26: Experiment Thermodynamics of the Dissolution of Borax
Your Name Lab Partner: Chemistry 1300 Instructor: Dr. Giarikos Laboratory Assistant: Date of Experiment:
Abstract:
The purpose of this experiment was to determine the thermodynamic properties of the changes in entropy, enthalpy and free energy, as well as the solubility product of borax as a function of temperature from the dissolution of borax in an aqueous solution. The thermodynamic properties of the reaction helped to determine the change in heat, disorder and spontaneity within the system. The hypothesis of this experiment was accepted on the basis that the acid- base titration of borax and HCl yielded the equation, y = -789.38x - 6.6469 (ln Ksp= [H/R][1/T] +[ S/ R]) from the graph of ln Ksp and 1/ T. Furthermore the values for S, H and G were calculated to be -55.3 J/mol K, 6.56 kJ/mol, and 23.0 kJ, respectively. 1
The purpose of this experiment was to determine the thermodynamic properties of the entropy, enthalpy and free energy, as well as the solubility product of borax as a function of temperature from the dissolution of borax in an aqueous solution. Thermodynamics is the study of heat and its transformations. The properties of thermodynamics are entropy, enthalpy and free energy. The properties of thermodynamics can be viewed in terms of spontaneity. Spontaneity is a spontaneous change of a system that occurs by itself under specific conditions, without input of energy from the surroundings. Entropy, S, is the tendency for the universe to move towards more disorder. If the value for entropy was negative, then the amount of disorder within a system would decrease, thus causing the reaction to be non-spontaneous. 1 Moreover, if the value for entropy was positive, then the amount of disorder would increase within a system which would cause the reaction to occur spontaneously. Enthalpy, H, is the total energy within a system in relation to work and heat. If the value of enthalpy is negative, then the reaction is exothermic. 1 Moreover, if the value of enthalpy is positive, then the reaction is endothermic. However, the magnitude of the enthalpy does not determine the spontaneity of a reaction. Gibbs free energy, G, is a measurement of spontaneity. If the value of free energy is negative, then the reaction is spontaneous. 2 Furthermore, if the value of free energy is positive, then the reaction is non-spontaneous. The free energy change of a chemical is proportional to the equilibrium constant according to the equation, G=-RT lnK. Moreover, free energy change of a chemical process is a function of enthalpy and entropy based on the equation, G= H TS. Furthermore, when the two free energy expressions were set equal to each other, the
2
equation, ln Ksp = - (H/R) (1/T) + (S/R) was utilized to determine the thermodynamic properties of a chemical system. It was hypothesized that the spontaneity of a thermodynamic reaction could be determined by the determination of entropy, enthalpy and free energy with the utilization of an acid-base titration. Moreover, the entropy, enthalpy and free energy were determined by the solubility product at varying temperatures. Materials and methods: Please refer to Experiment 26 on page 299-308 of Laboratory Manual for Principles of General Chemistry by J.A. Beran. The only deviation that was observed in this experiment was due to a procedural change in the molar concentration of HCl from 0.2 M to 0.25 M. Otherwise, there were no observable deviations in this experiment.
3
Results: Data:
Table 1: Standardized HCl Solution Trial 1 Mass of NaCO3 (g) Tared mass of Na2CO3 (g) Moles of NaCO3 (mol) Buret Reading, Intial (mL) Buret Reading, final (mL) Volume of HCl added (mL) Moles of HCl added (mol) Molar concentration of HCl (mol/L) Average molar concentration of HCl (mol/L) 0.199 1.89 x 10-3 4.50 8.80 4.30 1.08 x 10-3 0.250 Trial 2 0.199 0.199 1.87 x 10-3 9.50 13.6 4.10 1.03 x 10-3 0.251 0.259 0.201 1.90 x 10-3 0.00 x 100 4.50 4.50 1.13 x 10-3 0.276 Trial 3
Table 1 shows the standardization of HCl in solution and how much HCl must be titrated to determine the concentration of HCl.
