Question

a) The water is chlorinated to remove ammonia. If the end nitrogen product is nitrate, what minimum sodium hypochlorite concentration in mg/L is needed to remove 21 mg/L of ammonia "as N"?

b) If a 25% stoichiometric excess of hypochlorite is added to achieve the above ammonia oxidation, how much sodium bisulphite is then needed to remove the excess chlorine?

Answer 1

NH3 + 3NaClO - > 3NaNO3 + 3 HCl

21 mg/ l = conc of ammonia

conc of ammonia = 21/17 mmol/l

=1.23 mM

Fo1 1 mol H3 3 moles of NaOCl is required

So moles of NaOCl is required = 3 *1.24

=3.7 mMoles

= 276.1 mg/l

b.

If a 25% stoichiometric excess of hypochlorite is added, then sodium bisulphite is then needed to remove the excess chlorine

NaClO2 + 2Na2SO3 -> 2Na2SO4 + NaCl

moles of hypochlorite= .25 * 3.7

= 0.925 mM

moles of   sodium bisulphite = 2 * .925

= 1.85 mmol/ L

Mass of  sodium bisulphite = 233 mg/l