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?
To determine if the solution will be saturated, we need to compare the value of the ion product (Qsp) to the solubility product (Ksp) of Ca(OH)2. If Qsp is less than Ksp, then the solution is unsaturated. If Qsp is equal to Ksp, then the solution is saturated. If Qsp is greater than Ksp, then the solution is supersaturated.
First, let's calculate the moles of Ca(OH)2 that were added to the solution:
Given mass = 0.37 g
Molar mass of Ca(OH)2 = 40.08 + 2(16.00) + 2(1.01) = 74.10 g/mol
Moles of Ca(OH)2 = mass / molar mass = 0.37 g / 74.10 g/mol
Next, we need to convert the volume of water from milliliters to liters:
Given volume = 500 mL
Volume in liters = 500 mL / 1000 mL/L = 0.5 L
Now, we can calculate the concentration of Ca(OH)2 in the solution:
Concentration (mol/L) = moles / volume = (0.37 g / 74.10 g/mol) / 0.5 L
Using this concentration, we can determine the concentration of Ca^2+ and OH^- ions in the solution. Since Ca(OH)2 dissociates into one Ca^2+ ion and two OH^- ions, the concentrations can be calculated as follows:
Ca^2+ (mol/L) = Concentration (mol/L)
OH^- (mol/L) = 2 * Concentration (mol/L)
Now, we can calculate the ion product (Qsp) as the product of the concentrations of the ions:
Qsp = [Ca^2+] * [OH^-]
Finally, we can compare Qsp to the given Ksp value of Ca(OH)2 (Ksp = 6.5 x 10^-6). If Qsp is less than Ksp, the solution is unsaturated, and if Qsp is equal to or greater than Ksp, then the solution is saturated or supersaturated.
By following these steps, you can calculate the ion product (Qsp) and compare it to the solubility product (Ksp) to determine if the solution is saturated.