The standardization of same titrant during the determination of BOD5 of water sample was done separately for Day 1 and Day 5. Upon standardization, it turned out that the concentration of the titrant in Day 1 is higher than the concentration of the same batch of titrant in Day 5. How would you explain this discrepancy? A There were bubbles at the tip of the burette during titration. (B) Too much starch indicator was added in the conical flask. Sulfuric acid was added first before potassium iodide. D) The water used to prepare the Na₂S₂O3 solution was not boiled resulting to proliferation of bacteria.

The standardization of same titrant during the determination of BOD5 of water sample was done separately for Day 1 and Day 5. Upon standardization, it turned out that the concentration of the titrant in Day 1 is higher than the concentration of the same batch of titrant in Day 5. How would you explain this discrepancy? A There were bubbles at the tip of the burette during titration. (B) Too much starch indicator was added in the conical flask. Sulfuric acid was added first before potassium iodide. D) The water used to prepare the Na₂S₂O3 solution was not boiled resulting to proliferation of bacteria.

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...

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10-27. A 10.231-g sample of window cleaner containing ammonia was diluted with 39.466 g of water. Then 4.373 g of solution were titrated with 14.22 mL of 0.106 3 M HCl to reach a bromocresol green end point. (a) What fraction of the 10.231-g sample of window cleaner is contained in the 4.373 g that were analyzed? (b) How many grams of NH3 (FM 17.031) were in the 4.373-g sample? (c) Find the weight percent of NH3 in the cleaner.
The sample is dissolved in a mixture containing 10ml of distilled water and 20 ml of dilute HCL.After dilution, add 50 ml of water and 1g of potassium bromide.The mixture is titrated with 0.1N NaNO2 solution . 4 drops of atropine are used as an indicator .At the last point of the titration the color change from red to yellow. Write the reaction that takes place in the nitrometer and the sulfatiazole sample indicator?
The concentration of ammonia in a cleaning product was determined by back titration.Firstly, 10.00 cm3 of the cleaning product was pipetted into a large conical flask,containing 250.00cm3 of 0.50 mol/l HCl to give Solution A.Following a period of reaction and shaking, 50.00cm3 of Solution A was removed anddiluted to 250 cm3 with water in a volumetric flask to give Solution B.20 cm3 samples of Solution B were titrated against 0.05 mol/l Na2CO3 solution, givingan average titre of 12.45 cm3. i) Write equations for the reactions that have taken place.ii) Determine the concentration of NH3 in the original cleaning product in mol/l,g/l, ppm, and % w/v.
In a Complexometric Titration Experiment, the preparation of the EDTA solution is as follows: In a 250 mL beaker, weigh 0.10 g MgCl2*6H2O and 1.0 g NaOH. Add approximately 200 mL distilled water and stir to dissolve. Add about 2.0 g of EDTA and transfer the solution quantitatively to a 1-L volumetric flask. Dilute to the mark with distilled water. What is the purpose of adding MgCl2*6H2O? Please show COMPLETE solution.
25.0 cm3 of sodium hydroxide solution is tritrated against 0.2 mol dm–3 hydrochloric acid until the indicator just changes colour. 27.5 cm3 of the acid are required to reach the end point.Phenolphthalein is used as an indicator in the above experiment. What will be its colour change at the end point? From pink to yellow From pink to colourless From yellow to colourless From yellow to pink
A typical protein contains 16.2 wt% nitrogen. A 0.500-mL aliquot of protein solution was digested, and the liberated NH3 was distilled into 10.00 mL of 0.021 40 M HCl. Unreacted HCl required 3.26 mL of 0.019 8 M NaOH for complete titration. Find the concentration of protein (mg protein/mL) in the original sample.
A mixture containing only KCl and NaBr is analyzed by the Mohr method. A 0.2125-g sample is dissolved in 25 mL of water and titrated to the Ag2CrO4 endpoint, requiring 25.63 mL of 0.1110 M AgNO3. A blank titration requires 1.15 mL of titrant to reach the same end point. Report the %w/w KCl and NaBr in the sample. KCl = 74.551 NaBr = 102.89
A mixture containing only KCl and NaBr is analyzed by the Mohr method. A 0.2125- g sample is dissolved in 25 mL of water and titrated to the Ag2CrO4 end point, requiring 25.63 mL of 0.1110 M AgNO3. A blank titration requires 1.15 mL of titrant to reach the same end point. Report the %w/w KCl and NaBr in the sample.KCl = 74.551 NaBr = 102.89
A 0.1475-M solution of Ba(OH)2 was used to titrate the acetic acid (60.05 g/mol) in a dilute aqueous solution. The following results were obtained. (See attached image)(a) Calculate the mean w/v percentage of acetic acid in the sample.(b) Calculate the standard deviation for the results. (Take note if n-1 should be used)(c) Calculate the 90% confidence interval for the mean.(d) (d) At the 90% confidence level, could any of the results be discarded?
5. A 300.0 mg sample containing Na,CO3, NaHCO3 and NaOH and inert material either alone or in some combination was dissolved and titrated with 0.1000 M HCI the titration required 24.41 mL to reach the phenolphthalein endpoint. And an additional 8.67 mL to reach the methyl red endpoint. Determine the composition of the sample and calculate the percent of each titrated component.
A procedure7 for determining halogens in organic compounds uses an argentometric titration. To 50 mL50 mL of anhydrous ether is added a carefully weighed sample (10−100 mg)(10−100 mg) of unknown, plus 2 mL2 mL of sodium dispersion and 1 mL1 mL of methanol. (Sodium dispersion is finely divided solid sodium suspended in oil. With methanol, it makes sodium methoxide, CH3O−Na+,CH3O−Na+, which attacks the organic compound, liberating halides.) Excess sodium is destroyed by slowly adding 2‑propanol, followed by 100 mL100 mL of water. (Sodium should not be treated directly with water because H2H2 can explode in the presence of O2O2: 2Na+2H2O⟶2NaOH+H2.)2Na+2H2O⟶2NaOH+H2.) This procedure gives a two‑phase mixture, with an ether layer floating on top of the aqueous layer that contains the halide salts. The aqueous layer is adjusted to pH 44 and titrated with Ag+,Ag+, using the electrodes in Figure 7‑5.How much 0.02916 M AgNO30.02916 M AgNO3 solution will be required to reach each equivalence…
Why is the use of high-precision volumetric material essential for titration?