Question

A 25.225 g sample of aqueous waste leaving a fertilizer manufacturer contains ammonia. The sample is diluted with 75.815 g of water. A 14.842 g aliquot of this solution is then titrated with 0.1068 M HCl . It required 29.99 mL of the HCl solution to reach the methyl red endpoint. Calculate the weight percent NH3 in the aqueous waste

Answer 1

1.43 (w/w %)

Step-by-step explanation:

HCl reacts with NH3 as follows:

HCl + NH3 → NH4+ + Cl-

1 mole of HCl reacts per mole of ammonia.

Mass of NH3 is obtained as follows:

Moles HCl:

0.02999L * (0.1068mol / L) = 3.203x10-3 moles HCl = Moles NH3

Mass NH3 in the aliquot:

3.203x10-3 moles NH3 * (17.031g / mol) = 0.0545g.

Mass of sample + water = 22.225g + 75.815g = 98.04g

Dilution factor: 98.04g / 14.842g = 6.6056

That means mass of NH3 in the sample is:

0.0545g * 6.6056 = 0.36g NH3

Weight percent is:

0.36g NH3 / 25.225g * 100

1.43 (w/w %)