QUESTION 2 Titration of a 100.0 mL aliquot of deep well water sample for total hardness requires 28.90 mL of 0.01150 M EDTA. The total hardness of the sample in ppm CaCO3 (MM=100.0875): O 332.6 170.0 166.3 330.0
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- WARIARTAM 2. 1.2589 g of a certain brand of milk powder required 25.89 mL of 0.01873M EDTA to titrate the calcium contained in the sample. (a) What is the percent calcium in the sample? (b) Would the true result be higher, lower, or not be affected if the milk powder sample has been slightly wet?A sample of water from a river was analyzed by titrating a 125 mL aliquot with 0.0210 M EDTA, consuming 22.52 mL. Express the hardness of the water in ppm of CaCO3.The mass percent of Fe in an unknown salt sample was determined using both volumetric and gra vimetric titration. Ten determinations using both methods were performed. Volumetric(% of Fe): 24.22, 23.86, 24.07, 24.49, 24.69, 23.71, 24.28, 24.11, 24.26, 24.10 Gravimetric (% of Fe): 24.35, 24.26, 24.10,24.19, 24.18, 24.32, 24.11, 24.33, 24.17, 24.12 a. Do the two methods give similar standard deviations? b. Do the two methods give similar results for the mass percent of Fe at the 95% confidence level?
- The ethyl acetate concentration in an alcoholic solution was determined by diluting a 10.00-mL sample to 100.00 mL. A 20.00-mL portion of the diluted solution was refluxed with 40.00 mL of 0.04672 M KOH:CH3COOC2H5 + OH- → CH3COO- + C2H5OHAfter cooling, the excess OH2 was back-titrated with 3.41 mL of 0.05042 M H2SO4. Calculate theamount of ethyl acetate (88.11 g/mol) in the original sample in gramsM 23 (H) Inorganic Analytical Chemistry pard My courses CHEM 23 (H) FS-AY:20-21 Week 4-5: Quantitative Analysis by Gravimetry and Titration Quiz 4 Qu 10 The 500.0-mg sample of impure Na2 CO3 (FM: 105.96) required 22.00 mL of the HCI standard solution (the obtained concentration of HCI from the previous question, problem 5: Titrimetry short answer type) for complete neutralization. ut of 1. Give the balanced chemical equation between the sample, Na2CO3 and the standard, HCI. 2. What is the mass (g) of Na, CO3 in the sample? 3. Calculate the % (w/w) purity of a 500.0-mg sample of impure Na2 CO3 Note: Type your solution here or upload its pic in Jpeg or pdf format. Another option is to send it to me via email (subject: Fam name-Quiz 4) or messenger, privately. Do not forget to box the final answer and write your name. В I 1.A 22.41 mL of NaOH solution is required to reach the phenolphthalein endpoint in the standardization with 0.276 g of solid, monoprotic primary standard acid (MW 128.33) in a 25.00 mL solution. 12.34 mLof the same NaOH solution was added to titrate 5 mL of vinegar solution to the endpoint. Calculate the mass of acetic acid in this vinegar solution (d = 1.005 g/mL).
- A 0.1093-g sample of impure Na2CO3 was analyzed by the Volhard Method. After adding 50.00 mL of 0.06911 M AgNO3, the sample was back titrated with 0.05781 M KSCN, requiring 27.36 mL to reach the end point. Report the purity of Na2CO3sample.If 5.85 mL of 0.00832 M EDTA solution were needed to titrate a 0.0225 L water sample to an Eriochrome Black T endpoint, then what was the concentration of Ca2+ ions (in M) in the sample? volume of EDTA needed for blank titration: 2.18 mL Ca2“(aq) + EDTA“(aq) CAEDTA²-(aq)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?..
- You took 25.00 mL of unknown water solution and titrated it with EDTA. The EDTA solution molarity was 0.01 M, and it took 16.50 mL to reach the end point. With the blank it took only 0.98 mL to reach end point. 15.52 mL of EDTA solution were used to titrate hardness that actually came from the unknown 1.552×10−4 moles of EDTA reacted with hardness-causing ions from the unknown sample 1.552×10−4 moles of hardness-causing ions were present in the unknown sample A) Assuming that the total hardness of water is due to CaCO3, how many grams CaCO3 does it correspond to? B) What is the total hardness of the unknown water in ppm CaCO3?i have hard time geting it stright even after I do the experiment can u help me with the one pls Material ChemicalsSealed NO2 tubes, KSCN solution (0.002 M), acidified Fe(NO3)3 solution (0.2 M), KSCN crystals, Fe(NO3)3 crystals, KOH (concentrated (8 M)), distilled waterApparatus:Bunsen burner, ring stand, wire gauze, matches, thermometer, dropper, test tubes, glass stirring rod, beaker tongs Procedure Part A: Fill a 600 mL beaker to the 3/4 mark with water. Bring to a boil using a Bunsen burner. In another 600 mL beaker, fill to the 1/2 mark with water; add ice in order to lower the temperature to about 0°C. Note the room temperature. Observe the NO2 thoroughly at room temperature. Place the sealed NO2 tube in the cold bath. Note your observations. Remove the test tube from the cold bath. Let the test tube stand until it reaches room temperature. Note your observations. Place the test tube in the hot bath. Note your observations. Remove the test tube from the hot bath. Let the test…Precipitimetry The chloride in a 4.321-g food sample was precipitated through the addition of 50.00 mL of a standard AgNO3 solution (10.00 mL AgNO3 = 11.22 mL KSCN). The precipitate was coated with nitrobenzene and the mixture was diluted to 250.0 mL. A 50.00 mL aliquot was taken from the diluted solution and required 4.56 mL back titration with a %3D standard KSCN solution (22.33 mL KSCN = 0.9758 g AgNO3). %3D Formula Masses: AgNO3 = 169.87; CI = 35.45 Calculate the following: 1. Molar concentration of KSCN solution M %3D 2. Molar concentration of AgNO3 solution M %3D 3. % (w/w) chloride in the original sample =