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

Question

You weigh out a piece of copper wire (AW=63.546):Weight of copper standard....................0.3951 g
Dissolve it in 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 98.6 %
Standard 1 67.2 %
Standard 2 43.9 %

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.................... 70.4 %

CALCULATE
a) absorbance of Standard 1 is .167
b) absorbance of Standard 2 is .351
c) molarity of Cu in Standard 1 is 0.0062175 M
d) molarity of Cu in Standard 2 is 0.012435 M
e) Ratio: absorbance/molarity Std 1 is26.860 M^-1
f) Ratio: absorbance/molarity Std 2 is 28.227 M^-1
g) Average ratio of absorbance/molarity 27.544 M^-1
h) Absorbance of unknown solution .146
Concentration of copper in unknown...
i) solution as measured in cuvet 0.0050036 M
j) in original solution is 0.050036 M
****k) in original solution ????? w/V%

*****all of the calculations for 'a' through 'j' are correct but i can't figure out 'k'; i have already checked a similar problem on chegg and the equation is wrong. any help in step-by-step format would be appreciated, as i would like to understand it!

Explanation / Answer

the blank tells me to subtract 1.1% away from the transmittances a) The absorbance of standard 1: log of T : log of 0.598 = 0.223 b) The absorbance of standard 2: log of T : log of 0.368 = 0.434 c) The molarity of Cu in Standard 1 : 0.2848 g Cu @ 63.55 g/mol = 0.004482 moles Cu 0.004482 moles Cu /100 ml = 0.04482 Molar stock solution 0.04482 Molar stock diluted 10ml / 100ml = 0.004482 Molar d) The molarity of Cu in Standard 2 : same stock solution 0.04482 Molar stock diluted 20ml / 100ml = 0.008963 Molar e) Ratio: absorbance/molarity std 1: 0.223 / 0.004482 = 49.8 abs/ M f) Ratio: absorbance/molarity std 2: 0.434 / 0.008963 = 48.4 abs /M g) Average ratio of absorbance/molarity : 49.1 abs/M h) Absorbance of unknown solution : log 0.660 = 0.180 Concentration of copper in unknown... i) solution as measured in cuvet : 0.180 abs @ 49.1 abs/M = 0.00368 Molar j) in original solution had been diluted 10X's: 0.0368 Molar k) in original solution: 0.0368 mol Cu @ 63.55 g/mol = 0.2336 grams Cu per litre 0.2336 grams Cu per litre = 0.02336 grams /100 ml w/V% = 0.0234 % i've got to run.... double check my math... email me with any questions... I'll double check it later I did the work with all the digits my calculator had, & only rounded off each final answer

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