Colligative Properties of Solutions: Freezing Point Depression

Transcription

1 E1 Colligative Properties of Solutions: Freezing Point Depression PURPOSE The experiment to be performed is divided into three sections: (a) In part 1, the FP of the pure solvent, cyclohexane, is determined by constructing a cooling curve. (b) In part 2 we determine the value of the molal freezing point depression constant, kf, of cyclohexane by measuring the freezing point depression (relative to the FP of pure cyclohexane determined in part 1) of a solution with a known molality of solute. (c) In part 3 the molality of cyclohexane solution with an unknown solute is found. This makes possible the calculation of the molar mass of the unknown. DISCUSSION Consider a hypothetical cooling curve which describes the changes in phase of a pure substance in which heat is lost at a steady rate at constant pressure (see figure 1). The Freezing Point of Pure Liquid: This experiment will focus on the region in which the cooling liquid approaches and achieves the freezing point (The boxed area on cooling curve in figure 1 is seen close up in figure 2). At the freezing point both liquid and solid are present. If the system becomes thermally insulated from its surroundings, that is, heat can neither enter or leave the system, a state of equilibrium will be established. Here the number of molecules moving from solid to liquid is the same as the number of molecules moving from liquid to solid. This means that the relative quantities of solid and liquid present will remain constant and the temperature will remain constant until all of the pure substance has solidified. Liquid Solid + Heat (1)

2 The Freezing Point of a Solution: Now consider the effect of a nonvolatile solute upon a freezing solution. Any added solute would dissolve in the liquid phase but will be excluded from the solid crystal lattice. This means that, in effect, the concentration of liquid solvent molecules has been diminished and the rate of liquid solvent molecules moving to solid will decrease. On the other hand, the molecules of solid solvent (which remains pure), continue to move from solid to liquid at the same rate. If the temperature is held at the normal freezing point of the pure solvent, the system is thrown out of equilibrium and liquid phase is formed at the expense of solid phase, to a point where only liquid solution is present. In order to reestablish freezing, the temperature of the system needs to be lowered. This causes heat to be removed from the system and restores an equilibrium in which solid is present. The solution, therefore, has a lower freezing temperature than the pure solvent and the freezing point is said to be depressed. Now as a solution freezes, solvent molecules are removed from the liquid solution and deposited on the solid. This increases the concentration of solute in the liquid solution and the freezing point declines further. A solution therefore does not have a sharply defined freezing point (FP). usually the freezing point of a solution is taken as the temperature at which solid solvent crystals first begin to appear (See figure 3). The FP depression is one of a set of physical properties of solutions (vapor pressure lowering, boiling point elevation, and osmotic pressure) known collectively as colligative properties. These properties are affected by the quantity of solute particles dissolved in the solution, regardless of their identity. The FP depression (or the change in temperature of freezing) is described quantitatively by the following equation: T f = -kf m. (2) T f = T f,solution T f,solvent. T f is negative because the temperature of the solution is lower than that of the pure solvent. Molality is moles of solute per kilogram of solvent. The concentration unit mol/kg is temperature independent, unlike the mol/l concentration unit, because volume changes with temperature but mass does not. molality = m (3) The magnitude of the freezing point change is proportional to the molality of the dissolved solute: T f [solute] = m. The proportionality constant, k f, is called the molal freezing point constant. It is specific to a particular solvent. k f is the molal freezing point depression constant and expresses the sensitivity of the solvent to having its FP depressed by an added solute. The units are C/m, which may be viewed as the number of degrees the FP is depressed per one molal of solute concentration. Each solvent has its own unique value for kf. Page 2

