According to the website of the Hungarian Central Bank, the silvery white Hungarian 2 forint coin is — Physical Chemistry — Kinetics Chemistry Question
Analysis of a Hungarian 2-Forint Coin
According to the website of the Hungarian Central Bank, the silvery white Hungarian 2 forint coin is composed of an alloy containing only copper and nickel. A curious chemist (who did not know that it is illegal to destroy money in Hungary) weighed a 2-Ft coin (3.1422 g) and dissolved it completely in concentrated nitric acid in about 4 hours under a fume hood. A brown gas was produced during this process and no other gaseous products were formed.
What are the chemical equations for the dissolution reactions?
Model Answer
Cu + 4 HNO3 → Cu(NO3)2 + 2 NO2 + 2 H2O
Ni + 4 HNO3 → Ni(NO3)2 + 2 NO2 + 2 H2O
Our hero diluted the solution to 100.00 cm3 in a volumetric flask. To determine the composition of the coin, he devised a clever plan. First, he prepared a Na2S2O3 solution by dissolving 6 g of Na2S2O3 5 H2O in 1.0 dm3 of water. Then he weighed 0.08590 g KIO3, dissolved it in water and prepared 100.00 cm3 stock solution in a volumetric flask. He measured 10.00 cm3 of this stock solution, and then added 5 cm3 20 % hydrochloric acid and 2 g solid KI. The solution turned brown immediately. Then he titrated this sample with the Na2S2O3 solution. In a number of parallel measurements the average for the equivalence point was 10.46 cm3.
Write down the equations of all the reactions that have taken place and determine the concentration of the Na2S2O3 solution. What could our hero have used as an indicator?
Model Answer
IO3− + 5 I− + 6 H+ → 3 I2 + 3 H2O
I2 + I− I3−
I2 + 2 S2O3 2− → 2 I− + S4O6 2−
I3− + 2 S2O3 2− → 3 I− + S4O6 2−
n(IO3−) = 0.0895 g / 214.00 g mol−1 10 cm3 / 100 cm3 = 4.01410−5 mol
n(I2) = 3 n(IO3−) = 1.204210−4 mol
n(S2O3 2−) = 2 n(I2) = 2.408410−4 mol
c(S2O3 2−) = 2.408410−4 mol / 0.01046 dm3 = 0.02302 mol dm−3
Starch solution.
When our hero began to wash up, he noticed that some white precipitate appeared in the first sample. He remembered clearly that he added more Na2S2O3 solution to this sample than was necessary to reach the end point.
What is the chemical equation of the process producing the precipitate?
Model Answer
S2O3 2− + H+ → HSO3− + S
Next, our hero returned to the greenish blue stock solution he prepared first. He measured 1.000 cm3 of this solution into a titration flask, added 20 cm3 of 5 % acetic acid and 2 g solid KI. He waited about 5 minutes. The solution became brown and a light-colored precipitate appeared.
What is the chemical equation of the process producing the colored species and the precipitate? Why did our hero have to wait? Why would it have been a mistake to wait hours rather than minutes?
Model Answer
2 Cu2+ + 4 I− → 2 CuI + I2
I2 + I− I3−
The redox reaction between Cu2+ and I− is not instantaneous. Under the described conditions, Five minutes are sufficient to ensure that the reaction is complete. Waiting hours would be a mistake, because oxygen in the air also slowly oxidizes I−.
Our hero then titrated the sample with his Na2S2O3 solution. The average for the equivalence point was 16.11 cm3. Now he could calculate the composition of the 2-Ft coin.
What is the mass percent composition of the coin?
Model Answer
n(S2O3 2−) = c V = 0.02302 mol dm−3 0.01611 dm3 = 3.70910−4 mol
n(Cu2+) (in 1.000 cm3 stock solution) = 2 n(I2) = n(S2O3 2−) = 3.70910−4 mol
m(Cu) = 3.70910−2 mol 63.55 g mol−1 = 2.357 g
The copper content of the 2-Ft coin is 2.357 g / 3.1422 g = 75.01 % by mass
As a good analytical chemist, he was not satisfied with one method and tried to determine the composition of the coin with complexometry. In this measurement he did not take into account the results obtained in the iodometric titration. First, he dissolved 3.6811 g Na2EDTA 2 H2O (M = 372.25 g mol-1) to make 1.0000 dm3 solution. Then he measured 0.2000 cm3 of the original greenish blue stock solution, added 20 cm3 of water and 2 cm3 of 25 % ammonia solution. The color of the solution became an intense violet.
Which species are responsible for this color? What is the purpose of the addition of ammonia?
Model Answer
The color of the ammine complexes of both Cu2+ and Ni2+ are violet. (In reality, the color of the Cu2+ ammin complex is much more intense.)
