Data: Room Temperature of Borax = 23˚C Kₛₚ of borax at room temperature trials First: 21 mL of HCl Second: 20 mL of HCl Third: 17.29 mL of HCl The second and third trial were my two best titrations to evaluate Kₛₚ Ice-water bath of Borax = 4˚C Kₛₚ of borax at ice temperature trials First: 7 mL of HCl Second: 5.19 mL of HCl Third: 6.89 mL of HCl The first and third trial were my two best titrations to evaluate Kₛₚ The Molarity of HCl = 0.1012 M The Volume of borax solution 10.00 mL Question: Calculations for the two values for Kc and explain why you did the above calculations. You only need to explain the room temperature calculations – the ice bath calculations would have the same justification so you do not need to provide this. Do not simply tell me what you did. Explain the chemical principles that were involved in deciding to manipulate the numbers the way you did.
the free energy for a reaction can be related to the equilibrium
constant through the formula below.
K = e (-ΔG° / RT)
Therefore if Kc for a reaction is known, Go can be determined, or vice versa. Furthermore, if
you have the value for Go at two different temperatures, you can calculate H and S through
the familiar equation for Gibbs energy below, since you have two unknowns but also two
equations.
G = H – T S
In this lab you will be studying the solubility of borax (Na2B4O5(OH)4*8H2O), a slightly soluble
sodium salt, at two different temperatures. When solid borax is added to water, the
equilibrium below is established.
Na2B4O5(OH)4*8H2O (s) <--> 2 Na+ (aq) + B4O5(OH)42- (aq) + 8 H2O(l)
If you measure the concentrations for those substances that show up in the reaction quotient,
then the Kc for the reaction at that temperature can be calculated. In this lab, the
concentration of borate ion (B4O5(OH)42-) in solution will be measured by titration with standard
hydrochloric acid according to the equation below.
B4O5(OH)42- (aq) + 2 HCl (aq) + 3 H2O (l) <--> 4 H3BO3 (aq) + 2 Cl- (aq)
The concentrations of the other substances that appear in the reaction quotient can be
calculated from the borate concentration using stoichiometry.
Data:
- Room Temperature of Borax = 23˚C
- Kₛₚ of borax at room temperature trials
- First: 21 mL of HCl
- Second: 20 mL of HCl
- Third: 17.29 mL of HCl
- The second and third trial were my two best titrations to evaluate Kₛₚ
- Kₛₚ of borax at room temperature trials
- Ice-water bath of Borax = 4˚C
-
- Kₛₚ of borax at ice temperature trials
-
- First: 7 mL of HCl
- Second: 5.19 mL of HCl
- Third: 6.89 mL of HCl
- The first and third trial were my two best titrations to evaluate Kₛₚ
- The Molarity of HCl = 0.1012 M
- The Volume of borax solution 10.00 mL
-
- Kₛₚ of borax at ice temperature trials
-
Question: Calculations for the two values for Kc and explain why you did the above calculations. You only need to explain the room temperature calculations – the ice bath calculations would have the same justification so you do not need to provide this. Do not simply tell me what you did. Explain the chemical principles that were involved in deciding to manipulate the numbers the way you did.
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