Question

The following information is to be used for the next two questions. Molar absorptivity data for the cobalt and nickel complexes with 2,3-quinoxalinedithiol are ε_Co = 36400 and ε_Ni = 5520 at 510 nm and ε_Co = 1240 and ε _Ni = 17500 at 656 nm. A 0.445-g sample was dissolved and diluted to 250.0 mL. A 25.0-mL aliquot was treated to eliminate interferences; after addition of 2,3-quinoxalinedithiol, the volume was adjusted to 50.0 mL. This solution had an absorbance of 0.562 at 510 nm and 0.602 at 656 nm in a 1.00-cm cell.

Calculate the concentration of cobalt (in ppm) in the initial solution prepared by dissolving the sample.

Calculate the concentration of nickel (in ppm) in the initial solution prepared by dissolving the sample.

Answer 1

Both Co and Ni react with 2,3-quinoxalinedithiol in a 1:1 molar ratio; therefore, the concentrations of Co and Ni are equal to the concentrations of the complexes. Let the dilute solution contain x M Co and y M Ni (we choose molar concentrations since the absorptivity values are given in M^-1cm^-1) . Write down Beer’s law for the two complexes as

0.562 = (36400 M^-1cm^-1)*(x M)*(1.00 cm) + (5520 M^-1cm^-1)*(y M)*(1.00 cm) …..(1)

0.602 = (1240 M^-1cm^-1)*(x M)*(1.00 cm) + (17500 M^-1cm^-1)*(y M)*(1.00 cm) …..(2)

Multiply (1) by 1240 and (2) by 36400 and subtract and obtain

0.562*1240 – 0.602*36400 = (5520)*(1240)*y – (17500)*(36400)*y

====> -21215.92 = -630155200*y

====> y = 3.3668*10^-5

Put the value of y in (1) and obtain

0.562 = (36400)*x + (5520)*(3.3668*10^-5)

=====> 36400*x = 0.3595264

=====> x = 9.8771*10^-6

The concentrations of Co and Ni in the dilute solution are 9.8771*10^-6 M and 3.3668*10^-5 M respectively.

The dilute solution was prepared by taking 25.0 mL of the stock solution and diluting to 50.0 mL; the dilution factor is (50.0 mL)/(25.0 mL) = 2.0.

The concentrations are: Co = (9.8771*10^-6 M)*(2.0) = 1.97542*10^-5 M; Ni = (3.3668*10^-5 M)*(2.0) = 6.7336*10^-5 M.

The atomic masses of Co and Ni are 58.933 g/mol and 58.693 g/mol.

The concentrations (in ppm) are:

Co = (1.97542*10^-5 M)*(1 mol/L/1 M)*(58.933 g/1 mole)*(1000 mg/1 g)*(1 ppm/1 mg/L) = 1.164 ppm (ans).

Ni = (6.7336*10^-5 M)*(1 mol/L/1 M)*(58.693 g/1 mole)*(1000 mg/1 g)*(1 ppm/1 mg/L) = 3.952 ppm (ans).