Final answer:
The solubility of calcium hydroxide (Ca(OH)2) can be calculated using the solubility product constant (Ksp). The solubility is 0.789 g/L.
Step-by-step explanation:
The solubility of calcium hydroxide (Ca(OH)2) can be calculated using the solubility product constant (Ksp). The Ksp of calcium hydroxide is given as 5.02 x 10⁻⁶. The solubility of a compound is the amount of that compound that can dissolve in a given volume of solvent, usually water. In this case, we want to calculate the solubility in grams per liter (g/L).
To calculate the solubility, we need to use the molar mass of calcium hydroxide. Calcium has a molar mass of 40.08 g/mol, oxygen has a molar mass of 16.00 g/mol, and hydrogen has a molar mass of 1.01 g/mol. The molar mass of calcium hydroxide is therefore 40.08 + 2*(16.00 + 1.01) = 74.14 g/mol.
The solubility product constant (Ksp) expression for calcium hydroxide is:
Ksp = [Ca²+][OH-]²
Since the stoichiometry of the reaction is 1:1 for calcium and hydroxide ions, the concentration of calcium ions and hydroxide ions will be the same.
Setting the concentration of calcium ions and hydroxide ions as x, we can rewrite the Ksp expression as:
Ksp = x(2x)² = 4x³
Now we can substitute the value of Ksp into the expression and solve for x:
5.02 x 10⁻⁶ = 4x³
x³ = 5.02 x 10⁻⁶ / 4
x = (5.02 x 10⁻⁶ / 4)^(1/3)
x = 0.01064 M
Now we can calculate the solubility in grams per liter by multiplying the molar solubility by the molar mass:
Solubility = 0.10864 M * 74.14 g/mol = 0.789 g/L