thermodynamics-of-dissolution-of-borax-lab-report

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Studocu is not sponsored or endorsed by any college or university Thermodynamics of Dissolution of Borax lab report General Chemistry II (Ocean County College) Studocu is not sponsored or endorsed by any college or university Thermodynamics of Dissolution of Borax lab report General Chemistry II (Ocean County College) Downloaded by MF MFJ (malak.fdlou@gmail.com) lOMoARcPSD|7541396

CHEM 182 Thermo Dissolution of Borax Name: Thi Phuoc Loan Tran Date: 4/24/21 Sec. No. 04 Thermodynamics of the Dissolution of Borax Report Sheet Temperature (K) 313.15 318.15 323.15 328.15 333.15 Volume of HCl added (mL) (from class data) 4.40 4.25 6.90 9.90 10.35 1/T (K -1 ) 0.003193 0.003143 0.003095 0.003047 0.003002 Moles of HCl used 8.80× 8.50× 1.38× 1.98× 2.07× Moles of B 4 O 5 (OH) 4 2- 4.40× 4.25× 6.90× 9.90× 1.04× [B 4 O 5 (OH) 4 2- ] (M) 0.0880 0.0850 0.138 0.198 0.207 Molar solubility of borax (mol/L) 0.0880 0.0850 0.138 0.198 0.207 Solubility product, K sp 2.73× 2.46× 1.05× 3.11× 3.55× ln K sp -5.90 -6.01 -4.56 -3.47 -3.34 -ΔH°/R (from data plot) -15994 ΔS°/R (from data plot) 44.861 ΔH° (kJ/mol) 133 ΔS° (J/mol*K) 373 ΔG° (kJ), at 298 K 21.8 Downloaded by MF MFJ (malak.fdlou@gmail.com) lOMoARcPSD|7541396

CHEM 182 Thermo Dissolution of Borax Discussion/Conclusion The main goal of this experiment was to determine the solubility product of borax, the standard free energy (∆G°), standard enthalpy (∆H°), and standard entropy (∆S°) changes for the dissolution of borax in water. Borax, Na 2 B 4 O 5 (OH) 4 •8 H 2 O, dissolved slightly in water to give two sodium ions, a tetraborate ion, and eight molecules of water according to the equation: Na 2 B 4 O 5 (OH) 4 •8 H 2 O ( s )  2 Na + ( aq ) + B 4 O 5 (OH) 4 2- ( aq ) + 8 H 2 O ( l ) (1) The solubility product expression was: K sp = [Na + ] 2 [ B 4 O 5 (OH) 4 2- ] In order to determine the concentration of Na + and B 4 O 5 (OH) 4 2- , a titration of tetraborate with a standardized hydrochloric acid was conducted. Tetraborate was a weak base, so it could be titrated with a strong acid like HCl. Just as any other equilibrium constants, solubility product changes with temperature. So, in this experiment, five titrations were performed at five different temperatures to test the change in K sp as the temperature changing. The procedures of the five titrations were almost the same except for the temperatures that were maintained for each trial. A solution of 0.2 M HCl was slowly added into a known volume of borax samples until the yellow endpoint of the indicator bromocresol green. Tetraborate reacted with two hydrogen ions to produce four molecules of boric acid as the equation: B 4 O 5 (OH) 42- ( aq ) + 2 H 3 O + ( aq ) + H 2 O ( l )  4 H 3 BO 3 ( aq ) (2) The volumes of HCl added to reach equivalence points were recorded. Together with the known molarity, the number of moles of HCl were calculated as well as the number of moles of B 4 O 5 (OH) 42-. Noting that tetraborate reacted with HCl in a ratio of 1:2. Then the concentrations of tetraborate were calculated by divided by the volume of the borax samples, which was 5 mL. Once the concentrations of tetraborate were determined, the concentrations of sodium ion could be calculated using the stoichiometry of reaction (1). Having the concentrations of the two ions, the solubility product Ksp of borax at each temperature were calculated. It was concluded that as the temperature of the solution increased, the Ksp for borax increased. It was understandable because the increase in kinetic energy that came with higher temperatures facilitated the dissolving reaction by breaking apart the bonds between the solid and making it more soluble. When the solvation of borax was analyzed at several different temperatures, ∆G°, ∆H° and ∆S° could be determined. The relationship between free energy change ∆G° and the solubility product constant K sp was given by: ∆G° = –RT ln K. (3) The free energy change was also related to the enthalpy and entropy changes during the reaction: ∆G° = ∆H° – T∆S° (4) Downloaded by MF MFJ (malak.fdlou@gmail.com) lOMoARcPSD|7541396

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CHEM 182 Thermo Dissolution of Borax Combining these two equations (3) and (4) gave the general relationship between K, ∆H°, and ∆S°: –RT ln K = ∆H° – T∆S° (5) Dividing both sides by –RT gave the linear form of this relationship: ln K = –  H  R 1 T       S  R (6) This represented a linear equation of the form y = mx + b. In this case, y = ln K and x = 1/T; the slope m = –(∆H°/R), and y-intercept b = (∆S°/R). The Gas Constant, R, is equal to 8.314 J/mol·K. So, once knowing the slope and y-intercept, it would be able to determine ∆H° and ∆S°. The data collected of K sp at different temperatures T were used to plot a graph of ln K against 1/T 0 0 0 0 0 0 0 -7 -6 -5 -4 -3 -2 -1 0 f(x) = − 15993.87 x + 44.86 R² = 0.9 Thermodynamics of the Dissolu琀on of Borax 1/T (K-1) ln Ksp Downloaded by MF MFJ (malak.fdlou@gmail.com) lOMoARcPSD|7541396

