A 21.7 mL sample of 0.18 M KOH solution required 26.3 mL of aqueous acetic acid solu- tion in a titration experiment. Calculate the molarity of the acetic acid solution.

To solve this problem, we can use the concept of stoichiometry and the balanced chemical equation of the reaction between KOH and acetic acid.

The balanced chemical equation for the reaction is:
KOH + CH3COOH → CH3COOK + H2O

Based on this equation, we can see that the mole ratio between KOH and acetic acid is 1:1. This means that the number of moles of KOH used is equal to the number of moles of acetic acid in the reaction.

First, let's calculate the number of moles of KOH used:
Moles of KOH = Molarity of KOH × Volume of KOH (in L)

Given that the volume of KOH used is 21.7 mL, we need to convert it to liters by dividing by 1000:
Volume of KOH = 21.7 mL ÷ 1000 = 0.0217 L

Substituting the given values, we get:
Moles of KOH = 0.18 M × 0.0217 L = 0.00393 moles

Since the reaction is 1:1, the number of moles of acetic acid used is also 0.00393 moles.

Now, let's calculate the molarity of the acetic acid solution. We know that the volume of acetic acid used is 26.3 mL.

Volume of acetic acid = 26.3 mL ÷ 1000 = 0.0263 L

Molarity of acetic acid = Moles of acetic acid ÷ Volume of acetic acid = 0.00393 moles ÷ 0.0263 L

Performing the calculation, we find:
Molarity of acetic acid = 0.149 M

Therefore, the molarity of the acetic acid solution is 0.149 M.

To calculate the molarity of the acetic acid solution, you need to use the concept of stoichiometry. In a titration, the amount of a reactant (acid or base) is determined by reacting it with a solution of known concentration of another reactant (base or acid) until the reaction is complete.

In this case, the reaction is between KOH (potassium hydroxide) and acetic acid (CH3COOH) according to the balanced equation:
CH3COOH + KOH → CH3COOK + H2O

First, you need to find the number of moles of KOH used in the titration. To do this, you can use the molarity and volume of the KOH solution:
Moles of KOH = Molarity of KOH × Volume of KOH solution (in L)

Converting volume from mL to L:
Volume of KOH solution = 21.7 mL ÷ 1000 = 0.0217 L

Substituting the values, we get:
Moles of KOH = 0.18 M × 0.0217 L

Next, you need to determine the stoichiometric relationship between KOH and acetic acid. From the balanced equation, we see that the ratio of KOH to acetic acid is 1:1. Therefore, the moles of acetic acid used in the reaction will be the same as the moles of KOH:

Moles of acetic acid = Moles of KOH

Now, we can calculate the molarity of the acetic acid solution. We know that the moles of acetic acid is the same as the moles of KOH, and we have the volume of acetic acid solution:
Molarity of acetic acid solution = Moles of acetic acid ÷ Volume of acetic acid solution (in L)

Converting volume from mL to L:
Volume of acetic acid solution = 26.3 mL ÷ 1000 = 0.0263 L

Using the calculated moles of KOH:
Molarity of acetic acid solution = (0.18 M × 0.0217 L) ÷ 0.0263 L

Calculating this expression will give you the molarity of the acetic acid solution.