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Chemistry

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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.

  • Chemistry - ,

    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.

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