.02 + 0,72 ppm P2O5Mt mu ca Crystal Pepsi, phn tch nh m t trn, c gi tr mt quang l 0,565.Nng pht pho trong mu ban u ca Crystal Pepsi l bao nhiu ppm P? 13. EDTA to cc phc mu vi nhiu ion kim loi c th c dng nh lng chng trong phn tch o quang. H s hp th phn t ca phc EDTA ca Cu2+, Co2+, v Ni2+ ba bc sng c tm tt trong bng sau (tt c cc gi tr ca l trong M-1cm-1).kim loai462,9732,0378,7

Co2+15.82.113.11

Cu2+2.3295,27.73

Ni2+1.793.0313.5

S dng thng tin ny xc nh:(A) Nng ca Cu2+trong dung dch c gi tr mt quang 0,338 bc sng 732,0 nm.(B) Cc nng ca Cu2+v Co2+trong dung dch c gi tr mt quang 0,453 bc sng 732,0 nm v 0,107 bc sng 462,9 nm.(C) Cc nng ca Cu2+, Co2+, v Ni2+trong mt mu c gi tr mt quang 0,423 bc sng 732,0 nm, 0,184 bc sng 462,9 nm, v 0,291 bc sng 378,7 nm .Cho bit chiu di cuvet l 1,00 cm cho tt c cc php o.Bi gii: a) bc sng =732,0 nm ta c: A= CCu2+ .Cu2+.l= CCu2+.95,2 = 0,338 => CCu2+=0,338/95,2= 3,55 . 10-3(M)Vy nng ca Cu2+ l 3,55 . 10-3(M) b) =732,0 nm: A= Cu2+.l. CCu2+ + CCo2+. Cu2+.l= 0,453=> 95,2 CCu2+ + 2,11 CCo2+ = 0,453 (1) =462,9 nm, A= Cu2+.l. CCu2+ + CCo2+. Cu2+.l= 0,107 => 2,32 CCu2+ + 15,8 CCo2+ = 0,453 (2)T (1) v (2), gii h pt ta c CCu2+ = 4,62.10-3 (M) CCo2+= 6,09.10-3 (M) c) = 732,0 nm: A= Cu2+.l. CCu2+ + CCo2+. Cu2+.l + CNi2+. Ni2+.l= 0,423=> 95,2 CCu2+ + 2,11 CCo2+ + 3,03 CNi2+= 0,453 (1) =462,9 nm: A= Cu2+.l. CCu2+ + CCo2+. Cu2+.l + CNi2+. Ni2+.l= 0,184=> 2,32CCu2+ + 15,8 CCo2+ + 1,79 CNi2+= 0,453 (2) =378,7 nm: A= Cu2+.l. CCu2+ + CCo2+. Cu2+.l + CNi2+. Ni2+.l= 0,291=> 7,73CCu2+ + 3,11 CCo2+ + 13,5 CNi2+= 0,291 (3) T (1), (2), (3) gii h pt ta c CCu2+ = 3,69.10-3 (M) CCo2+= 9,14.10-3 (M) CNi2+= 0,0173 (M)14.Nng ca phenol trong mt mu nc c xc nh bng cch tch phenol khi cc tp cht khng bay hi bng cch chng ct hi nc, sau phn ng vi 4 aminoantipyrine v K3Fe(CN)6ti pH 7,9 to thnh mt cht nhum mu antipyrine.Mt tiu chun phenol vi nng 4,00 ppm c mt quang l 0,424 bc sng 460 nm s dng mt vuvet vi l= 1,00 cm.Mt mu nc c chng ct, ly 50,00 ml sn phm ct ny vo trong mt bnh nh mc100-mL v pha long n vch bng nc ct. Mt quang ca dung dch l 0,394.Nng ca phenol (phn triu) trong mu nc l bao nhiu (ppm)?Bi gii: i vi mu phenol tiu chun: A= .l.C m l=1 cm, C= 4ppm, A= 0,424=> = A/(l.C)= 0,424/4= 0,106i vi mu nc: A=0,394, l=1cm, C=?, = 0,106Ta c: A= .l.C => C= A/(.l)= 0,394/0,106 =3,72 (ppm)V 50,00 ml mu nc c pha long thnh 100ml nn nng ca phenol trong mu nc trc khi pha long l Cphenol=2.C= 3,72.2=7,44 (ppm)16. Trong cc phng php so mu DPD cho d clo t do ang c bo co l mg Cl2/ L, sc mnh oxy ha ca clo t do chuyn i amin khng mu N, N-diethyl-p-phenylenediamine nhum mu hp th mnh m hn phm vi bc sng 440-580 nm.Phn tch ca mt tp hp cc tiu chun hiu chun cho cc kt qu nh sau.mg Cl2/ Lhp th