CHEM 182 Thermo Dissolution of Borax The graph gave the best-fit line y = -15994x + 44.861 with the correlation coefficient R 2 =0.8989. Knowing the slope and the y-intercept of the equation, the standard enthalpy change ∆H° and standard entropy change ∆S° were determined to be 133 kJ/mol and 373 J/mol ∙ K, respectively. The value of enthalpy was positive, which indicated the dissolution of borax in water was an endothermic reaction. Moreover, the entropy value was also positive, which indicated the disorder of the system increased as the reaction occurred. The literature values for enthalpy and entropy of the dissolution of borax in water are 110 kJ/mol and 380 J/mol ∙ K, which gave the percent error to be 20.9% and 1.84%. The percent error of entropy value was less than 10%, which indicated high accuracy. However, the percent error of enthalpy value was more than 10%, which meant there were errors during the experiment that reduce the accuracy of the data collected. Possible errors included improper heating the calibrated test tubes. If the solutions in the test tubes were not heated entirely, it would affect the solubility product K sp of the sample. So, make sure that the level of water in the beaker outside the test tubes was equal to the level inside the test tubes so that the entire solution was heated up. Also make sure that the temperature was maintained during the experiment so that the borax would not crystallize. Another source of error would be some of the borax solid was inadvertently transferred to the flasks for titration. This would require more amount of HCl used to titrate the borax. Greater amount of HCl would lead to greater concentrations of Na + and B 4 O 5 (OH) 4 2- as well as the K sp . An effective way to minimize most of the errors was to perform the experiment multiple times. Based on the values of ∆H° and ∆S° that were determined for the dissolution of borax, at 75.0 ° C or 75.0 ° C + 273.15 = 348.15 K, the value of K sp would be: ln K = –  H  R 1 T       S  R ln K = - × ln K = -1.08 K = e -1.08 K = 0.340 Based on the values of ∆H° and ∆S° that were determined for the dissolution of borax, at 85.0 ° C or 85.0 ° C + 273.15 = 358.15 K, the solubility of borax in g/L would be: ln K = –  H  R 1 T       S  R ln K = - × ln K = 0.198 K = e 0.198 Downloaded by MF MFJ (malak.fdlou@gmail.com) lOMoARcPSD|7541396

CHEM 182 Thermo Dissolution of Borax K = 1.22 Na 2 B 4 O 5 (OH) 4 •8 H 2 O ( s )  2 Na + ( aq ) + B 4 O 5 (OH) 4 2- ( aq ) + 8 H 2 O ( l ) K sp = [Na + ] 2 [ B 4 O 5 (OH) 4 2- ] = (2s) 2 .s = 4s 3 = 1.22 s = ∛ = 0.673 M Solubility of borax at 85.0 ° C = × = 267 g/L If a little more than 5 mL of saturated solution were transferred to the corresponding calibrated test tube and subsequently titrated with the standardized hydrochloric acid solution, it would require greater amount of HCl to be used in the titration. Since the amount of HCl was greater than expected, the calculated concentrations of Na + and B 4 O 5 (OH) 4 2- would also greater than their expected values. Therefore, the calculated molar solubility of borax for that sample would also too high than the expected value. If the saturated solution of borax were diluted with more than 25 mL of deionized water, this dilution would not affect the calculated moles of B 4 O 5 (OH) 4 2- in the saturated solution. Because more deionized water would not change the number of moles of borax. The number of moles of borax would remain the same since the mass was not changing. The dilution would make the molarity lower, but the moles of borax was unaffected as well as the calculated moles of B 4 O 5 (OH) 4 2- . Judy’s data plot had a lesser slope than Nancy’s, assuming Judy’s slope was -100 and Nancy’s slope was -150. The plot of ln K against 1/T gave a straight line with the slope m = –(∆H°/R). Since the value of R is constant, 8.314 J/mol·K, Judy, ∆H° = -m × 8.314 = -(-100) × 8.314 = 831.4 J/mol Nancy, ∆H° = -m × 8.314 = -(-150) × 8.314 = 1247.1 J/mol The numerical value we were getting from the plot would be negative like the slope of Judy and Nancy, so the larger the slope the larger the value of ∆H°. So, Nancy would have the more positive ∆H° for the dissolution of borax. The equilibrium constant is the free energy change in the reaction: △ G°= -RT ln K. The free energy change is also written as: △ G°= △ H°-T △ S° to account for the enthalpy and entropy changes during the reaction. Both equations are combined and divided by the negative rate times temperature, and the relationship to the equilibrium constant becomes: ln K = - △ H°/RT + △ S°/R where T is in Kelvin and R = 8.314 J/Kmol The equilibrium constant is the free energy change in the reaction: △ G°= -RT ln K. The free energy change is also written as: △ G°= △ H°-T △ S° to account for the enthalpy and entropy changes during the reaction. Both equations are combined and divided by the negative rate times temperature, and the relationship to the equilibrium constant becomes: ln K = - △ H°/RT + △ S°/R where T is in Kelvin and R = 8.314 J/Kmo Downloaded by MF MFJ (malak.fdlou@gmail.com) lOMoARcPSD|7541396

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