A 10.00-mL aliquot of vinegar requires 16.95 mL of the 0.4874 M standardized NaOH solution to reach the end point of the titration. Demonstrate how to calculate the molarity of the vinegar solution (HC2H3O2). Show complete work below. 0.8261 M.

0.8261 M.

Explanation:

We'll begin by writing the balanced equation for the reaction. This is given below:

HC2H3O2 + NaOH —> NaC2H3O2 + H2O

From the balanced equation above, we obtained the following:

Mole ratio of the acid, HC2H3O2 (nA) = 1

Mole ratio of the base, NaOH (nB) = 1

Data obtained from the question include the following:

Volume of acid, HC2H3O2 (Va) = 10 mL

Molarity of acid, HC2H3O2 (Ma) =..?

Volume of base, NaOH (Vb) = 16.95 mL Molarity of base, NaOH (Mb) = 0.4874 M

Finally, we shall determine the molarity of the acid solution, as follow:

MaVa/MbVb = nA/nB

Ma x 10 / 0.4874 x 16.95 = 1

Cross multiply

Ma x 10 = 0.4874 x 16.95

Divide both side by 10

Ma = (0.4874 x 16.95) /10

Ma = 0.8261 M.

Therefore, the molarity of the vinegar solution (HC2H3O2) is 0.8261 M.

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