Enthalpy & Entropy of Aqueous Borax Solution Aims and Objectives
1. To measure the enthalpy (∆H) and entropy (∆S) of an aqueous solution of borax (sodium tetraborate decahydrate) from the temperature dependence of the equilibrium constant for the dissolution reaction of borax in water. 2. To study a system of an aqueous aqueo us solution of borax. 3. To become acquainted with the changes in free energy, enthalpy, and entropy and their relation to the equilibrium constant for a chemical reaction. Introduction Borax, most commonly known as sodium borate, sodium tetra borate, or disodium tetra borate, is an important boron compound, a mineral, and a salt of boric acid. Powdered borax is white, consisting of soft colorless crystals that dissolve easily in water. Borax is obtained as tincal, Na2B4O5(OH)4.8H2O and kernite, Na2B4O7.4H2O.
Borax has a wide variety of uses. It is a component of many detergents, cosmetics, and enamel glazes. It is also used to make buffer solutions in biochemistry, as a fire retardant, as an anti-fungal compound for fiberglass, for fiberglass, as a flux in metallurgy, neutron-capture shields for radioactive sources, a texturing agent in cooking, co oking, and as a precursor for other boron compounds. Enthalpy is a measure of the total energy of a thermodynamic system. It includes the internal energy, which is the energy required to create a system, and the amount of energy required to make room for it by displacing its environment and establishing its volume and pressure.
Enthalpy is a thermodynamic potential. It is a state function and an extensive quantity. The unit of measurement for enthalpy in the International System of Units (SI) is the joule, but other historical, conventional units are still in use, such as the British thermal unit and the calorie. Saturation is the point at which a solution of a substance can dissolve no more of that substance and additional amounts of it will appear as a separate phase. This point of maximum concentration, the saturation point, depends on the temperature and pressure of the solution as well as the chemical nature of the substances involved. This can be used in the process of re-crystallization of re-crystallization to purify a chemical: it is dissolved to the point of saturation in hot solvent, then as the solvent cools and the solubility decreases, excess solute precipitates. Impurities, being present in much lower concentration, do not saturate the solvent and so remain dissolved in the liquid. If a change in conditions (e.g. cooling) means that the concentration is actually higher than the saturation point, the solution has become supersaturated.
A spontaneous process is the time-evolution of a system in which it releases free energy (usually as heat) and moves to a lower, more thermodynamically stable energy state. The
sign convention of changes in free energy follows the general convention for thermodynamic measurements, in which a release of free energy from the system corresponds to a negative change in free energy, but a positive change for the surroundings. A spontaneous process is capable of proceeding in a given direction, as written or described, without needing to be driven by an outside source of energy. The term is used to refer to macro processes in which entropy increases; such as a smell diffusing in a room, ice melting in lukewarm water, salt dissolving in water, and iron rusting. The laws of thermodynamics govern the direction of a spontaneous process, ensuring that if a sufficiently large number of individual interactions (like atoms colliding) are involved then the direction will always be in the direction of increased entropy (since entropy increase is a statistical phenomenon).
The total enthalpy, H , of a system cannot be measured directly. Thus, change in enthalpy, Δ H , is a more useful quantity than its absolute value. The change Δ H is positive in endothermic reactions, and negative in heat-releasing exothermic processes. Δ H of a system is equal to the sum of non-mechanical work done on it and the heat supplied to it.
