Winkler Method
Chemistry: IB Diploma Higher Level, topic 6.6.2
The Winkler Method
The Winkler method is a redox titration process used to determine the concentration of dissolved oxygen in water samples.
The Winkler method
- The Winkler method can be used to determine the biological oxygen demand (BOD) of a water sample.
- The titration relies on a multi-step process, shown below:
- Step 1: 2Mn2+(aq) + 4OH-(aq) + O2(g) → 2MnO(2)(s) + 2H2O(l)
- Step 2: MnO2(s) + 2I-(aq) + 4H+(aq) → Mn2+(aq) + I2(aq) +2H2O(l)
- Step 3: 2S2O32-(aq) + I2(aq) → S4O62-(aq) + 2I-(aq)
- The endpoint of the titration is observed by the disappearance of the orange-brown iodine colour, which produces the colourless iodide ions.
Stoichiometry
- Considering the stoichiometry of the overall process, we can calculate the moles of oxygen in the initial water sample:
- O2 : MnO2 : I2 : S2O32- = 1 : 2 : 2 : 4
- The important ratio to consider is O2 : S2O32- = 1 : 4
- The concentration of oxygen in the analysed water sample is typically calculated in mg dm-3, which is the same as ppm (parts per million).
Sample question
- A 500.0 cm3 sample of river water was analysed to determine the dissolved oxygen concentration by adding Mn2+ in basic solution, followed by acidified KI. This solution required 13.50 cm3 of 0.0420 mol dm-3 of Na2S2O3(aq) to completely react with the iodine produced.
- Use the information and the equations for the Winkler process to determine the dissolved oxygen content of the water sample in mg dm-3 (ppm).
Answer
- No. of moles of S2O32-:
- n(S2O32-) = CV = 0.0420 mol dm-3 × 0.01350 dm3 = 5.67 ×10-4 mol
- No. of moles of O2 (1:4 stoichiometric ratio):
- n(O2) = 5.67 × 10-4 mol ÷ 4 = 1.4175 × 10-4 mol
- Mass of O2:
- m(O2) = 1.4175 × 10-4 mol × 31.998 g mol-1 = 4.536 ×10-3 g
- Convert to mg dm-3:
- mg dm-3 = 4.536 mg ÷ 0.5000 dm3 = 9.07 mg dm-3
Test yourself on Winkler Method
What is the endpoint of the titration in the Winkler method?
What is the correct stoichiometric ratio of O2 : S2O32- in the Winkler method?