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 εCo = 36400 and εNi = 5520 at 510 nm and εCo = 1240 and ε Ni = 17500 at 656 nm. A 0.556-g sample was dissolved and diluted to 150.0 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 0.866 at 510 nm and 0.218 at 656 nm in a 1.00-cm cell.

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

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

Answer 1

Here we have following two euations which have two unknown:

A (510) = εCo(510)b[Co2+] + ε Ni (510)b[Ni 2+]

A (656) = εCo(656)b[Co2+] + ε Ni (656 )b[Ni 2+]

Therefore;

A (510) = εCo(510)b[Co2+] + ε Ni (510)b[Ni 2+]

0.866 = 36400 [Co2+] + 5520 [Ni 2+]

So,

[Co2+ ] = 0.866 - 5520 [Ni 2+] / 36400

And A (656) = εCo(656)b[Co2+] + ε Ni (656 )b[Ni 2+]

0.218 = 1240 [Co2+] + 17500 [Ni 2+]

Now we put [Co2+] in the above equation:

0.218 = 1240[ 0.866 - 5520 [Ni 2+] / 36400 ] + 17500 [Ni 2+]

[Ni 2+] = [0.218 – 1240(0.866)/36400] / 17500 – 1240 (5520) /36400]

=    1.086*10^-5

Then;

[Co2+ ] = 0.866 - 5520 [Ni 2+] / 36400

[Co2+ ] = 0.866 - 5520 (1.086*10^-5 )/ 36400

= 2.21*10^-5

  This is the concentration in the cell.

50 ml = 0.050 L

0. 556 g = 5.56*10^-4 kg

0.050 L * 2.21*10^-5 moles Co 2+ / L * 58.9932 g / mol * milli/0.001 50 ml /25 ml

                                5.56 *10^-4 kg

= 234 .5 ppm Co2+

0.050 L * 1.086*10^-5 moles Ni2+ / L * 58.69 g / mol * milli/0.001 50 ml /25 ml

                                5.56 *10^-4 kg

= 114.64 ppm Ni2+