2. One of the common methods used to determine phosphorus in urine isto treat urine with molybdenum(VI) resulting in a complex compoundof 12- molybdophosphate and then reduced using ascorbic acid produces a blue-colored compound called molybdenum blue (max = 650 nm). A patient's urine sample collected over 24 hours produced 1122 mL of urine. A total of 1 mL of the sample was treated with Mo(VI) and ascorbic acid and diluted until it reached a volume of 50 mL. A calibration curve is also prepared by treating 1 mL of standard phosphate solution in the same way. The absorbance of the standard solution and the sample was then measured at a wavelength of 650 nm and produced the following data: Solution Absorbance at λ 650 nm Phosphate 1 ppm Phosphate 2 ppm Phosphate 3 ppm Phosphate 4 ppm Urine sample 0,230 0,436 0,638 0,848 0,518 a. Create a calibration curve for the data and determine the slope, intercept, and R values of 2! Determine the concentration of phosphates found in urine samples in units of ppm and mM! b. Determine the mass of phosphate excreted by the patient for 24 hours?

2. One of the common methods used to determine phosphorus in urine isto treat urine with molybdenum(VI) resulting in a complex compoundof 12- molybdophosphate and then reduced using ascorbic acid produces a blue-colored compound called molybdenum blue (max = 650 nm). A patient's urine sample collected over 24 hours produced 1122 mL of urine. A total of 1 mL of the sample was treated with Mo(VI) and ascorbic acid and diluted until it reached a volume of 50 mL. A calibration curve is also prepared by treating 1 mL of standard phosphate solution in the same way. The absorbance of the standard solution and the sample was then measured at a wavelength of 650 nm and produced the following data: Solution Absorbance at λ 650 nm Phosphate 1 ppm Phosphate 2 ppm Phosphate 3 ppm Phosphate 4 ppm Urine sample 0,230 0,436 0,638 0,848 0,518 a. Create a calibration curve for the data and determine the slope, intercept, and R values of 2! Determine the concentration of phosphates found in urine samples in units of ppm and mM! b. Determine the mass of phosphate excreted by the patient for 24 hours?

Fundamentals Of Analytical Chemistry

9th Edition

ISBN:9781285640686

Author:Skoog

Publisher:Skoog

Chapter24: Introduction To Spectrochemical Methods

Section: Chapter Questions

Problem 24.28QAP

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Given the Information that Fe3++Y4FeYKi=1.01025Cu2++Y4CuY2Ki=6.31018 and the further information that, among the several reactants and products, only CuY2- absorbs radiationat 750 nm, describe how Cu(II) could be used as an indicator for the photometric titration of Fe(III) with H2Y2-. Reaction:+ Fe3++ H2Y2- FeY- + 2H+.
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A standard solution was put through appropriate dilutions to give the concentrations of iron shown in the table that follows. The iron(II)-1,10-phenanthroline complex was then formed in 25.0-mL aliquots of these solutions, following which each was diluted to 50.0 mL. The following absorbances (1.00-cm cells) were recorded at 510 nm: Calculate the concentration, in ppm, of a sample with an abosrbance of 0.829. Fe(II) concentration (ppm) A510 4.00 0.160 10.0 0.390 16.0 0.630 24.0 0.950 32.0 1.260 40.0 1.580
One common way to determine phosphorous in urine is to treat the sample after removing the protein with molybdenum (VI) and then reducing the resulting 12-molybdophosphate complex with ascorbic acid to give an intense blue-coloured species called molybdenum blue. The absorbance of molybdenum blue can be measured at 650 nm. A 24-hour urine sample was collected, and the patient produced 1122 mL in 24 hours. A 1.00 mL aliquot of the sample was treated with Mo (VI) and ascorbic acid and diluted to a volume of 50.00 mL A calibration curve was prepared by treating 1.00 mL aliquot of phosphate standard solution in the same manner as the urine sample. The absorbances of the standards and the urine sample were obtained at 650 nm and the following results were obtained. ppm, P aBSORBANCE (A) 1.003.00 0.230 2.00 0.436 3.00 0.638 4.00 0.848 Urine sample 0.518 ffind the slope and intercept, then graph it.
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