A 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.

A 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.

Fundamentals Of Analytical Chemistry

9th Edition

ISBN:9781285640686

Author:Skoog

Publisher:Skoog

Chapter16: Applications Of Neutralization Titrations

Section: Chapter Questions

Problem 16.18QAP

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A sample of anhydrous NaHCO3 (FM = 84.007) is suspected to be contaminated with either NaOH (FM = 39.997) or Na2CO3 (FM = 105.989). 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 double flask method. If the sample requires 0.45 mL of standard 0.0125 N HCl to reach the phenolphthalein endpoint and 35.95 mL to reach the bromcresol green endpoint, calculate the percentage composition (in %w/w) of all the basic components in the sample.
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 sample of anhydrous NaHCO3 (FM = 84.007) is suspected to be contaminated with eitherNaOH (FM = 39.997) or Na2CO3 (FM = 105.989). 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 double flask method. If the sample requires 0.45 mL of standard 0.0125 N HCl to reach the phenolphthalein endpoint and 35.95 mL to reach the bromcresol green endpoint, calculate the percentage composition (in %w/w) of all the basic components in the sample.
If an instrument gives a response of 1240 for a standard containing 8 ppm of a substance, how much if this substance is in a sample that gives a response of 1705? Are any assumptions needed?
If exactly 200.0 mL of an aqueous solution that is 1.037E3 ppb Pb(NO3)2 (331.2 g/mol) is mixed with 100.0 mL of an aqueous solution that is 3.090E9 fM KI (166.00 g/mol), how many micrograms of solid PbI2 (461.01 g/mol, 6.16 g/mL) are formed? Assume complete reaction.
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In another chromatography experiment, pure chemical “A” leaves a yellow spot with an Rf = 0.49 and pure chemical “B” leaves a red spot with an Rf = 0 .68. An unknown mixture when chromatographed is found to contain two components, a blue spot (Rf = 0.49) and a yellow spot (Rf = 0.66). Does the unknown mixture contain any of the known pure chemicals’ “A” or “B”?
5.00 mL of stock solution is diluted to 25.00 mL, producing solution ALPHA. 10.00 mL of solution ALPHA is diluted to 25.00 mL, resulting in solution BETA. 10.00 mL of solution BETA is then diluted to 25.00 mL, producing solution GAMMA. dilution factor for ALPHA from stock solution = 0.167 dilution factor for BETA from ALPHA solution = 0.0476 part c and d?
(a) During the analysis of water sample by argentometric titration, results obtained are as follows: Experiment1 2 3 4 Volume 15.5, 15.2 ,15.1 ,15.4 A)Calculate average deviation and relative average deviation for the given data. (b)Calculate the molecular weight of an unknown acid if 8.5 g of it is dissolved in 200.0 ml of water and requires 50.0 ml of 1.5 M sodium hydroxide for complete neutralization .
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1. What is the MW of the analyte?2. What is the meq of the analyte?3. What will be the weight (in grams) of the analyte if the standardization is to be quartered?4. What will be Normality of the solution if 7.5ml of the titrant was consumed in the standardization?5. What is the average Normality if two or more trials were conducted with the values of 0.102N and 0.105N?