A 50.00-mL aliquot of solution containing 0.358 of MgSO4 (FM 120.37) in 0.500 L required 38.08 mL of EDTA solution for titration. How many milligrams of CaCO3 (FM 100.09) will react with 1.00 mL of this EDTA solution?
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A 50.00-mL aliquot of solution containing 0.358 of MgSO4 (FM 120.37) in 0.500 L required 38.08 mL of EDTA solution for titration. How many milligrams of CaCO3 (FM 100.09) will react with 1.00 mL of this EDTA solution?
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- Sketch a photometric titration curve for the titration of Sn2+ with MnO4 . What color radiation should be used for this titration? Explain.A50.0 mL aliquot of solution contairing 0.450 g of MgSO4 [FM 120.27) in 0.500 L required 376 mL of EDTA solution for titration. How many miligrams of CaCO3 (FM 106.09) will react with 100 mL of this EDTA solution?A 50.00 mL sample of hard water is titrated with 0.0102 M EDTA solution (volume of blank titration= 1.5 mL) and the volume of EDTA required is 22.44mL. What is the water hardness in ppm CaCO3?
- Standardization of EDTA was done using MgSO4 standard wherein a 50-mL aliquot of solution obtained from 0.480 g MgSO4 in 500 mL needed 39.4 mL of the EDTA solution to reach the endpoint. Determine how many milligrams of CaCO3 will react per mL of this EDTA solution.2. Titration of Ca²+ and Mg²+ in a 50mL sample of hard water required 23.65mL of 0.01205M EDTA. A second 50mL aliquot was made strongly basic with NaOH to precipitate Mg2+ as Mg(OH)2. The supernatant liquid was titrated with 14.53mL of the EDTA solution. Calculate a. concentration in ppm of CaCO3 in the sample. b. concentration in ppm of MgCO3 in the sample.A 50.00 ml aliquot of a solution containing Ca2+ and Mg2+ was buffered at pH 10 and titrated with 0.0499 M EDTA. The endpoint volume was 40.17 ml. A second aliquot of the same mixture was made strongly basic by the addition of NaOH – this causes the Mg2+ to precipitate as Mg(OH)2. The solution was then titrated with the 0.0499 M EDTA and the endpoint volume was found to be 34.70 ml.
- A 50.00-mL aliquot of solution containing 0.310 of MgSO4 (FM 120.37) in 0.500 L required 37.77 mL of EDTA solution for titration. How many milligrams of CaCO3 (FM 100.09) will react with 1.00 mL of this EDTA solution? Answer in 3 significant figures. Do not include the unit. need asapA 47.6 mL aliquot from a 0.5 L solution that contains 0.45g of MnSO4 (MW is 151 g/mol) required 41.9 mL of an EDTA solution to reach the titration endpoint. What mass of CaCO3 (MW is 100.09 g/mol) will react with 1.58 mL of the EDTA solution?A 50.00 mL aliquot of standard CaCO3 (0.3834g CaCO3 per liter) consumed 42.35 mL of EDTA solution. The same EDTA standard solution was used for the titration of a 3.00 mL water sample. This required 30.56 mL of the titrant. Calculate the molar concentration of EDTA standard solution. What is the total hardness of water sample (then after solving explain how it come up to the final answer.)
- A 15.00 mL urine specimen containing Ca2+ and Mg2+ was diluted to 2.000 L. A 10.00 mL sample was taken and buffered to a pH of 10 and required 27.32 mL of 0.003960 M EDTA to react with 'all' the Ca²+ and Mg2+. A second 10.00 mL sample was taken and all the Ca²+ was precipitated as Ca(C204)(s). The Ca(C2O4) (s) precipitate was then taken and dissolved in 18.00 mL of water, buffered to a pH of 10, and required 12.21 mL of 0.003960 M EDTA solution to react with all the Ca2+. Calculate the concentration of Ca²+ and Mg2+ in the original urine specimen.A 50.00 ml aliquot of a solution containing Ca2+ and Mg2+ was buffered at pH 10 and titrated with 0.0493 M EDTA. The endpoint volume was 42.15 ml. A second aliquot of the same mixture was made strongly basic by the addition of NaOH – this causes the Mg2+ to precipitate as Mg(OH)2. The solution was then titrated with the 0.0493 M EDTA and the endpoint volume was found to be 35.65 ml. Calculate the molar concentration of MgA solution is prepared by mixed 50.0 mL of 0.0400 M Ni2+ with 50.00 mL of 0.0400 M EDTA.Calculate the concentration of Ni 2+ after mixing. The mixture is buffered to a pH of 3.0. KNiY =4.2x1018, alpha4 =2.5x10 -11.