Question

A 25.199 g sample of aqueous waste leaving a fertilizer manufacturer contains ammonia. The sample is diluted with 78.055 g of water. A 13.740 g aliquot of this solution is then titrated with 0.1085 M HCl. It required 32.65 mL of the HCl solution to reach the methyl red endpoint. Calculate the weight percent NH3 in the aqueous waste.

Answer 1

1 mol of HCl consumes 1 mol of ammonia to form ammonium chloride

Hence 1 mol of HCL= 1 mol of Ammonia

36.5 g of HCL = 17 g of Ammonia ( based of the mol wt)

HCl consumed in the titration = 32. 65 mL of strength 0.1085 M

                            Calculation for      Wt of HCl in 32.65 mL HCl of 0.1085 M

                              32.65X 0.1085 X 36.5/1000 = 0.129 g

Total weight of the sample = 25.199 + 78.055

                                         =103.254 g

HCl consumed for 13.74 aloquot is 0.129 g

HCl required for 103.254 = 0.129 X 103.254/13.74

                                     =0.9694 g

36.5 g of HCl=17 g of Ammonia

0.9694 g HCl =0.4515 g of ammonia

103.254 g of sample =0.4515 g of ammonia

100 g of sample =0.43 g

Hence ammonia con is 0.43 %