0.000.000

0.500,270

1.000,543

1.500,813

2.001,084

Mt mu ly t mt ngun cung cp nc cng cng c phn tch xc nh clo d min ph, cho hp th ca 0,113.Phn d clo t do cho cc mu trong mg Cl l g2/ L?Bi gii: Theo phng php lp ng chun Gi phng trnh y= bx+a l phng trnh tuyn tnh v mi quan h gia nng o v mt quang AGi xtb: nng trung bnh ytb: gi tr trung bnh ca mt quang Khi : a= ytb - bxtb b= tng[(xi-xtb)(yi-ytb)]/tng(xi-xtb)2Ta c: xtb= (0+0,5+ 1+1,5+2) /5= 1ytb= (0+ 0,27 + 0,543 + 0,813 +1,084)/5 = 0,542=> b= [(-1)(0-0,542)+(-0,5)(0,27-0,542)+(0,5)(0,813-0,542)+(1,084-0,542)]/[2+0,52] = 0,5422=> a= 0,542 0,5422.1= -2.10-4Do phng trnh cn tm l y= 0,5422x - 2.10-4Khi A= y =0,113 => x= 0,209 Vy nng ca Cl trong mu l 0,209 mgCl2/l17. Lin v Brown m t mt phng php nh lng cho methanol da trn tc ng ca n trn quang ph nhn thy c ca methylene blue.23Trong s vng mt ca methanol, quang ph nhn thy c cho methylene xanh cho thy hai di hp th ni bt tm ti khong 610 nm v 660 nm, tng ng vi cc monomer v dimer, tng ng.Trong s hin din ca methanol, cng ca di hp th ca dimer gim, trong khi tng monomer.i vi nng methanol gia 0 v 30% v / v, t l hp th 663 nm,A663, cho rng ti 610 nm,A610, l mt hm tuyn tnh ca lng methanol.S dng cc d liu tiu chun sau y, xc nh% v / v methanol trong mt mu mA610l 0,75 vA663l 1,07.% V / v methanolA663/A610

0.01,21

5.01,29

10.01.42

15.01,52

20.01,62

25.01.74

30.01,84

Bi gii: Theo phng php lp ng chun Gi phng trnh y= bx+a l phng trnh tuyn tnh v mi quan h gia nng o v mt quang AGi xtb: nng trung bnh ytb: gi tr trung bnh ca mt quang Khi : a= ytb - bxtb b= tng[(xi-xtb)(yi-ytb)]/tng(xi-xtb)2Ta c: xtb= (0+5+ 10+15+20+25+30) /7= 15ytb= (1,21+ 1,29 + 1,42 + 1,52 +1,62 + 1,74 + 1,84)/7 = 1,52 (-15)(1,21-1,52)+(-10)(1,29-1,52)+(-5)(1,42-1,52)+5(1,62-1,52)+10(1,74-1,52) +15(1,84-1,52)= 14,95 (152 + 102 + 52 + 52 +102 +152)= 700=> b= 14,95/700 = 0,0214=> a= 1,52 0,0214.15= 1,199Do phng trnh cn tm l y= 0,0214x + 1,199Khi A663/A610 =y =1,07/0,75= 1,4267 => x= (1,4267-1,199)/ 0,0214=10,6 Vy nng ca methanol trong mu l 10,6 % V / v methanol

