Question

P23-27: Molar absorptivity data for the cobalt and nickel complexes with 2,3-
quinoxalinedithiol are εCo = 36,400 and εNi = 5520 at 510 nm and εCo = 1240 and εNi = 17,500 at 656 nm. A 0.425-g sample was dissolved and diluted to 50.0 mL. A 25.0-mL aliquot was treated to eliminate interferences; after addition of 2,3quinoxalinedithiol, the volume was adjusted to 50.0 mL. This solution had an absorbance of 0.446 at 510 nm and 0.326 at 656 nm in a 1.00-cm cell. Calculate the parts per million of cobalt and nickel in the sample.

Answer 1

At 510 nm Co absorbs strongly, According to Beer's law:

Absorption, A = εcl

or, 0.446 = 36400 M^­-1.cm^-1 * c * 1 cm

or, concentration, c = 1.23*10^-5 M

So, the concentration of Co in the final 50 mL solution = 1.23*10^­-5 M

Amount of Co in the final 50 mL solution = 1.23*10^­-5 moles * 50 mL /1000 mL = 6.13*10^-7 moles

This amount of Co is also present in the 25 mL aliquot taken.

And amount of Co present in the initial 50 mL solution (made from dissolution of 0.425 g sample)

= 6.13*10^-7 moles * 2

= 1.23*10^-6 moles

= (1.23*10^-6 moles * molar mass of Co) g

= 1.23*10^-6 * 58.93 g

= 7.25 * 10^-5 g

Co present in the sample

= (7.25 * 10^-5 g / 0.425 g) * 10⁶

= 170.55 ppm

At 656 nm Ni absorbs strongly, According to Beer's law:

Absorption, A = εcl

or, 0.326= 17500 M^­-1.cm^-1 * c * 1 cm

or, concentration, c = 1.86*10^-5 M

So, the concentration of Ni in the final 50 mL solution = 1.86*10^­-5 M

Amount of Ni in the final 50 mL solution = 1.86*10^­-5 moles * 50 mL /1000 mL = 9.31*10^-7 moles

This amount of Ni is also present in the 25 mL aliquot taken.

And amount of Ni present in the initial 50 mL solution (made from dissolution of 0.425 g sample)

= 9.31*10^-7 moles * 2

= 1.86*10^-6 moles

= (1.86*10^-6 moles * molar mass of Ni) g

= 1.86*10^-6 * 58.69 g

= 1.09 * 10^-4 g

Ni present in the sample

= (1.09 * 10^-4 g / 0.425 g) * 10⁶

= 256.47 ppm