4-Partition Coefficient

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Apr 3, 2024

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Mixtures: Determining the Partition Coefficient of a Compound Introduction A. Mixture Review: • Mixtures come in two varieties, and are defined by two characteristics: (1) homogeneous (a) variable composition (b) continuous properties (2) heterogeneous (a) variable composition (b) discontinuous properties • Variable composition: The amount of the individual components can vary between mixtures. You could make a homogeneous mixture consisting of 1 gram of NaCl in 20 g of H 2 O, or 2 g of NaCl in 15 g of H 2 O. This concept can also be applied to heterogeneous mixtures. So, the composition of a mixture is defined not by the identity of its components, but the identity AND amount of its components. • Continuous/discontinuous properties: If you were to use a pipet to withdraw part of a homogeneous mixture, the density of the sample would be the same no matter where in the mixture you put the tip of the pipet, i.e. the density would be uniform/continuous. However, in a heterogeneous mixture, the density of the sample could vary depending on your placement of the pipet tip. B. Inter-particle interactions For one substance to dissolve in another, they must have attraction forces between the particles. All attractions are based on the presence or absence of a net dipole moment. The important factors are the (i) shape of the molecule, and (2) polarity of the bonds. Consider some examples (other interactions exist in this mixture). (1) Acetic acid (C 2 H 4 O 2 or HC 2 H 3 O 2 or CH 3 COOH or HAc ) dissolves in water Water is polar (from information you learned in the first semester, you should be able to explain this with no difficulty: water is bent, and its bonds are polar, thus, it has a net dipole moment). HAc has both polar and nonpolar parts: In addition, it ionizes when it dissolves in water: HC 2 H 3 O 2 𝐻 2 𝑂 → C 2 H 3 O 2 - (aq) + H + (aq). (HAc does not dissociate 100 %. You will learn more about this later in the semester). Thus, acetic acid dissolves in water due to several inter-particle interactions between the two molecules: ion- dipole (H + interacting with water), dipole-dipole (the partially positive C in HAc interacting with water), hydrogen bonding (the partially negative oxygen in HAc interacting with water, or the partially positive hydrogen in HAc interacting with water), and dipole-induced dipole (the CH 3 part of HAc interacting with water). (2) Acetic acid dissolves in tert-butyl methyl ether (C 5 H 12 O or (CH 3 ) 3 COCH 3 or TBME). TBME is considered to be a mostly nonpolar substance even though it has both polar and nonpolar parts:

CH 3  H 3 C-O-C-CH 3  CH 3 Because the geometry of the C-O-C bond is bent and the bonds are polar a net dipole moment results. On the other hand, the CH 3 - groups are relatively nonpolar: even though the bonds are polar, the tetrahedral geometry causes the overall net dipole to be very small. One important factor is that the tert-butyl group CH 3  -C-CH 3  CH 3 is large so that it partially blocks the polar C-O- C (blocking due to size and shape is called “steric hindrance”). Thus, it is possible for TBME to have dipole-dipole interactions with a solute, but they are not strong since the tert-butyl group partially blocks the other dipole from getting close to the dipole in TBME. This is referred to as steric hindrance, and is an important concept you will study further in organic chemistry. When acetic acid dissolves in TBME it does not dissociate (ionize), thus, no ion-dipole attractions are possible with this pair of molecules. The intermolecular interactions between TBME and HAc are induced dipole-induced dipole, dipole-induced dipole, dipole-dipole, and hydrogen bonding (think about which parts of each molecule must be interacting for each listed interaction). (3) Water and TBME form a heterogeneous mixture (with exceptions) When large amounts of these two substances are mixed, they do not dissolve to a large degree, thus, rather than a homogeneous mixture, these substances form a heterogeneous mixture. However, in Discussion Question 1 you will discuss the attraction forces that are the basis for a small amount of water to dissolve in TBME (or a small amount of TBME to dissolve in water). C. Partitioning of a solute between two solvents To review: HAc dissolves in both water and TBME to form homogeneous mixtures, but water and TBME form a heterogeneous mixture. Thus, when all three are placed in the same container and shaken, some of the HAc will dissolve in the water, and some of it will also dissolve in the TBME. That is, the HAc will have partitioned between the two liquids. A ratio of the amount of solute in each solution is called the partition (distribution) coefficient ? . The units can be moles solute per mL solvent, moles solute per gram solvent, grams solute per mL solvent, etc., although many ? values given in the literature have units that are ambiguous or absent. In this exercise, you will use moles per mL. ? = 𝑐??𝑐????𝑎?𝑖?? ?? ??𝑙??? 𝑖? ??𝑙???? 2 𝑐??𝑐????𝑎?𝑖?? ?? ??𝑙??? 𝑖? ??𝑙???? 1 = ???𝑒? ?????𝑒 ?𝐿 ????𝑒?? 2 ???𝑒? ?????𝑒 ?𝐿 ????𝑒?? 1 (1) D. Extraction of a solute from a solution Suppose that a chemist wanted to remove a solute from a solvent. One method for accomplishing this goal would be to add another solvent that would more favorably dissolve the solute. Thus, the solute could be transferred from one solvent to another. In general, it is more effective to extract a solute with several small aliquots (portions) rather than one large aliquot.

