Question

2.The treated effluent from a wastewater treatment plant contains ammonium (NH4+) at 76.0 mg NH4+-N/L and nitrate at 22.0 mg NO3--N/L. Convert these concentrations to mg NH4+/L and mg NO3-/L.

For the data provided in question 2, express the ammonium and nitrate concentrations in (a) molar units, and (b) ppmm NH4+ and ppmm NO3-.

Answer 1

76 mg NH4+-N/L means that concentration of ‘N’ atom within ammonium cation is 76 mg/ L

Atomic mass of ‘N’ ~ 14

Molecular mass of NH4+ ~ 18

Therefore,

If 1 gm/ mol of ‘N’ is 76 mg/ L è 1 gm/ mol of NH4+ will be 76 * (18/14) = 97.71 mg/ L

97.71 mg/ L of NH4+ in moles = 97.71 X 10^-3/ 18 = 5.43 X 10^-3

In dilute solutions, 1ppm = 1 mg/L

Therefore, 97.71 mg/ L = 97.71 ppm

Similarly,

22 mg NO3--N/L means that concentration of ‘N’ atom within nitrate anion is 22 mg/ L

Atomic mass of ‘N’ ~ 14

Molecular mass of NO3- ~ 62

Therefore,

If 1 gm/ mol of ‘N’ is 22 mg/ L è 1 gm/ mol of NO3- will be 22 * (62/14) = 336.57 mg/ L

336.57 mg/ L of NO3- in moles = 336.57 X 10^-3/ 62 = 5.43 X 10^-3

In dilute solutions, 1ppm = 1 mg/L

Therefore, 97.71 mg/ L = 336.57 ppm