A 10.00-g sample containing an analyte was transferred to a 250-mL volumetric flask and diluted to volume. When a 10.00-mL aliquot of the resulting solution was diluted to 25.00 mL it was found to give a signal of o.235 (arbitrary units). A second 10.00-mL aliquot was spiked with 10.0o mL of a 1.00- ppm standard solution of the analyte and diluted to 25.0o mL. The signal for the spiked sample was found to be o.502. Calculate the weight percent of analyte in the original sample. 1010

A 10.00-g sample containing an analyte was transferred to a 250-mL volumetric flask and diluted to volume. When a 10.00-mL aliquot of the resulting solution was diluted to 25.00 mL it was found to give a signal of o.235 (arbitrary units). A second 10.00-mL aliquot was spiked with 10.0o mL of a 1.00- ppm standard solution of the analyte and diluted to 25.0o mL. The signal for the spiked sample was found to be o.502. Calculate the weight percent of analyte in the original sample. 1010

Principles of Instrumental Analysis

7th Edition

ISBN:9781305577213

Author:Douglas A. Skoog, F. James Holler, Stanley R. Crouch

Publisher:Douglas A. Skoog, F. James Holler, Stanley R. Crouch

Chapter1: Introduction

Section: Chapter Questions

Problem 1.9QAP

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The chromium in an aqueous sample was determined by pipetting 10.0 ml. of the unknown into each of five 50.0-mL volumetric flasks. Various volumes of a standard containing 12.2 ppm Cr were added to the flasks, following which the solutions were diluted to volume. Unknown,mLStandard, mLAbsorbancc 10.00.00.201 10.010.0 0.292 10.020.0 0.378 10.030.0 0.467 10.040.0 0.554 (a) Plot the data using a spreadsheet. (b) Determine an equation for the relationship between absorbance and volume of standard. (c) Calculate the statistics for the least-squares relationship in (b). (d) I)ctcrmine the conccnt ration oÍCr in ppm in the sample. (e) Find the standard deviation of the result in (d).
The acid-base indicator HIn undergoes the following reaction in dilute aqueous solution: HIncolor1H++Incolor2 The following absorbance data were obtained for a 5.00 I0-4 M solution of HIn in 0.1 M NaOH and 0.1 M HC1. Measurements were made at wavelengths of 485 nm and 625 nm with 1.00-cm cells. 0.1 M NaOH A485 = 0.075 A625 = 0.904 0.1 M HC1 A485 = 0.487 A625 = 0.181 In the NaOH solution, essentially all of the indicator is present as In-; in the acidic solution, it is essentially all in the form of HIn. (a) Calculate molar absorptivities for In- and HIn at 485 and 625 nm. (b) Calculate the acid dissociation constant for the indicator ¡fa pH 5.00 buffer containing a small amount of the indicator exhibits an absorbance of 0.567 at 485 nm and 0.395 at 625 nm (1.00-cm cells). (c) What is the pH of a solution containing a small amount of the indicator that exhibits an absorbance of0.492 at 485 nm and 0.245 at 635 nm (1.00-cm cells)? (d) A 25.00-mL aliquot of a solution of purified weak organic acid HX required exactly 24.20 mL of a standard solution of a strong base to reach a phenolphthalein end point. When exactly 12.10 mL of the base was added to a second 25.00-mL aliquot of the acid, which contained a small amount of the Indicator under consideration, the absorbance was found to be 0.333 at 485 nm and 0.655 at 625 nm (1.00-cmcells). Calculate the pH of the solution and Ka for the weak acid. (e) What would be the absorbance of a solution at 485 and 625 nm (1.50-cm cells) that was 2.00 10-4 M in the indicator and was buffered to a pH of 6.000?
Chloroform is used as an internal standard in the determination of the pesticide deltamethrin. To calibrate the method a mixture of 0.500 mM of chloroform and 0.800 mM of deltamethrin which produced signals of 15.3 (w.a.) and 10.1 (w.a.) for chloroform and deltamethrin, respectively. To prepare the sample, 10.00 mL of a sample of unknown concentration of deltamethrin and 10.10 μL of pure chloroform (FW = 119.39 g/mol, ρcloroformo = 1.484 g/L) is homogenized and diluted with water to the 100.00 mL mark in a volumetric flask. After preparing the sample, the observed signals of 30.0 (a.u.) and 20.7 (a.u.) for chloroform and deltamethrin, respectively, are observed. 1. Determine the response factor (K) for the method. 2.Determine the millimolar (mM) concentration of chloroform added to the sample. Remember that the final volume of the solution is 100.00 mL. 3.Determine the millimolar (mM) concentration of deltamethrin in the sample. 4.If the concentration of deltamethrin should not…
