You weigh out a piece of copper wire (AW=63.546): Weight of coppers standard0.28

Question

You weigh out a piece of copper wire (AW=63.546): Weight of coppers standard0.2828 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 I. 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 covet for all measurements, the following %Ts were obtained at 625 nm: 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 covet as for the standards. T unknown 41.0% You are given a solid UNKNOWN, containing copper. weighing 1.2556 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 covet and spectrometer as before gave little or no transmission (

Explanation / Answer

Absorbance = 2 - log(%T)

Molarity of stock standard solution = 0.2828 g/63.546 g/mol x 0.1 L = 0.0445 M

First part

a) Absorbance of standard 1 = (2 - log(50.6)) - (2 - log(99.7)) = 0.294

b) Absorbance of standard 2 = (2 - log(25.1)) - (2 - log(99.7)) = 0.599

c) Molarity of Standard 1 = 0.0445 M x 10 ml/100 ml = 4.45 x 10^-3

d) Molarity of Standard 2 = 0.0445 M x 20 ml/100 ml = 8.90 x 10^-3

e) ratio Absorbance/molarity standard 1 = 0.294/4.45 x 10^-3 = 66.07

f) ratio Absorbance/molarity standard 2 = 0.599/8.90 x 10^-3 = 67.30

g) average Absorbance/molarity = 66.70

h) Absorbance of unknown = (2 - log(41)) - (2 - log(99.7)) = 0.386

concentration of Cu in unknown

i) concentration of Cu as measured in cuvet = 0.386/66.70 = 5.79 x 10^-3 M

j) concentration of Cu in original solution = 5.79 x 10^-3 x 100/10 = 0.058 M

k) concentration of Cu in original solution = 0.058 M x 0.010 L x 63.546 g/mol/10 = 0.37% in 10 ml solution

For unknown solid

l) Absorbance of diluted solution = (2 - log(52.7)) - (2 - log(99.7)) = 0.277

m) Cu concentration in diluted solution = 0.277/66.70 = 4.153 x 10^-3 M

n) Cu concentration in original solution = 4.153 x 10^-3 M x 100 ml/9 ml = 0.046 M

o) percent Cu in original solid = 0.046 M x 0.1 L x 63.546 g/mol x 100/1.2556 g = 23.28% Cu

Related Questions

Navigate

• Browse (All)
• Browse Y
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.