Borax Analysis

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Johns Hopkins University *

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182

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Chemistry

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Oct 30, 2023

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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) 0 0 0 0 0 0 0 0 1 2 3 4 5 6 7 8 f(x) = 11707.6 x − 34.57 R² = 0.64 Ln(Ksp) Ln(Ksp) Linear (Ln(Ksp)) Combining these two equations (3) and (4) gave the general relationship between K, ∆H°, and ∆S°: –RTlnK=∆H°–T∆S° (5) Dividing both sides by –RT gave the linear form of this relationship: H 1 S ln K = –   RTR 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
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