Anna M Michałowska-Kaczmarczyk^{1} and Tadeusz Michałowski^{2*}

^{1}Department of Oncology, The University Hospital in Cracow, 31-501 Cracow, Poland

^{2}Faculty of Engineering and Chemical Technology, Technical University of Cracow, 31-155 Cracow, Poland

***Address for Correspondence:** Tadeusz Michałowski, Faculty of Engineering and Chemical Technology, Technical University of Cracow, 31-155 Cracow, Poland, Tel: +48126282035; +48 12 628 20 00; +048699 320; Email: michalot@o2.pl

**Dates:** **Submitted:** 16 May 2019; **Approved:** 10 June 2019; **Published:** 11 June 2019

**How to cite this article:** Michałowska-Kaczmarczyk AM, Michałowski T. Stoichiometric approach to redox back titrations in ethanol analyses. Ann Adv Chem. 2019; 3: 001-006. DOI: 10.29328/journal.aac.1001017

**Copyright:** © 2019 Michałowska-Kaczmarczyk AM, et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

**Keywords:** Ethanol analysis; Redox titration; Back titration; Stoichiometry

This article refers to calculations involved with determination of ethanol, analyzed according to redox back titration principle. A quantitative reasoning, based on logical sequence of statements, is presented for derivation of the formulas required to calculate the results of chemical analyses according to stoichiometric principles. The titrations are considered as two-step analytical procedures. This way, one can gain an insight into a classical redox titration and get a knowledge on the advantages of back titrations.

The literature designed for analytical laboratory purposes provides more or less detailed description of analytical methods. The description of quantitative nature is often preceded by preliminary information on the analytics. However, the question of the calculation of the results of analysis is commonly presented in the form of a ready formula, into which the data obtained from the experiments are inserted. The knowledge on the derivation of the related formula, based on stoichiometric reaction notations, is then presented as a “dry” information, for assimilation and memorization. This approach can be justified in the case of simple analyses. It is debatable, however, whether the same approach should be applied to more complex analytical procedures.

This paper concerns calculations involved with determination of ethanol contents, taken as interesting examples of classical, titrimetric analysis. Importance of classical determination of this analyte is revealed by the facts that: (1) a variety of analytical procedures were suggested for the ethanol determination [1-11]; (2) the international ethanol reference solution is prepared by the dichromate oxidation method [12].

Ethanol is a component of various alcoholic beverages, which include also other reducing components, considered as interferents in a suitable sample taken for analysis. Therefore, prior to the determination of ethanol in these beverages with use of suitable oxidizing agents, a separation of ethanol by distillation of a sample before the analysis is necessary. This way, less volatile interferents in the sample matrix are eliminated before the titrimetric procedure, realized according to back titration principle.

Titrimetry is one of the oldest analytical methods, frequently applied in analytical practice [13-16]. However, when the rate of reaction between analyte and reagent is slow, or when the standard solution lacks stability, or when the end point cannot be detected in a simple way – direct titration is not possible and resort is made to back-titration. Back-titration is a process in which the excess of a standard solution used to consume an analyte is determined by titration with a second standard solution.

**Analytics – preliminary data**

For clarity of presentation, all volumes are assumed to be expressed in mL units, and all concentrations in mol L^{–1} units. Reagents of at least p.a. quality, and water with a conductivity ≤ 10 μS cm^{–1} should be used during the experiments. For determination of ethanol concentration, standard/ized solutions of potassium dichromate (K_{2}Cr_{2}O_{7}), sodium thiosulfate (Na_{2}S_{2}O_{3}5H_{2}O), ammonium iron(II) sulfate ((NH_{4})_{2}Fe(SO_{4})_{2}6H_{2}O) and potassium permanganate (KMnO4) are required, while sulfuric acid (H_{2}SO_{4}), phosphoric acid (H_{3}PO_{4}) and potassium iodide (KI) solution are added in excess.

End-points of titrations can be determined either potentiometrically or visually. Determination of the end-point of iodine (I_{2}_{3}^{–1}) titration with sodium thiosulfate can be made more sensitive by adding starch solution as indicator. Phenanthroline or diphenylamine can be used as indicator to determine the end-point of titration in determination of potassium dichromate with ammonium iron (II) sulfate.

**Stoichiometric calculations**

Let V_{S} mL of wine sample (with unknown mass m_{et} of ethanol) be distilled and the distillate is collected in a receiving flask (V_{f} mL). After complete distillation of ethanol, the receiving flask is filled up to the mark with water and mixed; C_{0} mol/L ethanol solution is thus obtained. Subsequently, V1 mL of the standard K_{2}Cr_{2}O_{7} (C_{1} mol/L) solution is placed in a glass-stoppered flask, and a portion of concentrated H_{2}SO_{4} (1.84 g/mL) is added. The mixture is agitated and cooled to room temperature. Then V_{0} mL of the C_{0} mol/L ethanol solution taken from the receiving flask is added into this mixture; K_{2}Cr_{2}O_{7} is in due excess towards ethanol in this mixture. The flask is stoppered tightly and left to stand for ca. 15 min; it prevents the escape of acetaldehyde, as the intermediate product of this reaction [17]. Further analysis can be performed according to three options, see a flowchart in figure 1.

