Answere in detail please. An alloy of chromel containing Ni, Fe, and cr was anal

Question

Answere in detail please.

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 HNO_3 and diluted to 250.0 mL in a volumetric 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 murex end point. A second 50.00-mL aliquot was treated with hexamethylenetetramine to mask Cr, and required 35.43 mL of 0.05831 M EDTA 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 Cu^2+. Report the weight percents of Ni, Fe, and Cr in the alloy. Attach and, including representative chemical reactions:

Explanation / Answer

Answer:

First step to calculate the molarities of Ni, Fe and Cr.

Here, pyrophosphate mask Fe and Cr. The EDTA solution required for Ni is 26.14ml 0.05831 M

MEDTAVEDTA=MNiVNi------------ Formula

0.05831 X 26.14 = MNi X 50

MNi=0.03048 M

Here, hexamethylenetetramine to mask Cr. The EDTA solution required for Ni and Fe   is 35.43ml 0.05831 M

MEDTAVEDTA=MNiVNi + MFeVFe--------------Formula

0.05831 X 35.43 = 0.03048 X 50 + MFe X 50

MFe= 0.01083 M

EDTA Back titrated to murexide end point 6.21ml 0.06316 M Cu2+

From above, To calculate unreacted EDTA

MEDTAVunreacted EDTA=MCuVCu

0.05831 X Vunreacted EDTA = 0.06316 X 6.21

Vunreacted EDTA =6.72 ml

Volume of EDTA reacted with FE, Ni and Cr = 50-6.72 = 43.27ml.

MNiVNi + MFeVFe + MCrVCr = MEDTAVEDTA

0.03048 X 50 + 0.01083 X50 + MCr X 50 = 0.05831 X 43.27

MCr =0.00915M

After calculating the molarities, determine the weight of each metal in the sample

Wt of Ni=(0.03048 X 250 X 58.693)/1000 =0.4472 g.

Wt of Fe=(0.01048 X 250 X 55.84)/1000 =0.1463 g.

Wt of Cr=(0.00915 X 250 X 51.99)/1000 =0.1189 g.

Now calculate the percentage of each metal in the sample.

% Ni = (0.4472/0.7176)x 100 = 62.71 %
% Fe = (0.1463/0.7176)x 100 = 20.38 %
% Cr = (0.4472/0.7176)x 100 = 16.56 %

Chemical Reactions:

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