Question

23.495 g sample of aqueous waste leaving a fertilizer manufacturer contains ammonia. The sample is diluted with 72.311 g of water. A 12.949 g aliquot of this solution is then titrated with 0.1080 M HCl . It required 32.27 mL of the HCl solution to reach the methyl red endpoint. Calculate the weight percent NH3 in the aqueous waste.

Answer 1

1.86% NH₃

Step-by-step explanation:

The reaction that takes place is:

• HCl(aq) + NH₃(aq) → NH₄Cl(aq)

We calculate the moles of HCl that reacted, using the volume used and the concentration:

• 32.27 mL ⇒ 32.27/1000 = 0.03227 L

• 0.1080 M * 0.03227 L = 3.4852x10⁻³ mol HCl

The moles of HCl are equal to the moles of NH₃, so now we calculate the mass of NH₃ that was titrated, using its molecular weight:

• 3.4852x10⁻³ mol NH₃ * 17 g/mol = 0.0592 g NH₃

The weight percent NH₃ in the aliquot (and thus in the diluted sample) is:

• 0.0592 / 12.949 * 100% = 0.4575%

Now we calculate the total mass of NH₃ in the diluted sample:

Diluted sample total mass = Aqueous waste Mass + Water mass = 23.495 + 72.311 = 95.806 g

• 0.4575% * 95.806 g = 0.4383 g NH₃

Finally we calculate the weight percent NH₃ in the original sample of aqueous waste:

• 0.4383 g NH₃ / 23.495 g * 100% = 1.86% NH₃