Ammonium (NH^+_4) is a very troublesome water pollutant because it can have mult

Question

Ammonium (NH^+_4) is a very troublesome water pollutant because it can have multiple bad impacts if discharged to surface water. As we discussed in class, NH^+_4 will be nitrified by aerobic bacteria according to the following reaction (same as in course notes): NH^+_4 + 2O_2 rightarrow NO^-_3 + 2H^+ + H_2O The manure from a confined animal feeding operation was spilled into a river so that the concentration of NH^+_4 after mixing was 10 mgN/L. If all of the NH^+_4 reacts by nitrification according to the reaction above, how much dissolved oxygen and alkalinity will be consumed (in mg/L of O_2 or of CaCO_3, as appropriate).

Explanation / Answer

The reaction is NH4+ +2O2 --------> NO3-+ 2H+ +H2O,

The concentration of NH4+ is given after mixing in river is 10 mg N/L .,

HERE we will first calculate the moles of nitrogen from gms of nitrogen , first we will convert mg to gms by dividing mg/1000=10/1000= 0.01 grams of nitrogen per L,

THEN WE CAN CALCULATE THE MOLES of nitrogen= 0.01/14= 7.14*10-4 ( N= mol.wt =14) or it can be written as,

0.000714 moles,

from the reaction above , for one mole of nitrogen to nitrified it is required to use 2 moles of oxygen molecule.

we have calculated the moles of Nitrogen = 0.000714 *2 = 0.001428 moles of O2,,

The two moles of oxygen molecule will be dissolved , here we can calculate the quantity of dissloved oxygen as,

0.001428 moles of O2*32 = 0.04569 grams of oxygen is dissolved in per L (32 = MOL WT. OF O2)

WE CAN CONVERT THE grams to miligrams by , 0.04569*1000= 45.69 mg/ L O2

Also we can calculate the CaCO3 mg/L , By the following reaction,

2 NH4OH + CaCO3 ---------> Ca(OH)2 + (NH4)2 CO3 ,

because the Nitrogen in the form of NH4+ , Dissolved in water, it forms NH4OH and it reacts with calcium carbonate by the above reaction to give ammonium carbonate and calcium hydroxide,

we have moles of Nitrogen atom= 0.000714 , here 2 moles of ammonium hydroxide gives one mole of calcium hydroxide means, for one mole os N atom, it is required 0.5 moles of calcium carbonate,

i.e. 0.000714/2= 0.000357 moles of calcium carbonate is required,

mol wt. of caco3=100.08,

0.000357*100.08= 0.0357 grams caco3,

we will convert it to miligrams = 0.0357*1000= 35.7285 mg/L of CaCO3.

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