Ca(OH)2 has a Ksp of 6.5 x 10^-6. If 0.37 g of Ca(OH)2 is added to 500 mL of water and the mixture is allowed to come to equilibrium, will the solution be saturated? pls help me I stayed up all last night trying to get this one question :(
........Ca(OH)2 ==> Ca^2+ + 2OH^-
I.......solid........0.......0
C.......solid........x.......2x
E.......solid........x.......2x
Ksp = 6.5E-6 = (Ca^2+)(OH^-)^2
6.5E-6 = (x)(2x)^2 = 4x^3
Solve for x = [Ca(OH)2] in mols/L
Convert that to grams/500 mL.
g = mols x molar mass and that divided by 2 = g/500. Compare with the 0.37 g in the problem. If 0.37 g is larger than x the soln is not saturated. If 0.37 g is smaller it is saturated.
Thank you so much!
To determine if the solution will be saturated, we need to compare the value of the ion product (Q) with the solubility product constant (Ksp) of Ca(OH)2.
The molar mass of Ca(OH)2 is 74.09 g/mol (40.08 g/mol for Ca, and 2 × 1.01 g/mol + 16.00 g/mol for OH).
First, calculate the number of moles of Ca(OH)2:
0.37 g / 74.09 g/mol ≈ 0.00499 mol
Calculate the concentration of Ca(OH)2 in the solution:
Concentration (mol/L) = moles / volume (L)
Since the volume is given as 500 mL, convert it to liters:
500 mL / 1000 = 0.5 L
Concentration = 0.00499 mol / 0.5 L ≈ 0.00998 M
Now, let's calculate the ion product (Q):
Q = [Ca2+][OH-]^2
Since Ca(OH)2 dissociates into one Ca2+ ion and two OH- ions, the concentration of Ca2+ will be half the concentration of Ca(OH)2 added, and the concentration of OH- will be twice the concentration of Ca(OH)2 added.
[Ca2+] = 0.00998 M / 2 = 0.00499 M
[OH-] = 2 * 0.00998 M = 0.01996 M
Q = 0.00499 M * (0.01996 M)^2 ≈ 1.99512 x 10^-6
Now, compare Q with the Ksp value of Ca(OH)2 (6.5 x 10^-6).
Since Q (1.99512 x 10^-6) is less than Ksp (6.5 x 10^-6), the solution will not be saturated.
Therefore, the answer is that the solution will not be saturated after adding 0.37 g of Ca(OH)2 to 500 mL of water and allowing it to come to equilibrium.