A 18.0 mL sample of vinegar, which is an aqueous solution of acetic acid, CH3COO

Question

A 18.0 mL sample of vinegar, which is an aqueous solution of acetic acid, CH3COOH, requires 24.5 mL of 0.600 M NaOH to reach the endpoint in a titration.

CH3COOH(aq) + NaOH(aq) CH3COONa(aq) + H2O(l)

What is the molarity of the acetic acid solution?

Problem 10.86 Part A A 18.0 mL sample of vinegar, which is an aqueous solution of acetic acid, CH COOH requires 24.5 mL of 0.600 MNaOH to reach the endpoint in a titration What is the molarity of the acetic acid solution? CH3 COOH(aq) + NaOH(aq) CH&COONala;)+H2O) molarity= Submit My Answers Give Up

Explanation / Answer

Number of moles of NaOH = molarity of NaOH * volume of solution in L

Number of moles of NaOH = 0.600 * 0.0245 = 0.0147 mole

From the balanced equation we can say that

1 mole of NaOH requires 1 mole of CH3COOH so

0.0147 mole of NaOH will require 0.0147 mole of CH3COOH

Therefore, the number of moles of CH3COOH are 0.0147

Molarity of acetic acid (CH3COOH) = number of moles of acetic acid / volume of solution in L

Moalrity of acetic acid = 0.0147 / 0.018 = 0.817 M

Therefore, the molarity of acetic acid is 0.817 M

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