Molar absorptivity data for the cobalt and nickel complexes with 2,3-quinoxalinedithiol are = 36400 and = 5520 at 510 nm and = 1240 and = 17500 at 656 nm. A 0.411-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 0.483 at 510 nm and 0.367 at 656 nm in a 1.00-cm cell. Calculate the concentration in parts per million of cobalt and nickel in the sample.

Check the solution

Explanation:

Absorbance of solution = Absorbance due to Co + absorbance due to Ni

Absorbance = molar absorptivity x concentration x path length

with path length = 1 cm

molar absorptivity at 510 nm,

For Co = 36400 M-1.cm-1

For Ni = 5520 M-1.cm-1

molar absorptivity at 656 nm,

For Co = 1240 M-1.cm-1

For Ni = 17500 M-1.cm-1

let [Co] and [Ni] represent the concentration of Co and Ni in diluted solution

then,

0.650 = 36400[Co] + 5520[Ni]   at 510 nm

0.588 = 1240[Co] + 17500[Ni]   at 656 nm

Solving the two equations,

[Co] = 1.3 x 10^-5 M

[Ni] = 3.3 x 10^-5 M

Concentration in 25 ml aliquot,

[Co] = 1.3 x 10^-5 M x 50 ml/25 ml = 2.6 x 10^-5 M

[Ni] = 3.3 x 10^-5 x 50 ml/25 ml = 6.6 x 10^-5 M

concentration (ppm) in original solution prepared by dissolving sample,

[Co] = 2.6 x 10^-5 M x 250 ml x 58.93 g/mol x 1000/25 ml = 15.322 ppm

[Ni] = 6.6 x 10^-5 M x 250 ml x 58.69 g/mol x 1000/25 ml = 38.735 ppm

the answer

explanation:

the capillary tube was too close to the bottom of the beaker.

the ruler may have moved.

water got into the capillary tube.

the temperature was not allowed to equilibrate in the 2-4 minutes.

answer: the enthalpy of the overall chemical reaction is +87.88 kj or +88 kj.

according to hess's law, the overall enthalpy of a reaction is the difference between the sum of enthalpy values of all the products and the sum of enthalpy values of all the reactants.

δh(reaction) = ∑nδh(products) - ∑nδh(reactants)

δh(reaction) = [δh +δh] - [δh]

δh(reaction) = [1(-287.02) + 1(0) ] - [1(-374.9)] kj = +87.88 kj

here, δh (pcl3) = -287.02 kj, δh(cl2) = 0, δh(pcl5) = -374.9 kj

therefore, enthalpy of the overall reaction is +87.88 kj or +88 kj.