Table 2: Standardized HCl Solution (Formulas) Trial 1 Mass of NaCO3 Tared mass of Na2CO3 (g) Moles of NaCO3 (mol) Buret Reading, Intial (mL) Buret Reading, final (mL) Volume of HCl added (mL) Moles of HCl added (mL) Molar concentration of HCl (mol/L) 0.199 =B4/105.99 4.50 8.80 =B7-B6 =B8/1000*0.25 =B9/(B8/1000) Trial 2 =(20/1000)*(0.25)*(1/2)*(105.99) 0.199 =C4/105.99 9.50 13.6 =C7-C6 =C8/1000*0.25 =C9/(C8/1000) 0.201 =D4/105.99 0.00 x 100 4.50 =D7-D6 =D8/1000*0.25 =D9/(D8/1000) Trial 3
4
Average molar conc. of HCl (mol/L)
=AVERAGE(B10:D10)
Table 2 shows the formulas to determine the standardization of HCl in solution and how much HCl must be titrated to determine the concentration of HCl.
Table 3: Preparation of Borax Solution Sample Number Volume of sample (mL) Temperature of sample (C) Trial 1 5.00 55.0 Trial 2 5.00 48.0 Trial 3 5.00 38.0 Trial 4 5.00 29.0 Trial 5 5.00 24.0
Table 3 shows at what temperature and volume the Borax solution was prepared at.
Table 4: Analysis of Borax Test Solutions Sample Number Buret Reading, Intial (mL) Buret Reading, final (mL) Volume of HCl added (mL) Trial 1 17.40 23.70 6.30 Trial 2 0.00 x 100 6.20 6.20 Trial 3 6.20 11.50 5.30 Trial 4 11.50 17.40 5.90 Trial 5 0.00 x 100 5.80 5.80
Table 4 shows how much HCl was needed to determine the amount of Borax in the test solution.
Table 5: Analysis of Borax Test Solutions (formula) Sample Number Buret Reading, Intial (mL) Buret Reading, final (mL) Volume of HCl added (mL) Trial 1 17.40 23.70 =B6-B5 Trial 2 0.00 x 100 6.20 =C6-C5 Trial 3 6.20 11.50 =D6-D5 Trial 4 11.50 17.40 =E6-E5 Trial 5 0.00 x 100 5.80 =F6-F5
Table 5 shows the formulas of how much HCl was needed to determine the amount of Borax in the test solution.
5
Table 6: Data Analysis Sample Number Temperature (K) 1/T (K-1) Moles of HCl used (mol)
Moles of B4O5(OH)42-(mol) [B4O5(OH)42-] (mol/L)
Trial 1 3.28 x 102 K 3.05 x 10-3 K 1.58 x 10-3 7.90 x 10-4 3.16 x 10-2 3.16 x 10-2 1.26 x 10-4 -8.98
Trial 2 3.21 x 102 3.12 x 10-3 1.55 x 10-3 7.75 x 10-4 3.10 x 10-2 3.10 x 10-2 1.19 x 10-4 -9.04
Trial 3 3.11 x 102 3.22 x 10-3 1.33 x 10-3 6.65 x 10-4 2.66 x 10-2 2.66 x 10-2 7.50 x 10-5 -9.50 -7.89 x 102
Trial 4 3.02 x 102 3.31 x 10-3 1.48 x 10-3 7.40 x 10-4 2.96 x 10-2 2.96 x 10-2 1.04 x 10-4 -9.17
Trial 5 2.97 x 102 3.37 x 10-3 1.45 x 10-3 7.25 x 10-4 2.90 x 10-2 2.90 x 10-2 9.80 x 10-5 -9.23
Molar solubility of borax(mol/L) Solubility product, Ksp ln Ksp -H/R (from data plot) (kJ/mol) S/R(from data plot) (J/mol K) H (kJ/mol) S (J/mol K) G (kJ), at 298 K
-6.65
6.56 -55.3 23.0
Table 6 shows the analysis of the entropy, enthalpy and spontaneity of the thermo chemical reaction.
