41 9.6 Neutralization and Titration

Learning Objectives

By the end of this section, you will be able to:

  • Describe the fundamental aspects of titrations analysis.
  • Perform stoichiometric calculations using typical titration data.

In the 18th century, the strength (actually the concentration) of vinegar samples was determined by noting the amount of potassium carbonate, K2CO3, which had to be added, a little at a time, before bubbling ceased. The greater the weight of potassium carbonate added to reach the point where the bubbling ended, the more concentrated the vinegar.

We now know that the effervescence that occurred during this process was due to reaction with acetic acid, CH3CO2H, the compound primarily responsible for the odor and taste of vinegar. Acetic acid reacts with potassium carbonate according to the following equation:

[latex]2\text{CH}_3 \text{CO}_2 \text{H}(aq) + \text{K}_2\text{CO}_3(s) \longrightarrow \text{KCH}_3 \text{CO}_3(aq) + \text{CO}_2(g) + \text{H}_2 \text{O}(l)[/latex]

The bubbling was due to the production of CO2.

The test of vinegar with potassium carbonate is one type of quantitative analysis—the determination of the amount or concentration of a substance in a sample. In the analysis of vinegar, the concentration of the solute (acetic acid) was determined from the amount of reactant that combined with the solute present in a known volume of the solution. In other types of chemical analyses, the amount of a substance present in a sample is determined by measuring the amount of product that results.


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 (Figure 1) 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). 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 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.

Two pictures are shown. In a, a person is shown pouring a liquid from a small beaker into a buret. The person is wearing goggles and gloves as she transfers the solution into the buret. In b, a close up view of the markings on the side of the buret is shown. The markings for 10, 15, and 20 are clearly shown with horizontal rings printed on the buret. Between each of these whole number markings, half markings are also clearly shown with horizontal line segment markings.
Figure 1. (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. (credit a: modification of work by Mark Blaser and Matt Evans; credit b: modification of work by Mark Blaser and Matt Evans)

Example 1

Titration Analysis
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:

[latex]\text{HCl}(aq) + \text{NaOH}(aq) \longrightarrow \text{NaCl}(aq) + \text{H}_2\text{O}(l)[/latex]

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 modules of this chapter 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:

This figure shows four rectangles. The first is shaded lavender and is labeled, “Volume of N a O H.” This rectangle is followed by an arrow pointing right which is labeled, “Molar concentration,” to a second rectangle. This second rectangle is shaded pink and is labeled, “Moles of N a O H.” This rectangle is followed by an arrow pointing right which is labeled, “Stoichiometric factor,” to a third rectangle which is shaded pink and is labeled, “Moles of H C l.” This rectangle is followed by an arrow labeled, “Solution volume,” which points right to a fourth rectangle. This fourth rectangle is shaded lavender and is labeled, “Concentration of H C l.”

The molar amount of HCl is calculated to be:

[latex]35.23 \;\rule[0.5ex]{4.5em}{0.1ex}\hspace{-4.5em}\text{mL NaOH} \times \frac{1 \;\rule[0.5ex]{0.5em}{0.1ex}\hspace{-0.5em}\text{L}}{1000 \rule[0.5ex]{2em}{0.1ex}\hspace{-2em}\;\text{mL}} \times \frac{0.250 \;\rule[0.5ex]{3.5em}{0.1ex}\hspace{-3.5em}\text{mol NaOH}}{1 \;\rule[0.5ex]{0.5em}{0.1ex}\hspace{-0.5em}\text{L}} \times \frac{1 \;\text{mol HCl}}{1 \;\rule[0.5ex]{3.5em}{0.1ex}\hspace{-3.5em}\text{mol NaOH}} = \\ 8.81 \times 10^{-3} \;\text{mol HCl}[/latex]

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

[latex]\begin{array}{r @{{}={}} l} M & = \frac{\text{mol HCl}}{\text{L solution}} \\[1em] & = \frac{8.81 \times 10^{-3} \;\text{mol HCl}}{50.00 \;\text{mL} \times \frac{1 \;\text{L}}{1000 \;\text{mL}}} \\[1em] & = 0.176 \;M \end{array}[/latex]


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:

[latex]M = \frac{\text{mol solute}}{\text{L solution}} \times \frac{\frac{10^3 \;\text{mmol}}{\text{mol}}}{\frac{10^3 \;\text{mL}}{\text{L}}} = \frac{\text{mmol solute}}{\text{mL solution}}[/latex]


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

[latex]\frac{35.23 \;\text{mL NaOH} \times \;\frac{0.250 \;\text{mmol NaOH}}{\text{mL NaOH}} \times \frac{1 \;\text{mmol HCl}}{1 \;\text{mmol NaOH}}}{50.00 \;\text{mL solution}} = 0.176 \;M \;\text{HCl}[/latex]


Check Your Learning
A 20.00-mL sample of aqueous oxalic acid, H2C2O4, was titrated with a 0.09113-M solution of potassium permanganate.

[latex]{2\text{MnO}_4}^{-}(aq) + 5\text{H}_2 \text{C}_2 \text{O}_4(aq) + 6\text{H}^{+}(aq) \longrightarrow 10\text{CO}_2(g) + 2\text{Mn}^{2+}(aq) + 8\text{H}_2 \text{O}(l)[/latex]


A volume of 23.24 mL was required to reach the end point. What is the oxalic acid molarity?


0.2648 M


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College of Western Idaho General Chemistry: CHEM 111 & 112 Copyright © 2016 by Rice University is licensed under a Creative Commons Attribution 4.0 International License, except where otherwise noted.

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