18. Nng ca barbital barbiturat trong mu mu c xc nh bng cch chit 3.00 mL mu vi 15 ml CHCl3.Cc chloroform, m by gi c nhng barbital, c chit xut vi 10,0 ml NaOH 0,45 M (pH13).Mt mu 3,00 ml dung dch nc chit xut c t trong mt cuvet 1,00 cm v mt quang o c l 0,115. pH ca cc mu trong cc t bo hp thu c sau c iu chnh xung cn khong 10 bng cch thm 0,5 ml dung dch 16% w / v NH4Cl, cho mt quang l 0.023.Khi 3,00 ml dung dch barbital tiu chun vi mt nng 3 mg / 100 mL c thc hin thng qua cc th tc tng t, mt quang pH 13 l 0,295 v mt quang pH= 10 l 0.002.Bo co barbital mg / 100 ml trong mu.19. Jones v Thatcher pht trin mt phng php quang ph phn tch vin thuc gim au c cha aspirin, phenacetin, v caffeine.24Cc mu c ha tan trong CHCl3v chit vi dung dch NaHCO3 loi b cc aspirin.Sau khi khai thc xong, chloroform c chuyn giao cho mt 250-mL bnh nh mc v pha long n khi lng vi CHCl3.Mt phn 2,00 ml dung dch ny c pha long vi khi lng trong mt 200-mL bnh nh mc vi CHCl3. hp th ca cc gii php cui cng c o bc sng 250 nm v 275 nm, m ti cc cht hp th, trong ppm-1cm-1, cafein v phenacetin lmt250mt275

caffeine0,01310,0485

phenacetin0,07020,0159

Aspirin c xc nh bng cch trung ha cc NaHCO3trong dung dch nc v gii nn aspirin vo CHCl3.Cc cht chit xut kt hp c pha long n 500 ml trong bnh nh mc.Mt phn 20,00 ml dung dch c t trong mt 100-mL bnh nh mc v pha long n khi lng vi CHCl3. hp th ca dung dch ny c o 277 nm, ni hp th ca aspirin l 0,00682 ppm-1cm-1.Mt vin thuc gim au c iu tr bng phng php ny c pht hin l c hp th ca 0,466 250 nm, 275 nm ti 0,164 v 0,600 277 nm khi s dng mt t bo vi mt pathlength 1,00 cm.Bo co mg aspirin, caffeine, v phenacetin trong vin thuc gim au.Bi gii:C1 l nng ca caffeine C2 l nng ca phenacetin = 250 nm, ta c: 0,0131. C1 + 0,0702 . C2 = 0,466 = 275 nm, ta c: 0,0485. C1 + 0,0159. C2 = 0,164Gii h 2 pt trn => C1= 1,284 ppmC2= 6,399 ppmT 250 ml dd mu, ly 2ml dd ny pha long thnh 200ml, tc l pha long 100 ln => nng ca caffeine trong mu l 1,284.100 = 128,4 ppmnng ca phenacetin trong mu: 6,399.100 = 639,9 ppm m caffeine = 128,4. 0,25 = 32,1 mg m phenacetin = 639,9. 0,25 = 160 mgGi C3 l nng ca Aspirin sau khi pha long xong = 277 nm, ta c: 0,00682. C3= 0,6 => C3= 87,98 ppmT 500ml dd mu, ly 20ml pha long thnh 100ml, tc l pha long gp 5 ln nng ca Aspirin trong mu: 87,98 .5 = 439,9 ppm m Aspirin= 439,9 . 0,5= 219,95 mg

20. Nng SO2trong mu khng kh c xc nh bng ccphng php p-rosaniline.SO2c thu thp trong 10,00 ml dung dch HgCl42-, n to thnh Hg(SO3)22-, bng cch ht khng kh qua dung dch ny trong 75 pht tc 1,6 L / pht.Sau khi thmp-rosaniline v formaldehyde, dung dch mu c pha long n 25 ml trong bnh nh mc. Mt quang c o 569 nm trong cuvet 1 cm, t gi tr 0,485.Mt mu chun c chun b bng cch ly 1,00 ml dung dch chun cha tng ng vi 15,00 ppm SO2trong mu khng kh. Mt quang dung dch chun ny c tm thy l 0,181. Tnh nng SO2 trong khng kh theo ppm.Mt ca khng kh l 1,18 g / lt.Bi gii: Achun=.l.C => = Achun/l.C = 0,181/15= 0,012ppmKhi A=0,485 ta c C= A/l.= 0,485/0,012= 40,417 ppmTa c: Vkk= 120 . 1,18= 120 (l)mkk= 120. 1,18 = 141,6 g23. Cc danh mc bng cht hp th phn t sau cho hp Arsenazo ng v bari.27 ngh cc bc sng thch hp phn tch hn hp ng v bari s dng phc Arsenzao ca h.Bc sng (nm)Cu(M -1cm-1)Ba(M -1cm-1)

