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Animal fats and vegetable oils are triesters formed from the reaction between glycerol \((1,2,3\) -propanetriol \()\) and three long-chain fatty acids. One of the methods used to characterize a fat or an oil is a determination of its saponification number. When treated with boiling aqueous \(\mathrm{KOH},\) an ester saponifies into the parent alcohol and fatty acids (as carboxylate ions). The saponification number is the number of milligrams of KOH required to saponify 1.000 gram of the fat or the oil. In a typical analysis a \(2.085-\mathrm{g}\) sample of butter is added to \(25.00 \mathrm{~mL}\) of \(0.5131 \mathrm{M} \mathrm{KOH}\). After saponification is complete the excess \(\mathrm{KOH}\) is back titrated with \(10.26 \mathrm{~mL}\) of \(0.5000 \mathrm{M} \mathrm{HCl}\). What is the saponification number for this sample of butter?

Short Answer

Expert verified
The saponification number for this butter sample is 207.1 mg/g.

Step by step solution

01

Calculate Moles of HCl Used

We start by calculating the moles of hydrochloric acid (HCl) used in the back titration. Using the volume and molarity of HCl, the moles are:\[moles \ HCl = 10.26 \ mL \times 0.5000 \ M = 0.00513 \ mol\ HCl\]
02

Calculate Moles of Excess KOH

Since 1 mole of HCl reacts with 1 mole of KOH in the titration, the moles of excess KOH that were not used in the saponification are equal to the moles of HCl:\[moles \ KOH (excess) = 0.00513 \ mol\]
03

Calculate Initial Moles of KOH Added

We determine the initial moles of KOH added to the reaction using its molarity and volume:\[moles \ KOH (initial) = 25.00 \ mL \times 0.5131 \ M = 0.0128275 \ mol\]
04

Calculate Moles of KOH Used in Saponification

To find out how many moles of KOH were used in saponifying the butter, subtract the excess moles of KOH from the initial moles:\[moles \ KOH (saponification) = 0.0128275 \ mol - 0.00513 \ mol = 0.0076975 \ mol\]
05

Calculate Saponification Number

The saponification number is the number of milligrams of KOH needed to saponify 1 gram of fat. First, convert the moles of KOH used to mass in milligrams:\[mass \ KOH = 0.0076975 \ mol \times 56.11 \ g/mol \times 1000 \ mg/g = 431.81 \ mg\]Now, calculate the saponification number, which is the mass of KOH per gram of butter:\[\text{Saponification number} = \frac{431.81 \ mg}{2.085 \ g} = 207.083 \frac{mg}{g}\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Triesters
Triesters are a type of ester that forms when glycerol, a molecule with three hydroxyl groups, reacts with three fatty acids. Imagine a backbone of glycerol holding three long chains of fatty acids like a fork holding spaghetti. This combination creates a structure known as a triester. In the world of chemistry, understanding the structure of triesters is crucial because it determines many properties of fats and oils. These compounds are typically found in both plant-based oils and animal fats.
To simplify, triesters are like a triangle structure with glycerol at one corner and each of the fatty acids contributing to the other corners. This unique shape and the bonds formed between the glycerol and the fatty acids enable a variety of reactions, including saponification, which breaks these bonds to release glycerol and fatty acids. Knowing about triesters helps us understand how fats are broken down and used in different chemical processes, such as making soaps.
Back Titration
Back titration is a clever technique used when a direct titration is not possible or convenient. It involves adding an excess amount of a standard reagent to the reactant, allowing the reaction to proceed to completion, and then determining the leftover amount of the reagent through titration.
In our example, after the butter sample is treated with excess KOH for saponification, not all of the KOH is used. The remaining KOH is instead measured by reacting it with hydrochloric acid (HCl) through a second reaction, known as back titration. By knowing how much HCl is used to neutralize the leftover KOH, we can backtrack to find out how much KOH participated in the saponification. This method is particularly useful when dealing with complex reactions since it allows for precise calculations of reactants and products.
Glycerol
Glycerol, scientifically known as 1,2,3-propanetriol, is an important component in creating triesters. It is a simple, sweet, and non-toxic alcohol that plays a key role in various biological and chemical processes. Glycerol is the foundation upon which fatty acids attach to form triesters, which are the core of oils and fats.
Once a triester undergoes saponification, glycerol is one of the products released. The release of glycerol is crucial because it severs the link that binds the fatty acids, allowing them to react further, such as forming soap. Beyond chemistry labs, glycerol is also used in pharmaceutical formulations, food industries, and cosmetic products because of its moisturizing properties.
Fatty Acids
Fatty acids are long chains of carbon atoms bonded with hydrogen atoms, terminating in a carboxyl group. In triesters, three fatty acids react with glycerol to form an ester linkage. These chains can vary in length and saturation level, impacting the properties of the fats and oils they form.
Saponification, which refers to hydrolyzing the ester bond, frees the fatty acids from the glycerol backbone. As these fatty acids are released, they transform into carboxylate ions useful in creating soaps. The properties of these fatty acids are significant: they influence the melting point, hardness, and even the nutritional aspects of the fats. Understanding their role not only helps in scientific processes but also in an endless range of products we use daily.

