Example:

The absorbance values of a mixture of K₂Cr₂O₇ and KMnO4 at 440 nm and 545 nm using a cell of 1 cm path length were found to be 0.405 and 0.712 respectively. The absorbance values of pure solutions of K₂Cr₂O₇ (0.001 M) and KMnO₄ (0.0002 M) in similar conditions were as follows:

A (Cr,440 nm)   =  0.374,   A (Cr, 545 nm)  =  0.009

A (Mn, 440nm) = 0.019,   A (Mn, 545 nm)  =  0.475

Using the given data determine the concentration of Cr2O₇^2- ions and MnO₄^-   ions in the mixture.

Solution: Using the given absorbance values for the pure solutions, we can calculate the molar absorptivities as for Cr and Mn at 440 and 545 nm as follows:

0.374 = ε_{Cr, 440} × 1.0 × 1.0 × 10^-3       =>       ε_{Cr, 440} = 374

0.009 = ε_{Cr, 545} × 1.0 × 1.0 × 10^-3        =>       ε_{Cr, 545}   = 9

0.019 = ε_{Mn, 440}  × 1.0 × 2.0 × 10 ^-4     =>       ε_Mn,440  =  95

0.475 = ε_Mn,545  × 1.0 × 2.0 × 10^-4       =>       ε_Mn,545  =  2.38 × 103

For the mixture we can write the simultaneous equations as

 A₄₄₀ = εCr, 440 [Cr2O₇^2-] + ε_{Mn, 440}  [MnO₄^-  ]       

 A₅₄₅ = εCr, 545 [Cr2O₇^2-] + ε_{Mn, 545}  [MnO₄^-  ]       

Substituting the values, we get

0.405=374 [Cr2O₇^2-] + 95  [MnO₄^-  ]

0.712=9 [Cr2O₇^2-] + 2.38 x 10³ [MnO₄^-  ]               

Solving simultaneous equations, we get

[Cr2O₇^2-] = 1.0 × 10^-3 mol dm^-3

[MnO₄^- ] = 2.95 × 10^-4   mol dm^-3