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Calculations:Trial 1:Moles HCl: (0.250 x 0.0042) = 0.0011 mols of HClMoles of Borax: 0.0011mols HCl ( 1mol borax/2mols HCl) = 5.3 x 10-4mols borax5.3 x 10-4mols borax/0.0050L borax solution used = 0.011M boraxKsp= 4(0.011M)3= 0.0046 M3∆G= -(8.314J/mol∙K) x (303K) x ln(0.0046M3) = 13557 J/molRefer to figure 1:lnKsp=-11444(1/T)+31.8∆H (enthalpy) = -11444/8.314= -13542.1 J∆S (entropy) = 31.8/8.314= 3.83 J/K
Table 1:Trial 1 Borax TitrationGroupHCl (M)Temp (K)Volume Borax used (L)Volume of HCl used (L)Moles HCl (mols)Mole Borax (mols)Concentration of Borax (M)Ksp M310.2503080.0050.00330.000830.000410.0830.002220.2502980.0050.00600.00150.000750.150.01430.2503030.00480.00160.000400.000200.0420.0002940.2502980.0050.00200.000500.000250.0500.0005050.2503130.0050.01050.00260.00130.26250.07260.2502950.0050.00400.00100.000500.100.004070.2503030.0050.00420.00110.000530.110.004680.2503180.0050.0030.000700.000350.0700.001490.2502950.00510.00250.000630.000310.0610.0009Table 2: Trial 2 Borax Titration and averages of dataGroupTemp (K)Volume of HCl used (L)Moles HCl (mols)Volume Borax used (L)Mole Borax (mols)Conc. of Borax (M)Ksp M3Average Ksp M3∆G (J/mol) average13080.00310.000780.0050.000390.0780.00190.002115845.622980.00400.00100.0050.000500.100.00400.008811740.533030.00160.000400.00490.000200.0410.000270.0002820602.342980.00110.000280.0050.000140.0288.3E-050.0002920167.853130.01080.00270.0050.00140.270.0790.0766721.962950.00260.000650.0050.000330.0650.00110.002514647.073030.00430.00110.0050.000540.110.00500.004813450.183180.00600.00150.0050.000750.150.0140.007412958.692950.001750.000440.00510.000220.0430.000320.0006218122.8Figure
Conclusion:The purpose of this lab was the study of a sparingly salt. The solubility information was determined at various temperatures. The following thermodynamic quantities were determined the change in standard enthalpy, standard entropy , and standard free energy. The equation for this lab yields the following reaction when the borate ion was titrated with HCl:B4O5(OH)4-2(aq)+ 2HCl + 3H2O ↔ 4B(OH)3(aq)+ 2Cl-The equation for the equilibrium of borax is as follows:Na2B4O7∙10H2O(s)↔ 2Na+(aq)+ B4O5(OH)4-2(aq)+ 8H2O(l)This equation is written as an equilibrium process. The eight water molecules from the hydrated salt will be lost in the reaction medium.It is required that during the experiment there is always some solid borax remaining in the sample mixture before removed to analyze. This being the case the solid borax concentration can