595119007100

600155007200

60718.3007400

61119.3006900

61419.3007000

620178007100

62616.3008400

635109009900

641750010500

645530010000

65035008600

65522006600

65819006500

66515003900

67015002800

68018001500

26. Cc hc lng php ca mt kim loi-phi t phc tp, MLn, c xc nh bng phng php bin th lin tc. Nhiu dung dch c chun b trong nng tng hp ca M v L c gi khng i 5.15 10-4M. hp th ca cc dung dch c o bc sng m ch phc kim loi-phi t hp th.S dng cc d liu sau y, xc nh cng thc ca phc kim loi-phi t.phn mol Mmole phn LMt quang

1.00.00,001

0.90.10,126

0.80.20,260

0.70.30,389

0.60,40,515

0.50.50,642

0,40.60,775

0.30.70,771

0.20.80,513

0.10.90,253

0.01.00.000

33. Hobbins bo co cc d liu hiu chun sau y cho ngn la phn tch hp th nguyn t s dng pht pho.mg P / LMt quang

21300,048

42600,110

64000,173

85300,230

xc nh tinh khit ca mu Na2HPO4, mt mu 2,469 g c ha tan v pha long n khi lng trong mt 100-mL bnh nh mc.Phn tch cc gii php kt qu cho hp th ca 0,135. tinh khit ca Na2HPO4l bao nhiu?Bi gii: Theo phng php lp ng chun Gi phng trnh y= bx+a l phng trnh tuyn tnh v mi quan h gia nng o v mt quang AGi xtb: nng trung bnh ytb: gi tr trung bnh ca mt quang Khi : a= ytb - bxtb b= tng[(xi-xtb)(yi-ytb)]/tng(xi-xtb)2Ta c: xtb= (2130+4260+ 6400+8530) /4= 5330ytb= (0,084+ 0,11 + 0,173 + 0,23)/4 = 0,14=>b=[(2130-5330)(0,084-0,14)+(4260-5330)(0,11-0,14)+(6400-5330)(0,173-0,14)+(8530-5330)(0,23-0,14)]/[(2130-5330)2+(4260-5330)2+(6400-5330)2+(8530-5330)] b= 2,86 .10-5=> a= 0,14 2,86 .10-5.5330= -0,0123Do phng trnh cn tm l y= 2,86 .10-5x - 0,0123Khi A= y =0,135 => x= 5150,35 mgP/l m P = 5150,35 . 0,1= 515,035 mg m Na2HPO4= 515,035 . 142/31 = 2359,2 mg tinh khit ca Na2HPO4 l 2359,2/ 2469= 95,55 %

34. Bonert v Pohl bo co kt qu phn tch hp th nguyn t ca mt s kim loi trong h thng treo caustic sn xut trong sn xut soda bi qu trnh ammonia-soda.30(A) Nng ca Cu c xc nh bng axit ha mt mu 200 ml dung dch xt vi 20 ml dung dch HNO3, thm 1 ml dung dch 27% w / v H2O2, v un si trong 30 pht.Dung dch thu c pha long n 500 ml, lc v phn tch bng cch hp th nguyn t ngn la bng cch s dng cc tiu chun ph hp vi ma trn.Cc kt qu phn tch cho mt in hnh c th hin trong bng di y.dung dchmg Cu / LMt quang

Trng0.0000,007

tiu chun 10,2000,014

tiu chun 20,5000,036

tiu chun 31.0000,072

tiu chun 42.0000,146

mu0,027

Xc nh nng ca Cu trong vic nh ch n da.Bi gii: Theo phng php lp ng chun Gi phng trnh y= bx+a l phng trnh tuyn tnh v mi quan h gia nng o v mt quang AGi xtb: nng trung bnh ytb: gi tr trung bnh ca mt quang Khi : a= ytb - bxtb b= tng[(xi-xtb)(yi-ytb)]/tng(xi-xtb)2Ta c: xtb= (0+0,2+ 0,5+1+2) /5= 0,74ytb= (0,007+0,014+0,036 +0,072 + 0,146)/5 = 0,055=>b=[(0-0,74)(0,007-0,055)+(0,2-0,74)(0,014-0,055)+(0,5-0,74)(0,036-0,055)+(1-0.74)(0,072-0,055)+(2-0,74)(0,146-0,055)]/[(0-0,74)2+(0,2-0,74)2+(0,5-0,74)2+(1-0.74)2+(2-0,74)2] b= 0,071=> a= 0,055 0,071.0,74 = 2,46 .10-3

