Problem 13-5 new book: A 25 ml sample of river water was titrated with 0.001 M K2Cr2O7 (potassium dichromate) and required 8.3 ml to reach the end point. What is the chemical oxygen demand (COD) in milligrams of O2 per liter of sample.
Answer:
First, the chemical oxygen demand (COD) is determined by adding dichromate (Cr2O7-2) to the water. It gets reduced (by compounds that would theoretically react with O2 in the water) to Cr+3.
Cr2O7-2 +14 H+ +6e- à 2Cr+ +7H2O
Cr+3 is then an indicator of the amount of dichromate reacted
Note that 6 electrons (6e-) are required for the reaction and the number of electrons are what is carrying out the reduction of Cr2O7-2.
Next, we know from the text, that oxygen is reduced (or demanded) in water by the following reaction.
O2 + 4H+ 4e- à 2H2O
Since 6 electrons are required to reduce dichromate to Cr+3 and only 4 electrons are needed to reduce O2 to water, 6/4 times the moles of Cr2O7-2 reacted will be equivalent to the moles of oxygen (COD) reduced in the water if the same contaminants in water were reducing both.
Now back to the problem: From the concentration and volume of K2Cr2O7 we can calculate the number of moles of dichromate reacted:
0.001 moles/liter x 0.0083 liters = 8.3x 10-6 moles, right?
The equivalent O2 demand is 8.3x 10-6 x 6/4 moles.
Putting this is grams, we have 8.3x 10-6 x 6/4 moles x 32 grams/mole for O2.
Since this number of grams of O2 is in a 25 ml sample, divide by 0.025liters and the COD is 16 mg/liter.
Your book talks about some of the short comings of this test, but for this problem we are way above the BOD for unpolluted surface waters of 0.7 mg O2/liter and almost twice the equilibrium saturated O2 in water of 8.7 mg/liter (see below). Your book says the BOD for sewage can be several hundreds of mg/liter of O2.
Note that the Henry’s law constant give for oxygen in water in your book is 1.3 x10-3 moles per liter per atmosphere. Unfortunately when we covered Henry’s law before, I used the convention of atmospheres liters/mole {ie. KH = p/Cwsat) }; Your book uses the opposite convention, so Baird and Cann have KH book = Cwsat/p.
For oxygen this is:
KH book = O2 in the water phase/partial pressure of O2 in the atmosphere.
Your book gives a KH book of O2 = 1.3 x10-3 moles/liter water/atmosphere at 25oC.
If O2 has a partial pressure of 0.21 atmospheres (21%) you can then, in a crisis situation, calculate the saturated equilibrium concentration of O2 in water, which after converting to milligrams is 8.7 mg O2 /liter of water or 8.7 ppm.