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Most popular questions from this chapter

Using a ladder diagram, explain why the presence of dissolved \(\mathrm{CO}_{2}\) leads to a determinate error for the standardization of \(\mathrm{NaOH}\) if the end point's \(\mathrm{pH}\) is between \(6-10\), but no determinate error if the end point's \(\mathrm{pH}\) is less than 6.

An acid-base titration can be used to determine an analyte's equivalent weight, but it can not be used to determine its formula weight. Explain why.

Explain why it is not possible for a sample of water to simultaneously have \(\mathrm{OH}^{-}\) and \(\mathrm{HCO}_{3}^{-}\) as sources of alkalinity.

The amount of \(\mathrm{Cr}^{3+}\) in an inorganic salt is determined by a redox titration. A portion of sample that contains approximately \(0.25 \mathrm{~g}\) of \(\mathrm{Cr}^{3+}\) is accurately weighed and dissolved in \(50 \mathrm{~mL}\) of \(\mathrm{H}_{2} \mathrm{O} .\) The \(\mathrm{Cr}^{3+}\) is oxidized to \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\) by adding \(20 \mathrm{~mL}\) of \(0.1 \mathrm{M} \mathrm{AgNO}_{3}\), which serves as a catalyst, and \(50 \mathrm{~mL}\) of \(10 \% \mathrm{w} / \mathrm{v}\left(\mathrm{NH}_{4}\right)_{2} \mathrm{~S}_{2} \mathrm{O}_{8}\), which serves as the oxidizing agent. After the reaction is complete, the resulting solution is boiled for 20 minutes to destroy the excess \(\mathrm{S}_{2} \mathrm{O}_{8}^{2-},\) cooled to room temperature, and diluted to \(250 \mathrm{~mL}\) in a volumetric flask. A \(50-m L\) portion of the resulting solution is transferred to an Erlenmeyer flask, treated with \(50 \mathrm{~mL}\) of a standard solution of \(\mathrm{Fe}^{2+}\), and acidified with \(200 \mathrm{~mL}\) of \(1 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}\), reducing the \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\) to \(\mathrm{Cr}^{3+}\). The excess \(\mathrm{Fe}^{2+}\) is then determined by a back titration with a standard solution of \(\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\) using an appropriate indicator. The results are reported as \(\% \mathrm{w} / \mathrm{w} \mathrm{Cr}^{3+}\). (a) There are several places in the procedure where a reagent's volume is specified (see italicized text). Which of these measurements must be made using a volumetric pipet. (b) Excess peroxydisulfate, \(\mathrm{S}_{2} \mathrm{O}_{8}^{2-}\) is destroyed by boiling the solution. What is the effect on the reported \(\% \mathrm{w} / \mathrm{w} \mathrm{Cr}^{3+}\) if some of the \(\mathrm{S}_{2} \mathrm{O}_{8}^{2-}\) is not destroyed during this step? (c) Solutions of \(\mathrm{Fe}^{2+}\) undergo slow air oxidation to \(\mathrm{Fe}^{3+}\). What is the effect on the reported \(\% \mathrm{w} / \mathrm{w} \mathrm{Cr}^{3+}\) if the standard solution of \(\mathrm{Fe}^{2+}\) is inadvertently allowed to be partially oxidized?

A quantitative analysis for aniline \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}, K_{\mathrm{b}}=3.94 \times 10^{-10}\right)\) is carried out by an acid-base titration using glacial acetic acid as the solvent and \(\mathrm{HClO}_{4}\) as the titrant. A known volume of sample that contains \(3-4 \mathrm{mmol}\) of aniline is transferred to a \(250-\mathrm{mL}\) Erlenmeyer flask and diluted to approximately \(75 \mathrm{~mL}\) with glacial acetic acid. Two drops of a methyl violet indicator are added, and the solution is titrated with previously standardized \(0.1000 \mathrm{M} \mathrm{HClO}_{4}\) (prepared in glacial acetic acid using anhydrous \(\mathrm{HClO}_{4}\) ) until the end point is reached. Results are reported as parts per million aniline. (a) Explain why this titration is conducted using glacial acetic acid as the solvent instead of using water. (b) One problem with using glacial acetic acid as solvent is its relatively high coefficient of thermal expansion of \(0.11 \% /{ }^{\circ} \mathrm{C}\). For example, \(100.00 \mathrm{~mL}\) of glacial acetic acid at \(25^{\circ} \mathrm{C}\) occupies \(100.22 \mathrm{~mL}\) at \(27^{\circ} \mathrm{C}\). What is the effect on the reported concentration of aniline if the standardization of \(\mathrm{HClO}_{4}\) is conducted at a temperature that is lower than that for the analysis of the unknown? (c) The procedure calls for a sample that contains \(3-4\) mmoles of aniline. Why is this requirement necessary?

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