6
Figure 1: 1/T vs. ln Ksp
-8.9 0.003 -9 0.0031 0.0032 0.0033 0.0034
-9.1
-9.2 ln Ksp y = -789.38x - 6.6469 R = 0.2626 -9.3 Linear (Figure 1: 1/T vs. ln Ksp)
-9.4
-9.5
-9.6
1/T (K-1 )
Figure 1 showed the relationship between the solubility product and temperatures, which can help, determine the enthalpy and entropy.
7
Table 7: Data Analysis (Formula) Sample Number Temperature (K) 1/T (K-1) Moles of HCl used (mol)
Moles of B4O5(OH)42-(mol) [B4O5(OH)42-] (mol/L)
Trial 1 =B4+273 =1/B8 =B7/1000*(0.2 5) =B11(1/2) =B12/0.025 =B13 =4*(B14)^3 =LN(B15)
Trial 2 =C4+273 =1/C8 =C7/1000*(0.2 5) =C11(1/2) =C12/0.025 =C13 =4*(C14)^3 =LN(C15)
Trial 3 =D4+273 =1/D8 =D7/1000*(0.2 5) =D11(1/2) =D12/0.025 =D13 =4*(D14)^3 =LN(D15) -7.89 x 102
Trial 4 =E4+273 =1/E8 =E7/1000*(0.2 5) =E11(1/2) =E12/0.025 =E13 =4*(E14)^3 =LN(E15)
Trial 5 =F4+273 =1/F8 =F7/1000*(0.2 5) =F11(1/2) =F12/0.025 =F13 =4*(F14)^3 =LN(F15)
Molar solubility of borax(mol/L) Solubility product, Ksp ln Ksp -H/R (from data plot) (kJ/mol) S/R(from data plot) (J/mol K) H (kJ/mol) S (J/mol K) G (kJ), at 298 K
-6.65 =-B18*0.008314 =B19/1000*8.314 =B20-298*B22
Table 7 shows the formulas to determine the entropy, enthalpy and spontaneity of the thermo chemical reaction. Calculations: (Make sure you write complete formulas first) 1. Mass of NaCO3 = Volume of HCL in mL x Molraity x mole ratio x Formula Mass= 20 mL HCl (10-3 L/1mL)(0.25 mol/1L)(1 mol Na2CO3 /2 mol HCl) (105.99g Na2CO3/1 mol Na2CO3) = 0.199 g Na2CO3 2. Moles of NaCO3 (mol)= Mass of NaCO3 (1 mol Na2CO3/105.99 g Na2CO3) = 0.1999 g (1 mol Na2CO3/105.99 g Na2CO3) =1.89 x 10-3 mol Na2CO3 3. Volume of HCl added (mL) = Buret Reading, final (mL) - Buret Reading, intial (mL) = 8.8 mL 4.5 mL = 4.3 mL 4. Moles of HCl added (mol) = Volume of HCl added (mL)(1 L/1000 mL) *0.25 M HCl = (4.3 mL)(1L /1000 mL) *0.25 M HCl = 1.08 x 10-3 mol 5. Molar concentration of HCl (mol/L) = Moles of HCl added (mol) / Volume of HCl added (mL) (1L /1000 mL)= 1.08 x 10-3 mol / 4.3 mL (1L/ 1000 mL) = 0.25 M HCl 8
6. Average molar concentration of HCl (mol/L) = (Trial 1 + Trial 2 + Trial 3)/ 3 = 0.25 M + 0.251 M + 0.276 M / 3 = 0.259 M HCl 7. Volume of HCl added= Buret Reading, final (mL) - Buret Reading, (mL) intial = 23.7 mL 17.4 mL = 6.3 mL 8. Temperature(K) = Temperature (C) + 273 K = 55.0 C + 273 K = 328 K 9. 1/ T (K-1) = 1 / 328 K = 3.05 x 10-3 K-1 10. Moles of HCl used (mol) = Volume of HCl added (mL) (1 L/ 1000mL) * 0.25 M HCl = 6.3 mL (1 L/1000 mL) * 0.25 M HCl = 1.58 x 10-3 mol 11. Moles of B4O5 (OH)4-2 = Moles of HCl used (mol) ( 1 mol B4O5 (OH)4-2 / 2 mol HCl)= 1.58 x 10-3 mols HCl ( 1 mol B4O5 (OH)4-2 / 2 mol HCl)= 7.90 x 10-4 mol 12. [B4O5(OH)42-] (mol/L) = Moles of B4O5 (OH)4-2 / 0.025 L of B4O5 (OH)4-2 = 7.90 x 10-4 mol/ 0.025 L of B4O5 (OH)4-2 = 3.16 x 10-2 mol/L 13. Molar solubility of borax= [B4O5(OH)42-] (mol/L) = 3.16 x 10-2 mol/L 14. Solubility product, Ksp = 4 * (Molar solubility of borax) 3 = 4 * (3.16 x 10-2 mol/L)3 = 1.26 x 10-4 15. ln Ksp = ln (1.26 x 10-4) = -8.98 16. H (kJ/mol) = (H / 8.314 x 10-3 kJ/ mol K) = -7.89 x 102 = 6.56 kJ/ mol K 17. S (J/mol K)= (S (J/mol K) / 8.314 J/mol K) = -6.65 = -55.3 J/mol K 18. G (kJ), at 298 K = H (kJ/mol) (298 K)S (J/mol K)= 6.56 kJ/ mol K 298K *(-55.3 J/mol K/ 1000 J) = 23.0 kJ
