Question

Could someone help me with this question?

A vinegar sample was analyzed and found to be 5.88% acetic acid by volume. The density of pure acetic acid is 1.409g/ml. What volume of 0.200 M NaOH would be required to titrate a 2 ml aliquot of this vinegar.

Answer 1

5.88% v/v means 5.88 mL acetic acid/100 mL solution. m = v*d; therefore 5.88 mL x 1.409 g/mL = grams acetic acid.
Convert grams to moles.
moles = grams/molar mass.
Convert to M = moles/L of soln.

Then write the equation.
CH3COOH + NaOH ==> CH3COONa + H2O
moles CH3COOH frm above.
moles NaOH must be the same since the reaction is 1 mole CH3COOH to 1 mole NaOH.
moles CH3COOH = M x L
moles NaOH = M x L.
M x L = M x L
You have M CH3COOH and L CH3COOH and M NaOH. Solve for L NaOH.