Question

A 20.399 g sample of aqueous waste leaving a fertilizer manufacturer contains ammonia. The sample is diluted with 78.161 g of water. A 14.000 g aliquot of this solution is then titrated with 0.1076 M HCl. It required 28.62 mL of the HCl solution to reach the methyl red endpoint. Calculate the weight percent NH3 in the aqueous waste.

Answer 1

HCl + NH3 -----> NH4Cl

moles of HCl = molarity *volume in L

                   = 0.1076*0.02862 = 0.003079 moles

from balanced equation

1 mole of HCl react with moles of NH3

0.003079 moles of HCl react with 0.003079 moles of NH3

there solution is 14gm

mass of solution = 20.399+78.161 = 98.56gm

moles of NH3 in the mass =0.003079*98.56/14 = 0.043 moles

mass of NH3 = no of moles * gram molar mass

                    = 0.043*17 = 0.73gm in 98.56

Mass%NH3 = 0.73*100/98.56   = 0.74%maas/mass