A student prepares a solution for titration by adding a 10.00 mL aliquot of a so
ID: 1042766 • Letter: A
Question
A student prepares a solution for titration by adding a 10.00 mL aliquot of a solution saturated with Ca(OH)2 at 82.9°C to 25.0 mL of deionized water. This solution is titrated with a standardized 0.01252 M HCl solution. The equivalence point is observed after 17.94 mL of the HCl solution has been added.
1. Calculate the number of moles of hydroxide ion present in the aliquot of saturated Ca(OH)2 solution.
2. Calculate the solubility of Ca(OH)2 at 82.9°C.
3. Calculate the Ksp of Ca(OH)2 at 82.9°C.
4. Calculate the ?G°soln for the dissolution of Ca(OH)2 at 82.9°C.
Explanation / Answer
1)
moles of HCl = 0.01252 x 17.94 / 1000
= 2.246 x 10^-4 mol
Ca(OH)2 + 2 HCl ----------------> CaCl2 + 2 H2O
1 mol Ca(OH)2 --------------> 2 mol HCl
?? Ca(OH)2 -------------> 2.246 x 10^-4 mol HCl
moles of Ca(OH)2 = 1.123 x 10^-4 mol
concentration of Ca(OH)2 = 1.123 x 10^-4
moles of [OH-] ion present = 2.246 x 10^-4 mol
2)
solubility of Ca(OH)2 = 4.49 x 10^-3 M
3)
Ca(OH)2 --------------> Ca+2 + 2 OH-
Ksp = [Ca+2][OH-]^2
= (4.49 x 10^-3) (8.98 x 10^-3)^2
Ksp = 3.62 x 10^-7
4)
?G° = - RT ln Ksp
= - 8.314 x 10^-3 x 356.05 x ln (3.62 x 10^-7)
?G° = 43.9 kJ/mol
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