1. In the Thermodynamics of Borax Dissolution experiment, a student quickly transferred a 5.00 mL sample of the saturated borax solution into an Erlenmyer flasks using a pipet and pipet pump after the borax solution had cooled to 35oC. The student rinses the pipet with warm ddH2O, and add the rinsing to the flask. Then the student titrates the solution with a 0.500 M HCl solution. 12.00 mL of the HCl solution was used to reach the endpoint of the titration. Calculate the value of Ksp for borax at 35oC. You can use the following steps to do the calculation. Show your work.
a. Calculate the number of moles of HCl that were added during the titration.
b. Use following equation to relate the number of moles of HCl to the number of moles of tetraborate ion in the 5.00 mL sample.
B4O5(OH)42- (aq) + 2H+ (aq) + 3H2O (l) 4H3BO3 (aq)
c. Calculate the concentration of tetraborate ions in the 5.00 mL sample.
d. Use the concentrations of tetraborate ions to calculate the equilibrium constant (Ksp) at 35oC.
Na2B4O5(OH)4·8H2O (s) 2Na+ (aq) + B4O5(OH)42-(aq) + 8H2O (l)
Ksp = [Na+]2[B4O5(OH)42-]
[Na+] = 2[B4O5(OH)42-]
Ksp = 4[B4O5(OH)42-]3
a. To calculate the number of moles of HCl that were added during the titration, we can use the equation:
Number of moles of HCl = Molarity of HCl x Volume of HCl
Since the volume of HCl used is given as 12.00 mL and the molarity of HCl is given as 0.500 M, we can substitute these values into the equation:
Number of moles of HCl = 0.500 M x 0.01200 L
b. Now, to relate the number of moles of HCl to the number of moles of tetraborate ion in the 5.00 mL sample, we can use the balanced equation provided:
B4O5(OH)42- (aq) + 2H+ (aq) + 3H2O (l) → 4H3BO3 (aq)
From the equation, we can see that the stoichiometric ratio between HCl and tetraborate ion is 2:1. So, the number of moles of tetraborate ion in the 5.00 mL sample would be half the number of moles of HCl added.
Number of moles of tetraborate ion = (Number of moles of HCl) / 2
c. To calculate the concentration of tetraborate ions in the 5.00 mL sample, we need to convert the number of moles to concentration. Since the volume of the sample is given as 5.00 mL, we can use the equation:
Concentration of tetraborate ions = (Number of moles of tetraborate ions) / (Volume of sample in liters)
Concentration of tetraborate ions = (Number of moles of tetraborate ions) / 0.00500 L
d. Finally, to calculate the equilibrium constant (Ksp) at 35oC, we use the equation:
Ksp = 4[B4O5(OH)42-]3
Now that we have the concentration of tetraborate ions in the 5.00 mL sample, we can substitute this value into the equation to calculate Ksp.
a. To calculate the number of moles of HCl added during the titration, we need to use the volume and concentration of the HCl solution.
Given:
Volume of HCl solution = 12.00 mL
Concentration of HCl solution = 0.500 M
First, convert the volume of HCl solution to liters:
12.00 mL = 12.00 mL * (1 L / 1000 mL) = 0.01200 L
Next, calculate the number of moles of HCl using the formula:
moles of HCl = volume of HCl solution (in L) * concentration of HCl solution
moles of HCl = 0.01200 L * 0.500 M = 0.00600 moles
Therefore, the number of moles of HCl added during the titration is 0.00600 moles.
b. The equation B4O5(OH)42- (aq) + 2H+ (aq) + 3H2O (l) → 4H3BO3 (aq) shows the stoichiometry between HCl and the tetraborate ion.
From the equation, we can see that 1 mole of B4O5(OH)42- reacts with 2 moles of HCl.
Since we have determined that the number of moles of HCl is 0.00600 moles, the number of moles of tetraborate ions in the 5.00 mL sample is half of that:
moles of tetraborate ions = 0.00600 moles / 2 = 0.00300 moles
Therefore, the number of moles of tetraborate ions in the 5.00 mL sample is 0.00300 moles.
c. To calculate the concentration of tetraborate ions in the 5.00 mL sample, we need to use the volume of the sample.
Given:
Volume of sample = 5.00 mL = 0.00500 L
Concentration is calculated by dividing the number of moles by the volume:
concentration of tetraborate ions = moles of tetraborate ions / volume of sample
concentration of tetraborate ions = 0.00300 moles / 0.00500 L = 0.600 M
Therefore, the concentration of tetraborate ions in the 5.00 mL sample is 0.600 M.
d. The equilibrium constant (Ksp) at 35°C can be calculated using the concentrations of tetraborate ions.
The balanced equation for the dissolution of Na2B4O5(OH)4·8H2O is:
Na2B4O5(OH)4·8H2O (s) → 2Na+ (aq) + B4O5(OH)42- (aq) + 8H2O (l)
From the equation, we can see that the concentration of Na+ ions is twice the concentration of tetraborate ions, so [Na+] = 2[B4O5(OH)42-].
Therefore, the equilibrium constant Ksp = 4[B4O5(OH)42-]³.
Given that the concentration of tetraborate ions is 0.600 M, we can substitute this value into the equation:
Ksp = 4 * (0.600 M)³ = 0.864
Therefore, the value of Ksp for borax at 35°C is 0.864.
a. To calculate the number of moles of HCl that were added during the titration, we can use the formula:
Number of moles = concentration x volume
Since the concentration of the HCl solution is given as 0.500 M and the volume used is 12.00 mL (0.01200 L), we can substitute these values into the formula:
Number of moles of HCl = 0.500 M x 0.01200 L
b. The equation provided relates the number of moles of HCl to the number of moles of tetraborate ion (B4O5(OH)42-). From the balanced equation, we can see that 1 mole of B4O5(OH)42- reacts with 2 moles of H+.
So, the number of moles of B4O5(OH)42- can be calculated using the stoichiometry of the reaction:
Number of moles of B4O5(OH)42- = (Number of moles of HCl) / 2
c. To calculate the concentration of tetraborate ions in the 5.00 mL sample, we need to know the volume of the sample taken.
Since 5.00 mL of the saturated borax solution was transferred, we can assume that the volume of the sample is indeed 5.00 mL.
Concentration of tetraborate ions = (Number of moles of B4O5(OH)42-) / (Volume of the sample in liters)
d. Finally, to calculate the equilibrium constant (Ksp) at 35oC, we need to know the concentrations of tetraborate ions.
From the balanced equation, we can see that the ratio of [Na+] to [B4O5(OH)42-] is 2:1. So, the concentration of tetraborate ions is half the concentration of sodium ions.
Concentration of tetraborate ions = (Concentration of sodium ions) / 2
Substitute this value into the equation for Ksp:
Ksp = 4 x ([Concentration of tetraborate ions])^3
a).500Mx.012L=.006mole
b) ..5*.006=.003moles
c) .003mole/.005Liter=.0M
d)4*.6^2=.90=Ksp