The biochemical oxygen demand (BOD) is the capacity of organic and biological matter in natural water to consume oxygen. It is determined experimentally by measuring the oxygen concentration in a sealed sample before and after a 5-day incubation period. The standard titration procedure reacts the oxygen in the sample with manganese (II) sulfate in basic solution. Manganese (IV) oxide is precipitated, which oxidizes iodide ion to diatomic iodine. The concentration of iodine is determined by titration with a standard solution of sodium thiosulfate. The reactions are:
2 Mn2+(aq) + 4 OH-(aq) + O2(aq) -> 2 MnO2(s) + 2 H2O(l)
MnO¬2(s) + 4 H+(aq) + 2 I-(aq) -> Mn2+(aq) + I2(aq) + 2 H2O(l)
I2(aq) + 2 S2O32-(aq) -> S4O62-(aq) + 2 I-(aq)
On the first day and the fifth day, 10.00 mL aliquots of the water sample were subjected to the above series of reactions. Of the 1.00 x 10-3 M solution of Na2S2O3, 10.15 mL were required for the titration on day 1 and 2.40 mL were required on day 5 to titrate the I2 produced. Calculate the BOD of this natural water sample in milligrams per liter.© BrainMass Inc. brainmass.com October 25, 2018, 5:18 am ad1c9bdddf
On day 1, we used 10.15mL of 0.001M thiosulfate to titrate I2. We should find the number of moles here:
0.001mol/L * 0.01015L = 1.1015x10^-5 moles of thiosulfate required
We can change this amount to moles of I2, and then change I2 to MnO2 using the second equation. ...
If we take aliquots of water samples and test them through titration with manganese sulfate, how do we calculate he biochemical oxygen demand of the system? The math for this problem is explained step-by-step.
Biological oxygen demand and water hardness
1.) Two determinations of the dissolved oxygen content of a 200 mL sample of water, made five days apart, gave 5.3 ppm and 2.2 ppm of dissolved O2 respectively. What is the biological oxygen demand of the sample? The following formula was given to us:
BOD= mg dissolved 02 consumed/ Liter of water
but I'm not sure how to convert ppm to mg dissolved 02 consumed
2.) A 25.00 mL sample of a standard solution containing 1.0 g of CaCO3/L required 20.20 mL of EDTA to fully complex the Ca present. Calculate the volume of EDTA stoichiometrically equivalent to 1.0 mg of CaCo3View Full Posting Details