Question

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.

Answer 1

find the concentration: \(M_1V_1=M_2V_2\)

then find the moles (n): \(Molarity=\dfrac{n_{solute}}{L_{solution}}\)

then the mass: \(n=\dfrac{mass}{Molar\;mass}\)

then: mass percent=\(\dfrac{mass \;of\;HCl_{(s)}}{mass\;of\;sample}*100%\%\)