Discussion: The purpose of this experiment was to determine the thermodynamic properties of the changes in entropy, enthalpy and free energy, as well as the solubility product of borax as a function of temperature for the reaction of borax in an aqueous solution. An acid- base titration was utilized for this experiment because the B4O5(OH)42- ion was the conjugate base of the weak acid, boric acid. Moreover, the B4O5(OH)42- ion accepted two protons from the strong acid, HCl which was represented by the equation, B4O5(OH)42- (aq) + 2 H+ (aq) + 3 H2O(l) 4 H3BO3(aq). Figure 1 and table 6 was used to determine the thermodynamic properties of enthalpy, entropy and free energy for borax from the graph of ln Ksp vs. 1/T. This graph was analyzed by the general equation of y= mx + b, where the slope was equivalent to -H/R, and the yintercept was S/R which resulted in the overall equation, ln Ksp= (-H/R)(1/T) +( S/R). In 9
addition, free energy, G, was calculated with the determination of entropy and enthalpy. Moreover, since the data was collected at standard state, the equation utilized was, G=H TS. In table 6 trial 3 the data collected had a decreased value because heat was lost to the surroundings, which resulted in a lower solubility product. The decreased solubility product resulted from the dissipation of heat from the system to the surroundings. Thus, as borax was titrated with HCl, fewer milliliters of titrant was needed to reach an equivalence point (this is the point at which the weak conjugate base and strong acid were neutralized, thus resulting in a color change from clear to a light blue). Table 6, illustrated the entropy, S,(the tendency for the universe to move towards more disorder) within a system which, resulted in a negative value. This negative value resulted from the dissociation of the slightly soluble crystalline salt, Na2B4O5(OH)4 8 H2O. This dissociation was represented by the equation, Na2B4O5(OH)4 8 H2O(s) 2 Na+(aq) + B4O5(OH)42- (aq) + 8 H2O (l), which illustrated how the reaction went from ordered to more disordered. Furthermore, the value for H (the amount of heat gained or lost for a reaction to proceed) was determined to be positive, which illustrated that the reaction was an endothermic reaction. This reaction was considered to be endothermic, which meant that heat was applied to the system for borax to dissociate into an aqueous solution. In addition, the value for G(Gibbs free energy determined the spontaneity of a reaction) was calculated to have a positive value based on the calculations of entropy and enthalpy. Therefore, the reaction of solid borax was considered to be non-spontaneous because the values for S and G were negative and positive, respectively. Table 3, illustrated that the temperature for the test solutions for borax were maintained at temperatures that were less than 61C. This was significant because borax was stable at
10
temperatures lower than 61C, otherwise with higher temperatures the B4O5(OH)42- ion would dehydrate since Borax was an anhydrous salt.
