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7.13: Titration Calculations

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    The manufacture of soap requires a number of chemistry techniques. One necessary piece of information is the saponification number. This is the amount of base needed to hydrolyze a certain amount of fat to produce the free fatty acids that are an essential part of the final product. The fat is heated with a known amount of base (usually \(\ce{NaOH}\) or \(\ce{KOH}\)). After hydrolysis is complete, the leftover base is titrated to determine how much was needed to hydrolyze the fat sample.

    Titration Calculations

    At the equivalence point in a neutralization, the moles of acid are equal to the moles of base.

    \[\text{moles acid} = \text{moles base}\nonumber \]

    Recall that the molarity \(\left( \text{M} \right)\) of a solution is defined as the moles of the solute divided by the liters of solution \(\left( \text{L} \right)\). So the moles of solute are therefore equal to the molarity of a solution multiplied by the volume in liters.

    \[\text{moles solute} = \text{M} \times \text{L}\nonumber \]

    We can then set the moles of acid equal to the moles of base.

    \[\text{M}_A \times \text{V}_A = \text{M}_B \times \text{V}_B\nonumber \]

    \(\text{M}_A\) is the molarity of the acid, while \(\text{M}_B\) is the molarity of the base. \(\text{V}_A\) and \(\text{V}_B\) are the volumes of the acid and base, respectively.

    Suppose that a titration is performed and \(20.70 \: \text{mL}\) of \(0.500 \: \text{M} \: \ce{NaOH}\) is required to reach the end point when titrated against \(15.00 \: \text{mL}\) of \(\ce{HCl}\) of unknown concentration. The above equation can be used to solve for the molarity of the acid.

    \[\text{M}_A = \frac{\text{M}_B \times \text{V}_B}{\text{V}_A} = \frac{0.500 \: \text{M} \times 20.70 \: \text{mL}}{15.00 \: \text{mL}} = 0.690 \: \text{M}\nonumber \]

    The higher molarity of the acid compared to the base in this case means that a smaller volume of the acid is required to reach the equivalence point.

    The above equation works only for neutralizations in which there is a 1:1 ratio between the acid and the base. The example below demonstrates the technique to solve a titration problem for a titration of sulfuric acid with sodium hydroxide.

    Example \(\PageIndex{1}\)

    In a titration of sulfuric acid against sodium hydroxide, \(32.20 \: \text{mL}\) of \(0.250 \: \text{M} \: \ce{NaOH}\) is required to neutralize \(26.60 \: \text{mL}\) of \(\ce{H_2SO_4}\). Calculate the molarity of the sulfuric acid.

    \[\ce{H_2SO_4} \left( aq \right) + 2 \ce{NaOH} \left( aq \right) \rightarrow \ce{Na_2SO_4} \left( aq \right) + 2 \ce{H_2O} \left( l \right)\nonumber \]

    Step 1: List the known values and plan the problem.
    • Molarity \(\ce{NaOH} = 0.250 \: \text{M}\)
    • Volume \(\ce{NaOH} = 32.20 \: \text{mL}\)
    • Volume \(\ce{H_2SO_4} = 26.60 \: \text{mL}\)

    First determine the moles of \(\ce{NaOH}\) in the reaction. From the mole ratio, calculate the moles of \(\ce{H_2SO_4}\) that reacted. Finally, divide the moles of \(\ce{H_2SO_4}\) by its volume to get the molarity.

    Step 2: Solve.

    \[\begin{align*} &\text{mol} \: \ce{NaOH} = \text{M} \times \text{L} = 0.250 \: \text{M} \times 0.03220 \: \text{L} = 8.05 \times 10^{-3} \: \text{mol} \: \ce{NaOH} \\ &8.05 \times 10^{-3} \: \text{mol} \: \ce{NaOH} \times \frac{1 \: \text{mol} \: \ce{H_2SO_4}}{2 \: \text{mol} \: \ce{NaOH}} = 4.03 \times 10^{-3} \: \text{mol} \: \ce{H_2SO_4} \\ &\frac{4.03 \times 10^{-3} \: \text{mol} \: \ce{H_2SO_4}}{0.02660 \: \text{L}} = 0.151 \: \text{M} \: \ce{H_2SO_4} \end{align*}\nonumber \]

    Step 3: Think about your result.

    The volume of \(\ce{H_2SO_4}\) required is smaller than the volume of \(\ce{NaOH}\) because of the two hydrogen ions contributed by each molecule.


    • The process of calculating concentration from titration data is described and illustrated.

    7.13: Titration Calculations is shared under a CC BY-NC license and was authored, remixed, and/or curated by LibreTexts.

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