Ammonia is needed to adjust the pH to a suitable value to ensure practically complete complex formation with EDTA.
The equivalence point was 10.21 cm3 as calculated from the average of a few parallel experiments.
Did this experiment confirm the earlier conclusion about the composition of the coin?
Model Answer
n(EDTA) = 3.6811 g / 372.25 g mol−1 10.21 cm3 / 1000.00 cm3 = 1.01010−4 mol
n(Cu) + n(Ni) = 1.01010−4 mol 100.0 cm3 / 0.2000 cm3 = 0.05048 mol
From the mass of the coin:
M(Cu) n(Cu) + M(Ni) n(Ni) = 3.1422 g
Solving the simultaneous equations for n(Cu) and n(Ni):
n(Ni) = 0.0136 mol ⇒ m(Ni) = 0.796 g
n(Cu) = 0.0369 mol ⇒ m(Cu) = 2.35 g
This result agrees with the composition calculated earlier based on the iodometric titration.
Our hero was still not satisfied and also began to suspect that he made an error when he weighed the coin, so he turned on the old spectrophotometer in the lab. The lab he worked in was very well maintained so he found recently prepared and standardized 0.1024 mol dm–3 CuCl2 and 0.1192 mol dm–3 NiCl2 solutions in the lab. First, he measured the absorbance spectrum of the CuCl2 solution using a 1.000 cm quartz cell and made notes of the absorbance values at a few wavelengths he thought suitable:
Then he measured the absorbances of the NiCl2 solution at the same wavelengths in the same cell:
He diluted 5.000 cm3 of his original greenish blue stock solution to 25.00 cm3 in a volumetric flask and measured the absorbances. He obtained readings of 1.061 at 815 nm and 0.1583 at 395 nm.
Why did he dilute the solution? What is the composition of the coin based on these spectrophotometric data alone?
Model Answer
The undiluted stock solution would give absorbance values higher than 2.0, which cannot be measured reliably.
The molar absorption coefficients for Cu2+:
ε(260 nm) = 6.687 dm3 mol−1 cm−1
ε(395 nm) = 0.107 dm3 mol−1 cm−1
ε(720 nm) = 9.076 dm3 mol−1 cm−1
ε(815 nm) = 13.95 dm3 mol−1 cm−1
The molar absorption coefficients for Ni2+:
ε(260 nm) = 0.501 dm3 mol−1 cm−1
ε(395 nm) = 5.617 dm3 mol−1 cm−1
ε(720 nm) = 2.517 dm3 mol−1 cm−1
ε(815 nm) = 0.9916 dm3 mol−1 cm−1
The concentrations of the diluted stock solution can be obtained by solving the following simultaneous equations:
A(815 nm) = [ε(815 nm,Cu) c(Cu) + ε(815 nm,Ni) c(Ni)] 1.000 cm
A(395 nm) = [ε(395 nm,Cu) c(Cu) + ε(395 nm,Ni) c(Ni)] 1.000 cm
c(Cu) = 0.07418 mol dm–3 and c(Ni) = 0.02677 mol dm–3 for the diluted solution
The concentrations of the stock solution are five times greater.
c(Cu) = 0.3709 mol dm–3 and c(Ni) = 0.1338 mol dm–3
For the total volume of the stock solution (100.0 cm3):
n(Cu) = 0.03709 mol and n(Ni) = 0.01338 mol
This composition is in agreement with the titration results.
Next, he measured the absorbance at 720 nm and obtained 0.7405.
Is this value in agreement with the previous conclusions?
Model Answer
The expected absorbance value at 720 nm:
A(720 nm) = [ε(720 nm,Cu) c(Cu) + ε(720 nm,Ni) c(Ni)] 1.000 cm = 0.7404
This is in agreement with the measured value.
Finally, he tuned the instrument to 260 nm. He was surprised to see a reading of 6.000.
What was his expected reading?
Model Answer
The expected absorbance value at 260 nm:
A(260 nm) = [ε(260 nm,Cu) c(Cu) + ε(260 nm,Ni) c(Ni)] 1.000 cm = 0.5093
This does not agree with experimental observations.
He decided to measure the absorbance at this wavelength in a smaller, 1.00 mm quartz cell as well. Again, he obtained a reading of 6.000.
Suggest a possible explanation for this finding and a method to confirm it using chemicals and equipment that have already been used by our hero.
Model Answer
The spectrophotometer reading of 6.000 means that practically no light goes through the sample. This is unchanged in a shorter cell.
The molar absorbances of Cu2+ and Ni2+ were measured using CuCl2 and NiCl2 solutions. Nitric acid was used to dissolve the coin, so the concentration of the nitrate ion is high in the stock solution. The observations can be explained if the nitrate ion absorbed at 260 nm. This can be confirmed by recording the UV-VIS spectrum of a sample of dilute nitric acid.