Question

A 24.201 g sample of aqueous waste leaving a fertilizer manufacturer contains ammonia. The sample is diluted with 72.053 g of water. A 13.774 g aliquot of this solution is then titrated with 0.1045 M HCl. It required 28.02 mL of the HCl solution to reach the methyl red endpoint. Calculate the weight percent NH3 in the aqueous waste.

Answer 1

Answer:

1.44%

Explanation:

Initially, there is 24.201g of the waste. Let this waste contain 'x' mols of  ammonia($NH_{3}$). Now, this waste is dissolved in 72.053g of water, thus making the weight of the solution to 24.201+72.053 = 96.254g. Now, from this, 13.774g is taken. Note that, when the aqueous waste is dissolved in water, the ammonia particles are uniformly distributed, i.e, the 'x' mols of ammonia is uniformly present in the 96.254g of the solution. Hence, when we take 13.774g of the solution, only a fraction of the ammonia particles is taken. This fraction is equal to $\frac{13.774}{96.254}$ of x, which is equal to 0.1431 times 'x'.

For the HCl solution, 28.02mL of 0.1045M solution contains 28.02x0.1045 = 2.9281mmols of HCl is present in it.

The basic titration reaction that occurs is : $HCl + NH_{3}$→$NH_{4}Cl$  , i.e, one mol of ammonia requires one mol of HCl for neutralization. Therefore, for the above solution of HCl containing 2.9281mmols, same amount of ammonia, i.e, 2.9281mmols is required for complete neutralization.

Therefore, 0.1431x = 2.9281 mmols ⇒ x = 20.4619 mmols.

The molecular weight of ammonia is 17g/mol. Therefore, the weight of 20.4619mmols of ammonia has a weight(w) = 20.4619 x $10^{-3}$ x 17 = 0.3478 g.

Therefore the weight of ammonia in the initial aqueous waste is 0.3478 g. The total weight of the waste is 24.201g, hence, the percent weight of ammonia is given by $\frac{0.3478}{24.201}$×100 = 1.44%