3 So expanding equation 2: T f,solution T f,solvent = T f = - k f (4) The minus sign in the last expression in Eq. 4 result in a positive kf because T is negative while concentration is positive. Freezing point depression provides a convenient way to determine the molar mass of an unknown substance. A solution containing a known mass of solute per mass of solvent is prepared and its freezing point is measured. The number of moles of solute can be determined by rearranging Eq. 4 to solve for moles of solute, n solute : T f = - k f n solute = (5) Because the mass of unknown solute is known (measured on the balance) and the number of moles has been calculated (by Eq. 5 ), you can determine the molar mass of the unknown solute using the equation below. M m = The freezing point apparatus is shown in figs. 4 below. 1. Carefully adjust the height of the thermometer so the entire bulb of the thermometer is immersed in the liquid. 2. Carefully rotate the split stopper so the graduations on the thermometer are visible through the split. Make sure the utility clamp is located so that you can easily read the thermometer. 3. Prepare an ice-water slurry. Fill your 250-mL beaker about ½ full with ice, add enough water to fill the spaces, and stir with stirring rod. The temperature should be within a degree or so of 0 C. Procedure Part 1. Freezing point of pure cyclohexane: 4. Place a clean, dry 25 x 250 mm test tube into a 250 ml Erlenmeyer flask with the tube up-right. Put both pieces of glassware on a top loading centigram balance and tare it (this sets the balance to read 0.00 g). 5. Pour about 20 ml of cyclohexane into the test tube and record the mass to the nearest ±0.01 g in your notebook (information may be transferred to your report form, when available). Page 3

4 6. We will construct one cooling curve to observe experimentally the equilibrium between the liquid and solid solvent at the freezing(melting) point and to determine the temperature at which this takes place. Assemble the freezing point apparatus and transfer the test tube to the ice-water bath making sure that the cyclohexane is below the surface of the ice-water bath, add more ice if needed. Stir the pure cyclohexane solvent with your stirring loop. When the temperature is approximately 15 C begin to take temperature readings and record them in the table every 15 seconds while continually stirring the cyclohexane with the stirring loop in an up and down smooth continuous motion. Continue to stir and take temperature readings every 15 seconds until most of the cyclohexane solvent has solidified (~ 10 data points after solid cyclohexane has formed). Plot the data using Excel and record the temperature of freezing of the pure solvent, T f, in your notebook. Allow the cyclohexane solvent to warm up and liquefy so that it can be use in the first trial of part 2. Part 2. Determination of kf for Cyclohexane: In this part of the experiment, the molal freezing point depression constant, kf, of cyclohexane will be determined experimentally. The freezing temperature will be found for a solution using a known mass of cyclohexane as the solvent and a measured mass of para-dichlorobenzene as the solute. The freezing point depression, T f, of the solution is found by calculating the difference between the freezing temperature of the solution and the freezing temperature of the pure solvent found in part 1. After the molality of the solution, m, is calculated from the known masses of solute and solvent, the k f may be calculated: k f = T f / m Procedure Part 2. Freezing point of known solution: 7. Using a weighing boat mass out approximately 0.5 g (±0.01 g) of para-dichlorobenzene (M m = g/mol). Remove the stopper and attached thermometer and stirrer (avoiding loss of cyclohexane), just far enough to add the sample of p-dichlorobenzene. Stir the mixture to dissolve the solute completely. To your ice-water bath, in the 250 ml beaker, add about 10 ml (~7 g) of NaCl and stir to dissolve. Place your freezing point apparatus into the ice-water bath. When the temperature is at C, start taking temperature-time data every 15 seconds until most of the solution has solidified (again ~10 data points after first sign of solid). 8. Allow solution to warm and collect time-temperature data again using the same solution for a second determination. When you have finished the measurement, pour the solution into the waste cyclohexane bottle located under the fume hood. Do not pour it into the sink! Remove the last trace of solution by rinsing the test tube with few ml cyclohexane in the fume hood and warm in your hand to dry. Also clean with paper towel your thermometer/stirrer/rubber stopper apparatus. When a solute is added to the solvent the shape of the cooling curve changes so that we don t see a clear horizontal plateau as the example shows in Figure 5. In this case we must draw a trend line through the data points corresponding to the cooling of the liquid and a trend line through the data points corresponding to the freezing of the solvent. The temperature at the point where those two lines intersect is the freezing point of the solution. This temperature can be obtained from the two trendlines. Figure 5. Time vs. temp. data for Solution Page 4

