50.0-mL sample containing Ni2+ was treated with 25.0 mL of 0.0500 M EDTA to complex all the Ni2+ and leave excess EDTA in solution. The excess EDTA was then back-titrated, requiring 5.00 mL of 0.0500 M Zn2+. What was the molar concentration of Ni2+ in the original solution?
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A 50.0-mL sample containing Ni2+ was treated with 25.0 mL of 0.0500 M EDTA to complex all the Ni2+ and leave excess EDTA in solution. The excess EDTA was then back-titrated, requiring 5.00 mL of 0.0500 M Zn2+. What was the molar concentration of Ni2+ in the original solution?
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- A 100.0 mL sample containing Zn2+ was treated with 50.0 mL of 0.0500 M EDTA to complex all the Zn2+ and leave excess EDTA in solution. The excess EDTA was then back titrated, requiring 10.00 mL of 0.0500 M Mg2+. What was the concentration of Zn2+ in the original solution?A 25.00-mL sample containing Fe3+ was treated with 10.00 mL of 0.036 7 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 0.0461 M Mg2+. What was the concentration of Fe3+ in the original solution?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+?
- A 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)?A 100, 0mL containing Zn^ 2+ was treated with 50.0 of 0.0500 M EDTA complex all the Zn^ 2+ and leave excess EDTA in solution. The excess EDTA was then back titrated, requiring 10.00 ml of 0.0500 M Mg^ 2+ was the concentration of Zp^ 2+ original solution ? (Marks)A 50.0-mL sample containing Ni21 was treated with 25.0 mL of 0.050 0 M EDTA to complex all the Ni21 and leave excess EDTA in solution. How many millimoles of EDTA are contained in 25.0 mL of 0.050 0 M EDTA?
- A 25 ml solution of unkown containing Fe+3 and Ca+2 required 16.06 ml of 0.051 M EDTA for complete the reaction. A 50 ml of the same of the unkown was treated with NH4F to protect Fe+3. then the Ca+2 was reduced and masked by addition thiourea. upon addition of 25 ml of 0.051 M EDTA, the Fe+3 was liberated fron its floride complex and formed EDTA complex . the excess ETDA required 19.77 ml of 0.01883 M pb+2 to reach the end point by using xylenol orange indicator. Find the concentarion of Ca+2 in the unkown sample?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 50.0-mL solution containing Ni21 and Zn21 was treated with 25.0 mL of 0.045 2 M EDTA to bind all the metal. The excess unreacted EDTA required 12.4 mL of 0.012 3 M Mg21 for complete reaction. An excess of the reagent2,3-dimercapto-1-propanol was then added to displace the EDTA from zinc. Another 29.2 mL of Mg21 were required for reaction with the liberated EDTA. Calculate the molarities of Ni21 and Zn21 in the original solution.
- A 10.0 mL sample containing Fe³⁺ and Cu²⁺ was diluted to 100.0 mL solution. A 20.0 ml aliquot required 25.0 ml of 0.0500 M EDTA for complete titration. A separate 20.0 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 0.0500 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 end point, 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 sampleThe Cr plating on a surface that measured 3.00 cm x 4.00 cm was dissolved in HCl. The pH was suitably adjusted, and 10.19 mL of 0.01441 M EDTA was then introduced. The excess reagent required a 2.91-mL back-titration with 0.007171 M Cu2+. Calculate the average weight (in mg) of Cr (51.9961 g/mol) on each square centimeter of surface.b. EDTA cannot be used as a primary standard always. When can EDTA be used as a primarystandard ? Give reason why EDTA is used in complexometric titrations.