2Mn2+(aq)? +?? 4OH-?(aq) + O2?(aq) → 2MnO2?(s)?? + 2H2O (l)
MnO2?(s)??+? 2I?(aq) + 4H+?(aq) ?→ Mn2+(aq) + I2(aq) + 2H2O(l)
2S2O32?(aq) + I2?(aq)?→ S4O62?(aq) + 2I?(aq)
1 mole of O2?→ 2 moles of MnO2?→ 2 moles of I2?→ 4 moles of S2O32?
A sample of lake water was analysed using the Winkler Method. The size of the sample was 600 cm3?and the following table shows the results of a titration of the liberated iodine against 0.0500 mol dm-3?sodium thiosulfate solution in the final step of the analysis:Determine the dissolved oxygen content of the water sample in g dm-3
Answer:
Step 1:?Determine the average volume of? delivered
Average volume Na2S2O3??=? (18.35 + 18.30 + 18.40)?÷ 3
Average volume Na2S2O3??= 18.35 cm3
Step 2: Determine the number of moles of sodium thiosulfate reacted
Moles of Na2S2O3?= volume in dm3?x concentration = 0.01835 dm3?x 0.0500 mol dm-3= 9.175 x 10-4?mol
Step 3:?Determine the moles of oxygen reacted
1 mol of O2?is equivalent to 4 mol of Na2S2O3
Therefore, amount of oxygen in the sample is = (9.175 x 10-4)?÷ 4? = 2.294 x 10-4??mol
Step 4:?Calculate the concentration of the dissolved oxygen
Concentration = mol?÷?volume in dm3?= (2.294 x 10-4??mol)?÷?0.600 dm3= 3.823 x 10-4??mol dm-3
Convert to g dm-3?= 3.823 x 10-4?mol dm-3??x 32.00 g mol-1?=?0.01223 g dm-3
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