21.18: Titration Calculations
- Page ID
- 53948
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)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.
\[\ce{H_2SO_4} \left( aq \right) + 2 \ce{NaOH} \left( aq \right) \rightarrow \ce{Na_2SO_4} \left( aq \right) + 2 \ce{H_2O} \left( l \right)\nonumber \]
Solution
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 \(\ce{H_2SO_4} = 26.60 \: \text{mL}\)
Unknown
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.
Summary
- The process of calculating concentration from titration data is described and illustrated.