Question 5: A 200.00-mL sample containing possible mixture of NaOH, Na2CO3, and/or NaHCO3 is diluted to form 250.00 mL solution. 50.00 mL aliquot of this solution is titrated with 0.2011 M HC1, requiring 22.20 mL to reach the phenolphthalein endpoint. A second 50.00 mL aliquot requires 49.80 mL of the titrant to reach the methyl orange endpoint. Determine the composition and the concentration of the solute in the original solution.

Question 5: A 200.00-mL sample containing possible mixture of NaOH, Na2CO3, and/or NaHCO3 is diluted to form 250.00 mL solution. 50.00 mL aliquot of this solution is titrated with 0.2011 M HC1, requiring 22.20 mL to reach the phenolphthalein endpoint. A second 50.00 mL aliquot requires 49.80 mL of the titrant to reach the methyl orange endpoint. Determine the composition and the concentration of the solute in the original solution.

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

See similar textbooks

Similar questions

An unknown sample of Cu2+gave an absorbance of 0.262. Then 1.00 mL of solution containing 100 ppm (ug/mL)Cu2+was mixed with 95.0 mL of the unknown and the mixture was diluted to 100 mL in a volumetric flask with deionized water. The absorbanceof the new solution was 0.500. a.Denoting the initial unknown concentrationas [Cu2+]i,write an expression for the final concentrationafter dilution [Cu2+] b.Find [Cu2+]in the unknown.
0.1500 g of an unknown sample was passed through a column of Amberlite IR-120(+) resin and titrated with 0.2050 M NaOH. If 9.75 mL of titrant was required to reach phenolphthalein endpoint, what is the molecular weight of the unknown substance?
A 0.3005g sample containing Na2CO3 and NaOH requires 11.5 mL of 0.1055 M HCl to reach the phenolphthalein end point. Methyl orange is then added and 7.9 mL more of 0.1055 M HCl is required to reach the end point. Calculate the % composition of Na2CO3 and NaOH in the sample. a. 5.15% NaOH and 29.55% Na2CO3 b. 5.05% NaOH and 29.55% Na2CO3 c. 5.15% NaOH and 29.40% Na2CO3 d. 5.05% NaOH and 29.40% Na2CO3
A 0.500-g sample containing BaBr2 (FM = 297.14) is dissolved in 50 mL of distilled water. Titrating with 0.0400 M AgNO3 requires 25.00 mL to reach the Mohr endpoint. A blank titration requires 0.50 mL to reach the same endpoint. Calculate the %w/w BaBr2 in the sample. Please provide a complete solution.
mix 500 ml of 0.1 M HCl solution with 300 ml of 2 M Ba (OH) 2 solution a) indicate the environment of the solution after mixing b) What color will the detectors, phenophthalein and litmus get?
Compound X was dissolved in a 5-mL volumetric flask. A 1.00-mL aliquot of the compound was placed in a 10-mL volumetric flask and diluted to the mark. The solution was read in a UV-Vis spectrophotometer and an absorbance of 0.427 in a 1.000-cm cuvette was observed at 340 nm. Solve the following question with complete solution Data for Compound X: Molecular mass = 292.16 g/mol Molar absorptivity (ɛ) at 340 nm= 6130 M -1 cm -1 1. How many milligrams of compound X were used to make the 5-mL solution?
To which sample solution the spike will be add? the distillate>130C the mass standard the distilled <130C
Five white, 500-mg uncoated ascorbic acid (AA) tablets with an average weight of 0.6152-g were pulverized in a mortar. A sample of the powdered ascorbic acid weighing 0.4700-g was placed in an iodine flask and was dissolved in 50-mL H2SO4 then 5-g of KBr was added to the resulting solution. The solution was titrated with 47.81-mL of 0.09640 N STD. KBrO3 to reach a faint yellow endpoint then 3-g KI and 5-mL Starch TS. The blue color solution is then titrated with 2.73-mL of 0.09123 N STD. Na2S2O3 to reach the disappearance of the blue iodostarch complex. MW: KBrO3 = 167.0 ; KIO3 = 214.0 ; Na2S2O3 = 158.11 ; C6H8O6 = 176.12 Compute the milligrams of pure AA per tablet from the assay. 293.3 mg 502.5 mg None of the choices 383.9 mg
From a 10-mL sample, a 1-mL aliquot was taken and diluted to 100mL. From this, a 5-mL aliquot was taken and diluted to 20mL.The final 20mL was found to have a concentration of 0.004M analyte X.What is the concentration of analyte X in the 10-mL sample?
Titration of a 0.824 g of 99.99% KHP (204.23 g/mol) with phenolphthalein required 18.3 mL of NaOH (40.00 g/mol) solution to reach the end point. The same titrant was used to analyze an impure acetic acid (CH3COOH, 60.06 g/mol) solution. A 10.0 mL aliquot of the sample required 12.9 mL of the titrant. What substance served as the primary standard?Which served as the indicator in the titration?What is the color at the end point?What is the concentration of the titrant?What is the molar concentration of the acetic acid solution?
A Na2S2O3 titrant was standardized by taking 50.0 mL of 6.7863 x 10-3 M standard KIO3 solution, acidifying, and adding KI crystals. Titration of this solution required 9.80mL to reach the starch endpoint. What is the concentration of the Na2S2O3 titrant?
A 0.951-g sample containing (NH4)2C2O4 and compounds was dissolved in water and made alkaline with KOH. The liberated ammonia was distilled into 50.00 mL of 0.1007 N H2SO4. The excess H2SO4 was back titrated with 11.13 mL of 0.1214 N NaOH. Calculate the percentage of nitrogen in the sample.