Question

The Winkler method for dissolved oxygen in water is based on the rapid oxidation of solid Mn(OH)₂ to Mn(OH)₃ in alkaline medium. When acidified, the Mn (III) readily releases iodine from iodide. A 240.-mL water sample, in a stoppered vessel, was treated with 1.10 mL of a concentrated solution of NaI and NaOH and 1.10 mL of a manganese(II) solution. Oxidation of the Mn(OH)₂ was complete in about 1 min. The precipitates were then dissolved by addition of 2.20 mL of concentrated H₂SO₄, whereupon an amount of iodine equivalent to the Mn(OH)₃ (and hence to the dissolved O₂) was liberated. A 24.5-mL aliquot (of the 244 mL) was titrated with 14.2 mL of 0.00861 M thiosulfate. Calculate the mass in milligrams of O₂ per milliliter sample. Assume that the concentrated reagents are O₂ free and take their dilutions of the sample into account.

Answer 1

Write down the equations taking place:

Mn²⁺ (aq) + 2 OH^- (aq) ------> Mn(OH)₂ (s) …..(1)

2 Mn(OH)₂ (s) + O₂ (aq) + 2 H₂O (l) -------> 2 Mn(OH)₃ (s) …..(2)

2 Mn(OH)₃ (s) + 2 I^- (aq) + 4 H⁺ (aq) ------> 2 Mn²⁺ (aq) + I₂ (aq) + 6 H₂O (l) …..(3)

I₂ (aq) + 2 S₂O₃^2- (aq) -------> S₄O₆ (aq) + 2 I^- (aq) …..(4)

Therefore, moles I₂ titrated = 2*moles of thiosulfate added (as per the balanced stoichiometric reaction above).

Moles thiosulfate added = (volume of thiosulfate added in L)*(concentration of thiosulfate in mol/L) = (14.2 mL)*(1 L/1000 mL)*(0.00861 mol/L) = 1.22262*10^-4 mol.

Moles I₂ titrated = (1.22262*10^-4 mol thiosulfate)*(1 mole I₂/2 mole thiosulfate) = 6.1131*10^-5 mole.

Now, moles I₂ titrated = 2*moles of Mn(OH)₃ present = moles of O₂ present (refer to equations 3 and 2 above).

Therefore, moles I₂ titrated = moles of O₂ present in the sample = 6.1131*10^-5 mol.

6.1131*10^-5 mol O₂ is present in 24.5 mL aliquot of the sample solution. The original volume of the sample solution taken = 240 mL(we took this volume of sample solution).

Therefore, moles O₂ present in 240 mL sample solution = (6.1131*10^-5 mol)*(240 mL/24.5 mL) = 5.9883*10^-4 mol.

Molar mass of O₂ = 32 g/mol.

Therefore, mass of O₂ present in the original sample = (5.9883*10^-4 mol)*(32 g/1 mol) = 0.01916 g.

240 mL sample contains 0.01916 g O₂. Therefore, 1 mL sample contains (0.01916 g/240 mL)*(1 mL) = 7.9833*10^-5 g = 0.07983 mg ≈ 0.08 mg

Therefore, concentration of dissolved O₂ = 0.08 mg/mL (ans).