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If 74 L of 0.643 M KOH is required to completely neutralize 33.1 L of a CH3COOH solution, what is the molarity of the acetic acid solution?

User Stchang
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In this question, we have to determine the molar concentration, or molarity of a CH3COOH solution.

We have as informations:

74 Liters of 0.643 M of KOH

33.1 Liters of CH3COOH

Although we have a situation of weak acid + strong base, the question provides informations considering a strong acid + strong base situation, so I will consider CH3COOH as a strong acid, even though IT IS NOT A STRONG ACID.

In order to find the molarity, we can use the Titration formula:

MaVa = MbVb

Ma = molarity of the acid

Va = volume of the acid

Mb = molarity of the base

Vb = volume of the base

Ma * 33.1 = 0.643 * 74

33.1Ma = 47.58

Ma = 1.44 M

The concentration of this acid, based on the information provided, will be 1.44 M

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