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
Aa.81.
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
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 0.03676 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 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 0.3674 g powdered milk sample was analyzed for calcium by igniting in a crucible at 1000°C, the ash dissolved in dilute HCI, and the solution buffered to pH 10. A titration with an EDTA solution required 15.62 mL of EDTA. The EDTA was standardized by titrating a 10.00 mL aliquot of a solution containing 500.0 mg Zn/L. This titration required 9.81 mL of EDTA. What is the mg% of calcium in the powdered milk?
The concentration of EDTA is 2.497mM
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
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.
C- Al is determined by titrating with EDTA:- Al3 +H2Y2 AlY +2H* 0.5 mg sample required 20.5 ml. EDTA for titration. The EDTA was standardized by titrating 25 ml. of 0.01M CaCl2 solution, requiring 30 ml. EDTA. Calculate Al2O3 in the sample.
Chromel is an alloy composed of nickel, iron, and chromium. A 0.6418-g sample was dissolved and diluted to 250.0 mL. When a 50.00-mL aliquot of 0.05173 M EDTA was mixed with an equal volume of the diluted sample, all three ions were chelated, and a 5.27-mL back-titration with 0.06139 M copper(II) was required. The chromium in a second 50.0-mL aliquot was masked through the addition of hexamethylenetetramine; titration of the Fe and Ni required 35.81 mL of 0.05173 M EDTA. Iron and chromium were masked with pyrophosphate in a third 50.0-mL aliquot, and the nickel was titrated with 25.77 mL of the EDTA solution. Calculate the percentages of nickel, chromium, and iron in the alloy. Percentage of nickel = % Percentage of iron = Percentage of chromium = % %

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