Problem : You weigh out a piece of copper wire (AW=63.546): Weight of copper sta

Question

Problem:
You weigh out a piece of copper wire (AW=63.546):
Weight of copper standard....................0.2785 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:

Next:
You pipetted 10 mL of UNKNOWN into a 100 mL volumetric flask. The required amount of ammonia was added and the mixture diluted to volume. The %T of this solution was measured at 625 nm, using the same spectrometer and cuvet as for the standards.
%T unknown.................... 38.8 %

Finally:
You are given a solid UNKNOWN, containing copper, weighing........ 1.0791 g.
You dissolve this in concentrated Nitric Acid, HNO3, transfer quantitatively to a 100 mL volumetric flask, add the required amount of Ammonia, NH3, dilute to volume and mix thoroughly.
This resulting solution is too concentrated and the resulting %T measured in the same cuvet and spectrometer as before gave little or no transmission (<10%).
You are told to pipet.................... 17.00 mL
into a 100 mL volumetric flask, add the required amount of ammonia, dilute to volume and mix.
Spectroscopic measurement of this solution gave %T... 30.8 %
CALCULATE:
l) Absorbance of the diluted solution....................._____________________
m) The copper concentration in the diluted solution...._____________________ M
n) The copper concentration in the original solution..._____________________ M
o) The Per Cent copper in the original solid UNKNOWN..._____________________ % Cu

Explanation / Answer

Determination of Cu in a sample

Absorbance = 2 - log(%T)

Absorbance of blank = 2 - log(98.7) = 0.006

Absorbance of standard 1 = 2 - log(48.4) = 0.315 - 0.006 = 0.309

Absorbance of standard 2 = 2 - log(21.7) = 0.664 - 0.006 = 0.658

Absorbance of unknown solution = 2 - log(30.8) = 0.511 - 0.006 = 0.505

molarity of Cu stock solution = 0.2785/63.546 x 0.1 = 0.044 M

Calculations

a) Absorbance of standard 1 = 0.309

b) Absorbance of standard 2 = 0.658

c) molarity of standard 1 = 0.044 M x 10 ml/100 ml = 0.0044 M

d) molarity of standard 2 = 0.044 M x 20 ml/100 ml = 0.0088 M

e) ratio absorbance/molarity for standard 1 = 0.309/0.0044 = 70.23

f) ratio absorbance/molarity for standard 2 = 0.658/0.0088 = 74.77

g) average ratio absorbance/molarity = 72.50

h) absorbance of unknown solution = 0.505

concentration of copper in unknown

i) solution as measured in cuvet = 0.505/72.50 = 0.007 M

j) in original solution = 0.007 x 100/17 = 0.0412 M

k) in original solution = 0.0412 M x 63.546 g x 100/mol/100 ml = 2.62%

l) absorbance of diluted solution = 0.505

m) Copper concentration in diluted solution = 0.007 M

n) copper concentration i original solution = 0.0412 M

o) percent copper in original solid unknown = 0.262 x 100/1.0791 = 24.25%

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