Calculate the molarity of an acetic acid solution. you take 20.0 ml of the acetic acid solution and it takes 33.7 ml of 0.28m naoh to titrate it to the end point (barely pink with phenolphthalein).

The Molarity of acetic acid (CH3COOH) = 0.4718 M

Explanation:

At the end point, the moles of NaOH is the same as the moles of the acetic acid because of the reaction and their corresponding coefficient (Stoichiometry).

Reaction: CH3COOH + NaOH → CH3COONa + H2O

molarity (M) = moles (n) divided by Volume (V) in litres

the volume should be converted to Liter 33.7 ml = 0.0337 L and 20 ml = 0.02 L

mole of NaOH at end point = (0.28 M) x (0.0337 L) = 9.436 x 10^-3 mole = mole of acetic acid.

Molarity of acetic acid solution = (9.436 x 10^-3)/(0.02) = 0.4718 M

16

explanation:

the marks go by 4s