It\'s long but any help at all would be nice. You weigh out a piece of copper wi

Question

It's long but any help at all would be nice.

You weigh out a piece of copper wire (AW=63.546):
Weight of copper standard....................0.3645 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.................... 55.1 %

Finally:
You are given a solid UNKNOWN, containing copper, weighing........ 0.8070 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.................... 25.00 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... 37.3 %
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

Part A.

wt. of copper sample = 0.3645 g

moles of Cu = g/molar mass = 0.3645/63.546 = 5.74 x 10^-3 mols

Molarity of Cu = moles/L = 5.74 x 10^-3/0.1 = 0.0574 M

Std. 1 : 10 ml diluted to 100 ml

Molarity = 0.0574 x 10/100 = 5.74 x 10^-3 M

std. 2 : 20 ml diluted to 100 ml

Molarity = 0.0574 x 20/100 = 0.0115 M

Absorbance = 2 - log(%T)

Calculations:

a) Abs. of std. 1 = (2-log(51.9)) - (2-log(99.7) = 0.2835

b) Abs. of std. 2 = (2-log(27.3)) - (2-log(99.7) = 0.5625

c) Molarity of Cu in std. 1 = 5.74 x 10^-3 M

d) Molarity of Cu in std. 1 = 0.0115 M

e) Ratio Abs/Molarity std 1. = 0.2835/5.74 x 10^-3 = 49.39

f) Ratio Abs/Molarity std 2. = 0.5625/0.0115 = 48.91

g) Average ratio Abs/Molarity = 48.65

h) Abs. of unknown = (2-log(55.1)) - (2-log(99.7) = 0.2575

concentration of Cu in unknown

(i) as measured in diluted solution = 0.2575 x 5.74 x 10^-3/0.2835 = 5.21 x 10^-3 M

(ii) Cu in original solution = 5.21 x 10^-3 x 0.1/0.01 = 0.0521 M

(iii) in original solution = (5.21 x 10^-3 x 63.546/10) x 100 = 3.31 w/V%

Part B.

wt. of Cu sample = 0.8070 g

moles = 0.8070/63.546 = 0.127 mols

Dissolved in 100 ml

Molarity = 0.127/0.1 = 1.27 M

Dilution 25 ml to 100 ml

Molarity = 25 x 1.27/100 = 0.3175 M

%T = 37.3%

Calculations:

l) Abs of diluted solution = 2-log(37.3) = 0.4283 - 0.0013 = 0.427

m) Cu conc. in diluted solution = 0.427 x 5.74 x 10^-3/0.2835 = 8.64 x 10^-3 M

n) Cu conc. in original solution = 4 x 8.64 x 10^-3 = 0.03456 M

o) % of Cu in original solution = (0.03456 x 0.01 x 63.546/0.8070) x 100 = 2.72%

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