In Option 1, the excess of K_{2}Cr_{2}O_{7} is reduced by adding an excess of potassium iodide (KI) solution. Iodine formed is subsequently determined in redox titration with standard solution of Na_{2}S_{2}O_{3} (C_{2}, V_{2}).

The total number of mmoles of K_{2}Cr_{2}O_{7} added is equal to C_{1}·V_{1}, while the number of mmoles (n(Cr)) of K_{2}Cr_{2}O_{7} consumed in the reaction with ethanol is calculated from the stoichiometry of the reaction

2Cr_{2}O7^{–2} + 16H^{+1} + 3C_{2H5}OH = 4Cr^{+3} + 3CH_{3}COOH + 11H_{2}O (1)

2

n(Cr)

3

C_{0}·V_{0}

i.e. n(Cr) = $\frac{2}{3}$
•C_{0}•V_{0}. The excess of K_{2}Cr_{2}O_{7}, i.e., C_{1}•V_{1} – $\frac{2}{3}$
•C_{0}•V_{0}, is treated with an excess of KI solution, according to the reaction

Cr_{2}O_{7}^{–2} + 14H^{+1} + (6, 9)I^{–1} = 2Cr^{+3} + 3(I_{2}, I_{2}^{–1}) + 7H_{2}O (2)

1

C_{1}•V_{1} – $\frac{2}{3}$
•C_{0}•V_{0}

3

n(I_{2})

An excess of I^{–1} from KI increases the dissolution of the solid iodine; it is readily soluble in KI solution. The number of mmoles, n(I_{2}), of the oxidized iodine species formed in the reactions (2) is equal to

n(I_{2}) = 3•C_{1}•V_{1} – 2•C_{0}•V_{0} (3)

It is quantitatively determined in redox titration with V_{2} mL of standardized Na_{2}S_{2}O_{3} (C_{2}) solution. The reactions occurred here are as follows:

2S_{2}O_{3}^{–2} + (I_{23}^{-1}) = S_{4}O_{6}^{–2} + 2I^{–1} (4)

2

C_{2}•V_{2}

1

n(I_{2})

At the end of the titration, when nearly all iodine (I_{2}, I_{2}^{-1}) is consumed in reaction (4), freshly prepared starch solution (1%) is added. The thiosulfate solution is added until its color turns from dark blue into a clear green blue. Thus the number of mmoles of iodine consumed in reaction (4) is

n(I_{2}) = $\frac{1}{2}$
C_{2}•V_{2} (5)

After comparing the right sides of equations (3) and (5), the following equation is obtained

${\text{C}}_{0}\cdot {\text{V}}_{0}=\frac{1}{4}\cdot (6\cdot {\text{C}}_{1}\cdot {\text{V}}_{1}-{\text{C}}_{2}\cdot {\text{V}}_{2})$ (6)

Mass m_{0} [g] of ethanol (molar mass M = 46.07 g moL^{–1}) in the aliquot V_{0} of the solution taken for analysis from the flask (V_{f}) is then calculated by inserting the relation C_{0}•V_{0} = 10^{3}•m_{0} /M, into equation (6). Then we get

${\text{m}}_{0}={10}^{-3}\cdot \frac{\text{M}}{4}\cdot (6\cdot {\text{C}}_{1}\cdot {\text{V}}_{1}-{\text{C}}_{2}\cdot {\text{V}}_{2})=1.152\cdot {10}^{-2}\cdot (6\cdot {\text{C}}_{1}\cdot {\text{V}}_{1}-{\text{C}}_{2}\cdot {\text{V}}_{2})$ (7)

The mass m_{et} of ethanol in V_{f} mL of the solution is m_{et} = m_{0}∙V_{f}/V_{0}. It is assumed that the same mass of ethanol, (i.e., m_{et}) was contained in V_{S} mL of the sample tested. Then the percentage content of ethanol (p, $\text{\%}\frac{\text{m}}{\text{V}}$
) in the sample tested is as follows

$\text{p}=1.152\cdot (6\cdot {\text{C}}_{1}\cdot {\text{V}}_{1}-{\text{C}}_{2}\cdot {\text{V}}_{2})\cdot \frac{{\text{V}}_{\text{f}}}{{\text{V}}_{0}\cdot {\text{V}}_{\text{S}}}\left(\text{\%},\frac{\text{m}}{\text{V}}\right)$ (8)

Second option (Option 2) of the method is a titration of the excess of dichromate with V_{3} mL of standard solution of (NH_{4})_{2}Fe(SO_{4})_{2} (C3). This titration proceeds according to the reaction

Cr_{2}O_{7}^{–2} + 6Fe^{+2} + 14H^{+1} = 6Fe^{+3} + 2Cr^{+3} + 7H_{2}O (9)

1

C_{1}• V_{1} – •C_{0}•V_{0}

6

C_{3}•V_{3}

From this proportion we get: 6•C_{1}• V_{1} – 4•C_{0}•V_{0} = C_{3}•V_{3}, and then:

$\text{p}=1.152\cdot (6\cdot {\text{C}}_{1}\cdot {\text{V}}_{1}-{\text{C}}_{3}\cdot {\text{V}}_{3})\cdot \frac{{\text{V}}_{\text{f}}}{{\text{V}}_{0}\cdot {\text{V}}_{\text{S}}}\text{}\left(\text{\%},\frac{\text{m}}{\text{V}}\right)$ (10)

In this titration, when the solution is almost clear-green, o-phenanthroline is added as indicator, that at the end point changes the color of the solution from blue-green to brown [18,19].

