Question

Chromel is an alloy composed of nickel, iron, and chromium. A 0.6445-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 5.23-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 38.95 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 25.17 mL of the EDTA solution. Calculate the percentages of nickel, chromium, and iron in the alloy.

Answer 1

solution: 0.6445 g Ni, Fe Cr rightarrow 250.0 ml for back 5.23 ml cu^+ 2 0.06139 M exequiued titration Aliquot 50 ml NI, Fe Cr + 50 ml 0.05173 M EDTA [NI-EDTA] [Fe-EDTA] [Cr-EDTA] MV_eu = MV_unreactcel EDTA 0.06139 M times 5.23 ml = 0.05173M times V V_unreacted EDTA = 0.06139 times 5.23/0.05173 = 0.3210697/0.05173 V_unreacted EDTA = 6.20 ml volume of EDTA reacted with Ni, Fe and Cr = 50.0ml -6.20 ml = 43.8 ml (MV)_Ni + (MV)_Fe + (MV)_Cr = (MV)_EDTA (1) (MV)_Ni + (MV)_Fe + (MV)_Cr = 0.05173 times 43.8 For titration 38.95 ml EDTA = 0.05173 m 50 ml Ni, Fe Cr atiquot MV_EDTA = MV_ni + MV_Fe (2) 0.05173 M times 38.95 ml = M 50 Ni + M 50 Fe for Ni 25 17 ml of EDTA 0.05173 M requirment 50 ml Ni, Fe Cr aliquot MV_EDTA = MV_Ni (3) 0.05173 times 25.17 = m times 50 1.3020441 = m times 50 M = 1.3020441/50 M_Ni = 0.0260 M Put the values of M_Ni is equation (2) to get the value of M_Fe so we bave 0.05173 M times 38.95 ml = 0.0260 M times 50 + M 50 Fe