Question

A 22.683 g sample of aqueous waste leaving a fertilizer manufacturer contains ammonia. The sample is diluted with 77.686 g of water. A 14.695 g aliquot of this solution is then titrated with 0.1055 M HCl. It required 32.13 mL of the HCl solution to reach the methyl red endpoint. Calculate the weight percent NH3 in the aqueous waste.

Answer 1

Solution) Total mass of solution = 22.683+77.686 = 100.369 grams

NH3 + HCl -------> NH4Cl

1 mole NH3 require 1 mole HCl.

Moles of HCl = 0.1055*32.13/1000 = 0.0034 moles

So, moles of NH3 = moles of HCl = 0.0034 moles

Mass of NH3 = moles * molar mass = 0.0034*17 = 0.0578 grams

It means 14.695 grams solution contain 0.0578 grams of NH3

So, 100.369 grams will contain 0.0578*100.369/14.695 = 0.389 grams of NH3

mass % of NH3 = 0.389*100/22.683 = 1.715 % ...... Answer