Question

A wastewater treatment plant does secondary treatment consisting of BOD oxidation and nitrification. What is the total nitrogen concentration in the wastewater effluent if the wastewater contains 2 mg/L NH4+ and 10 mg/L NO3−. Answer in mg/L as N.

Answer 1

Concentration of nitrogen =------ mg /L as 'N'

Given-

2 mg/L as NH₄⁺

10 mg /L as NO₃^-

as per information

# Molar mass of NH₄⁺ = 18.03 g/mol

#Molar mass of NO₃^-  = 62 g/mol

We have to calculate number of moles of ammonium and nitrate ion in mentioned solution -

Number of moles of NH₄⁺ present in 1 Lit of solution =

= (mass of NH₄⁺/ Molar mass of NH₄⁺)

=(2 .0 mg /18.03 g/mol)

=0.111 mmol of NH₄⁺

In Similar way moles of nitrate =

=(Mass of NO₃^- /Molar mass of NO₃^-)

= (10 mg /62 g/mol)

=0.161 mmol of NO₃^-

Total moles as NH₄⁺   and NO₃^- = 0.111 mmol + 0.161 mmol

=0.272 mmol

Now we can convert total nitrogen containing per liter =

= total moles x molar mass  of nitrogen

= 0.272 mmol x 14 g/mol)

= 3.8 mg

Form this it is clear that, one liter of effluent contains 3.8 mg N ,

So, concentration effluent solution = 3.8 mg /L as N.