Final answer:
The molar solubility of calcium hydroxide (Ca(OH)2) in water can be calculated by using its solubility product constant (Ksp). The molar solubility 's' is found by solving the equation 5.02 x 10^-6 = 4s^3 for 's', which results in approximately 1.11 x 10^-2 M.
Step-by-step explanation:
To calculate the molar solubility of calcium hydroxide (Ca(OH)2), we must first write out the balanced dissolution equation and its corresponding equilibrium expression using the solubility product constant (Ksp). The dissolution of calcium hydroxide in water can be represented by the following equation:
Ca(OH)2 (s) ⇌ Ca2+ (aq) + 2OH- (aq)
If we let 's' represent the molar solubility of Ca(OH)2 in water, then at equilibrium the concentration of Ca2+ will be 's' and the concentration of OH- will be '2s'. The Ksp expression for this equilibrium is:
Ksp = [Ca2+][OH-]2
Substituting the equilibrium concentrations into the Ksp expression gives us:
Ksp = s(2s)2 = 4s3
Given that the Ksp of Ca(OH)2 is 5.02 x 10-6, we solve for 's' using the following equation:
5.02 x 10-6 = 4s3
To find 's', we take the cube root of both sides:
s = √3(5.02 x 10-6 / 4)
s ≈ 1.11 x 10-2 M
Therefore, the molar solubility of Ca(OH)2 in water is approximately 1.11 x 10-2 M.