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

we are given with the titration with 0.100 M NaOH obtained at 15.00 mL. The moles equivalent of HCl here is equal to 0.1*15x10^-3 or 0.0015 moles. This is equivalent to 0.055 grams HCl. the percent purity then is equal to 0.055/0.5725 or 9.55%.

Answer 2

9.55%

Step-by-step explanation:

First thing's first, we write out the balanced chemical equation;

HCl (aq) + NaOH (aq) --> NaCl (aq) + H2O (l)

Now convert 15.00mL into L by dividing by 1000.

15.00 mL * (1L/1000mL) = 0.01500L

Next we multiply the number of liters by the molar concentration of NaOH to get the number of moles in the amount of titrated solution.

Number of moles (Mol) = Molar Concentration (Mol/L) * Volume (L)

0.01500L * 0.1000 mol/L = 1.50*10^-3 mol NaOH

The endpoint of the reaction will be approximately the equivalence point, thus at the endpoint:

The equivalence point is the point in a titration where the amount of titrant added is enough to completely neutralize the analyte solution. The moles of titrant (standard solution) equal the moles of the solution with unknown concentration.

Hence;

mol HCl = mol NaOH

From the reaction;

1 mol of HCl reacts with 1 mol of NaoH

Now we know that we have reacted equal amounts of HCl and NaOH, so we also know the number of moles of HCl that were reacted in the equation: 1.50*10^-3 mol

Mass (g) = Number of moles (mol) * Molar mass (g/mol)

1.50*10^-3 mol * 36.46 g/mol = 0.05469 g HCl

Now if we divide the number of grams of HCl present by the total mass of the original sample, we will get the percent by mass of hydrochloric acid in Lysol.

(0.05469g/0.5725g) * 100= 0.0955 * 100 = 9.55%