Analytical chem

Cu+ reacts with neocuproine to form the coloured complex (neocuproine)2Cu+, with an absorption maximum at 454 nm. Neocuproine is particularly useful because it reacts with few other metals. The copper complex is soluble in 3-methyl-1-butanol, an organic solvent that does not dissolve appreciably in water. If (neocuproine)2Cu+ is present, virtually of it goes into the organic phase. For the purpose of this problem, assume that all of the coloured complex transfers into the organic phase. Supposed that the following procedure is carried out:
 A rock containing copper is pulverized, and all metals are extracted from it using strong acid. The acidic solution is neutralized with base and made up to 250.0 mL in flask A.
 Next, 15.00 mL of the solution is transferred to flask B and treated with 15.00 mL of reducing agent to convert Cu2+ to Cu+. Then 15.00 mL of buffer are added so that the pH is suitable for complex formation with neocuproine.
 25.00 mL of this solution are withdrawn and placed in flask C. To the flask are added 15.00 mL of an aqueous solution containing neocuproine and 30.00 mL of 3-methyl-1-butanol. After the mixture has been shaken well and the phases allowed to separate, all (neocuproine)2Cu+ is in the organic phase.
 A few milliliters of the organic layer are withdrawn, and the absorbance at 454 nm is measured in a 1.00-cm cell. A blank carried through the same procedure gives an absorbance of 0.072.
a. Suppose that the rock contained 3.00 mg of Cu. What will be the concentration of Cu (mol/L) in the organic phase?
b. If the molar absorptivity of (neocuproine)2Cu+ is 7.90 x 103 M-1 cm-1, what will be the observed absorbance?
c. Another rock is analyzed in the same manner and found to give a final absorbance of 0.753 (uncorrected for the blank). How many milligrams of Cu are in the rock?

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asked by panchi
  1. If I read this right, you have 3 mg Cu initially, you place it in 250 mL, withdraw 15, add 15 more of something else and take 25 of that 30. So you should have 0.003/63.55 mol = ? initially and at the end it will be
    (0.003/63.55) x (15/250) x (25/30) = ? and that is in 30 mL of the solvent so ?/0.030L = approx 8E-5M for part a.
    A = e*b*c
    A = 7.90E3*1.00*about 8E-5M will give you the A with no blank. Add 0.072 to that for the reading for part b. I think that is about 0.632 and add 0.072 for the final of about 0.704. But note that the 8E-5 is an estimated value so the numbers won't be exactly as above.

    c. A uncorrd = 0.753
    A corrected = 0.753-0.072 = 0.681
    Here is what I would do for part c. I would forget all of those dilution steps and extraction steps and recognize that everything done to the standard was done to the sample, also. Then I would compute a new "corrected" molar absorptivity(call it k). It won't be the molar absorptivity but it will be k when c is in mg and it will work the problem.
    A = k*b*c
    0.681 = e*b*3 mg Cu
    k = 0.632/3 = 0.210 and use that for the new sample.
    A = k*b*c
    0.681 = 0.210*1*c
    c = about 3.24 mg (still estimated).
    You can do it another way.
    3 mg Cu x (0.681/0.632) = about 3.24 mg.

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