Often a chemist will want to calculate the amount of solute in each solvent by using the partit ion coefficient. You’ll be doing this sort of problem for part (4) of the calculations section. As an example, consider the solute A that has k = 12 for solvents B and C (A dissolves in both B and C, but it is more soluble in B). Suppose that you have a sample with 10.0 total moles of A per 1.00 L of C, and you want to transfer A from C into B by performing 2 extractions with 0.330 L of B. To calculate the amount of A in B after the first extraction, you solve the following for x: 12 = 𝑥 ???𝑒 ? 0.330 𝐿 ? (10.0−𝑥) ???𝑒 ? 1.00 𝐿 ? . Solving for x yields a value of 8.0 moles of A in B, and (10.0 moles – 8.0 moles) = 2.0 moles of A in C. For the second extraction , another portion of the 2.0 mol of A left in C will be transferred into B…thus one can set up a similar equation and solve for a new x: 12 = 𝑥 ???𝑒 ? 0.330 𝐿 ? (2.0???𝑒 ?−𝑥 ???𝑒 ?) 1.00 𝐿 ? . Solving this equation yields x = 1.6 moles A transferred to B, and thus 0.40 moles of A would remain in C. E. Application of the extraction process to the mixture you will investigate In this lab, you will determine the value of ? for HAc in a H 2 O/TBME mixture, and then use that value in further calculations to make predictions regarding extents of extraction with various volumes: ? = 𝑐??𝑐????𝑎?𝑖?? ?? ?????? 𝑇??𝐸 𝑐??𝑐????𝑎?𝑖?? ?? ?????? 𝑖? (𝑎?)?ℎ𝑎?? = ( ????? 𝐻?𝑐 𝑖? 𝑇??𝐸 ?? 𝑇??𝐸 ) ( ????? 𝐻?𝑐 𝑖? (𝑎?)?ℎ𝑎?? ?? (𝑎?)?ℎ𝑎?? ) which when using the properties of fractions can be brought into the form ? = ( ????? 𝐻?𝑐 𝑖? 𝑇??𝐸 ?? ??????? 𝑇??𝐸 )( ?? ??????? (𝑎?)?ℎ𝑎?? ????? 𝐻?𝑐 𝑖? (𝑎?)?ℎ𝑎?? ). In order to find the ? value, you need to perform an experiment to determine how much HAc is in H 2 O, and how much is in TBME (the volumes of each will be known via experimental data. Specifically, you will determine the amount of acetic acid in the aqueous phase by titration with aqueous sodium hydroxide: HC 2 H 3 O 2(aq) + NaOH (aq) → H 2 O (l) + NaC 2 H 3 O 2(aq) This is an acid/base neutralization reaction, which you will analyze by titrating to a faint pink endpoint while using phenolphthalein as an indicator. Before you come to class, copy the table below in your lab notebook, then fill in the blanks using data from part B of the procedure: mL TBME mL HAc mL NaOH Ctt  V Vi Vf  V Vi Vf  V 1 2

F. Standardization of a solution As you recall from instruction in the first semester course, the concentrations of solutions purchased from a manufacturer are often neither precise nor accurate. Thus, you need to standardize the solution; that is, you must determine the concentration precisely and accurately. This is often done by reaction with another chemical in a reaction with a known stoichiometry. In this exercise, you will standardize acetic acid by titration with a known concentration of aqueous sodium hydroxide. Notice that this is the same reaction for the analysis. Before you come to class, copy the table below in your lab notebook. You will fill in the blanks using your data from part A of the procedure during your lab period. mL HAc mL NaOH titration Vi Vf  V Vi Vf  V 1 2 Procedure Review the important techniques in your General Lab Practices Handout that you will need for this exercise. A. Standardization of Acetic acid HAc (aq) Prepare two microburets. Obtain about 10 mL of HAc (CAUTION: caustic, pungent odor) to be used in one microburet, and obtain about 10 mL standardized NaOH (aq) to be used in the other microburet. In a clean 10 mL Erlenmeyer flask, dispensed exactly 0.9 mL of HAc, add about 3 mL of dH 2 O and two drops of phenolphthalein indicator, then titrate to the endpoint with standardized NaOH (aq). Repeat the titration for exactly 0.7 mL of HAc. B. Determination of 𝒌 Obtain 2 conical test tubes (ctt). Use a microburet to place exactly 1 mL of HAc into one of the ctts. In a fume hood, pipet exactly 1 mL of TBME (CAUTION: flammable. Keep away from open flames) into the ctt that contains the HAc. Firmly seat a cork into the ctt opening and hold it with your thumb. Shake the mixture for about 1 minute, uncork the ctt then place it in the test tube rack to allow the mixture to separate into two layers (place the cork in the opening at an angle). When the layers have separated, use a disposable (Pasteur) pipet to carefully transfer the bottom (aqueous) layer into a 10 mL Erlenmeyer flask. (DO NOT WITHDRAW LIQUID FROM THE TOP LAYER OR FROM THE INTERFACE BETWEEN THE LAYERS) Add about 3 mL of dH 2 O and two drops of phenolphthalein and titrate to the endpoint with NaOH (aq) . Repeat using 0.75 mL HAc in a ctt with 1 mL TBME. C. Start working on the calculations With any remaining time, start to work through the calculations, making sure to direct any relevant questions to your instructor.

WORKSHEET - Calculations 1. Calculate the standard molarity of HAc (use your data from part A): Use the  V values, the molarity of NaOH (aq) on the bottle label, and the balanced reaction equation to calculate the standard molarity of the HAc. Do this for each trial. Take an average. Use this average molarity in subsequent calculations.

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