Compounds of sulfur in its lower oxidation states are present in many industrial wastes (metallurgy, — Physical Chemistry — Electrochemistry Chemistry Question
Sulfur determination
Compounds of sulfur in its lower oxidation states are present in many industrial wastes (metallurgy, production of paper, chemical) and are dangerous ecotoxicants. The prevalent forms of sulfur in lower oxidation states in solutions are S2–, SO3 2– and S2O3 2– ions. Their content can be determined by redox titration under different conditions.
To a 20.00 cm3 sample containing S2–, SO3 2– and S2O3 2– an excess of ZnCO3 suspended in water was added. Upon completion of the reaction the solution was filtered into a 50.00 cm3 volumetric flask and diluted to the mark. To 20.00 cm3 of the filtrate an excess of aqueous formaldehyde was added. The mixture was acidified with acetic acid and titrated with 5.20 cm3 of iodine standard solution (c = 0.01000 mol dm–3).
a) Write down the net ionic equations of the reactions taking place during the analysis.
b) Which ion, S2–, SO3 2– or S2O3 2–, can be determined by this method?
c) Calculate the concentration of this ion in ppm in the initial solution.
Model Answer
a) ZnCO3(s) + S2– → ZnS(s) + CO3 2–
SO3 2– + CH2O + H+ → CH2(OH)SO3 –
2 S2O3 2– + I2 → S4O6 2– + 2 I–
b) S2O3 2–
c) n(S2O3 2–) = 2 × 5.20 × 0.01000 = 0.104 mmol (in 20.00 cm3 of the filtrate)
c(S2O3 2–) = 0.104 / 20.00 × 50.00 / 20.00 = 0.01300 mol dm-3 (in the initial) = 0.01300 × 112.13 g dm-3 = 1.46 g dm-3 (1460 ppm)
A 20.00 cm3 of the sample of the iodine solution (c = 0.01000 mol dm–3) was acidified with acetic acid and then combined with 15.00 cm3 of the filtrate above. The mixture was titrated with 6.43 cm3 of the standard sodium thiosulfate solution with a concentration of 0.01000 mol dm–3.
a) Write down the net ionic equations of the reactions taking place during the analysis.
b) Which ion, S2–, SO3 2– or S2O3 2–, can be determined by this method taking into account the result of the previous experiment?
c) Calculate the concentration of this ion in ppm in the initial solution.
Model Answer
a) 2 S2O3 2– + I2 → S4O6 2– + 2 I–
SO3 2– + I2 + H2O → SO4 2– + 2 H+ + 2 I–
b) SO3 2–
c) n(I2)initial = 20.00 × 0.01000 = 0.2000 mmol
n(I2)excessive = 0.5 × 6.43 × 0.01000 = 0.0322 mmol
n(SO3 2–) + 0.5 n(S2O3 2–) = 0.2000 – 0.03215 = 0.1679 mmol (in 15.00 cm3 of the filtrate)
n(SO3 2–) = 0.1679 – 0.5 × 0.1040 / 20.00 × 15.00 = 0.1289 mmol (in 15.00 cm3 of the filtrate)
c(SO3 2–) = 0.1289 / 15.00 × 50.00 / 20.00 = 0.0215 mol dm-3 (in the initial) = 0.0215 × 80.07 g dm-3 = 1.720 g dm-3 (1720 ppm)
A 10.00 cm3 sample of iodine solution (0.05000 mol dm–3) was acidified with acetic acid and then 10.00 cm3 of the original sample containing S2–, SO3 2– and S2O3 2– were added. The mixture was titrated with 4.12 cm3 of the standard sodium thiosulfate solution with a concentration of 0.05000 mol dm–3.
a) Write down the net ionic equations of the reactions taking place during the analysis.
b) Which ion, S2–, SO3 2– or S2O3 2–, can be determined by this method taking into account the results of two previous determinations?
c) Calculate the concentration of this ion in ppm in the initial solution.
Model Answer
a) 2 S2O3 2– + I2 → S4O6 2– + 2 I–
SO3 2– + I2 + H2O → SO4 2– + 2 H+ + 2 I–
S2– + I2 → S + 2 I–
b) S2–
c) n(I2)initial = 10.00 × 0.05000 = 0.5000 mmol
n(I2)excessive = 0.5 × 4.12 × 0.05000 = 0.103 mmol
n(S2–) + n(SO3 2–) + 0.5 n(S2O3 2–) = 0.5000 – 0.1030 = 0.3970 mmol (in 10.00 cm3 of the initial)
n(S2–) = 0.3970 – 10.00 × 0.02148 – 10.00 × 0.5 × 0.01300 = 0.1172 mmol (in 10.00 cm3 of the initial)
c(S2–) = 0.1172 / 10.00 = 0.01172 mol dm–3 = 0.01172 × 32.07 g dm–3 = 0.376 g dm–3 (376 ppm)