Theory
Borax dissociates in water to form sodium and borate ions and water. This reaction occurs by the following chemical equation: Na2B4O7 • 10H2O(s) ⇔ 2Na(aq) + B4O5(OH)42(aq) + 8H2O(l )
Free energy change of a chemical process is directly proportional to the equilibrium constant of the chemical process. This can be deduced by the following equation: G = RT ln K =
S
Where R= gas constant T= absolute temperature of reaction in Kelvin , k= equilibrium constant
Making ln K the subject, we obtain : Ln K = (-
R
S/R
In this form the equation is now linear and 1/T and lnK can be used as (x, y) points on a graph. The slope of such a graph is related to the change in enthalpy and the change in entropy is related to the yintercept:
ln K sp =
ln K sp =
y
=
-H
+
RT
T S RT
-H
1
R
T
m
+
S
R x
+
However; the equilibrium constant used here is the solubility constant. The solubility constant is used because equilibrium is established in a saturated solution a t a specific temperature. Solubility is highly dependent on temperature. The solubility constant is obtained by the following equation: Na2B4O7 • 10H2O(s) ⇔ 2Na(aq) + B4O5(OH)42(aq) + 8H2O(l ) K sp = [Na]2 [B4O5(OH)42] 2-
Since there are two sodium ions produced for each borate ion [B4O5(OH)4 ] then [Na] = 2 [B4O5(OH)42] 2- 3
And the K sp can be rewritten as: K sp = 4[B4O5 (OH)4 ]
Hence the k sp can be calculated for and used in equation (1) in order to calculate the changes in entropy and enthalpy of the reaction. The concentration of borate can easily found by a simple acid-base titration of the borax solution with HCl. The simplicity of the titration is due to th e fact that the borate ion is a weak base. The endpoint of the titration is signaled by the colour change of bromo cresol green indication +
This is the equation of the titration reaction: B4O5(OH)4 + 2H + 3H2O → 4B(OH)
3
experimental procedure will specifically state that you are to make sure there is always some solid borax remaining in the sample mixture before you remove some of it to analyze. That being the case, the 'concentration' of solid borax can be assumed constant, and the equilibrium expression can be simplified: RT
ln Ksp = = H TS
• A table of free energy values (G) with the temperatures those values correspond to should be
compiled. This is most conveniently done using equation (A), with values for Ksp and the temperatures for which they are valid. • A graph of ln Ksp vs. 1/T should be made, the slope of which is related to change in enthalpy. • The same graph has a relationship between change in entropy and its y-intercept
The experiment requires that the solubility of borax be found at various temperature values. Samples of saturated borax solution are collected at no less than 5 different temperatures, four above room temperature, and one close to or at room temperature. These samples are then warmed (if necessary) to re-dissolve any precipitated borax, and titrated to the yellow bromocresol green endpoint with standardized aqueous hydrochloric acid.
.
Chemicals and equipment
Distilled Water
Digital analytical Balance
0.500 HCl
125 ml Erlenmeyer flask
30g Borax
5 Test tubes
Bromo cresol Indicator
Beakers
Wash Bottle
Water bath
Burette
Laboratory Thermometer
Volumetric pipette
Heating Apparatus
Procedure
1) 5 clean test tubes were labeled and 5.00ml of water was transferred into each test tube with a volumetric pipette. 2) The exact level of the water in the test tube is marked after which it is poured out and the test tube is dried. 3) Approximately 30g of borax is dissolved in a 100mL beaker in 50mL of distilled water. o 4) The mixture is then heated gently to 65 C. A little more borax is added to the solution until excess solid was present and the solution is completely saturated. o 5) The solution is allowed to cool to approximately 60 C and was poured into the test tube #1 until the level of the solution reached the exact 5.00mL mark on the test tube. The o exact temperature of the solution was noted at 60 C. o 6) The parent solution is then allowed to cool to approximately 50 C and yet again exactly 5.00mL of it is poured into test tube #2. The exact temperature of the solution is noted as well. o o o 7) The process is repeated at temperatures of: 40 C, 30 C, and 20 C. the various cooled solutions were poured into test tubes #3, #4 and #5 respectively. 8) The contents of test tube #1 were transferred into a clean 125mL Erlenmeyer flask. Several portions of distilled water was added to the borax solution in order to keep the solution completely dissolved. 9) Water is added to the crystallized solid remains in the test tube, and is heated gently until the borax dissolved and then transferred to the flask. 10) 3 drops of Bromo cresol green indicator is add ed to the flask. 11) A clean 50mL burette is filled with 0.5M HCl solution. 12) A few mL of the acid was drained into an empty beaker and 1 drop of the bromocresol green was added to the acid in order to verify the colour of the indicator in an acidic solution. 13) The content of the beaker was titrated to the end point with the HCl solution. 14) Steps 8 through 13 were done to the borax solutions in the other test tubes.
