Question

A pond’s water contains 25.0 mg/l of algae.

Find the theoretical carbonaceous oxygen demand (CBOD) of water in the pond

Find the total theoretical (carbonaceous plus nitrogenous ) oxygen demand (BOD) on water in the pond

Note: algae can be represented by the chemical formula C₆H₁₅O₆N.

Theoretically oxygen in pond can be consumed following equations 1 and 2.

C₆H₁₅O₆N + 6O₂ → 6CO₂ + 6H₂O + NH₃                     (1)

NH₃ + 2O₂ → NO₃^- + H⁺ + H₂O                                               (2)

Answer 1

Let's consider 1L of the pond's water

Given the mass of algae = 25.0 mg = 25.0 mg x(1 g / 1000 mg) = 0.0250 g

Molecular mass of algae (C6H15O6N) = 197.18 g/mol

Moles of algae(C6H15O6N) = mass / molecular mass = 0.0250 g / 197.18 g/mol = 1.268x10^-4 mol

The given chemical reaction for the oxidation of the organic compound is

C6H15O6N + 6O2 → 6CO2 + 6H2O + NH3

1 mol 6 mol 6 mol 6 mol 1 mol

Hence moles of oxygen required for the oxidation of 1.268x10^-4 mol of C6H15O6N

= (1.268x10^-4 mol C6H15O6N) x (6 mol O2 / 1 mol C6H15O6N) = 7.61x10^-4 mol O2 /L

Molecular mass of O2 = 32 g/mol

Hence mass of O2 required = 7.61x10^-4 mol x 32 g O2 / mol = 0.0243 g O2 / L = 24.3 mg / L

Hence  theoretical carbonaceous oxygen demand (CBOD) of water in the pond = 24.3 mg / L (answer)

Nitrogenous Oxygen demand:

1 mol of C6H15O6N forms 1 mol of NH2.

Hence moles of NH3 formed by 1.268x10^-4 mol C6H15O6N

= (1.268x10^-4 mol C6H15O6N) x ( 1mol NH3 / 1mol C6H15O6N) =1.268x10^-4 mol NH3

Now NH3 undergoes oxidation through the following reaction

NH3 + 2O2 → NO3- + H+ + H2O

1 mol 2 mol

Hence moles of O2 required by 1.268x10^-4 mol NH3 moles of NH3

= 1.268x10^-4 mol NH3 x ( 2 mol O2 / 1 mol NH3) = 2.536x10^-4 mol O2.

Molecular mass of O2 = 32 g/mol

Hence mass of O2 required = 2.536x10^-4 mol x 32 g O2 / mol = 0.00811 g O2 / L = 8.11 mg / L

Hence  theoretical nitrogenus oxygen demand of water in the pond = 8.11 mg / L

Hence total theoretical (carbonaceous plus nitrogenous ) oxygen demand (BOD) on water in the pond

= 24.3 mg / L + 8.11 mg/L

= 32.4 mg/L (answer)