Question

A 25.888 g sample of aqueous waste leaving a fertilizer manufacturer contains ammonia. The sample is diluted with 73.464 g of water. A 10.762 g aliquot of this solution is then titrated with 0.1039 M HCl . It required 31.89 mL of the HCl solution to reach the methyl red endpoint. Calculate the weight percent NH3 in the aqueous waste.

Answer 1

Answer:

Weight % of NH₃ in the aqueous waste  = 2.001 %

Explanation:

The chemical equation for the reaction

$\\\\NH_3} + HCl -----> NH_4Cl$

Moles of HCl = Molarity × Volume

= 0.1039 × 31.89 mL × $\frac{1 \ L}{1000 \ mL}$

= 0.0033 mole

Total mass of original sample = 25.888 g + 73.464 g

= 99.352 g

Total HCl taken for assay = $\frac{10.762 \ g}{99.352 \ g}$

= 0.1083 g

Moles of NH₃ = $\frac{0.0033 \ mol}{0.1083}$

= 0.03047 moles

Mass of NH₃ = number of  moles × molar mass

Mass of NH₃ = 0.03047 moles × 17 g

Mass of NH₃  = 0.51799

Weight % of NH₃  = $\frac{0.51799 \ g}{25.888 \ g} * 100%$%

Weight % of NH₃ in the aqueous waste  = 2.001 %