If 0.370 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?
no
To determine if the solution will be saturated, we first need to calculate the solubility of Ca(OH)2 in water. The solubility of a compound is defined as the maximum amount of that compound that can dissolve in a given amount of solvent at a specific temperature.
To calculate the solubility, we need to know the solubility product constant (Ksp) of Ca(OH)2. The Ksp is a constant that represents the product of the concentrations of the ions in a saturated solution, raised to the power of their respective stoichiometric coefficients.
For Ca(OH)2, the balanced equation for its dissociation is:
Ca(OH)2 (s) -> Ca2+ (aq) + 2OH- (aq)
The Ksp expression for this reaction is:
Ksp = [Ca2+][OH-]^2
The value of Ksp for Ca(OH)2 is approximately 5.5 x 10^-6 at 25°C.
Now, we can calculate the solubility of Ca(OH)2 in water using the Ksp value. Let's assume that x moles of Ca(OH)2 dissolve in 500 mL of water:
Ca(OH)2 (s) -> Ca2+ (aq) + 2OH- (aq)
Initial concentration: 0
Change: +x, +2x, +x
Equilibrium concentration: x, 2x, x
Substituting these concentrations into the Ksp expression:
Ksp = (x)(2x)^2 = 4x^3
Now we can solve for x:
4x^3 = 5.5 x 10^-6
x^3 = (5.5 x 10^-6) / 4
x^3 ≈ 1.375 x 10^-6
Taking the cube root of both sides:
x ≈ 1.11 x 10^-2
The solubility of Ca(OH)2 is approximately 1.11 x 10^-2 moles per liter (mol/L) or 11.1 mmol/L.
Now, let's compare this solubility with the amount of Ca(OH)2 added to the solution (0.370 g). We need to convert the mass to moles using the molar mass of Ca(OH)2:
Molar mass of Ca(OH)2 = 40.08 g/mol (Ca) + 2(16.00 g/mol) + 2(1.01 g/mol) = 74.09 g/mol
Number of moles of Ca(OH)2 added:
0.370 g / 74.09 g/mol ≈ 0.00499 mol
So, 0.00499 mol of Ca(OH)2 is added to a 500 mL (0.5 L) solution.
Since the solubility of Ca(OH)2 is approximately 1.11 x 10^-2 mol/L, and we added 0.00499 mol to 0.5 L of water, the resulting concentration would be:
Concentration of Ca(OH)2 in the solution = 0.00499 mol / 0.5 L ≈ 0.00998 mol/L
Comparing this concentration to the solubility of Ca(OH)2 (1.11 x 10^-2 mol/L), we can see that the concentration of Ca(OH)2 in the solution is lower than its solubility. Therefore, the solution will not be saturated.
In conclusion, if 0.370 g of Ca(OH)2 is added to 500 mL of water, the resulting solution will not be saturated with Ca(OH)2.