Results Table of Results 5-
o Temperature / C
Volume B4O5(OH)4 /ml
Volume HCl/ml
60
5.0
26.40
50
5.0
12.60
40
5.0
7.2
30
5.0
4.8
20
5.0
3.0
o
-1
Temp ( C)
T (Kelvin)
1/T (Kelvin )
[Oxalate]
Ksp
In(Ksp)
60
333
3.003 x 10
-3
1.32
9.20
2.22
50
323
3.096 x 10
-3
0.63
1.000188
0.00
40
313
3.195 x 10
-3
0.36
0.187
-1.68
30
303
3.300 x 10-3
0.24
0.055
-2.90
20
293
3.413 x 10
-3
0.15
0.0135
-4.31
o
1) At T = 20 C = 273 + 20 = 293K -1 -1 -3 -1 T = 293 = 3.413 × 10 K
o
4) At T = 50 C = 273 + 20 = 323K -1 -1 -3 -1 T = 323 = 3.096 × 10 K
o
2) At T = 30 C = 273 + 20 = 303K -1 -1 -3 -1 T = 303 = 3.00 × 10 K o
3) At T= 40 C =273+20=313K -1
-1
T = 313
-3
o
5) At T = 60 C = 273 + 20 = 333K -1 -1 -3 -1 T = 333 = 3.003 × 10 K -1
= 3.195 × 10 K
Calculation From the equation: -
2-
B4O5(OH)4
(aq)
+ 2HCl (aq) + 3H2O(l) ----------> 4B(OH)3 (aq) + 2Cl
1) Vol (HCl) = 26.40ml 1 mol Borate
2 mol HCl
(aq)
+
N(Borate) = 0.5 x n(H ) +
+
+
N(H ) = [H ] x Volume (H ) = 0.5 x 26.40/1000 = 0.0132M Hence n(Borate) = 0.5 × 0.0135 -3
= 6.6 x 10 mol C= n/v 5
-3
Concentration of borax, [B4O5(OH)4 ] = 6.75 x 10 /0.005 = 1.32M 3
3
K sp = 4 [Borate] = 4(1.32) = 9.20 Ln (K sp) = ln(9.20) = 2.22
2) Vol (HCl) = 12.60 ml 1 mol Borate
2 mol HCl +
N (Borate) = 0.5 x n(H ) +
+
+
N (H ) = [H ] x Volume (H ) = 0.5 x 12.60/1000 -3
= 6.3 x 10 M Hence N (Borate) = 0.5 × 0.0135 = 3.15 x 10-3 mol C= n/v 5
-3
Concentration of borax, [B4O5 (OH)4 ] = 3.15 x 10 /0.005 = 0.63M 3
3
K sp = 4 [Borate] = 4(0.63)
= 1.000188 Ln (K sp) = ln(1.00) = 1.88 x 10-4 3) Vol (HCl) = 7.20 ml 1 mol Borate
2 mol HCl +
N (Borate) = 0.5 x n(H ) +
+
+
N (H ) = [H ] x Volume (H ) = 0.5 x 7.20/1000 = 3.6 x 10-3M Hence -3
N (Borate) = 0.5 × 3.6 x 10 = 1.8 x 10-3 mol C= n/v
5
-3
Concentration of borax, [ B4O5 (OH)4 ] = 6.75 x 10 /0.005 = 0.36M 3
3
K sp = 4 [Borate] = 4(0.36) = 0.187 Ln (K sp) = ln(0.187) = -1.68
4) Vol (HCl) = 4.80 ml 1 mol Borate
2 mol HCl +
N (Borate) = 0.5 x n(H ) +
+
+
N (H ) = [H ] x Volume (H ) = 0.5 x 4.80/1000 = 2.4 x 10-3M Hence
-3
N (Borate) = 0.5 × 2.4 x 10 M = 1.2 x 10-3 mol C= n/v 5
-3
Concentration of borax, [ B4O5 (OH)4 ] = 1.2 x 10 /0.005 = 0.24M 3
3
K sp = 4 [Borate] = 4(0.24) = 0.055 Ln (K sp) = ln(0.055) = -2.90
5) Vol (HCl) = 3.00 ml 1 mol Borate
2 mol HCl +
N (Borate) = 0.5 x n(H ) +
+
+
N (H ) = [H ] x Volume (H ) = 0.5 x 3.0/1000 = 1.5 x 10-3M Hence -3
N (Borate) = 0.5 × 1.5 x 10 -4
= 7.5 x 10 mol C= n/v 5
-4
Concentration of borax, [ B4O5 (OH)4 ] = 7.5 x 10 /0.005 = 0.15M 3
3