The third option (Option 3) of ethanol contents determination is based on the reaction of the dichromate excess with standard (NH_{4})_{2}Fe(SO_{4})_{2} (C_{4}, V_{4}) solution, added in excess; this excess is titrated with V_{5} of standard KMnO_{4} (C_{5}) solution. From the proportion

1

C_{1}•V_{1} – $\frac{2}{3}$
•C_{0}•V_{0}

6

n(Fe)

resulting from reaction (9) we get

n(Fe) = 6•C_{1}• V_{1} – 4•C_{0}•V_{0} (11)

From the reaction

MnO_{4}^{–1} + 5Fe^{+2} + 8H^{+1} = 5Fe^{+3} + Mn^{+2} + 4H_{2}O (12)

1

C_{5}•V_{5}

5

C_{4}•V_{4} – n(Fe)

we get

C_{4}•V_{4} – n(Fe) = 5•C_{5}•V_{5} (13)

From (11) and (13)

C_{0} • V_{0} = $\frac{2}{3}$
(6 • C_{1} • V_{1} – C_{4} • V_{4} + 5 • C<5 • V_{5}) (14)

and the content of ethanol in the analyzed sample is calculated from the formula:

$\text{p}=1.152\cdot \left(6\cdot {\text{C}}_{1}\cdot {\text{V}}_{1}-{\text{C}}_{4}\cdot {\text{V}}_{4}+5\cdot {\text{C}}_{5}\cdot {\text{V}}_{5}\right)\cdot \frac{{\text{V}}_{\text{f}}}{{\text{V}}_{0}\cdot {\text{V}}_{\text{S}}}\left(\%,\frac{m}{V}\right)$ (15)

Historically, the Option 3, involving three standard(ised) solutions, was the first one, introduced by Nicloux in 1896 (cit. in [20]).

A note on the principle of stoichiometric excess

The limiting reagent can be explained as the reactant that requires more moles to react than are available in the experiment. The term “stoichiometric excess” refers to the reagent, which remains in excess in the reaction mixture, after the reaction is completed. The minor component (in stoichiometric sense) defines the quantity (the number of mmoles) of the product formed.

An excess of the reagent can play different roles. For example, an excess of H_{2}SO_{4} assures the right course of a chemical reaction; an excess of iodide increases dissolution of iodine, I_{2}, moderately soluble in aqueous media.

To apply the reasonable/moderate quantities of the corresponding reagents, the stoichiometry of a chemical reaction should be considered before starting the experiment. In other words, stoichiometric amount of a reactant/product should be adequate for the reaction to proceed. Such a procedure is valid for the systems where the reaction leads to completion, as in the analysis specified in this paper. In general case, in particular when this requirement is not fulfilled, a use of a more advantageous simulating approach, which involves all prior physicochemical knowledge about the system tested, is suggested; its use was illustrated in the authors’ papers cited in [13-16].

This paper is an illustration how to teach complex classical analytical procedures, referred to determination of ethanol contents. These determinations, based on redox back titrations, are presented from the viewpoint of the stoichiometric calculations involved. Combination of different redox titrations for determination of the same analyte, in which prior students’ knowledge is engaged, is presented. The principle of stoichiometric excess is considered. Importance of classical analysis is highlighted.

Stoichiometry is an essential part of chemistry education; it is the science aiming to study how compounds react one with the other. It is easy to find students who are competent in calculations at particular steps of an analytical procedure, but it is difficult to find a large number of students able to link all these steps in a logical whole. Starting from this viewpoint, we consider the derivations presented in this paper as not trivial ones. As a matter of fact, a solving of stoichiometric problems encountered some problems for many students attending in introductory courses of chemistry. Certainly, the description of experimental procedure (analytical measurement) has here a prominent place. We must help the students to develop a systematic approach to solve such problems; this is the main aim of this paper. The drawback of use memorized formulas to solve problems is that they are shortcuts only, and thus a systematic reasoning is not involved in a didactics. The method presented here helps the students to calculate the results obtained from titration data in a systematic way.

The example considered is important in chemical analysis. In particular, quantitative determination of ethanol from blood samples is used for scientific research made in the forensic laboratory. An international reference solution of ethanol is prepared according to the dichromate oxidation method. Quantitative determination of ethanol in wine and liquors may be also carried out by the dichromate method. And last but not least, the costs of the related analyses are relatively low.

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