Question

A 23.130 g sample of aqueous waste leaving a fertilizer manufacturer contains ammonia. The sample is diluted with 80.347 g of water. A 13.078 g aliquot of this solution is then titrated with 0.1061 M HCl. It required 29.58 mL of the HCl solution to reach the methyl red endpoint. Calculate the weight percent NH3 in the aqueous waste.

Answer 1

Balanced chemical equation is:

HCl + NH3 ---> NH4Cl

lets calculate the mol of HCl

volume , V = 29.58 mL

= 2.958*10^-2 L

use:

number of mol,

n = Molarity * Volume

= 0.1061*2.958*10^-2

= 3.138*10^-3 mol

According to balanced equation

mol of NH3 reacted = (1/1)* moles of HCl

= (1/1)*3.138*10^-3

= 3.138*10^-3 mol

This is number of moles of NH3

Molar mass of NH3,

MM = 1*MM(N) + 3*MM(H)

= 1*14.01 + 3*1.008

= 17.034 g/mol

use:

mass of NH3,

m = number of mol * molar mass

= 3.138*10^-3 mol * 17.03 g/mol

= 5.346*10^-2 g

Now use:

Mass % of NH3 = mass of NH3 * 100 / mass of mixture

= (5.346*10^-2)*100 / 13.078

= 0.409 %

Answer: 0.409 %