Consider thé following statement: Statement I: Matrix is equal to blank plus the analyte. Statement II: In 250-mL water sample, 50-mL is subjected for analysis. The 250-mL is the aliquot. Statement IlI: When the technical grade chemical is used to prepare solution, a primary standard is used to standardized and determine the actual concentration of solution. Which of the following option is correct?
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- A Medical Technology student was given a capsule of a multivitamins and she was asked to determine the % by mass (w/w) of ascorbic acid present in the capsule. The student analyzed the 2.001 g sample using volumetric titration. The following data was generated in the analysis: KIO3 + 5KI + 6H+ → 3I2 + 6K+ + 3H2O C6H8O6 + I2 → C6H6O6 + 2I- + 2H+ Table 1. Standardization of KIO3 Molarity of Ascorbic Acid Standard Solution 0.03542 M Volume of Ascorbic Acid 25.00 mL Volume of KIO3 8.70 mL Molarity of KIO3 ______________M Table 2. Determination of Ascorbic Acid Concnetration Initial burette reading, KIO3 0.00 mL Final burette reading, KIO3 33.60 mL Volume consumed, KIO3 33.60 mL MM of Ascorbic Acid 176.12 g/mole choices 7.30% 30.1% 33.6% 32.5%00 mL of a diprotic acid primary standard solution was accurately prepared to a concentration of 0.1431 M. Three samples of this primary standard solution were used as samples in a titration to standardize an aqueous solution of sodium hydroxide, NaOH, which would be used as a titrant. Using the following table of data for the titration of the primary standard acid with NaOH, calculate the average concentration of NaOH. Trial # Volume of primary standard Initial titrant volume Final titrant volume 1 10.00 mL 8.21 mL 27.22 mL 2 10.00 mL 27.22 mL 46.23 mL 3 10.00 mL 30.28 mL 49.29 mL 0.1506 M 0.0753 M 0.0376 M 0.1431 M 0.0526 MPS. Further values required for the solvings are give in the various situations below. (ANSWER) Situation: A community in a mountainous area of Bohol uses water collected from a nearby natural spring. A sample was submitted to a laboratory for the analysis of its total hardness. Required: SHOW YOUR COMPLETE CALCULATIONS. BASED ON THE IMAGE PROVIDED BELOW FOR THIS QUESTION: Calculate the amount of titrant used in each trial to reach endpoint. Report total hardness of the sample as mean ±sd. a. 250.0 mL of 500.0 ppm of CaCO3 solution from a primary standard (assume solvent is distilled water only). Answer : Mass of CaCO3 = 0.125 g b, The EDTA solution was standardized by titrating it with a 25.0 mL aliquot of the CaCO3 solution. How much of the titrant was consumed. Answer: Volume of EDTA consumed = 12.405 g c. Calculate the average titer (mg CaCO3/mL EDTA). Mass of CaCO3 in 25 mL CaCO3 solution: 0.0125 g Answer: 1.008 mg CaCO3/ mL EDTA
- Sources of Error Determine the relationship between the observed/apparent value (EX) VERSUS that of the true value (ET) for the quantity being sought by writing either <, >, or = on the space provided TOPIC: Measured mass of the precipitate 1. Filter paper was dried prior to filtration. EX _____ ET TOPIC: Standardization of Titrant 2. Distilled water was not equilibrated to room temperature before the preparation of NaOH titrant. EX ______ ET TOPIC: Determination of Molar Concentration of each component (Double Indicator Titration) 3. No blank correction EX ______ ETCalculate the gravimetric factor of the following. 2 Fe3O4 is sought(Analyte), 3 Fe2O3 is weighed (Precipitate) 2 Fe is sought(Analyte), 1Fe2O3 is weighed (Precipitate)Calculate the gravimetric factor of the following. 2 Fe3O4 is sought(Analyte), 3 Fe2O3 is weighed (Precipitate)
- Mr. Clean recently bought a laboratory-grade sodium carbonate from a chemical company known as Brand X. He was supposed to use it in the production of detergents. Unfortunately, he was scammed by the company. He suspected that he purchased a crude sodium carbonate so he tasked the Quality Assurance Department to determine the components of the purchased chemical. The chemist assigned to analyze the sample used double indicator method. For the standardization of HCl titrant, 0.1025 g Na2CO3 of 99.5% purity (FW: 106.00) required 8.20 mL of the titrant to reach the phenolphthalein endpoint. FW: NaOH (40.00), NaHCO3 (84.01), Na2CO3 (106.00) a. What is the molarity of the titrant? The chemist obtained a 3.150 g sample and dissolved it in distilled water to produce a 50.0 mL solution. An aliquot of 10.00 mL was obtained and diluted in a 100.0 mL volumetric flask. A 50.00-mL aliquot of the diluted sample was taken and it required 25.70 mL of titrant for the methyl orange endpoint, while…If 500mL of 0.10M Ca2+ is mixed with 500mL of 0.10M SO42-, what mass of calcium sulfate will precipitate? Ksp for CaSO4 is 2.40 • 10-5. express answer to 3 significant figures. And include units.Aspirin powder = 0.8110g MW of Aspirin = 180g.mol-1 Volume of 0.5N HCl consumed in back titration = 23.50mL Volume of 0.5N HCl consumed in blank titration = 44.50mL Percent purity (USP/NF) = Aspirin tablets contain NLT 90.0% and NMT 110.0% of the labeled amount of aspirin (C9H8O4) What is the calculated weight (in grams) of pure aspirin?..
- Sources of Error Determine the relationship between the observed/apparent value (EX) VERSUS that of the true value (ET) for the quantity being sought by writing either <, >, or = on the space provided TOPIC: Measured mass of the precipitate Question 4 Overignition which causes the conversion of BaSO4 precipitate to BaO. EX ___ ET Question 5 Precipitate was not washed thoroughly. Ex ___ ET TOPIC: Standardization of Titrant Question 6 Adding insufficient amount of titrant to reach acceptable endpoint color: EX ___ ETHow much amount (in grams) do you need to prepare the following solutions 500.0 mL 0.1000 M stock EDTA solution from Na2H2EDTA•2H2O (FW=372.24) and MgCl2•6H2O crystals 100.0 mL 0.0500 M stock Ca2+ solution from pure CaCO3 (FW=100.09) and concentrated HCl 250 mL 1.0 M NH3-NH4+ pH 10 buffer solution from NH4Cl and NH3A 500ml solution of NaOH was made using 2g of NaOH(s) Three trials of titration were made with using KHP(s) as the acid dissolved with about 25ml of deionized water and 4 drops of phenolphthalein indicator. Slowly adding the NaOH solution until the clear solution had turned pink which would give us our end point and allow us to find the NaOH molarity by equivalence point. Trial 1: - 0.484g of KHP were used - initial volume of buret containing NaOH solution was 0.0 ml - final volume of buret was 23.60ml Trial 2: - 0.485g of KHP were used - initial volume was 0.0ml - final volume was 24.00ml Trial 3: - 0.486g of KHP - initial volume was 0.0ml - final volume was 23.80ml The molarity of NaOH was found by using the moles of KHP(as at equivalence, both solutions are balanced in moles) divided by the total volume of NaOH used to neutralize the solution. Giving us 0.100M for trial 1, 0.099M for trial 2, and 0.100M for trial 3. Make a rough sketch of a titration curve that…