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

Question

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

Weight of copper standard....................0.2802 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: Blank 99.7 % Standard 1 49.3 % Standard 2 23.8 %

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.................... 39.7 % CALCULATE

a) The absorbance of Standard 1 ______________

b) The absorbance of Standard 2 ______________

c) The molarity of Cu in Standard 1 ______________ M

d) The molarity of Cu in Standard 2 ______________ M

e) Ratio: absorbance/molarity Std 1 ______________

f) Ratio: absorbance/molarity Std 2 ______________

g) Average ratio of absorbance/molarity ______________

h) Absorbance of unknown solution ______________

Concentration of copper in unknown...

i) solution as measured in cuvet ______________ M

j) in original solution ______________ M

k) in original solution ______________ w/V%

Finally: You are given a solid UNKNOWN, containing copper, weighing........ 1.2296 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.................... 8.00 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... 53.3 % 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 by absorbance

a) Absorbance of Standard 1 = (2 - log(49.3)) - (2 - log(99.7)) = 0.306

b) Absorbance of Standard 2 = (2 - log(23.8)) - (2 - log(99.7)) = 0.622

c) concentration of Cu in initial 100 ml solution = 0.2802 g/63.546 g/mol x 0.1 L = 0.0441 M

Molarity of Cu in standard 1 = 0.0441 M x 10 ml/100 ml = 0.00441 M

d) Molarity of Cu in standard 2 = 0.0441 M x 20 ml/100 ml = 0.00882 M

e) Ratio absorbance/molarity Std 1 = 0.306/0.00441 = 69.39

f) Ratio absorbance/molarity Std 1 = 0.622/0.00882 =70.52

g) average ratio of absorbance/molarity = 69.955

h) Absorbance of unknown = (2 - log(39.7)) - (2 - log(99.7)) = 0.400

concentration of Cu in unknown (in 10 ml) = 0.4 x 10/69.955 = 0.0572 M

i) solution as measured in cuvette = 0.00572 M

j) original solution = 0.0572 M

k) in original solution w/v% = 0.0572 M x 10 ml x 63.546 g x 100/mol/10 x 1000 = 0.363%

Unknown solid = 1.2296 g

l) Absorbance = (2 - log(53.3)) - (2 - log(99.7)) = 0.272

m) Concentration of Cu in diluted solution = 0.272/69.955 = 0.0039 M

n) Concentration of Cu in original solution = 0.0039 M x 100 ml/8 ml = 0.049 M

o) % Cu in original solid = 0.049 M x 0.100 L x 63.546 g/mol x 100/1.2296 g = 25.194%

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