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? (4).
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- 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 25.0 mL aliquot of 0.0700M EDTA was added to a 32.0 mL solution containing an unknown concentration of V3+. All of the V3+ present in the solution formed a complex with EDTA, leaving an excess of EDTA in solution. This solution was back- titrated with a 0.0380 M Ga+ solution until all of the EDTA reacted, requiring 15.0 mL of the Ga+ solution. What was the original concentration of the V³+ solution? [V3+] = MA 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 25.0 mL aliquot of 0.0570 M EDTA was added to a 59.0 mL solution containing an unknown concentration of V³+. All of the V³ + present in the solution formed a complex with EDTA, leaving an excess of EDTA in solution. This solution was back- titrated with a 0.0480 M Ga³ + solution until all of the EDTA reacted, requiring 14.0 mL of the Ga³ + solution. What was the original concentration of the V3+ solution? [V³ t] = M The end point of the Zn²+-EDTA titration was observed after 15.50 mL of 0.0500 M EDTA solution was dispensed. Determine the number of moles of zinc ion present in the sample. number of moles of zinc ion = molA 25.0 mL aliquot of 0.0580 M EDTA was added to a 47.0 mL solution containing an unknown concentration of V³+. All of the V3+ present in the solution formed a complex with EDTA, leaving an excess of EDTA in solution. This solution was back- 3+ 3+ titrated with a 0.0420 M Ga³+ solution until all of the EDTA reacted, requiring 11.0 mL of the Ga³+ solution. What was the original concentration of the V3+ solution? [V3+]= MA 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?
- 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)The total concentration of Ca²+ and Mg²+ in a sample of hard water was determined by titrating a 0.100-L sample of the water with a solution of EDTA. The EDTA chelates the two cations: Mg+ + [EDTA]+ Ca2+ + [EDTA]+ [Mg(EDTA)]²- [Ca(EDTA)]?- It requires 31.5 ml. of 0.0104 M[EDTA]* solution to reach the end point in the titration. A second 0.100-L sample was then treated with sulfate ion to precipitate Ca?+ as calcium sulfate. The Mg2+ was then titrated with 18.7 mL of 0.0104 M [EDTA]*. Calculate the concentrations of Mg²+ and Ca2 in the hard water in mg/L.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 25.0 mL sample containing Fe3+ was treated with 10.00 mL of EDTA0.0367 mol / L to complex all the iron and leave an excess of EDTA insolution. To the excess of EDTA was added 2.47 mL of Mg2+ 0.0461 mol / Lwhich guaranteed all EDTA consumption. What was the concentration of Fe3+ in the original solution?A 50.00 mL solution containing Ni2+ and Fe2+ was treated EDTA to bind all the metal ions. After back titration, the amount of EDTA used is 2.500 mmol. In another 50.00 mL solution was added pyrophosphate to mask the Fe2+ ions, and the solution required 25.00 mL of 0.04500 M EDTA. Calculate the ppm Fe (55.85 g/mol) in the solution.3. A 50.00 mL solution containing Ni2+ and Fe2+ was treated EDTA to bind all the metal ions. After back-titration, the amount of EDTA used is 2.500 mmol. In another 50.00 mL solution was added pyrophosphate to mask the Fe2+ ions, and the solution required 25.00 mL of 0.04500 M EDTA. Calculate the ppm Fe (55.85 g/mol) in the solution.