Molar absorptivity data for the cobalt and nickel complexes with 2, 3-quinoxalin

Question

Molar absorptivity data for the cobalt and nickel complexes with 2, 3-quinoxalinedithiol are epsilon C_0 = 36400 and epsilon Ni = 5520 at 510 nm and epsilon co = 1240 and epsilon_Ni = 17500 at 656 nm. A 0.467-g sample was dissolved and diluted to 50.0 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 0.488 at 510 nm and 0.360 at 656 nm in a 1.00-cm cell. Calculate the concentration in parts per million of cobalt and nickel in the sample Concentration of cobalt = ppm Concentration of nickel = ppm

Explanation / Answer

As per Beer's law we have

A = E C L

A = absorbance, E = absorptivity coefficient, C = concentration and L = path length

So at 510 nm

we have

0.488 = ECO x CCO + ENi + CNi (L is 1)

0.488 = 36400 x CCO + 5520 x CNi -(1)

Similarly At 656 nm we have

0.36 = 1240 x CCO + 17500 x CNi - (2)

Solving for 1 and 2 by multiplying (1) by 1240 and (2) by 36400

605 = 45136000 CCO + 6844800 CNi

13104 = 45136000 CCO + 637000000 CNi

subtraction the the first from the second

12499 = 630155200 CNi

CNi = 1.98 x 10-5 M Nickel concentration

CCO is got by substituting CNi in (1)

0.488 = 36400 x CCO + 5520 x 1.98 x 10-5   

CCO = 1.039 x 10-5M so Cobalt concentration

Co will be 0.61 ppm and Ni will be 1.16 ppm

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