A solution using 0.1500 g of CaCO3 unknown is made up in a 100.00 mL volumetric flask. If a 25.00 mL aliquot of this solution was titrated and 35.83 mL of .004373M EDTA was necessary to achieve the endpoint, what is the CaO%?
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A solution using 0.1500 g of CaCO3 unknown is made up in a 100.00 mL volumetric flask. If a 25.00 mL aliquot of this solution was titrated and 35.83 mL of .004373M EDTA was necessary to achieve the endpoint, what is the CaO%?
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- An aqueous solution of 0.2100 g of a mixture containing potassium cyanide and potassium chloride required 14.56 mL of 0.1000 M silver nitrate to produce a faint permanent turbidity; 30.00 mL more of the silver nitrate was added (an excess) and the precipitate of AgCl and AgCN filtered off from the solution. The filtrate and washings were titrated with 13.06 mL 0.1000 M thiocyanate solution. Calculate the percentages of KCl and KCN in the sample.The arsenic in a 1.010 g sample of a pesticide was converted to H3AsO4 by suitable treatment. The acid was then neutralized, and exactly 40.00 mL of 0.06222M AgNO3 was added to precipitate the arsenic quantitatively as Ag3AsO4. The excess Ag in the filtrate and in the washings from theprecipitate was titrated with 10.76 mL of 0.1000MKSCN; the reaction wasAg+ + SCN-→ AgSCN(s) Calculate the percent AsO3 in the sample.The arsenic in a 1.010 g sample of a pesticide was converted to H3AsO4 by suitable treatment. The acid was then neutralized, and exactly 40.00 mL of 0.06222M AgNO3 was added to precipitate the arsenic quantitatively as Ag3AsO4. The excess Ag in the filtrate and in the washings from the precipitate was titrated with 10.76 mL of 0.1000M KSCN; the reaction was Ag+ + SCN-→ AgSCN(s) Calculate the percent AsO3 in the sample.
- If 1.000 ml. of a solution of KMn04 is equiva- lent to 0.1000 millimole of NaCHO2 (sodium formate) in the fol- lowing titration: 3CHO2- + 2MnO- + H2O -> 3CO2 + 2MnO2 + 5OH-, what is the value of the KMnO4 in terms of grams of CaO in the volumetric method for calcium in which that element is precipitated as CaC2 4 .H2O and the precipitate is filtered, dis- solved in dilute H2S04 , and the oxalate titrated with permanganate?A. 2.00 ml sample of hydrogen peroxidesolution required 8.5ml of a permanganatesolution in titration. If each ml of thepermanganate solution is equivalent to0.007295g of Fe, what % w/v of H2O2 was inthe sampleThe arsenic in a 1.010 g sample of a pesticide was converted to H3ASO4 by suitable treatment. The acid was then neutralized, and exactly 40.00 mL of 0.06222M AgNO3 was added to precipitate the arsenic quantitatively as Ag;AsO4. The excess Ag in the filtrate and in the washings from the precipitate was titrated with 10.76 mL of 0.1000MKSCN; the reaction was Ag* + SCN→ AgSCN(s) Calculate the percent AsO3 in the sample.
- A 4.912-g sample of a petroleum product was burnedin a tube furnace, and the SO2produced was collectedin 3% H2O2.Reaction:SO2(g)+H2O2→H2SO4A 25.00-mL portion of 0.00873 M NaOH was introducedinto the solution of H2SO4, following whichthe excess base was back-titrated with 15.17 mL of0.01102 M HCl. Calculate the sulfur concentrationin the sample in parts per million.6mL of 125mM EDTA was added to 244mL of water. The resulting solution was added to 219mL of water and the pH was adjusted to 8.0 using 6mL of NaOH. A further 25mL of water was added. What is the concentration of the final solution (in micromolar)? MW of EDTA is 292.24, MW of NaOH is 39.99The bismuth in 0.7405 g of an alloy was precipitated as BIOCI and separated from the solution by filtration. The washed precipitate was dissolved in nitric acid and treated with 10.00 mL of 0.1498 M AgNO3, causing the precipitation of AgCl. The excess AgNO; required 12.92 mL of 0.1008 M KSCN for titration. Calculate the % Bi in the sample. 6. (Answer : %Bi = 5.54%)
- A sample of Al;(SO.); weighing 7.52-g was dissolved in enough water and diluted to 250.0-mL. 100.0-mL of the dilution was transferred to another flask, 10.0-mL of 0.050M EDTA, 20-ml of buffer solution, 50-mL of alcohol and 2.0-ml dithizone TS were added. The resulting solution required 8.7-mL of 0.031M ZnSO, to reach the endpoint. (Note: Each ml of 0.050M EDTA is equivalent to 16.66-mg Al:(SO.);•18H;0). Compute for the percent purity of the sample.A 48.0048.00 mL aliquot from a 0.4500.450 L solution that contains 0.3700.370 g of MnSO4MnSO4 (MW=151.00MW=151.00 g/mol) required 35.835.8 mL of an EDTAEDTA solution to reach the end point in a titration. What mass, in milligrams, of CaCO3CaCO3 ( MW=100.09 MW=100.09 g/mol) will react with 1.011.01 mL of the EDTAEDTA solution?A salt sample was analyzed for its purity. A 0.5000g sample was dissolved in water and diluted to 250.0mL. From this solution, 25.00mL was taken for analysis. To this aliquot, 20.00mL of a 0.08735M AgNO; solution was added to precipitate the chloride. The excess AgNO required 12.45mL of a 0.09473M KSCN solution for back-titration. Calculate the purity of the salt sample as percent by mass of KCI (74.55 g/mol). (Answer: 84.63% KCI)