# A solution of sodium hydroxide was prepared by dissolving 4.0 g of sodium hydroxide, NaOH(s), in 250 ml of water. It was found that 20.0 mL of the sodium hydroxide solution neutralizes 25.0 mL of vinegar.  Determine the concentration of acetic acid, CH3COOH(aq), in the sample of vinegar.  Assume that acetic acid is the only acidic substance in the vinegar. First determine the molarity of the sodium hydroxide solution.

4 g in 250 mL = 16 g/liter

molar mass of NaOH = 40 g/mole so you have 16/40 = 0.4 moles /L = 0.4 M solution.

Now if you take the produce of molarity * volume you get the number...

First determine the molarity of the sodium hydroxide solution.

4 g in 250 mL = 16 g/liter

molar mass of NaOH = 40 g/mole so you have 16/40 = 0.4 moles /L = 0.4 M solution.

Now if you take the produce of molarity * volume you get the number of moles of a substance in that volume.

If you write a balanced chemical equation for the reaction of NaOH + acetic acid you get:

NaOH + CH3COOH -->  CH3CooNa  + HOH

so for every mole of NaOH you neutralize one mole of acetic acid.

Since mL * M of the NaOH = mL * M for the acetic acid you can solve for the molarity of the acetic acid.

20 mL * 0.4 M = 25 mL * M of acetic acid

M = 20/25 * 0.4 = 0.32 M solution of acetic acid.

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