If you assume that there is 5% acetic acid in vinegar, how much vinegar should you weigh out so that the endpoint requires 25.0 mL of 0.186 M NaOH?

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  1. Is that 5% w/w or some other unit? I will assume w/w. That means 5g CH3COOH in 100 g soln. Convert 5g to mols. That's 5/molar mass = about 0.083 mols. How many mols do you need? That equals to mols NaOH = M x L = 0.025 x 0.186 = about 0.0046. So how much of the 0.083 mol vinegar to weigh. It is
    0.083mols x (?g/100g) = 0.0046
    Solve for ?g.

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  2. 5.58 g

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