In: Chemistry
This question is for a laboratory experiment titled "Determination of the Solubility Product Constant of Calcium Hydroxide" where a saturated solution of Ca(OH)2 in 0.02523M NaOH was titrated with HCl.
Part 2 Data - Saturated Solution of Ca(OH)2 in 0.02523M NaOH:
Volume of Ca(OH)2/NaOH aliquot: 25.00mL
Concentration of standard HCl: 0.1342M
Indicator Used: Bromothymol Blue
Average volume of HCl to reach end point: 10.26mL
a. Calculate the TOTAL [OH-] in the saturated solution of Ca(OH)2 in sodium hydroxide for the solution assigned (Ca(OH)2 in 0.02523M NaOH).
b. Calculate the [OH-] that comes from the dissolution of Ca(OH)2. The total [OH-] (calculated above) is the sum of the [OH-] from the NaOH and the [OH-] from Ca(OH)2.
c. Calculate the solubility of Ca(OH)2 in the NaOH solution (in mol/L).
d. Calculate the experimental Ksp of Ca(OH)2 for the saturated solution of Ca(OH)2 in NaOH.
Please explain steps/show equations! Thank you.
Part a) Calculate the total [OH-] in the saturated solution of Ca(OH)2 in sodium hydroxide for the solution assigned (Ca(OH)2 in 0.02523M NaOH).
This is an example of a strong acid-strong base reaction
H+ + OH- H2O
At the equivalence point,
Moles of H+ = Moles of OH-
[OH-] = [(0.1342) x (10.26)]/(25) = 0.05508 M
Part b) Calculate the [OH-] that comes from the dissolution of Ca(OH)2. The total [OH-] (calculated above) is the sum of the [OH-] from the NaOH and the [OH-] from Ca(OH)2.
[OH-] from the dissolution of Ca(OH)2 = 0.05508 M - 0.0252 M = 0.0298 M.
Part c) Calculate the solubility of Ca(OH)2 in the NaOH solution (in mol/L)
Ca(OH)2 Ca2+ + 2OH-
Therefore,
[Ca2+] = 0.0298/2 = 0.0149 M
Part d) Calculate the experimental Ksp of Ca(OH)2 for the saturated solution of Ca(OH)2 in NaOH
Ksp = [Ca2+][OH-]2 = (0.0149)(0.055)2 = 4.5 x 10-5