Do phng trnh cn tm l y= 0,071x + 2,46.10-3Khi A= y =0,027 => x= 0,346 mgP/l T 200ml pha long n 500ml, tc l pha long gp 2,5 lnVy nng ca Cu trong vic nh ch n da l 0,346. 2,5= 0,865 ppm

(B) Vic xc nh Cr c thc hin bng cch axit ha mt mu 200 ml dung dch xt vi 20 ml dung dch HNO tp trung3, thm 0,2 g Na2SO3v un si trong 30 pht.Cc Cr c phn lp t cc mu bng cch thm 20 ml dung dch NH3, sn xut mt cht kt ta bao gm crom cng nh oxit khc.Kt ta c phn lp bng cch lc, ra sch, v chuyn giao cho mt cc thy tinh.Sau khi axit ha vi 10 ml dung dch HNO3, cc gii php c bc hi n kh.Phn d c ra gii trong mt s kt hp ca HNO3v HCl v bc hi n kh.Cui cng, phn d c ha tan trong 5 ml dung dch HCl, lc, pha long vi khi lng trong mt th tch bnh 50 ml, v phn tch bng cch hp th nguyn t s dng cc phng php b sung tiu chun.Cc kt qu hp th nguyn t c tm tt trong bng di y.mumg Crthm/ Lhp th

trng0,001

mu0,045

Ngoi tiu chun 10,2000,083

Ngoi tiu chun 20,5000,118

Ngoi tiu chun 31.0000,192

Bo co nng Cr trong vic nh ch n da.35. Quigley v Vernon bo co kt qu cho vic xc nh cc kim loi vi lng trong nc bin bng cch s dng mt l than ch nguyn t hp th quang ph v cc phng php b sung tiu chun.31Cc kim loi nh du ln u tin c tch t, ma trn nhiu mui phc tp ca h bi coprecipitating vi Fe3+.Trong mt phn tch in hnh mt phn 5.00 ml 2000 ppm Fe3+ c thm vo 1.00 L nc bin.PH c iu chnh s dng 9 NH4OH, v kt ta Fe (OH)3 yn qua m.Sau khi ly v ra kt ta, Fe (OH)3v kim loi coprecipitated c ha tan trong 2 ml tp trung HNO3v pha long n khi lng trong mt th tch bnh 50 ml. phn tch cho Mn2+, mt mu 1,00 ml dung dch ny c pha long n 100 ml trong bnh nh mc.Cc mu sau y c bm vo l graphite v phn tch.muhp th

2,5 ml mu + 2,5 ml 0 ppb Mn2+0,223

2,5 ml mu + 2,5 ml 2,5 ppb Mn2+0,294

2,5 ml mu + 2,5 ml 5,0 ppb Mn2+0,361

Bo co cc phn t Mn2+trong mu nc bin.36. Nng ca Na trong nguyn liu thc vt c th c xc nh bng pht x nguyn t ngn la.Cc vt liu c phn tch c chun b bng cch nghin, homogenizing, v sy 103oC.Mt v d v khong 4 g c chuyn giao cho mt ni thch anh v un trn mt tm nng char cc cht hu c.Cc mu c lm nng trong l mp 550oC trong vi gi.Sau khi lm lnh n nhit phng cn c ha tan bng cch thm 2 ml dung dch 1: 1 HNO3v bc hi n kh.Phn cn li c ra gii trong 10 ml 1: 9 HNO3, lc v pha long n 50 ml trong bnh nh mc.Cc d liu sau y thu c trong qu trnh phn tch tiu biu cho nng Na trong mt mu 4,0264 g cm yn mch.mumg Na / LCng pht x

