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:

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

Correct the concentration depending on the number of acidic or basic ions present:

**For acids:** Multiply by the number of *H*^{+}* ions*

**For bases:** Multiply by the number of *OH*^{–}*, H*^{–}*, O*^{2–}*, or NH*_{2}^{–}* ions*

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

- There are
**3 H**^{+}ions in H_{3}PO_{4}:

$\mathbf{3}\mathbf{\xb7}{\mathbf{M}}_{\mathbf{acid}}\left(\mathbf{31}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{mL}\right)\mathbf{=}\left(\mathbf{0}\mathbf{.}\mathbf{120}\mathbf{}\mathbf{M}\mathbf{}\mathbf{NaOH}\right)\left(\mathbf{26}\mathbf{.}\mathbf{43}\mathbf{}\mathbf{mL}\right)$

A 31.00 mL sample of an unknown H_{3}PO_{4} solution is titrated with a 0.120 M NaOH solution. The equivalence point is reached when 26.43 mL of NaOH solution is added. What is the concentration of the unknown H_{3}PO_{4} solution? The neutralization reaction is H_{3}PO_{4} (aq) + 3 NaOH(aq) → 3 H_{2}O(l) + Na_{3}PO_{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.