Question

Determine the carbonaceous and nitrogenous oxygen demand in mg/L for a 1 L solution containing 300 mg of a wastewater represented by the formula CN2H602 (N is converted to NH3 in the first step) Afterwards calculate the COD of the solution

Answer 1

Explanation:

It is given that in 1 liter of solution, there is 300 g of wastewater in $C_{9}N_{2}H_{6}O_{2}$.

Therefore, the reaction equation will be as follows.

           $C_{9}N_{2}H_{6}O_{2} + 8O_{2} \rightarrow 2NH_{3} + 9CO_{2}+ 0.H_{2}O$

For carbonaceous oxygen demand:

        Molecular weight of $C_{9}N_{2}H_{6}O_{2}$ = $9(12) + 2(14) + 6(1) + 2(16)$

                                           = 174 g/mol

or,                                        = $174 \times 10^{3}$ mg/mol   (as 1 g = 1000 mg)

So, total moles present in 300 mg of solution will be as follows.

                 Moles = $\frac{mass}{\text{molar mass}}$

                            = $\frac{300 mg}{174 \times 10^{3} mg/mol}$

                            = $1.724 \times 10^{-3} mol$

Hence, for 1 mol of carbonaceous oxygen demand is B mol.

Therefore, for $1.724 \times 10^{-3} mol$ oxygen required is calculated as follows.

                    $8 \times 1.724 \times 10^{-3} mol$

                      = 0.0138 mol

Weight of oxygen required (m) = $0.0138 mol \times 32 g/mol$

                                                   = 0.4414 g

or,                                                = 441.4 mg

This means COD (carbonaceous oxygen demand) is 441.4 mg.

For nitrogenous oxygen demand:

              $NH_{3} + 2O_{2} \rightarrow HNO_{3} + H_{2}O$

For 1 mol of $NH_{3}$ to convert into $HNO_{3}$, oxygen required is 2 mol.

For $2 \times 1.724 \times 10^{-3} mol$ of $NH_{3}$, oxygen is calculated as follows.

                  $2 \times 2 \times 1.724 \times 10^{-3} mol$

                      = 0.006896 mol $O_{2}$

Mass of $O_{2}$ required is 0.22 g = 220.6 mg

         NOD (Nitrogenous oxygen demand) = $\frac{220.6 mg O_{2}}{1 L}$

Therefore, calculate total biological oxygen demand (BOD) as follows.

                   = $\frac{(441.4 + 220.6) mg \text{O_{2}}}{1 L}$                              

                   = 662 mg $O_{2}$/L