Recall that at the ** equivalence point** of a titration:

$\overline{){\mathbf{moles}}{\mathbf{}}{\mathbf{acid}}{\mathbf{=}}{\mathbf{moles}}{\mathbf{}}{\mathbf{base}}}$

Also, recall that:

**moles = molarity × volume****moles = mass/ molar mass**

This means:

$\overline{){\mathbf{\left(}\mathbf{MV}\mathbf{\right)}}_{{\mathbf{acid}}}{\mathbf{=}}{\left(\mathbf{MV}\right)}_{{\mathbf{base}}}}$

Solving for **M _{acid}**:

${\mathbf{M}}_{\mathbf{acid}}\left(\mathbf{3}\mathbf{.}\mathbf{55}\mathbf{}\mathbf{mL}\right)\mathbf{=}\left(\mathbf{0}\mathbf{.}\mathbf{130}\mathbf{}\mathbf{M}\mathbf{\hspace{0.17em}}\mathbf{NaOH}\right)\left(\mathbf{43}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{mL}\right)$

${\mathbf{M}}_{\mathbf{acid}}\mathbf{=}\frac{(\mathbf{0}\mathbf{.}\mathbf{130}\mathbf{}\mathbf{M}\mathbf{\hspace{0.17em}}\mathbf{NaOH})(\mathbf{43}\mathbf{.}\mathbf{00}\mathbf{}\overline{)\mathbf{mL}})}{\mathbf{3}\mathbf{.}\mathbf{55}\mathbf{}\overline{)\mathbf{mL}}}$** = 1.575 M**

The distinctive odor of vinegar is due to acetic acid, CH_{3}COOH, which reacts with sodium hydroxide in the following fashion:

CH_{3}COOH(aq) + NaOH(aq)(l) + NaC_{2}H_{3}O_{2} (aq)

You may want to reference (Pages 149 - 153) Section 4.6 while completing this problem.

If 3.55 mL of vinegar needs 43.0 mL of 0.130 M NaOH to reach the equivalence point in a titration, how many grams of acetic acid are in a 1.60 qt sample of this vinegar?

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