Calculate the molar concentration (and uncertainty) of an EDTA solution of which 5.40(±0.05) mL was used to titrate 100.00(±0.03) mL sample containing 0.786(±0.005) mg primary standard CaCO₃. Consider the molecular weight [100.09 g/mol] of CaCO₃ to have a negligible uncertainty
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Calculate the molar concentration (and uncertainty) of an EDTA solution of which 5.40(±0.05) mL was used to titrate 100.00(±0.03) mL sample containing 0.786(±0.005) mg primary standard CaCO₃. Consider the molecular weight [100.09 g/mol] of CaCO₃ to have a negligible uncertainty
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- For an analysis of phthalates by GC-MS, the limits of detection and quantitation are 0.55 and 3.56 mg/g, respectively. Based on the results obtained below, how are the 3 samples reported? Sample Concentration (mg/g) A 4.50 B 0.055 C 0.563c. A solution containing 2.50 mM X (analyte) and 1.20 mM S (standard) gave peak areas of 2509 and 7132 respectively, in a chromatographic analysis. Then, 1.00 mL of 6.00 mM S was added to 5.00 mL of unknown X and the mixture was diluted to 10.0 mL. This solution gave peak areas of 4016 and 3138 for X and S, respectively. Calculate i. the response factor for the analyte ii. the concentration of S (mM) in the 10.0 mL of mixed solutioniii. the concentration of X (mM) in the 10.0 mL of mixed solutioniv. the concentration of X in the original unknown solutionA sample of anhydrous Na2CO3 (FM = 105.989) is suspected to be contaminated with either NaHCO3 (FM = 84.007) or NaOH (FM = 39.997) To verify the suspicion, a 0.7483 g sample was dissolved to prepare a 50.00 mL solution, and a 10.00 mL aliquot was taken to prepare a 100 mL solution, where 25.00 mL was analyzed using single flask method. If the sample requires 27.50 mL of standard 0.0125 N HCl to reach the phenolphthalein endpoint and another 28.40 mL to reach the bromcresol green endpoint, calculate the percentage composition (in %w/w) of all the basic components in the sample.
- An automatic titration was performed with the following results: Standardized NaOH Solution Concentration: 0.1191 M Volume of HA- product examined in each trial: 25.00 mL Volume of Standardized NaOH Titrant used to achieve the first equivalence point: Trial #1: 1.300 mL Trial #2: 1.137 mL Trial #3: 1.140 mL Calculate the %(w/v) for each trial, Average Concetration (%), Deviation (%), Standard Deviation (%), and RSD (%)A sample of anhydrous Na2CO3 (FM = 105.989) is suspected to be contaminated with eitherNaHCO3 (FM = 84.007) or NaOH (FM = 39.997) To verify the suspicion, a 0.7483 g sample was dissolved to prepare a 50.00 mL solution, and a 10.00 mL aliquot was taken to prepare a 100 mL solution, where 25.00 mL was analyzed using single flask method. If the sample requires 27.50 mL of standard 0.0125 N HCl to reach the phenolphthalein endpoint and another 28.40 mL to reach the bromcresol green endpoint, calculate the percentagecomposition (in %w/w) of all the basic components in the sample.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.
- Concentration EDTA (M): 1.94e-3 Trial #1 Trial #2 Trial #3 Volume of Water Sample titrated (mL): 25.00 25.00 25.00 Volume EDTA used (mL): 36.23 39.00 37.77 Calculate the average of the volume of EDTA used. Calculate the number of moles of EDTA solution required to titrate the water sample. The total hardness is due to one or a combination of Ca2+, Mg2+, and Fe2+ in your sample. It is convenient to express this hardness as though it was entirely due to Ca2+. Making this assumption, determine the number of moles of Ca2+ present in the bottled water sample. The total hardness and the information on Table 1 is always listed in parts-per-million (ppm) of CaCO3 (or mg CaCO3/KgH2O). Since the density of water is 1.0 g/mL, one ppm would be the same as the number of mg of CaCO3 per liter of water. Determine the number of moles of CaCO3 present in the bottled water, assuming all the Ca2+ combine with CO32−. Calculate the number of grams of CaCO3 present and convert to mg.…What is the result of the Analytical Method Validation of 0.1 N H2SO4 Standardization? Given: 50 mL of 1N sulphuric acid is used Added 460 mL water in volumetric flask 20 mL of prepared Sulphuric Acid and placed in volumetric flask Performed Secondary standardization using 0.1 N sodium hydroxide Trials in titration Trial 1 = 19.9 mL Trial 2 = 18 mL Trial 3 = 18.5 mL Trial 4 = 18.5 mL Calculate the concentration of the sulphuric acid and the result of the titration.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.
- 25 ml of pickled brine was taken from an apple pickled with vinegar and diluted to 500 ml, 25 ml of this was taken and acidity was determined with 0.2 N (Factor: 1.05) NaOH. In the main trials, 6.3 ml - 6.5 ml - 6.9 ml, respectively, 0.1 ml base was used in the analysis without using the sample.For example, how many ppm is its acidity?The ethyl acetate (CH3COOC2H5) concentration in an alcoholic solution was determined by diluting a 10.00-mL sample to 100.00 mL. A 20.00-mL aliquot of the diluted solution was refluxed with 40.00 mL of 0.04672 M KOH: ethyl acetate.JPG After cooling, the excess KOH was back-titrated with 3.85 mL of 0.04644 M H2SO4. Calculate the %(w/v) CH3COOC2H5 in the alcoholic solution. MM CH3COOC2H5: 88.11 MM NaOH: 40.00 MM H2SO4: 98.08plot of concentration (Ci , ppm) vs. absorbance (A) gave the following equation for the calibration curve: y = 0.051x + 0.003 ; where y = absorbance and x = concentration Ci. A solution of a 10.0 ppm standard gave the following absorbance for 5 replicate measurements: 0.512 0.551 0.449 0.495 0.507 Answer the following: What is the calibration sensitivity for this analysis: Calibratation sensitivity = ________ (report result to three decimal places with no