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?
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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?
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The water used to prepare the Na2S2O3 solution was not boiled resulting to proliferation of bacteria.
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Too much starch indicator was added in the conical flask.
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Sulfuric acid was added first before potassium iodide.
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There were bubbles at the tip of the burette during titration.
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- In a similar experiment, the mass of 3 PB tablets was measured to be 3.1823 g. The tablets were crushed, and 2.9801 g of the powder was transferred to a beaker and reacted with HCl. After filtration, the filtrate was transferred to a 50-mL volumetric flask and diluted with water. 10.00 mL of this stock solution was combined with 0.2 M Na3PO4. The resulting precipitate weighed 0.2425 g after drying. What is the mass of BSS per tablet in mg?Round your answer to 1 decimal place.1. A sample of an infusion was diluted 10 ml to 250 ml and then 10 ml to 200 ml. It was then analysed and was found to contain sodium at 0.789 mg/100 ml. Calculate the concentration of sodium in the original sample in %w/v. The sample was composed of a mixture of sodium lactate and sodium carbonate in equimolar amounts. Calculate the amount of sodium lactate and sodium carbonate in mg/10 ml of the sample (Na = 23, lactate = 89, carbonate = 60) 2. 0.641 g of a semi-synthetic alkaloid was dissolved in 25 ml of 1% w/v acetic acid and was analysed directly by HPLC. The solution was found to contain 1.42 mg/100 ml of an impurity. What is the level of impurity in % w/w and ppm? 3. Calculate the pH of a buffer system made by dissolving 1.2 g of acetic acid and 0.82 g of sodium acetate in 500 ml of distilled water (pKa of acetic acid = 4.7) 4. Convert the following concentrations into the required expression. Concentration given Expression required 0.5% NaCl molarity 1 mM of KBr…A piece of Gold weighing 12,359 Kg is suspected of being contaminated with Iron. To perform an instrumental analysis and To confirm whether or not it contains Fe, a portion of the sample (0.954 g) is taken from the piece and dissolved with 25 mL of aqua regia. Heats up For its complete dissolution, it is cooled and made up to 100 mL. A 10 mL aliquot is taken from this solution and made up to 50 mL. From This last solution is given the appropriate treatment to visualize Fe+2, for which the 1,10-phenanthroline reagent is added. (it forms a complex that is red in color) and is taken to a visible spectrophotometer and with a 12 mm cell a absorbance of 0.45. Previously, a calibration curve of Fe+2 was obtained under the same instrumental conditions obtaining the following data: (view table) Calculate the purity of the gold piece, assuming impurities only due to Fe.
- 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.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 = 102.891.50 mL of 0.15M Fe2(SO4)3 is diluted with deionized water to 25.00 mL in a volumetric flask. What is the concentration (in mg/mL) of iron (not Fe2(SO4)3) in the diluted sample? Please note that the units in this question are different than presented in the lab manual to avoid issues with exponents in a numerical answer. Of course if the units were in micrograms/g the results would be * 103 or (1000 times bigger).
- Q3/ A 0.80868 gm sample of a commercial phosphate detergent was ignited at a red heat to destroy the organic matter the residue was then taken up in hot HCl, which converted the P to H3PO4. The phosphate was precipitated as MgNH₂PO4.6H₂O by addition of Mg¹2 followed by aqueous NH3 after being filtered and washed. The precipitate was converted to Mg₂P₂O(222.57gm mol) by ignition at 1000 C this residue weighed represent 61.5% of the total sample mass. Calculate the percent P(30.974gm mol),O (16 gm/mol) and Mg (24.3 gm/mol) in the sample.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.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
- Use the following experimental titration data to calculate the concentration of the acid being analysed. Observations: The initial solution of acetic acid (HC2H3O2) is clear and colourless. A few drops of phenolphthalein indicator are added to each sample. A dilute solution of sodium hydroxide (concentration 2.50 x 10-4 mol/L) is used as the titrant. As the mixture reaches the endpoint, flashes of pink colour are seen and the titrant is added drop by drop. The endpoint is reached when one drop of titrant turns the mixture a pale pink colour that does not fade.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?up the 1. In a reaction involving the iodination of acetone, the following volumes were used to make reaction mixture: 10 mL 4.0 M acetone + 10mL 1.0 M HCl + 10 mL 0.0050 M 1₂ + 20 mL H₂O a. How many moles of acetone were in the reaction mixture? Recall that, for a component A, no. moles A = MA X V, where MA is the molarity of A and V is the volume in liters of the solution of A that was used. b. What was the molarity of acetone in the reaction mixture? The volume of the mixture was 50 mL, 0.050 L, and the number of moles of acetone was found in Part a. Again, MA = moles acetone no. moles A V of soln. in liters M acetone