Volumetric analysis question, need help!

[yabbaboo]

The conncentration of ethanoic acid (CH3C00H) in a vinegar sample was determined by volumetric analysis. 20.00ml of vinegar was transferred to a volumetric flask and water added to the total 250.0ml volume. 20.00ml aliquots of the diluted vinegar solution required an average titre of 19.82mL of standardised 0.102M NaOH solution for neutralisation.

How do I find the molar concentration of ethanoic acid in the vinegar sample?

So the total volume of the vinegar sample is 250.0mL.
c(CH3C00H)=?
M(CH3C00H) =52g/mol
n(CH3C00H)=?

Can I us this equation below? Or it can't be used because water is mixed with vinegar?
CH3C00H(aq) + NaOH(aq) --> NaCH3COO(aq) + H20(l)

Sorry if this seems an utterly dumb question....

[Nobby]

CH3C00H(aq) + NaOH(aq) --> NaCH3COO(aq) + H20(l)

n(NaOH)=0.01982x0.102
n(CH3COOH)=n(NaOH)
c(CH3COOH)=n(CH3COOH)/0.02000
                    =0.10132M
                    =0.101M

[Nobby]

CH3C00H(aq) + NaOH(aq) --> NaCH3COO(aq) + H20(l)

n(NaOH)=0.01982x0.102
n(CH3COOH)2=n(NaOH)
c(CH3COOH)2=n(CH3COOH)2/0.02000
c(CH3COOH)1=C(CH3COOH)2x250.0/20.00
                      =1.26M

[Shenz0r]

Rule of thumb: Always find the number of mol with the stuff that you can find first.
First write the equation.
1. CH3COOH (aq) + NaOH (aq) -> CH3COONa (aq) + H2O (l) (for some reason I've always seen ionic compounds with CH3COO that have the cation written at the back)
Next, find the number of mol of NaOH with the formula n = cv.
2. n(NaOH) = 0.102 x 0.01982 = 0.002022 mol
Next, find the number of mol of CH3COOH using the molar ratios.
3.n(CH3COOH)aliquot = 1/1 x 0.002022 = 0.002022 mol
Work out the concentration of the 20.00 mL CH3COOH aliquot by rearranging n=cv.
4. [CH3COOH] = n/v = 0.002022/0.02 = 0.101082 M
Use the dilution formula (c1v1=c2v2) to find out the original concentration of the 20.00 mL sample.
5. c1 = c2xv2/v1 = 0.101082 x 250/20 = 1.26M

Water is not really mixed with vinegar, but rather it dilutes its concentration so that it will be lower during the titration.