Heat of Reaction: Solubility method
Abstract: The experiment uses solubility method and Van Hoff’s equation in order to obtain
the solubility and integral heat of solution of oxalic acid from various levels of saturations. Meanwhile, in the experiment proper, during the preparation of the samples of saturated solution, initial solutions of higher saturation should be used in order to have a more approximate saturation of solution (lower than the initial) of temperature lower than the initial. With this, 5mL of samples are acquired, of various level of saturation (identified based on the temperature used in the preparation). Important reminder: when acquiring a sample from the prepared saturated solution, assuring that no solid crystals is included in the sample one wishes to obtain so that error in the results may be avoided. When enough and relevant datum are acquired throughout the experiment, the solubility and integral heat of solution of oxalic acid can then be attained using Solubility me thod and Vant Hoff’s equation. After then, a semi log plot of Log S vs. 1/T can be plotted.
Introduction: The main purpose of the experiment is to determine the Integral heat of solution of an organic acid (Oxalic acid) by the Solubility Method the Vant Hoff’s eq uation wherein the integral heat of solution is the enthalpy of a solution minus the sum of the enthalpies of its components. It refers to the heat of the process of dissolving rather than the physical mixture
that is formed.
Solubility is the property of a solid, liquid, or gaseous chemical substance called solute to dissolve in a liquid solvent to form a homogeneous solution of the solute in the solvent. The solubility of a substance fundamentally depends on the used solvent as well as on temperature and pressure. The extent of the solubility of a substance in a specific solvent is measured as the saturation concentration where adding more solute does not increase the concentration of the solution. Thus, in acquiring the solubility of the oxalic acid, the equivalent weight of oxalic acid from the saturated sample of the solution is determined. The Vant Hoff’s equation (also known as the Vukancic-Vukovic equation) in
chemical thermodynamics relates the change in temperature (T) to the change in the equilibrium constant (K) given the standard enthalpy change ( ΔH ΔHo) for the process. it follows that
Therefore, a plot of the natural logarithm of the equilibrium constant versus the reciprocal temperature gives a straight line. The slope of the line is equal to minus the standard enthalpy o
change divided by the gas constant,
- ΔH /R and
change divided by the gas constant,
ΔS /R .
o
the intercept is equal to the standard entropy
Differentiation of this expression yields the van 't
Hoff equation. Using the equation, the solubility S as a function of the reciprocal of the temperature T (1/T) plotted on a semi log plot. The slope of this line is equal to - ∆H/2.303R, so that the integral heat of solution is obtained.
Materials and Methods: Prepare 50ml of a saturated solution of oxalic acid at 50 oC so that at lower temperature, assurance of acquiring a saturated solution at desired temperature (lower than 50oC) is high. Cool the solution in constant baths of 35 oC, 30 oC, and 25 oC for 10 minutes of each interval. At these temperatures, crystallization occurs due to the lowering of the solubility of the solvent. Acquire 5mL of samples using a pipette with cotton at the tip of the pipette to prevent the inclusion of solid crystals, for the reason that it may cause to oversaturation of the sample solutions. Get the masses of these samples. Dilute each up to 25mL with distilled water and titrate to standardized NaOH with phenolphthalein as indicator. If necessary, carry another runs until constant solubility is obtained. Here, equilibrium is reached.
Results and Discussions: Table 1 – Preparation of sample solutions
Run
Temperature (oC)
Mass of empty flask (g)
Mass of flask and solution (g)
Mass of Solution (g)
1
35
28.79
33.85
5.06
2
30
28.90
34.01
5.11
3
25
28.83
33.98
5.15
The table 1 shows three (3) runs of samples prepared from various temperatures. From the obtained sample, masses can then be acquired. The mass of solution is the difference of the mass of the flask with the solution and the mass of the empty flask. Also, the mass of the solution is the sum of the mass of oxalic acid and water at equilibrium or saturation point. Table 2 – Titration of sample solutions Mass of
Run (sample)
Temperature (oC)
Volume of NaOH used (mL)
1
35
12.94
3.882 x 10
-3
7.763 x 10
-3
0.699
4.361
2
30
12.23
3.699 x 10
-3
7.397 x 10
-3
0.666
4.444
3
25
12.01
3.604 x 10
-3
7.208 x 10
-3
0.649
4.501
Moles of
Moles of NaoH
C2H2O4
C2H2O4
(g)
Mass of H2O (g)
Table 2 shows the values after titration of the samples with NaOH in order to get the masses of oxalic acid in various samples. By titration, at certain volumes of NaOH added to the samples, equilibrium is reached. Using the volumes of the base multiplied by its concentration, the moles of NaOH is acquired. Using stoichiometric ratio of the equation NaOH + 2 C2H2O4
Na2C2O4 + 2H2O, moles of C 2H2O4 is obtained. Hence, masses of C 2H2O4 is
obtained (molecular weight of C 2H2O4 is 90.04 g/mol). Afterwards, the masses of water in the samples are acquired knowing that it is equal to the difference of the total mass of samples and mass of C2H2O4. Solubility method is used here. Table 3 – Determination of Solubility
Run(sample)
Temperature (oC)
Solubility (g C2H2O4 / g H2O)
Solubility (g C2H2O4 /100 g H2O)
1
35
0.16032
16.032
2
30
0.14997
14.997
3
25
0.14421
14.421
Table 3 shows the solubility of each samples at each bath temperatures in different units; (g C 2H2O4 / g H2O) and (g C2H2O4 /100 g H2O). C2H2O4 served as the solute and H2O served as the solvent of the solution. Solubility of each solution depends on the temperature. As the temperature decreases, the solubility decreases. So solubility is proportional to temperature.
