Co and Ni can be analysed simultaneously by their reaction with 2, 3-quinoxalinedithiol. Molar absorptivity data at the wavelengths are:

Molar Absorptivity.ε

510 nm

656 nm

Co-Complex

36400

1240

Ni-Complex

5520

17500

A 0.519 g sample was dissolved and diluted to 50 ml. A 25 ml aliquot was treated to eliminate interferences; after addition of 2, 3-quinoxalinedithiol, the volume was adjusted to 50 ml. this solution had an absorbance of 0.477 at 510 nm and 0.219 at 656 nm in 1.00 cm cell. 

Calculate the parts per million of Co and Ni in the sample.

Answer:

Let x be the concentration of Co and y be the concentration of Ni

Given that

molar absorptivity of Co-complex at 510 nm = 36400

molar absorptivity of Ni -complex at 510 nm = 5520

molar absorptivity of Co-complex at 665 nm = 1240

molar absorptivity of Ni-complex at 665 nm = 17500

Absorbance at 510 nm = 0.477

absorbance at 665 nm = 0.219

From Beer-lambert's law we know that

Absorbance = molar absorptivity*concentration*length

Since length remains constant, we can take it to be equal to 1

Therefore the equations are

36400*x+5520*y = 0.477 --------(1)

1240*x+17500*y = 0.219 -------(2)

solving equations (1) and (2), we get

x = 1.13*10^-5 g/L = 11.3 microgram/L (since 1 microgram = 10^6 g)

y = 1.17*10^-5 g/L = 11.7 microgram/L

Therefore ppm of Co = 11.3 microgram/1000 g = 0.0113 microgram/g = 0.0113 ppm (since 1 microgram/g = 1 ppm)

ppm of Nickel = 11.7 microgram/1000g = 0.0117 ppm