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An alloy of chromel containing Ni, Fe, and Cr was analyzed by a complexation titration using EDTA as the titrant. A 0.7176-g sample of the alloy was dissolved in HNO3 and diluted to 250 mL in a volumetric flask. A 50.00-mL aliquot of the sample, treated with pyrophosphate to mask the Fe and Cr, required 26.14 mL of 0.05831 M EDTA to reach the murexide end point. A second 50.00-mL aliquot was treated with hexamethylenetetramine to mask the Cr. Titrating with 0.05831 M EDTA required 35.43 mL to reach the murexide end point. Finally, a third 50.00-mL aliquot was treated with 50.00 mL of 0.05831 M EDTA, and back titrated to the murexide end point with 6.21 mL of 0.06316 M Cu2+. Calculate the weight percentages of Ni, Fe, and Cr in the alloy. 

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Related questions

Chromel is an alloy composed of nickel, iron and chromium. A0.7176g sample of the alloy was dissolved in HNO3 and diluted to250ml in a volumetric flask. A 50ml aliquot of this sample, treatedwith pyrophosphate to mask Fe and Cr, required 26.24ml of 0.05831 MEDTA to reach the endpoint. A second 60ml aliquot was treated withhexamethylenetetra to mask the Cr. Titrating with 0.05831M EDTArequired 40.35ml to reach the end point. Finally, a third 60mlaliquot was treated with 50ml 0.05831 M EDTA and back titrated tothe end point with 7.21ml of 0.07316M Cu2+. what are the weights ofNi, Fe and Cr in the alloy

**Step by Step calculation would be appreciated

A student wishes to analyze a seawater sample to determine the amounts of Na+ , Ca2+ and Mg2+ present. She begins with 50.00 mL sample of seawater that she dilutes to 500.00 mL. She removes a 50.00 mL aliquot of the diluted solution, which she analyzes for total cations present by exchanging the cations for H+ on an ion exchange column and titrating the liberated acid with 0.5000 M NaOH. The amount of NaOH required to reach the endpoint was 22.48 mL.

The student then titrates another 50.00 mL aliquot of her diluted seawater solution with 15.31 mL of 0.0100 M EDTA solution at a pH of 10 to the Erio T endpoint.

The third titration, performed on a 50.00 mL aliquot of the diluted seawater solution involves precipitating Mg2+ as the hydroxide and titrating the Ca2+ remaining in solution with 0.010 M EDTA solution. Only 5.15 mL of EDTA were required to reach the end point.

Summary of titration data:

Original seawater = 50.00 mL; diluted seawater = 500.00 mL

Vol. of titration aliquot = 50.00 mL

First titration: [NaOH] = 0.5000 M; V NaOH = 22.48 mL

Second titration: [EDTA] = 0.0100 M; VEDTA = 15.31 mL

Third titration: [EDTA] = 0.0100 M; VEDTA = 5.15 mL

From the titration data, calculate the following (show work):

A). mmol of Ca2+ present in the third 50.00 mL titration aliquot - ans.__________

B). mmol of Ca2+ present in the original 50.00 mL seawater sample - ans.__________

C). mg of Ca2+ present in the original 50.00 mL seawater sample - ans.__________

D). mmol of Ca2+ and Mg2+ present in the second 50.00 mL titration aliquot - ans.__________

E). mmol of Mg2+ present in the 50.00 mL titration aliquot - ans.__________

F). mmol of Mg2+ present in in the original 50.00 mL seawater sample - ans.__________

G). mg of Mg2+ present in in the original 50.00 mL seawater sample - ans.__________

H). meq of Ca2+ and Mg2+ present in the 50.00 mL titration aliquot - ans.__________

I). total meq of Na+ , Ca2+ and Mg2+ present in the 50.00 mL titration aliquot - ans.__________

J). meq of Na+ in the 50.00 mL titration aliquot - ans.__________

K). meq of Na+ present in the original 50.00 mL seawater sample - ans.__________

L). mg of Na+ present in in the original 50.00 mL seawater sample - ans.__________

A 1.000-mL aliquot of a solution containing Cu2 and Ni2 is treated with 25.00 mL of a 0.04068 M EDTA solution. The solution is then back titrated with 0.02053 M Zn2 solution at a pH of 5. A volume of 16.47 mL of the Zn2 solution was needed to reach the xylenol orange end point. A 2.000-mL aliquot of the Cu2 and Ni2 solution is fed through an ion-exchange column that retains Ni2 . The Cu2 that passed through the column is treated with 25.00 mL 0.04068 M EDTA. This solution required 20.25 mL of 0.02053 M Zn2 for back titration. The Ni2 extracted from the column was treated witn 25.00 mL of 0.04068 M EDTA. How many milliliters of 0.02053 M Zn2 is required for the back titration of the Ni2 solution?

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