A 10.0 mL of vinegar, an aqueous solution of acetic acid (HC2H3O2), is titrated with .5062M NaOH, and 16.58mL is required to reach the equivalence point.

Question

A. What is the molarity of the acetic acid?
B. If the density of the vinegar is 1.006 g/cm^3, what is the mass percent of acetic acid in the vinegar?

Answer 1

Acetic acid is CH3COOH (HC2H3O2). Its the right end H that is acidic.
CH3COOH + NaOH ==> CH3COONa + HOH

mols NaOH required = M x L = ??
mols acetic acid must be the same since the reaction is 1 mol CH3COOH to 1 mol NaOH.
Molarity CH3OOH = # mols/L = mols/0.10 = xx.

For percent, convert mols CH3COOH from above to grams. Convert 10.0 mL to grams using the density to give the sample mass.
percent CH3COOH = [grams CH3COOH/mass sample]*100 = ??
Post your work if you get stuck.

Answer 2

Ok, so far this is what i've done.
A. to get moles of NaOH, i multiplied 0.5062M x 0.01658L=8.39x10^-3, then i divided this number by .010L, i got .8392M
B. (8.392x10^-3)x(60.052g HC2H3O2)=.5039g, after this part i got stuck.

Answer 3

You're almost home. You have grams and that is correct except I would have carried it out to another place to get 0.50399 which rounds to 0.5040 g. (The molarity I calculated as 0.83928 which rounds to 0.8393 but I didn't round and re-enter numbers in my calculator.

%acetic acid = [grams/mass sample]*100
You have grams from above.
To obtain mass of the sample,
mass = volume x density
10 cc x 1.006 g/cc = ??