5 Part 3. The Molar Mass of an Unknown Solid In this section, a measured mass of an unknown solute will be added to a fresh sample of cyclohexane solvent, whose mass is also known. The freezing point depression, T f, of this solution will be determined. Since kf has been obtained in part 2, the molality of the solution containing the unknown, m, may be calculated: m = - T f /kf. The molality, which is in units of moles solute per kg of solvent, may now be used to find the number of moles of unknown solute present, and, since the mass of the unknown was determined in the beginning, the molar mass may now be calculated. Watch your units. They insure a proper sequence of calculations. Procedure Part 3. Freezing point of unknown solution: 9. Repeat step 4, and 5 of the procedure above. 10. Using a weighing boat mass out approximately 0.5 g (±0.01 g) of an unknown solute provided and add it to the solvent in the test tube. Insert the thermometer/stirrer/rubber stopper apparatus. Stir the mixture to dissolve the solute completely. 11. Start collecting time vs. temperature data when the temperature is about ~15 C in 15 s intervals. Continue collecting data until you have ~ 10 data points past the temperature at which the first sign of a solid was observed. Record your data in your notebook. Allow solid to dissolve completely and warm the test tube and contents in your hand until it is about at 16 C. 12. Repeat step 11 for a second determination of the freezing temperature of your unknown solution.. When you have finished the measurement, pour the solution into the waste cyclohexane bottle located under the fume hood. Wipe clean your thermometer/stirrer/rubber stopper apparatus and return it to the cart. Clean out your test tube with a little soap, water, and a test tube brush and place it on the equipment cart. Page 5

6 REPORT 1 Colligative Properties of Solutions: T f Name Section: MW/TTH (circle) Date Partner s Name (if any) DATA Part 1. Temperature of freezing of the pure solvent, T f. Mass of (~20 ml) of cyclohexane used in part 1, and Trial 1/2 in part 2. g Time(s) / Temperature ( C) data for pure solvent: Part 2. Determination of known solution, T f(soln). Instructor: Trial 1 Trial 2 Mass of (~20 ml) of cyclohexane g same as in part 1 g Mass of para-dichlorobenzene g same as trial 1 g Time(s) / Temperature ( C) data Trial 1: Time(s) / Temperature ( C) data Trial 2: Part 3. Determination of unknown solution, T f(soln). Unknown# Trial 1 Trial 2 Mass of (~20 ml) of cyclohexane g same as trial 1 g Mass of unknown solute g same as trial 1 g Time(s) / Temperature ( C) data Trial 1: Page 6

7 Time(s) / Temperature ( C) data Trial 2: Calculations and Results Part 1. Temperature of freezing of cyclohexane, T f (from Excel plot, attach to your report) t C Part 2. Determination of k f Trial 1 Trial 2 Temperature of freezing of known cyclohexane/para-dichlorobenzene solution, T f(soln) C C (Attach Excel data and calculations.) T f = T f,solution T f,solvent (show calculation below) Avg T f(soln) = C C Molality of known solution (show calculation below) m Molality = n solute m solvent (kg) Average k f = - T f / m (show calculation below) Avg k f = /molal Part 3. Determination of the M mass of unknown solute Trial 1 Trial 2 Temperature of freezing of unknown solution, T f(soln) C C (Attach Excel data and calculations.) T f = T f,solution T f,solvent (show calculation below) Avg T f(soln) = C C molality = - T f /kf (show calculation below) m n solute = m(molality)(mass solvent (kg)) (show calculation below) mol M mass = (show calculation below) g/mol Unk# Page 7

8 Questions/Problems Problems 1. Cyclohexanol, C 6 H 11 OH, is sometimes used as the solvent in molecular weight determinations by freezing point depression. If g of benzoic acid, C 6 H 5 COOH, dissolved in g of cyclohexanol, lowered the freezing-point of pure cyclohexanol by 6.55 C, what is the molal freezing-point constant, k f, of this solvent? (show calculation below) /m 2. Since the freezing point of a solution depends on the relative number of particles, what would you calculate to be the freezing point of 0.1 m solutions in water of (a) NaCl, (b) BaCl2? Assume that these salts are 100% ionized in solution. {Compare your answers with the actual respective freezing points: C and C. The difference is due to the decreased activity of the ions (discussed in lecture). Because of the attractive forces between the positively and negatively charged ions, they do not move completely independently of one another.} (show calculation below) (a) C (b) C 3. How many grams of each of the following per kilogram of water in your car radiator are needed to give equal protection against freezing down to -10 C? (a) Methyl alcohol, CH 3 OH, b.p C. g (b) Ethylene glycol, C 2 H 4 (OH) 2, b.p C. g (c) In spite of higher cost, what advantage does ethylene glycol possess over methyl alcohol as a winter antifreeze and/or summer coolant? (show calculation below) Page 8