Question

1. When a 0.5725 g sample of Lysol toilet bowl cleaner was titrated with 0.100 M NaOH, an end point was obtained at 15.00 mL. Calculate the percent (by weight) of hydrochloric acid in the Lysol sample.

b. When a 3.529 g sample of Liquid-Plumr was titrated with 0.100 M HCl, two end points were obtained. The first end point occured at 15.0 mL, the second at 42.00 mL. Calculate the percentages of both NaOH and NaOCl in Liquid-Plumr.

Answer 1

1. moles of NaOH added = 0.1 M x 15 ml = 1.5 mmol

moles of HCl present would be = 1.5 mmol

mass of HCl = 1.5 mmol x 36.5 g/mol/1000 = 0.055 g

Mass % HCl in lysol = 0.055 x 100/0.5725 = 9.607%

2. Liquid-plumr has NaOH and NaOCl in it

moles of HCl at first end point = 0.1 M x 15 ml = 1.5 mmol = mols of NaOH in sample

mass of NaOH = 1.5 mmol x 40 g/mol/1000 = 0.06 g

Mass % NaOH in sample = 0.06 x 100/3.529 = 1.7%

moles of HCl at IInd end point = 0.1 M x (42 - 15)ml = 2.7 mmol = mols of NaOCl

mass of NaOCl in sample = 2.7 mmol x 74.44 g/mol/1000 = 0.201 g

Mass % NaOCl in sample = 0.201 x 100/3.529 = 5.70%