Cu2+ reacts with neocuproine to form the colored complex (neocuproine)2Cu+, with

Question

Cu2+ reacts with neocuproine to form the colored 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 (isoamyl alcohol), an organic solvent that does not dissolve appreciably in water. In other words, when isoamyl alcohol is added to water, a two-layered mixture results, with the denser water layer at the bottom. If (neocuproine)2Cu+ is present, virtually all of it goes into the organic phase. For the purpose of this problem, assume that the isoamyl alcohol does not dissolve in the water at all and that all of the colored complex will be in the organic phase. Suppose that the following procedure is carried out:

1. A rock containing copper is pulverized, and all metals are extracted from it with strong acid. The acidic solution is neutralized with base and made up to 250.0 mL in flask A.

2. Next, 10.00 mL of the solution are transferred to flask B and treated with 10.00 mL of reducing agent to convert Cu2+ to Cu+. Then 10.00mL of bugger are added so that the pH is suitable for complex formation with neocuproine.  

3. 15.00 mL of this solution are withdrawn and placed in flask C. To the flask are added 10.00 mL of an aqueous solution containing neocuproine and 20.00 mL of isoamyl alcohol. After the mixture has been shaken well and the phases allowed to separate, all (neocuproine)2Cu+ is in the organic phase.

4. A few millimeters of the upper layer are withdrawn, and the absorbance at 454 nm is measured in a 1.00-cm cell.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 at 0.056.

a)Suppose that the rock contained 1.00 mg of Cu. What will be the concentration of Cu (mol/L) in the isoamyl alcohol phase?

b) If the molar absorptivity of (neocuproine)2Cu+ is 7.90 x 103 M-1 cm-1, what will be the observed absorbance? Remember that a blank carried through the same procedure gave an absorbance of 0.056.

c) A rock is analyzed and found to give a final absorbance of 0.874 (uncorrected for the blank). How many mg of Cu are in the rock?

Explanation / Answer

c) Volume of 1.0*10-3 M NaSCN taken = 5.8 mL;.

Total volume of the solution = 60.0 mL (see its written diluted to and not diluted with).

a) We start with 1.00 mg of copper.

Molar mass of copper = 63.54 g mol-1.

Therefore, moles of copper present initially = (1.00 mg)/(63.54 g mol-1) = 0.0157 mmole.

Concentration of the original solution in flask A = (0.0157 mmole)/(250 mL) = 6.28*10-5 mol/L

10.00 mL of the solution in flask A was taken and after work-up, made upto 30.00 mL.

Concentration of solution in flask B = (10.00 mL)*(6.28*10-5 mol/L)/(30.00 mL) = 2.093*10-5 mol/L 2.10*10-5 mol/L.

15.00 mL of the solution in flask B was taken and finally made upto a volume of 45.00 mL.

Concentration of the solution in flask C = concentration of Cu in the isoamyl alcohol phase = (15.00 mL)*(2.10*10-5 mol/L)/(45.00 mL) = 7.00*10-6 (ans).

b) The absorbance of the solution can be obtained by using Beer’s law:

A = *(concentration of Cu complex)*(length of the solution through which light travelled)

Now note that Cu and (neocuproine)Cu2+ are formed in a 1:1 ratio, so that the concentration of (neocuproine)Cu2+ = 7.00*10-6 mol/L

Plug in values:

A = (7.90*103 M-1cm-1)*(7.00*10-6 M)*(1.00 cm) = 0.055

This is the absorbance due to the Cu complex; the observed absorbance will take into account the absorbance due to water which is 0.056. Therefore, the observed absorbance will be the sum total of the two.

Observed absorbance = (0.055 + 0.056) = 0.111 (ans).

c) Final absorbance = 0.874; absorbance due to blank = 0.056; corrected absorbance of the sample = (0.874 – 0.056) = 0.818

Plug in values:

0.818 = (7.90*103 M-1cm-1)*c*(1.00 cm)

====> c = (0.818)/(7.90*10-1 M-1) = 1.035*10-4 M

Find out the moles of the concentrated sample = (1.035*10-4 M)*(45.00 mL/15.00 mL)*(30.00 mL/10.00 mL)*(250 mL)*(1 L/1000 mL) = 2.32875*10-4 mole.

Mass of Cu in the sample = (2.32875*10-4 mole)*(63.54 g/mol) = 0.01479 g = (0.01479 g)*(1000 mg/1 g) = 14.79 mg 14.80 mg (ans).

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.