Page 347 - Modern Analytical Chemistry

1400-CH09 9/9/99 2:13 PM Page 330

330 Modern Analytical Chemistry

SOLUTION
Conservation of electron pairs for the three titrations requires that for
Titration 1: moles Ni = moles EDTA1 (Fe, Cr masked)

Titration 2: moles Ni + moles Fe = moles EDTA2 (Cr masked)

Titration 3: moles Ni + moles Fe + moles Cr + moles Cu = moles EDTA3
Note that the third titration is a back titration. Titration 1 can be used to
determine the amount of Ni in the alloy. Once the amount of Ni is known, the
amount of Fe can be determined from the results for titration 2. Finally,
titration 3 can be solved for the amount of Cr.

Titration 1
g Ni
= M EDTA1 ´V EDTA1
AW Ni

g Ni = M EDTA1 ´V EDTA1 ´ AW Ni

.
.
.
0.05831 M ´0 02614 L ´58 69 g / mol =0 08946 g Ni
Titration 2
Moles EDTA1 + moles Fe = moles EDTA2

g Fe
= moles EDTA2 - moles EDTA1
AW Fe

g Fe = (M EDTA2 ´V EDTA2 -M EDTA1 ´V EDTA1 ) ´
AW Fe
0
0
0
(.05831 M ´ 0.03543 L - .05831 M ´ .02614 L)
´55 .847 g/mol = .03025 g Fe
0
Titration 3
Moles EDTA2 + moles Cr + moles Cu = moles EDTA3

g Cr
= moles EDTA3 - moles EDTA2 - moles Cu
AW Cr

g Cr = (M EDTA3 ´V EDTA3 -M EDTA2 ´V EDTA2 -M Cu ´ V Cu ) AW Cr
´
(.05831 M ´ 0.05000 L - .05831 M ´ .03543 L
0
0
0
0
0
–.06316 M ´ .00621 L) ´ 51.996 g/mol = 0.02378 g Cr
Each of these titrations was conducted on a 50.00-mL aliquot of the original
250.0-mL sample. The mass of each analyte, therefore, must be corrected by
multiplying by a factor of 5. Thus, the grams of Ni, Fe, and Cr in the original
sample are