Question

A 0.6004 g sample of Ni/Cu condenser tubing was dissolved in acid and diluted to
100.0 mL via a volumetric flask. Titration of both cations in a 25.00 mL aliquot of this solution required 45.81 mL of 0.05285 M EDTA. Mercaptoacetic acid and NH3 were then introduced; production of the Cu complex with the former resulted in the release of an equivalent amount of EDTA, which required a 22.85 mL titration with 0.07238 M Mg2+. Calculate the percentages of Cu and Ni in the alloy.

Answer 1

Given mass of Ni is cu -0.6004 g. in 100 mL Amount of EDTA = - 24 2+ Amount of Nit cu = 45. 81 mk yo. 05285 mmol 2.4 210 mmol. Now, 2+ А мампk of мо“. А илм осврт A A Awaxwe of си. = 22.85 th x 0.07 238 mmol ML 1.6538 mmol. 2+ : Amount of cu= 1.6538 mmol. 24 weight of cu= 1.6538 x 63.5 ms mmol mwool 1.6538 x 63.5 mg. : present in 25 mL of the This weight of curt is solution a weight of cut in - 100 mL of the solution 1.6538x 63.5 mg x too me 25 mt 420 mg = 0.42 g. % cu : 0.42 g x 100 0.6004 g = 69.95.).

Now . Amosunt of Ni Total Amassons - Amarums of 27 ONI + ĐỆ NỞ, u" 27 - . (24210 - 1.6538) mimosd = 0.7672 must weight of Nikt = 0.7672 mynd x 58.7 mmg a 45.03 46 me weight of in 100 me Niht of a solution 1. the 45.0 346 me x too yht 25 yu = 180.13 mg 0.1801 g % Ni = 0.1801% X 160 0.60048 a 29.99 ./. - % Ni = 29.99.60 .6. cu = 69.95%