posted by shannan .
a 10 ml vinegar sample was completely neutralized by 22.5 0.2M NaOH solution . calculate the molarity and percent acetic acid in vinegar I keep getting 2.7% acetic acid is this correct? my friend got 27% and if I am wrong where do you think I messed up?
I don't believe you did.
Is that 22.5 mL of 0.2M NaOH?
mols NaOH = M x L = 0.0225 x 0.20 = 0.00450 mols.
mols acetic acid (HAc)in the vinegar = mols NaOH (Look at the coefficients in the balanced equation.)
grams HAc = mols x molar mass = 0.00450 x 60 = about 0.270g
%HAc = g HAc/100 mL
%HAc = 0.270/10 mL = 2.70 g/100 mL = 2.70%
M = mols/L soln. Can you handle that?