When a 0.5725 g sample of lysol toilet bowl cleaner was titrated with 0.1025 M NaOH an end point was obtained at 14.73 mL. Calculate the mass percent of hydrochloric acid in the lysol sample?

Answer 1

First write the balanced chemical equation.

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

Now convert 14.73mL into L by dividing by 1000.

14.73 mL * (1L/1000mL) = 0.01473L

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

0.01473L * 0.1025 mol/L = 1.51*10^-3 mol NaOH

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

mol HCl = mol NaOH

The coefficient of each reactant should match that of the balanced chemical equation. Both HCl and NaOH have a coefficient of 1 in the balanced chemical equation, thus it is a 1:1 exchange ratio.

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: 1.51*10^-3 mol

If we multiply the number of moles of HCl by the molar mass of HCl, we will get the number of grams of HCl present.

1.51*10^-3 mol * 36.46 g/mol = 0.0551 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.

0.0551g/0.5725g = 9.62% HCl

Answer 2

Hydrochloric acid has a molar mass of 36.45 g/mol.

0.1025(mmol/mL)*14.73(mL)*36.45(mg/mmol)*0.001(g/mg) / 0.5725(g) * 100% = 9.613% HCl

Note: all value were calculated in 4 signifcant digits according to the given figures.

However this does not suggest that the outcomes are as accurate; at least triplicate titrations on 3 different weights (amounts) of the same sample are needed for an accuracy of <0.1% of outcome values.