A metal element (M) forms a hexadentate complex with the ligand ethylenediammine

Question

A metal element (M) forms a hexadentate complex with the ligand ethylenediammine (en). Its
formula is M(en)32*. The metal ion and the ligand do not absorb in the visible region of
spectrophotometer and a 1.00 cm cuvette, the following absorbances were measured for various
concentrations of the metal and the ligand:
[M2+] M
1.0 x 10 5
2.0 x 10 5
3.0 x 10 5
4,0xl05
5.0 x 10'5
6.0 x 10 5
7.0 x 10"5
8.0 x 10's
[en]M
1.0 x 10"5
4.0 x 10 5
9.0 x 10's
6.0 x 10'5
10.0 x 10'5
18.0 x 10 5
25.0 x 10'5
20.0 x 10"5
Absorbance at 450 nm
0.03
0.12
0.27
0.18
0.30
0.54
0.63
0.60
a.) Calculate the absorbance (same as extinction) coefficient for the complex at 450 nm
b.) What would be the absorbance of a solution with [M2+] = 2.5 x 10"5 M and [en] = 7.0 x 10"5 M ?

Explanation / Answer

a.) 2.2*10^-5 b.) 0.56

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