There are 3 seperate questions. All questions must be answered and with detailed

Question

There are 3 seperate questions. All questions must be answered and with detailed solutions to get five stars.

At a pH of 4, the Cu-neocuproine complex formed in the liquid-liquid extraction experiment has a partition coefficient of about 6.1 when equilibrated between the acid aqueous solution and chloroform.   Calculate the concentration of Cu in the chloroform phase if 39.0 mL of a 0.26 M aqueous Cu solution buffered to pH 4 is equilibrated with 25.0 mL of chloroform.

A spectrophotometer was calibrated with a 8.128  10-4M (in Cu) Cu-neocuproine solution.  This calibration was carried out in a 1.00 cm cell, and yielded an absorbance of 0.2540 AU.  An unknown was then analyzed in this same spectrophotometer, but using a 2.50 cm cell. If the absorbance of the unknown was 0.3050 AU, what is the Cu concentration?

You are using the following procedure from your lab instructions to determine the copper content of a brass sample:

If you actually used a 22.3 mg sample of brass, and the resulting diluted solution was found to contain 9.399 mg/L of Cu, what is the %Cu in the sample?

Explanation / Answer

Total moles of Cu = volume x concentration

= 39/1000 x 0.28 = 0.01092 mol

Let a be the moles of Cu left in the aqueous phase after extraction

Moles of Cu in chloroform = 0.01092 - a

Partition coefficient = [Cu(chloroform]/[Cu(water)]

= (moles of Cu in chloroform/volume of chloroform)/(moles of Cu in water/volume of water)

= ((0.01092 - a)/0.012)/(a/0.039) = 6.5

((0.01092 - a)/0.012) = 6.5 x (a/0.039)

250a = 0.91

a = 0.00364

Moles of Cu in chloroform = 0.01092 - a = 0.00728 mol

Concentration of Cu in chloroform = moles/volume

= 0.00728/0.012 = 0.607 M

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