Chemical dosimeters are widely used for remote determination of high-level doses of radiation. Most — Physical Chemistry Chemistry Question
Chemical dosimeter
Chemical dosimeters are widely used for remote determination of high-level doses of radiation. Most common type of dosimeter is a ferrous-ferric sulfate dosimeter. When ionizing radiation passes through the solution, lots of products (radicals, ions, and molecules) are formed. Most of them can oxidize Fe2+ into Fe3+.
Choose the electronic configuration of Fe2+ cation:
А) 3d64s0 B) 3d54s0 C) 3d44s2 D) 3d54s1
Model Answer
А)
Write the balanced equations for oxidation of Fe2+ with the following particles:
А) H2O+
В) ОН–
С) H2O2
Model Answer
А) H2O+ + Fe2+ → H2O + Fe3+
B) ОН· + Fe2+ → OH– + Fe3+
C) H2O2 + Fe2+ → OH– + ОН· + Fe3+
The resulting solution is then titrated and the amounts of Fe(II) and Fe(III) are calculated. For these purpose, permanganatometric and iodometric titrations can be used.
Write down the balanced redox equation for the reaction between:
а) iron(II) sulfate and potassium permanganate in acidic media;
b) iron(III) sulfate and potassium iodide;
с) sodium thiosulfate and iodine.
Model Answer
a) 10 FeSO4 + 2 KMnO4 + 8 H2SO4 → 5 Fe2(SO4)3 + 2 MnSO4 + K2SO4 + 8 H2O.
b) Fe2(SO4)3 + 6 KI → 2 FeI2 + I2↓ + 3 K2SO4.
c) I2 + 2 Na2S2O3 → 2 NaI + Na2S4O6.
Calculate the concentrations of cations Fe2+ and/or Fe3+ in each of the following cases:
а) 12.30 cm3 of potassium permanganate solution (c = 0.1000 mol dm-3) were necessary to titrate 20.00 cm3 of Fe(II) solution.
b) 1.00 cm3 of solution containing Fe(III) was diluted up to 20.00 cm3 and an excess of potassium iodide solution was then added to the prepared solution. The iodine formed was titrated with 4.60 cm3 sodium thiosulfate solution (c = 0.0888 mol dm-3).
с) An aliquot 5.00 cm3 was titrated with 0.1000 mol dm-3 potassium permanganate solution, the average volume being equal 7.15 cm3. After that, an excess of potassium iodine solution was added to the flask. Titration of the resulting solution required 13.70 cm3 of sodium thiosulfate solution with its concentration of 0.4150 mol dm-3.
Model Answer
а) Concentration of iron(II) can be calculated from equation 15.3(а):
12.3 × 0.1000 × 5 = 20 * x
x = 0.3075 mol dm-3
b) Using equation 15.3(b) one can assume that the amount of iodine occurred after potassium iodide addition is two times smaller than the amount of iron(III). From equation 15.3(с) the amount of thiosulfate spent for iodine titration is two times greater than that of iodine. Therefore, concentration of iron in the initial aliquot is:
x = 0.0888 × 4.6,
x = 0.4085 mol dm-3
с) The initial amount of potassium permanganate is 7.15 cm3 × 0.1000 mol dm-3 = 0.7150 mmol. So, from stoichiometry of 15.3(а) the amount of substance of Fe2+ is:
n(Fe2+) = 7.15 cm3 × 0.1000 mol dm-3 × 5 = 3.5750 mmol.
At the endpoint of the redox titration all iron is in the Fe3+ form and therefore total iron is determined by titration with sodium thiosulfate.
n(Fe total) = 13.7 cm3 × 0.4150 mol dm-3 = 5.6855 mmol.
Then, the amount of iron(III) is:
n(Fe3+) = 5.6855 – 3.5750 = 2.1105 mmol.
The iron concentrations are:
c(Fe2+) = 3.5750 / 5 = 0.7150 mol dm-3,
c(Fe3+) = 2.1105 / 5 = 0.4221 mol dm-3.
Frequently, chemical dosimeters are used for measuring doses near nuclear reactors where large amounts of various radionuclides are accumulated.
Match mother and daughter radionuclide, indicate the type of decay (α or β–) in each case:
Mother radionuclide:
a) 60Co b) 90Sr c) 226Ra d) 137Cs e) 212Po
Daughter radionuclide:
1) 90Zr 2) 137Xe 3) 214Rn 4) 222Rn 5) 137Ba 6) 60Ni 7) 60Fe 8) 208Pb 9) 90Y
Model Answer
a–6–β–;
b–9–β–;
c–4–α;
d–5–β–;
e–8–α
Radioactivity A is directly proportional to the number of particles N of a substance J (A = λ N), where λ is the decay constant related to half-life by the equation λ = ln 2 / T1/2. Radioactivity is measured in becquerel (symbol Bq) units: 1 Bq is one decay per second.
Using values of half-lives, calculate (in GBq) radioactivity of samples containing:
a) 1.3141 g 226RaCl2
b) 1.0 mg 90Sr(NO3)2 and 0.5 mg 137CsNO3
T1/2(226Ra) = 1612 years, T1/2(90Sr) = 29 years, T1/2(137Cs) = 30 years.
Model Answer
a) First, calculate the number of 226Ra atoms:
NRa = NA·ν = NA·m/M = 6.02·10^23 · 1.3141 / (226+71) = 2.6636·10^21.
Then, the decay constant (half-life in seconds) can be determined:
λRa = ln 2 / (1612 × 365 × 24 × 3600) = 1.3635·10^–11 s^–1.
The radioactivity is:
ARa = λRa * NRa = 2.6636·10^21 × 1.3635·10^–11 = 36.32 GBq.
b) The same calculations for cesium-137 and strontium-90 give:
ACs = 1.5126·10^18 × 7.3265·10^–10 = 1.108 GBq
ASr = 2.8131·10^18 × 7.5792·10^–10 = 2.132 GBq
Total radioactivity is the sum of both values:
Atot = 3 GBq.
Explain why 226Ra is dangerous for humans?
Model Answer
The daughter radionuclide of 226Ra is 222Rn – noble gas that is radioactive and can easily penetrate different obstacles.
It is well known that radionuclide 64Cu decays to 64Ni (this is attributed to electron capture, process when nuclei absorb inner electron) and 64Zn (β– decay). Half-life for electron capture is 20.8 hours, for β– decay is 32.6 hours.
a) Calculate the average half-life of 64Cu.
b) How much time is required for radioactivity of a 64Cu sample to decrease by 10 times?
Model Answer
a) This process can be described as two competitive reactions of the first order. Then:
λ_total = λ_(e capture) + λ_(β)
ln 2 / T_1/2(total) = ln 2 / T_1/2(e capture) + ln 2 / T_1/2(β)
1 / T_1/2(total) = 1 / T_1/2(e capture) + 1 / T_1/2(β)
T_1/2(total) = (T_1/2(e capture) * T_1/2(β)) / (T_1/2(e capture) + T_1/2(β)) = (20.8 × 32.6) / (20.8 + 32.6) = 12.7 hours
b) The integrated equation of decay is as follows: A(t) = A0e^–λt. Taking into account that radioactivity decreased ten times, time is calculated as follows:
0.1 = e^–λt
ln 0.1 = –λ t
t = ln 10 / λ = 12.7 × ln 10 / ln 2 = 42.2 h.