25.00 mL of an acetic acid solution requires 23.86mL of a 0.1550M potassium hydroxide solution to reach the end point. What is the molarity of the acetic acid solution? (Hint: write the chemical reaction, and remember stoichiometry)

Answer 1

Write and balance the equation.
moles KOH = M x L
Use the equation to convert moles KOH to moles acetic acid.
Macetic acid = moles/L

Answer 2

C2H4O2 + KOH-->

I'm not quite sure how you balance the equation?

Answer 3

The empirical formula you have written tells you nothing about the molecule (at least not much except for the molar mass). If you write it as CH3COOH OR as HC2H3O2, it is easier to see that the H on the right most end of the CH3COOH is the H that ionizes or it's the left most H on HC2H3O2 so the ionization is
CH3COOH ==> H^+ + CH3COO^- or
HC2H3O2 ==> H^+ + C2H3O2^-
the C2H3O2 is the acetate ion (CH3COO^-). Frankly, I reduce it even further because I get tired of typing all that "other" stuff. I just call it HAc where the H^+ stands for hydrogen and the Ac^- stands for the acetate ion.
The neutralization reaction is
HAc + KOH ==> KAc + H2O.