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**.

This means:

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

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

${\mathbf{M}}_{\mathbf{acid}}\left(\mathbf{26}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{mL}\right)\mathbf{=}\left(\mathbf{0}\mathbf{.}\mathbf{2100}\mathbf{}\mathbf{M}\mathbf{\hspace{0.17em}}\mathbf{NaOH}\right)\left(\mathbf{21}\mathbf{.}\mathbf{92}\mathbf{}\mathbf{mL}\right)$

A 26.00-mL sample of an unknown HClO_{4} solution requires titration with 21.92 mL of 0.2100 M NaOH to reach the equivalence point. What is the concentration of the unknown HClO_{4} solution? The neutralization reaction is: HClO_{4}(aq) + NaOH(aq) → H_{2}O(l) + NaClO_{4}(aq)

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Our tutors have indicated that to solve this problem you will need to apply the Equivalence Point concept. If you need more Equivalence Point practice, you can also practice Equivalence Point practice problems.