Pre-Laboratory Assignment for Copper Analysis by Spectroscopy Before you can sta

Question

Pre-Laboratory Assignment for Copper Analysis by Spectroscopy Before you can start this experiment you must: 1. Solve the problem and enter the correct answers 2. Pass an 8 question on-screen quiz Problem: You weigh out a piece of copper wire (AW-63.546): Weight of copper standard...._.0.3223 g Dissolve it in a slight excess of concentrated nitric acid. Add to the solution the required amount of concentrated ammonia. Transfer this quantitatively to a 100 mL volumetric flask, dilute to volume, and mix thoroughly. 10.00 mL of this is pipetted into a 100 ml volumetric flask, the required amount of concentrated ammonia is added and diluted to volume. This is Standard 1 Standard 2 is made by pipetting 20 mL of the original solution into a 100 mL volumetric flask, adding the required amount of ammonia, NH3, and diluting to volume. Using the same cuvet for all measurements, the following %T's were obtained at 625 nm:

Explanation / Answer

Transmittance, T = I / I0
% Transmittance, %T = 100 T

Absorbance,

A = log10 I0/I
A = log10 1/T

Transmittance of the dilute solution

= 33.7/100

l. Absorbance of the diluted solution, A = log10 1/T

= 0.472

m. To calculate the concentration you need the value of the molar absorption coefficient () of the Cu complex. You can determine the value of from the standard solutions prepared above.

From Beer's law, A = cl

A graph of A vs c for the standard solutions give a slope of l. With the knowledge of l (length of the cell used in spectroscopy) you can get the value of .

Once you have , you can calculate the concentration of the unknown solution from the equation:

c = A/ l

= (0.472/ l) M

n. Cu concentration in the original solution

= {(0.472/ l)*100}/17 M

o. Amount of Cu in the concentrated solution

= {(0.472/ l)*100}/(17*10) moles              as 1000 mL solution contains {(0.472/ l)*100}/17 moles of Cu                                                                    so 100 mL solution contains {(0.472/ l)*100}/(17*10) moles of                                                                    Cu

= [{(0.472/ l)*100}/(17*10)] * 63.546 g Cu

% Cu in the original sample

= ([{(0.472/ l)*100}/(17*10)] * 63.546/ 1.2758) * 100

Transmittance, T = I / I0
% Transmittance, %T = 100 T

Absorbance,

A = log10 I0/I
A = log10 1/T

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