K sp = 4 [Borate] = 4(0.15) = 0.0135 Ln (K sp) = ln(0.0135) = -4.31
3 2 1 0 ) p s K ( n L
0
1
2
3
4
5
6
-1 -2 -3 y = -1.596x + 3.454 -4 -5
1/T
From the graph, the slope is 1.596. But the slope is =
R= 8.314 J/molK
= 8.314 x 1.596 = 13.269 J/molK
If the y-intercept is3.454 then:
= 3.454
= 8.314 x 3.454 = 28.72 J/molK
The two points are: -3
-3
(3.413 x 10 , -4.31) and (3.003 x 10 ,2.22) o
-3
o
-3
-4.31 = -ΔH (3.413 x 10 ) + ΔS....... (1) R R 2.22 = -ΔH (3.003 x 10 ) R
+ ΔS…... R
(2)
But R = 8.314 J/mol and multiplying both sides by the value of R o
-3
8.314(-4.31) = -ΔH (3.413 x 10 ) + ΔS…...
(1)
8.314(2.22)
o
-3
= -ΔH (3.003 x 10 ) + ΔS…...
(2)
(1) – (2) o
-54.29042= 0.00041 x ΔH o
→ ΔH = -132415.66 J/mol = -132.416 KJ/mol Substituting
o
ΔH into (1) -3
-35.83334 = -(3.413 x 10 × -132415.66) + ΔS ΔS
= 416.101 J/mol
Discussion
By theory we obtain and enthalpy value of 150kj. We got a little less than that. Indicating that the substances used in the experiment wasn’t pure. When the borax solution was heated to the specified temperatures, it took only a matter of seconds to begin recrystallizing. Hence more water was added to the already heated solution and heated again for a little while to prevent the borax from completely recrystallizing. Upon the addition of bromo cresol indicator to the borax solution, the colur of the solution within the Erlenmeyer flask changed from creamy to yellow. The mixture was then titrated with 0.5M HCl. The endpoint of the titration comes by like a flash. The colour change was from yellow to bluish green. It was necessary to add the HCl in little drops. At a temperature of 40oC the volume of the titrant required was 7.20ml. After the experiment was complete and the calculations made, it was deduced that the values of the changes in enthalpy and entropy of the sytem were: -132.416 KJ/mol and 416.101 J/mol respectively.
Sources of error Precautions
1. Lab coats, goggles, and disposable gloves were put on throughout the course of the experiments. 2. We ensured that the glassware were washed with samples of the solution before they were used
3. We ensured that the experiments were done at the right temperatures required. 4. Readings from the burette were taken from the bo ttom of the meniscus. 5. The burette was erected upright to ensure accu rate reading of the meniscus. 6. The volumetric flask was covered with the stopper immediately it was filled up to its mark with the solution. 7. Our hands were thoroughly washed before leaving the laboratory.
