4.6 Volumetric Analysis

Quantitative chemical analysis: is branch of chemistry that oversees the determination of the amount or percentage of one or more the components of a chemical compound.

Several physical methods can be included in the quantitative chemical analysis.

Such physical methods include precipitation, neutralization, oxidation, or, in general, the formation of a new product. The major types of strictly chemical methods are known as gravimetric analysis and volumetric, or titrimetric analysis. Some other physical methods include also measurement of some physical property such as density, refractive index, absorption or polarization of light, electromotive force, magnetic susceptibility.

Some Physical Methods:

Titration:

A method by which a compound with unknown concentration is determined by a neutralization reaction with another compound that has known concentration. Titration is also known as neutralization reaction of acid and base.

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

2CH₃CO₂H(aq) + K₂CO₃(s) ⟶ 2CH₃CO₂K(aq) + CO₂(g) + H₂O(l)

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.

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.

The setup of the titration is given below:

Figure 4.24 Titration Set up

Ref: www.commons.wikimedia.com

Figure 4.25 Titraion Lab Experiment

Reference: www.flickr.com

The buret holds chemical compound called titrant which has known concentration. The Erlenmeyer flask holds the unknown concentration of the compound which is known as analyte or titrant.

Two points are very important in the titration:

End point: A point by which the indicator in the analyte solution changes its color by an extra drop added of the titrant inside the buret.

Equivalent point: A point by which number of moles of the titrant solution compound inside the buret equals the number of moles of the analyte solution compound inside in Erlenmeyer flask.

The figure below shows the difference between the end point and equivalent point of the titration of HCl with NaOH.

Figure 4.26 Equivalence point & End Point

Ref: Wikimedia commons

Note that: the end point comes always after the equivalent point. One drop of the titrant signals the color change.

Figure 4.27 End point color change

Ref: commons.wikimedia.org/

Numerical problems are solved to find unknown concentration or volume of Acid/Base in neutralization reaction using titration data.

Example:

Practice Problem#1: 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: HCl(aq) + NaOH(aq) ⟶ NaCl(aq) + H2 O(l) What is the molarity of the HCl?

Solution 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.

The molar amount of HCl is calculated to be:

35.23 mL NaOH × 1 L   NaOH            × 0.250 mol

                                1000 ml NaOH           1 L             × 1 mol HCl

                                                                                       1 mol NaOH          =  8.81 × 10−3 mol HCl

Using the provided volume of HCl solution and the definition of molarity, the HCl concentration is: M = mol HCl L solution M = 8.81 × 10−3 mol HCl

                                                    50.00 mL × 1 L /1000 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 = mol solute L solution × 103 mmol mol 103 mL L = mmol solute mL solution Using this version of the molarity unit will shorten the calculation by eliminating two conversion factors: 35.23 mL NaOH × 0.250 mmol NaOH mL NaOH × 1 mmol HCl      ×  50.00 mL solution = 0.176 M HCl

                                1 mmol NaOH

Practice Problem#2: 2 KOH   +   H₂SO₄     à  K₂SO₄    +   2 H₂O

If 35.0 mL of 0.150 M KOH are used to neutralize 35.0 mL of H₂SO₄. Calculate the molarity of H₂SO₄.

Molarity of H₂SO₄ = [0.0350 L KOH] x [0.150 mol KOH /  L KOH] x [1 mol H₂SO₄ / 2 mol KOH] x [1 / 0.035 L H₂SO₄]  = 0.075 mol H₂SO₄ / L H₂SO₄ = 0.075 M = 0.075 molar

Try  this ACID BASE VIRTUAL LAB on https://.chemcollective.org

http://chemcollective.org/activities/autograded/124

Activity: Set up the lab and find the concentration of unknown monoprotic acid in this virtual lab.