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Titration Analysis

The described approach to measuring vinegar strength was an early version of the analytical technique known as titration analysis. A typical titration analysis involves the use of a buret (see image below) to make incremental additions of a solution containing a known concentration of some substance (the titrant) to a sample solution containing the substance whose concentration is to be measured (the analyte).


(a) A student fills a buret in preparation for a titration analysis. (b) A typical buret permits volume measurements to the nearest 0.1 mL. Image credit:- a: modification of work by Mark Blaser and Matt Evans; b: modification of work by Mark Blaser and Matt Evans

The titrant and analyte undergo a chemical reaction of known stoichiometry, and so measuring the volume of titrant solution required for complete reaction with the analyte (the equivalence point of the titration) allows calculation of the analyte concentration. The equivalence point of a titration may be detected visually if a distinct change in the appearance of the sample solution accompanies the completion of the reaction.

The halt of bubble formation in the classic vinegar analysis from the previous lesson is one such example, though, more commonly, special dyes called indicators are added to the sample solutions to impart a change in color at or very near the equivalence point of the titration.

Equivalence points may also be detected by measuring some solution property that changes in a predictable way during the course of the titration. Regardless of the approach taken to detect a titration’s equivalence point, the volume of titrant actually measured is called the end point.

Properly designed titration methods typically ensure that the difference between the equivalence and end points is negligible. Though any type of chemical reaction may serve as the basis for a titration analysis, the three described in this chapter (precipitation, acid-base, and redox) are most common. Additional details regarding titration analysis are provided in the chapter on acid-base equilibria.

Titration Analysis: Example

The end point in a titration of a 50.00-mL sample of aqueous HCl was reached by addition of 35.23 mL of 0.250 M NaOH titrant. The titration reaction is:

\(\mathrm{HCl}(aq) + \mathrm{NaOH}(aq) \longrightarrow \mathrm{NaCl}(aq) + \mathrm{H_2O}(l)\)

What is the molarity of the HCl?


As for all reaction stoichiometry calculations, the key issue is the relation between the molar amounts of the chemical species of interest as depicted in the balanced chemical equation. The approach outlined in previous lessons is followed, with additional considerations required, since the amounts of reactants provided and requested are expressed as solution concentrations.

For this exercise, the calculation will follow the following outlined steps:


Image credit: OpenStax, Chemistry

The molar amount of HCl is calculated to be:

\(\require{cancel}\mathrm{35.23 \; \cancel{mL \; NaOH}} × \cfrac{\mathrm{1 \; \cancel{L}}}{\mathrm{1000 \; \cancel{mL}}} × \cfrac{\mathrm{0.250 \; \cancel{mol \; NaOH}}}{\mathrm{1 \; \cancel{L}}} × \cfrac{\mathrm{1 \; \cancel{mol \; HCl}}}{\mathrm{1 \; \cancel{mol \; NaOH}}} = 8.81 × 10^{-3} \text{ mol HCl}\)

Using the provided volume of HCl solution and the definition of molarity, the HCl concentration is:

\(M = \cfrac{\text{mol HCl}}{\text{L solution}}\)

\(M = \cfrac{8.81 × 10^{-3} \text{ mol HCl}}{50.00 \text{ mL} × \frac{1 \text{ L}}{1000 \text{mL}}}\)

\(M = 0.176 M\)

Note: For these types of titration calculations, it is convenient to recognize that solution molarity is also equal to the number of millimoles of solute per milliliter of solution:

\(M = \cfrac{\text{mol solute}}{\text{L solution}} × \cfrac{\frac{10^3 \text{ mmol}}{\text{mol}}}{\frac{10^3 \text{ mL}}{\text{L}}} = \cfrac{\text{mmol solute}}{\text{mL solution}}\)

Using this version of the molarity unit will shorten the calculation by eliminating two conversion factors:

\(\cfrac{35.23\text{ mL NaOH} × \frac{0.250 \text{ mmol NaOH}}{\text{mL NaOH}} × \frac{1\text{ mmol HCl}}{1\text{ mmol NaOH}}}{50.00\text{ mL solution}} = 0.176 M \text{ HCl}\)

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