Table 1. Data on EDTA Standardization Weight of CaCO3 (g): 0.2003g Trial Volume of EDTA (mL) Molarity of EDTA (M) 1 8.60 2 8.50 3 8.55 Average Molarity of EDTA (M) Table 2. Data on Sample Analysis Volume of water sample (mL) : 10.00mL Trial Volume of EDTA (mL) ppm CaCO3 1 0.19 0.16 3 0.19 Average ppm CaCO3
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- DATA AND RESULTS: Titration Data (Standardization of EDTA) Example II Mass, g Caco, 0.0250 0.0251 0.0246 0.0254 Final reading, mL 25.00 24.91 24.87 25.31 Initial Reading, mL 0.00 0.02 0.02 0.02 Volume, mL EDTA 25.00 Molarity of EDTA, M 0.009992 B. Analysis Data Example II Volume of Sample, mL 25.00 24.90 25.10 25.0 Final reading, mL 27.46 25.08 25.06 24.99 Initial Reading, mL 0.00 0.02 0.02 0.02 Volume of EDTA, mL 27.46 Molarity of EDTA, M 0.009992 Ca" concentration, ppm 440 STATISTICAL EVALUATION OF RESULTS Standardization of EDTA Mean, M EDTA Median, M EDTA Standard Deviation Ci for u at 95% Prob. Level Reported value: M EDTA Ca" Concentration Mean, Ca" ppm Median, Ca" ppm Standard deviation Cl for u at 95% Prob. Level Reported value: Ca" ppmM with a volume of 0.05 Na2G03 if it is molarity of Q1 25.00 cm 3 while the acid titre value is 22.33 em 3, calculate the concentrationHGL is HGL in mol / dm.3. The eguation - a2GO3 reaction is (ag) +) (ag 2HGI 2NAGL (ag) + H2O (1) + GO2 (g) Na: 2.3. Gl = 35,5, G = 12.0 = 16. H = 1 * %3D %3D %3Dthe solution. 2. A standard solution of EDTA is prepared, and by titration each ml is found to complex with the Mg in 10.0ml of a solution containing 0.300g of MgCl₂ per liter. A 100 ml pipetful of a certain well water is found to require 8.60 ml of the standard EDTA. (With the knowledge that the conventional method of expressing water hardness in terms of ppm of CaCO3 regardless of the nature of the Carbons and anions actually present, what is the hardness of the well water?
- A 50.00 mL solution containing NaBr was treated with excessive AgNO3 to precipitate 0.214 6g of AgBr (FM 187.772). What was the Molarity of NaBr in the solution?A 10.00-mL solution containing Cl2 was treated with excess AgNO3 to precipitate 0.436 8 g of AgCl (FM 143.321). What was the molarity of Cl2 in the unknown?The sulfur content of an iron pyrite ore sample is determined by converting it to H2S gas, absorbing the H2S in 10.0mL of 0.00500 M I2, and then back-titrating the excess 12 with 0.00200 M Na2S203. If 2.6mL Na2S20zis required for the titration, how many milligrams of sulfur are contained in the sample? (Mwt of H2S is 34.08g/mol). Reactions: H2S + I2 -------> S+21 + 2H* I2+ 2S20,2- ---------> 21- + S,0,2- I () II !!
- Concentration of EDTA solution is 1.006 mg CaCO3/mL 10mL of unknown diluted with 40mL of distilled water 5mL of buffer and 5 drops of indicator what is the mg, ppm, and molarity for both trials?? Unknown titration with EDTA Trial 1 Trial 2 Volume of titrated 10mL 10mL Final Buret volume 4.1 4.5 Initial Buret Volume 25 25 Volume of EDTA solution used 20.9 20.5 mg of CaCO3 in sample ppm CaCO3 in sample Molarity of Ca2+1.5m 6A 200. mL solution that is 6.7 x 104 M in Cu²+ is mixed with 0.30 mole NH3. There is no change in volume as NH3 is bubbled in. What is the final [Cu]? Kf Cu(NH3)42 = 5.6 x 10¹¹ 124 . Kover= ksp kf 5." -9 Cu2++NH₂ CU(NH3) 4² 2+Example /A0.527g sample of mixture containing NazCO; and NaHCO; and inert impurities phenolphthalein endpoint and a total of 43.8mL to reach the modified methyl orange endpoint, what is the percent each of Na CO3 and NaHCO; in mixture? is titrated with 0.109M HCІ, requiring 15.7mL to reach the Solution:
- a primary standard of (COOK)2.H2O was prepared up to 0.05M. This primary standard was then titred against an unknown concentration of 50cm3 of KMnO4. An average titre value of 13 was obtained. The balanced equation is as follows: 5(COOK)2.H2O + 2KMnO4 + 8H2SO4 = 6K2SO4 +2MnSO4 + 10CO2 +18H2O. Find the concentration of the KMnO4.A 320.00 mL solution of 0.00290 M AB2 is added to a 260.00 mL solution of 0.00235 M C3D2. What is pQsp for A3D2?Experiment 28 Advance Study Assignment: Determination of the Hardness of Water 1. A 0.3946 g sample of CaCO, is dissolved in 12 M HCI and the resulting solution is diluted to 250.0 mL in a volumetric flask. a. How many moles of CACO, are used (formula mass = 100.1)? moles b. What is the molarity of the Ca²* in the 250 mL of solution? M c. How many moles of Ca" are in a 25.00-mL aliquot of the solution in Ib? moles .. moles 아 Cacos 드 Weight in gram formula mass 0-3946 1001 E3.942 x 10-3 mol moles of solute Molanity Volume of solution (L) 3.942 x 1o-3 x 1000 250 0.0158 M Imol Caco, produce I mol Cast ion 0.0158 M Moles off Ca²+ = Molarity x 0.0158 Vo lume lin L) 25 1000 3.94 X10-4mol.