A sample of body serum is to be analysed for sodium by flame emission spectroscopy. 1.00cm' aliquot of serum was pipetted into each of two 50.0cm' volumetric flasks. The first flask was diluted to volume with deionised water. The absorbance of this first solution was0.350. To the second flask 10.0 cm' of a 25.0 ppm sodium standard was added and the flask made up to volume with deionised water. The absorbance of this second solution was 0.720.(i) Calculate the concentration of the sodium in the body serum in mg dm-3.
A spectrophotometric method for the quantitative determination of the concentration ofPb2+ in blood yields an Ssamp of 0.133 for a 1 mL sample of blood that has been diluted to 6 mL. A second sample is spiked with 1 µL of a 1467 ppb Pb2+ standard and diluted to 6 mL, yielding an Sspike of 0.491. Determine the concentration of Pb2+ in the original sample of blood.
A second spectrophotometric method for the quantitative analysis of Pb2+ in blood yields an Ssamp of 0.712 for a 5.00 mL sample of blood. After spiking the blood sample with 5.00 μL of a 1560 ppb Pb2+ external standard, an Sspike of 1.546 is measured. What is the concentration of Pb2+ in the original sample of blood?
A standard solution was prepared containing 10.0 ppm of an analyte and 15.0 ppm of an internal standard. Analysis of the sample gave signals for the analyte and internal standard of 0.155 and 0.233 (arbitrary units), respectively. Sufficient internal standard was added to an unknown sample to make it 15.0 ppm in the internal standard. Analysis of the unknown sample yielded signals for the analyte and internal standard of 0.274 and 0.198, respectively. Calculate the concentration of analyte in the unknown sample.
A solution of Ba(OH)2 was standardized against0.1215 g of primary-standard-grade benzoic acid, C6H5COOH (122.12 g/mol). An end point was observedafter addition of 43.25 mL of base. (a) Calculate the molar concentration of the base. (b) Calculate the standard deviation of the molarconcentration if the standard deviation for themass measurement was ±0.3 mg and that for thevolume measurement was ±0.02 mL. (c) Assuming an error of ±0.3 mg in the mass measurement,calculate the absolute and relative systematicerror in the molar concentration.
Anionic surfactants in wastewater are measured spectrophotometrically as methylene blue active substances (MBAS). A volume of a water sample is first reacted with methylene blue, a cationic dye, in a basic buffered aqueous solution. The resulting solution is agitated after the addition of chloroform. The organic layer is then extracted, and the absorbance is read at 655 nm. A standard calibration curve is produced by plotting absorbance vs. concentration for the concentration range between 0 ppm to 2.50 ppm. The line had the following equation: y = 0.286x + 0.009 a. The limit of surfactant content in Philippine class B water is 300 ppb. If a 5 mL aliquot of a 50 mL water sample had an absorbance of 0.084 after MBAS analysis, is the sampled water body suitable for bathing? Support your answer
1 mL was taken from the sample filtrate and mixed with 13 mL pure water and 4 mL sulfomolybdic acid and 2 mL dilute SnCI2 solution by adding it, and after waiting for 15 minutes, the absorbance of the resulting solutions against pure water was read at 520 nm. if the function of the calibration graph obtained with standard phosphorus solutions of 0.5-2.5 mg/mL is y= 0.245x + 0.107 and the absorbance of the serum sample is 0.342, how many grams of phosphorus is the amount in the sample?
A 0.0200 gram blood sample was decomposed by a microwave digestion technique followed by dilution to 100.0 mL in a volumetric flask. Aliquots of the sample solution were treated with a lead complexing reagent and water as follows: Solution 1: 10.0 ml blood sample + 20.0 mL complexing agent + 30.0 mL H20. Solution 2: 10.0 ml blood sample + 20.0 mL complexing agent + 26.0 mL H20 + 4.00 mL of 78 ppb Pb2+ standard. The resulting solutions were analyzed by UV/Vis at 375 nm. Absorbance for solution 1 = 0.155 and for solution 2 = 0.216. Calculate the concentration of lead (ppb) in the original sample.
Internal standard (As , Cs ), analyte with known concentration (Ax , Cx ) In a chromatographic equipment, a solution containing 0.006727 X and 0.008331 M S give peak area of Ax= 2316 and As= 207. To analyze an unknown sample, 5.0mL of 0.008331M S was added to 5.0mL of X, and the mixture was diluted to 50.0mL. This mixture gave a chromatography spectrum with area Ax= 3206 and As= 200. Find Cx.