trng0.000.0

tiu chun 12.0090,3

tiu chun 24.00181

tiu chun 36.00272

tiu chun 48.00363

5 tiu chun10.00448

mu238

Xc nh lng ppm Na trong mu la mch.Bi gii: Theo phng php lp ng chun Gi phng trnh y= bx+a l phng trnh tuyn tnh v mi quan h gia nng o v cng pht xGi xtb: nng trung bnh ytb: gi tr trung bnh ca cng pht xKhi : a= ytb - bxtb b= tng[(xi-xtb)(yi-ytb)]/tng(xi-xtb)2Ta c: xtb= (0+2+4+ 6+8+10) /6= 5ytb= (0+ 90,3 + 181 + 272 +363+ 448)/6 = 225,71=>b=[(0-5)(0-225,71)+(2-5)(90,3-225,71)+(4-5)(181-225,71)+(6-5)(272-225,71)+(8-5)(363-225,71)+(10-5)(448-225,71)]/[ (0-5)2+(2-5)2+(4-5)2+(6-5)2+(8-5)2+(10-5)2] b= 3149,1/70= 44,987=> a= ytb - bxtb =225,71 44,987.5= 0,781Do phng trnh cn tm l y= 44,987x + 0,781Khi A= y =238 => x= 5,273 ppm =5,273 mg/lHm lng Na c trong mu la mch: 5,273 50/4,0264= 65,48 ppm37. Gluodenis m t vic s dng cc pht x nguyn t ICP phn tch mu bng ng thau cho Pb v Ni.Cc phn tch cho Pb s dng cc tiu chun bn ngoi chun b t cc mu ng thau c cha mt lng ch c bit n.Kt qu c th hin trong bng di y.% W / w Pbcng pht x

0.0004.29 104

0,01001,87 105

0,02003.20 105

0,06501,28 106

0,3506.22 106

0,7001,26 107

1.041,77 107

2.243,88 107

3.155.61 107

9.251,64 108

Cc% l g w / w Pb trong mu ng cung cp cho mt cng pht x ca 9,25 104?Bi gii: Theo phng php lp ng chun Gi phng trnh y= bx+a l phng trnh tuyn tnh v mi quan h gia nng o v cng pht x IGi xtb: nng trung bnh ytb: gi tr trung bnh ca cng pht x Khi : a= ytb - bxtb b= tng[(xi-xtb)(yi-ytb)]/tng(xi-xtb)2Ta c: xtb= (0+0,01+0,02+ 0,065+0,35+0,7+1,04+2.24+3,15+9,25)/10= 1,6825ytb=(4,29.104+1,87.105+3,20.105+1,28.106+6,22.106+1,26.107+1,77.107+ 3,88.107+ 5,61.107+ 1,64.108)/10= 2,97. 107Ta c: tng[(xi-xtb)(yi-ytb)]=(0,00-1,6825).(4,29.104-2,97. 107)+(0,01-1,6825).( 1,87.105-2,97. 107)+(0.02-1,6825).(3,20.105-2,97. 107)+( 0,065-1,6825)( 1,28.106-2,97. 107)+ (0,35-1,6825).( 6,22.106-2,97. 107)+( 0,7-1,6825)( 1,26.107-2,97. 107)+ ( 1,04-1,6825)( 1,77.107-2,97. 107)+( 2.24-1,6825)( 3,88.107-2,97. 107)+( 3,15-1,6825)( 5,61.107- 2,97. 107)+ (9,25-1,6825)( 1,64.108-2,97. 107)=tng(xi-xtb)2=(0,00-1,6825)2+(0,01-1,6825)2+ (0.02-1,6825)2+(0,065-1,6825)2+ (0,35-1,6825)2+( 0,7-1,6825)2+( 1,04-1,6825)2+( 2.24-1,6825)2+( 3,15-1,6825)2+(9,25-1,6825)2= => b= tng[(xi-xtb)(yi-ytb)]/tng(xi-xtb)2==> a= Do phng trnh cn tm l y= 2,86 .10-5x - 0,0123Khi I= y =9,25 .104 => x=

Cc phn tch cho Ni s dng mt tiu chun ni b.Kt qu cho mt hiu chun in hnh l chng trnh trong bng sau.% W / w Nit l cng pht thi

0.0000,00267

0,01400,00154

0,03300,00312

0,1300,120

0,2800,246

0,2800,247

0,5600,533

1.301.20

4,824.44

Cc% l g w / w Ni trong mt mu m t l cng pht x l 1.10 10-3?Bi gii: Theo phng php lp ng chun Gi phng trnh y= bx+a l phng trnh tuyn tnh v mi quan h gia nng o v t l cng pht xGi xtb: nng trung bnh ytb: gi tr trung bnh ca t l cng pht xKhi : a= ytb - bxtb b= tng[(xi-xtb)(yi-ytb)]/tng(xi-xtb)2Ta c: xtb= (0,014+0,033+0,13+0,28+0,28+0,56+1.30+4,82) /9= ytb= (0,00267+0,00154+0,00312+0,120+0,246+0,247+0,533+1.20+4.44)/9 = =>b =tng[(xi-xtb)(yi-ytb)]/tng(xi-xtb)2 b= 2,86 .10-5=> a = ytb - bxtb=Do phng trnh cn tm l y= Khi y =0,135 => x=