Conclusion: The hypothesis of this experiment was accepted because the graph of Ksp and 1/T yielded the equation, y = -789.38x - 6.6469. Thus, the values of S, H and G were calculated to be -55.3 J/mol K, 6.56 kJ/mol, and 23.0 kJ, respectively. Errors resulted in the handling of the test solutions causing either a loss or gain of heat, thus resulting in an inconsistent titration between borax and HCl. This would have caused a higher or lower calculation of S, H and G. In addition, the inaccuracy in pipetting the saturated solid borax, which would require an increased amount of HCl, needed to reach an equivalence point thus, causing a higher S, H and G. Improvements in this experiment could have been made by increasing the number of trials. In addition, the use of a centrifuge would have increased the precision of the data by limiting the amount of saturated solid borax that could be pipetted. Thus, the value of Ksp found could increase the precision of the calculations of S, H and G. Post Lab Questions: 1. Part A.1 No desiccators is available. The sodium carbonate is cooled to room temperature , but the humidity in the room is high. How will this affect the reported molar concentration of the hydrochloric acid solution in Part A.7... too high, too low, or unaffected? Explain. The humidity in the room causes the sodium carbonate to have condensation thus causing a higher mass of sodium carbonate. The molar concentration of the HCl would be too high because as the mass increases the molar concentration increases.
11
2. Part A.5 The endpoint in the titration is "overshot." a. Is the reported molar concentration of the hydrochloric acid solution too high or too low? Explain. If the endpoint is overshot then the reported amount of HCl added would be higher thus causing the molar concentration to be too low because as the grams of HCl are massed the more amount of titrated HCl would cause the molar concentration to be lowered. b. As a result of this poor titration technique, is the reported number of moles of B4O5(OH)4-2 in each of the analyses (part C.2) too high or too low? Explain. The moles of borax would be too high because the mole ratio of HCl to borax is 1:2. Therefore, the higher amount of HCl titrated would cause in a higher amount of moles of HCl added thus causing the moles of borax to increase. 3. Part B.2. The solid borax reagent is contaminated with a water-soluble substance that does not react with hydrochloric acid. How will this contamination affect the reported solubility product of borax? Explain. The borax would cause a diluted solution, which would require more hydrochloric acid to be titrated. The addition of more HCl would cause the solubility product to be higher because the molar solubility would be higher. 4. Part B.4 For the borax solution at 48C, no solid borax is present in the test tube. Five milliliters is transferred to the corresponding calibrated test tube and subsequently titrated with the standardized hydrochloric acid solution. How will this oversight in technique affect the reported Ksp of borax at 48C ... too high, too low or unaffected? Explain.
12
Since there was not a presence of borax in the test solution that meant that the solid borax dissociated more and that lower amounts of HCl would be needed to titrated the solution thus causing the Ksp to be too low because the molar solubility would be lowered. 5. Part B.5. A "little more" than 5 mL of a saturated solution is transferred to the corresponding calibrated test tube and subsequently titrated with the standardized HCl acid solution. How will this "generosity" affect the reported molar solubility of borax for that sample ... too high, too low or unaffected? Explain. The molar solutbility will be too high because more HCl will be required since there is more borax present the amount of HCl needed to reach an equivalence point will require more HCl. 6. Part C.2 The saturated solution of borax is diluted with " more than" 25 mL of deionized water. How does this dilution affect the reported number of moles of B4O5 (OH)42- in the saturated solution ... too high, too low or unaffected? Explain. The amount of borax that is diluted within solution would require more HCl to be titrated that would make the number of moles of borax to be too high. 7. Explain why the slope of a "hand drawn" straight line may be more representative of the data than a slope calculated from a least squares program (e.g., trendline of Excel). The hand drawn straight line is more representative because the points that are outliers could be eliminated versus where on a program the trendline cannot be extrapolated and take into consideration the outliers.
13
Works Cited
1
2
Beran, J.A. ,( 2008). Page 299-308 Laboratory Manual for Principles of General Chemistry Zumdahl, Steven. (2007). page 999-1007 Chemistry
14