i am completely lost, i confused myself.

i need to find the concentration of HC2H3O2

HC2H3O2 + NaOH ==> NaC2H3O2 + HOH

NaOH
C= 0.3 M
v= 0.25 L
n= 0.075 moles
M= 40.00g/mol
m= 3.00 g

NaC2H3O2
n= 0.075 moles
m= 0.4973 g(ammount of acetic acid in
solution)

HC2H302
v= 0.020 L
M= 60.06g/mol

thats all the information i have. i don't know how to calculate it. my friend said c1v1=c2v2 but i don't know what numbers id use

2 answers

You are confused because you have too much information piled in together. You need to think this through. And you don't say what unit the concentration should be in.
M(NaOH) x L(NaOH) = mols NaOH.
mols NaOH = mols acetic acid since the equation is 1 mol NaOH to 1 mol acetic acid.

Now M(acid) x L(acid) = mols. You know mols, you know L acid (is that the 0.020 L), calculate M acid. That will be the molarity of the acetic acid in that 20 mL sample. By the way, I can't believe you used 0.25 L of the NaOH. That's 250 mL which is a huge volume to be titrated with. That means you would have had to refill a 50 mL buret five times. (Of course this could be a problem.)
I'm getting ready to leave for the night so if there is a quick question, please type it in before I leave.
alright thank you i think i should be able to figure it out. and i know we only did 3 trials though