Acetic Acid Quality Control

Question

a 5.00 ml sample of toilet cleaner weighs 4.902 g. when titrated, those 5.00 ml of the household product need 25.50 ml of naoh 0.506 m for the end point to be reached. a) what is the density of the toilet cleaner? b) how many moles of hcl are contained in those 5.00 ml of household product? c) what mass of hcl is contained in those 5.00 ml of household product? d) what is the molarity of the toilet cleaner? e) what is the mass percentage of hcl in the toilet cleaner?

I could calculate:

a) 0.9804 g/ml

b) 0.1344 mols

c) ?

d) ?

e) ?

Need help please.

Answer 1

I'll need to start at the beginning and also be sure we have the same answers for a) and b).

Assuming the only titratable ingredient is HCl, we proceed as follows:

NaOH + HCl ==> H2O + NaCl ... neutralization reaction

a) density of cleaner = mass / volume = 4.902 g / 5.00 ml = 0.9804 g / ml

b) Find mols NaOH used then mols HCl

25.50 mls NaOH x 1 L / 1000 mls x 0.506 mol / L = 0.01290 mols NaOH

0.01290 mols NaOH x 1 mol HCl / mol NaOH = 0.01290 mols HCl

Not sure how you got 0.1344 mols HCl. I'm guessing you used 4.902 g, but that isn't ALL HCl. See (e).

c) mass of HCl:

0.01290 mols HCl x 36.5 g / mol = 0.471 g HCl

d) molarity of HCl in the cleaner:

0.01290 mols HCl / 5.00 mls x 1000 mls / L = 2.58 mols HCl/ L = 2.58 M

e) find mass % HCl

mass % HCl = mass HCl / total mass (x100%) = 0.471 g / 4.902 g (x100%) = 9.61% HCl by mass