Question

When a 0.5725 g sample of Lysol toilet bowl cleaner was titrated with 0.1025 M NaOH, an end point was obtained at 11.49 mL. Calculate the percent (by weight) of hydrochloric acid in the Lysol sample.

Answer 1

moles of NaOH= 0.01149 L x 0.1025 = xxx moles of HCl
xxx moles of HCl x molecular mass HCl= yyy g HCl
yyy g HCl x 100/0.5725 g sample= zzz % of HCl in the sample