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+] = 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 25.0 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+] = 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

Fundamentals Of Analytical Chemistry

9th Edition

ISBN:9781285640686

Author:Skoog

Publisher:Skoog

Chapter17: Complexation And Precipitation Reactions And Titrations

Section: Chapter Questions

Problem 17.31QAP

See similar textbooks

Similar questions

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?
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
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.
Calcium in powdered milk is determined by dry ashing a 3.00g sample then titrating the calcium with 12.10mL EDTA solution. The EDTA was standardized by titrating 10.00mL of a Zn solution by dissolving 0.632g Zn metal in acid and dilution to 1.00L (10.80 mL EDTA required for titration). What is the concentration of Ca in the powdered milk in ppm?
The Cl– content of a solution was analyzed via an EDTA titration. 2.00 g of AgNO3 was added to 25.00 mL of the sample solution. A titration with 0.096 53 M EDTA was then carried out, reaching an endpoint at 38.25 mL. What’s the concentration of Cl – (M)?
Calcium in powdered milk is determined by dry ashing a 1.46g sample and then titrating the calcium with EDTA solution, 12.1mL being required. The EDTA was standardized by titrating 10.4mL of a Zn solution prepared by dissolving 0.634g Zn metal in acid and diluting to 1L (10.83mL EDTA required for titration). What is the concentration (ppm) of calcium in the powdered milk?
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.
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 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+]= M
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.
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
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 = % %
A cyanide solution with a volume of 13.72 mL was treated with 25.00 mL of Ni2+ solution(which contained an excess of Ni(II) ions) to convert the cyanide to the complex ion tetracyanonickelate(II): 4CN- + Ni2+ → 6 Ni(CN)2- The excess Ni 2+ was then titrated with 10.15 mL of 0.01307 M EDTA. One mole of this reagent reacts with one mole of Ni(II) as follows: Ni(CN)42- does not react with EDTA. If 39.35 mL of EDTA was required to react with 30.10 mL of the original Ni 2+ solution, calculate the molarity of CN- in the 13.72 mL cyanide sample.