Question

A 21.466 g sample of aqueous waste leaving a fertilizer manufacturer contains ammonia. The sample is diluted with 70.668 g of water. A 11.619 g aliquot of this solution is then titrated with 0.1062 M HCl. It required 30.75 mL of the HCl solution to reach the methyl red endpoint. Calculate the weight percent NH3 in the aqueous waste.

Answer 1

We know that

No. mole of = molarity Volume of solution in liter

no mole of HCl = 0.1062M 0.03075L = 0.00326565 mole

reaction take place between HCl and NH₃ as follow

HCl_(aq) + NH_3(aq)  NH₄Cl_(aq)

1:1 HCl and NH₃ required then for 0.00326565 mole of HCl required 0.00326565 mole of NH₃

That mean 11.619 gm of aliquot contain 0.00326565 mole of NH₃

Total mass of NH₃ solution = 21.466gm +70.668 = 92.134 gm

11.619 gm of aliquot contain 0.00326565 mole of NH₃ then 92.134 gm contain

92.1340.00326565 / 11.619 = 0.02589 mole of NH₃

molar mass of NH₃ = 17.031 gm/mol that mean 1 mole of NH₃ = 17.031 gm then 0.02589 mole of NH₃ =

0.0258917.031 = 0.4409 gm of NH₃

92.134 gm of solution contain 0.4409 gm Ammonia

weight % of solute = mass of solute100 / mass of solution

weight % of NH₃ = 0.4409100/92.134 = 0.4785%