A 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 theV3+ 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 theoriginal concentration of the V3+ solution?[V3+]=M

Step 1: Here EDTA will make complex with v+3 and remaining EDTA will react with Ga+3 So the…

Answered: A 25.0 mL aliquot of 0.0580 M EDTA was…

Fundamentals Of Analytical Chemistry

9th Edition

ISBN:9781285640686

Author:Skoog

Publisher:Skoog

Chapter17: Complexation And Precipitation Reactions And Titrations

Section: Chapter Questions

Problem 17.29QAP

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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+] = M
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 = 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+?
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 25.0 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?
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The Cr metal was dissolved in HCl. The pH was suitably adjusted by adding ammnoniaammonium buffer solution and 15.00 mL of 0.01768 M EDTA were introduced. The excess reagent required a 4.30-mL back-titration with 0.008120 M Cu2+. Calculate mass of Cr metal.
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The concentration of a solution of EDTA was determined by standardizing against a solution of Ca2+ prepared from the primary standard CaCO3. A 0.4302g sample of CaCO3 was transferred to a 60mL volumetric flask, dissolved using a minimum of 6 M HCl solution, and diluted to volume. A 44mL portion of this solution was transferred into a 250-mL Erlenmeyer flask and the pH adjusted by adding 5 mL of a pH 10 NH3-NH4 L buffer containing a small amount of Mg2+ EDTA. After adding calmagite as a visual indicator, the solution was titrated with the EDTA, requiring 632mL to reach the end point. Calculate the molar concentration of the titrant
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