Table 4 – Data for the graph
Solubility, S (g C2H2O4 /100 g
Log S
Temperature, T (K)
1/T
16.032
1.205
308.15
3.25 x 10
-3
14.997
1.176
303.15
3.30 x 10
-3
14.421
1.159
298.15
3.35 x 10
-3
H2O)
Table 4 converts Solubility, S to log S and temperature, T(in Kelvin units) in to 1/T. These values are plotted in a semi log plot to obtain a straight line and determine the constant slope that is get using linear regressions. The graph of this is shown below. By linear regression or best fit method, Log S = -460 (1/ T) + 2.698 where -460 served as the slope of the line. Using the Vant Hoff’s equation, the slope served as -∆H/2.303R. So ∆H, integral heat of solution is obtained by equating it to the experimental slope. Since R is equal to 8.314J/mol so ∆H is equal to 8807.68 J/mol. This also indicate that the greater the Log S t he smaller the 1/T.
Therefore, the greater the Solubility (relative to log S) is the greater the temperature (inverse to 1/T) and vice versa. Since the solubility of a substance is a special case of the equilibrium constant, the Vant equation is applicable replacing K to S. The value obtained may vary slightly from the value of a more dilute solution. The heat of solution so calculated is approximately the average heat of solution over the temperature range studied, and it is the heat of solution at the saturation concentration.
Graph for Log S vs. 1 /T (K) 1.21 1.2 1.19 1.18 S g o L
1.17 Log S = -460 (1/ T) + 2.698
1.16 1.15 1.14 1.13 3 . 2 5 x 1 0- 3
3.30 x 10-3
3.35 x 10-3
1/T(K)
Sample Calculations: Run 1 Obtained: mass of flask = 28.79g, mass of flask with solution = 33.85g Total mass of sample solution = 33.85g – 28.97g = 5.06g Obtained: volume of NaOH solution used = 12.94mL Moles of NaOH = Volume NaOH used x molarity of solution = (0.01294L)(0.3M) = 3.882 x 10 Moles of C2H2O4 = 2 Moles of NaoH = 2(3.882 x 10 Mass of C2H2O4 = (7.763 x 10
-3
-3
moles) = 7.763 x 10-3 moles
moles)(90.04g/mol) = 0.699g
Mass of H2O = mass of solution – mass of C2H2O4 = 5.06g - 0.699g = 4.361g Solubility= mass solute/mass solvent = 0.699g/= 4.361g = 0.16032 g C 2H2O4 / g H2O Solubility=( 0.16032 g C 2H2O4 / g H2O)(100/100)= 16.032 g C2H2O4 /100 g H2O Obtained equation: Log S = -460 (1/ T) + 2.698 , slope =-460(best fit method)
-3
moles
∆H = -(slope)(2.303R) = -(-460)(2.303)(8.314) =8807.68 J/mol
Conclusion: The integral heat of solution of the oxalic acid-water solution is 8807.68 J/mol. It was determined from the Van’t equation with equating equilibrium constant to solubility of the
solution. The graph is seen above with equation, Log S = -460 (1/ T) + 2.698. The solubility of oxalic acids(g C2H2O4 /100 g H2O) are 16.032 at 35 oC , 14.997 at 30 oC and 14.421 at 25 oC. The lower the temperature is the lower its solubility. Reference: http://web.njit.edu/~grow/solubility/solubility.html http://en.wikipedia.org/wiki/Van_%27t_Hoff_equation http://dl.clackamas.edu/ch105-03/heatof.htm
