Question

Part A Calculate the molarity of an acetic acid solution if 24.9 mL of an HC2H2O2 solution is titrated with 22.0 mL of a 0.171 M KOH solution: HC2H2O2(aq)+KOH(aq) + H2O(1)+ KC2H2O2(aq) Express your answer with the appropriate units. Di μΑ ? molarity = Value Units Submit Request Answer

Answer 1

Answer and Explanation:

To calculate the molarity of the acetic acid solution, we need to use the information given in the titration reaction and the volumes of the solutions.

The balanced chemical equation for the reaction is:

HC2H2O2(aq) + KOH(aq) → H2O(l) + KC2H2O2(aq)

From the balanced equation, we can see that the stoichiometric ratio between HC2H2O2 and KOH is 1:1.

First, we need to calculate the number of moles of KOH used in the titration. To do this, we use the formula:

moles of KOH = Molarity of KOH * Volume of KOH solution

moles of KOH = 0.171 M * 0.022 L

moles of KOH = 0.003762 mol

Since the stoichiometric ratio between HC2H2O2 and KOH is 1:1, the number of moles of HC2H2O2 is also 0.003762 mol.

Next, we convert the volume of HC2H2O2 solution from milliliters to liters:

Volume of HC2H2O2 solution = 24.9 mL = 24.9 mL * (1 L / 1000 mL)

Volume of HC2H2O2 solution = 0.0249 L

Finally, we calculate the molarity of the acetic acid solution using the formula:

Molarity = moles of solute / volume of solution

Molarity = 0.003762 mol / 0.0249 L

Molarity = 0.151 M

Therefore, the molarity of the acetic acid solution is 0.151 M.