A 25.0 mL sample containing Fe3+ was treated with 10.00 mL of EDTA 0.0367 mol / L to complex all the iron and leave an excess of EDTA in solution. To the excess of EDTA was added 2.47 mL of Mg2+ 0.0461 mol / L which guaranteed all EDTA consumption. What was the concentration of Fe3+ in the original solution?
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A 25.0 mL sample containing Fe3+ was treated with 10.00 mL of EDTA
0.0367 mol / L to complex all the iron and leave an excess of EDTA in
solution. To the excess of EDTA was added 2.47 mL of Mg2+ 0.0461 mol / L
which guaranteed all EDTA consumption. What was the concentration of Fe3+ in the original solution?
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- A 50.0 mL sample containing Cd2+ and Mn2+ was treated with 65.1 mL of 0.0400 M EDTA. Titration of the excess unreacted EDTA required 15.0 mL of 0.0190 M Ca2+. The Cd2+ was displaced from EDTA by the addition of an excess of CN−. Titration of the newly freed EDTA required 22.7 mL of 0.0190 M Ca2+. What are the concentrations of Cd2+ and Mn2+in the original solution?A 0.9352g sample of ore containing Fe³+, Al³+ and Sr²+ was dissolved and made up to 500.00 mL. The analysis of metals was performed using complexation volumetry. Initially, an aliquot of 50.00 mL had its pH adjusted to 1.0 and titrated with a standard 0.03145 mol/L EDTA solution, requiring 6.95 mL to reach the end point. Subsequently, another 25.00 mL aliquot was buffered at pH=5 and titrated with the same EDTA solution, requiring 6.24 mL to reach the end point. Finally, a third aliquot of 25.00 mL was titrated at pH=11, requiring 11.10 mL of the same EDTA solution to complete the titration. Given the molar masses: Fe=55.845 g/mol; Al-26.982 g/mol and Sr-87.620 g/mol. a) Determine the percentage of each of the metals in the sample. b) Explain why the change in pH allows the determination of the three ions in this sample.An alloy containing Ni, Fe and Cr was analyzed by a complexation titration using EDTA as titrant. A 0.7176 g sample of the alloy was dissolved in HNO3 and diluted to 250 mL in a flask. A 50.00 mL aliquot of the sample, treated with pyrophosphate to mask Fe and Cr, required 26.14 mL of 0.05831 M EDTA to reach the murexide endpoint. A second 50.00 mL aliquot was treated with hexamethylenetetramine to mask Cr and titration with 0.05831 M EDTA required 35.43 mL to reach the murexide endpoint. Finally, a third 50.00 mL aliquot was treated with 50.00 mL of 0.05831 M EDTA and titrated back to the murexide endpoint with 6.21 mL of 0.06316 M Cu(II). the weight percentages of Ni, Fe and Cr in the alloy.
- To analyze the amount of iron (Fe; Mw = 55.85 g/mol) contained in an ore sample, the sample was digested with acid and diluted to 50 mL with water. This solution was then treated with 25.00 mL of 0.2922 M EDTA. The excess EDTA was back titrated with 6.47 mL of 0.0843 M Zn2+ to reach the equivalence point. How many grams of Fe contained in the ore sample?a 10 ml sample containing fe3+ and cu2+ was diluted to 100.0 mL solution. A 20 mL aliquot required 25 mL of 0.00500 M EDTA for complete titration. A separate 20 mL aliquot of the diluted sample was treated with NH4F to protect the Fe3+. Then the Cu2+ was treated with thiourea. Upon addition of 25.0 ml of 0.00500 M EDTA, the Fe3+ was liberated from its fluoride complex and formed an EDTA complex. The excess EDTA required 21.5 ml of 0.0180 M Pb2+ to reach an endpoint, using xylenol orange. (a) identify the masking agent (b) identify the auxiliary complexing agent (c) calculate the molar concentration of the fe3+ in the original sample (d) calculate the molar concentration of the cu2+ in the original sampleA 25.00-mL sample containing Ni2+ was treated with 25.00 mL of 0.5000 M EDTA to complex all the Ni2+ and leave excess EDTA in solution. The excess EDTA was then back titrated, requiring 7.11 mL of 0.450 M Zn2+. What was the concentration of Ni2+ in the original solution (keep 4 SF)?
- 0.4545 g CaCO3 was dissolved in HCl and the resulting solution was diluted to 0.25 L. Twenty-five mL aliquot of this solution required 35.2 mL of EDTA upon performing titrimetric analysis. Determine the molarity of EDTAA sample of steel weighing 2.00 g is analyzed for Cr (AW 52.0). The Cr is oxidized into chromate with alkaline permanganate and the excess permanganate is destroyed. A certain volume of 0.120 M FeSO4 is added to the acid solution and the excess is titrated with 0.0220 M KMnO4, requiring 31.0 mL. If the sample contained 0.50 % Cr, what volume (in mL) of FeSO4 was added?When I was a boy, I watched Uncle Wilbur measure the iron content of runoff from his banana ranch. He acidified a 25.0-mL sample with HNO3 and treated it with excess KSCN to form a red complex. (KSCN itself is colorless.) He then diluted the solution to 100.0 mL and put it in a variablepathlength cell. For comparison, he treated a 10.0-mL reference sample of 6.80X 1024 M Fe31 with HNO3 and KSCN and diluted it to 50.0 mL. The reference was placed in a cell with a 1.00-cm pathlength. Runoff had the same absorbance as the reference when the pathlength of the runoff cell was 2.48 cm. What was the concentration of iron in Uncle Wilbur’s runoff ?
- A 25.00 mL aliquot of a solution containing Cu2+ and Fe3+ was titrated with 17.08 mL of 0.05095 M EDTA. A second 25.00 mL aliquot of the Cu/Fe mixture was treated with NaF to form a stable iron-fluoride complex. This mixture was then titrated with EDTA and the endpoint volume was found to be 5.47 mL. Calculate the amounts of Cu2+ and Fe3+ in mg/L. Molar mass (g/mol): Fe = 55.85 and Cu = 63.55A 0.7352g sample of ore containing Fe3+, Al3+ and Sr2+ was dissolved and made up to 500.00 mL. The analysis of metals was performed by a chemistry using complexation volumetry. Initially, an aliquot of 50.00 mL had its pH adjusted to 1.0 and titrated with a standard solution of EDTA 0.02145 mol/L, requiring 6.95 mL to reach the end point. Subsequently, another 25.00 mL aliquot was buffered at pH=5 and titrated with the same EDTA solution, requiring 6.24 mL to reach the end point. Finally, a third aliquot of 25.00 mL was titrated at pH=11, requiring 11.10 mL of the same EDTA solution to complete the titration. Determine the percentage of each of the metals in the sample Given the molar masses: Fe=55.845 g/mol; Al=26.982 g/mol and Sr=87.620 g/mol.A 25.00 mL sample containing Fe3+ was treated with 10.00 mL of 0.03676 M EDTA to complex all the Fe3+ and leave excess EDTA in solution. The excess EDTA was then back-titrated, requiring 2.37 mL of 0.04615 M Mg2+. What was the concentration of Fe3+ in the original solution in ppm Fe3+?