Question

4) Calculate the solubility of calcium hydroxide, Ca(OH)2 (Ksp = 5.5 x 10-5) in grams per liter in:

a. Pure water
b. 0.10 M CaCl2

Answer 1

In pure H2O.
........Ca(OH)2 ==> Ca^2+ + 2OH^-
I.......solid........0........0
C.......solid........x........2x
E.......solid........x........2x

Ksp = (Ca^2+)(OH^-)^2
Substitute the E line into Ksp expression and solve for x = solubility in mols/L and convert to g/L.

In 0.10M CaCl2 which ionizes 100%.
Same as above except CaCl2 must be added in as follows:
.........CaCl2 = Ca^2+ + 2Cl^-
I.........0.1....0.......0
C........-0.1...0.1......0.1
E.........0......0.1.....0.1

Ksp still = (Ca^2+)(OH^-)
(Ca^2+) = 0.1 from CaCl2 and x from Ca(OH)2 for total of 0.1+x
(OH^-) = 2x
Substitute into Ksp expression and solve for x = solubility in M = mols/L and convert to g/L.
grams = mols x molar mass.

Answer 2

0.027