Ca(OH)2 has a Ksp of 6.5×10−6.
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?
You can do this two ways.
1. Determine the solubility of Ca(OH)2 from the Ksp and compare with 0.379 g.
2. Determine Qsp, using the 0.370 g, and compare with Ksp. If Qsp>or=Ksp, it is saturated. I'll show you how to do 1.
mols Ca(OH)2 in 0.370 g = 0.370/molar mass Ca(OH)2 = estimated 0.005 M.
........Ca(OH)2 ==> Ca^2+ + 2OH^-
I.......solid.......0........0
C.......solid.......x.......2x
E.......solid.......x.......2x
Ksp = 6.5E-6 = (Ca^2+)(OH^-)^2
6.5E-6 = (x)(2x) = 4x^3 and I get approx
x^3 = approx 1.6E-6 or x = approx 0.012 M
Convert to mols by mols = M x L = approx 0.012M x 0.5L = 0.006 mols. Then convert to grams. g = mols x molar mass = approx 0.006 x 74 = approx 0.4 g but the solution has only 0.370 so it isn't saturated.
Post your work if you get confused.
The other method is quicker.
Well, if the solution is feeling a little "salty," then it might be saturated! But in terms of chemistry, let's crunch some numbers.
To determine if the solution will be saturated, we need to compare the ion concentrations to the Ksp. But first, let's find the moles of Ca(OH)2.
The molar mass of Ca(OH)2 is 74.093 g/mol, so 0.370 g is equal to:
0.370 g / 74.093 g/mol = 0.00499 mol
Since Ca(OH)2 dissociates into Ca2+ and 2 OH-, the concentration of Ca2+ ions will be twice the concentration of Ca(OH)2. So, the concentration of Ca2+ ions is:
2 * 0.00499 mol / 0.500 L = 0.01996 M
Now, let's calculate the OH- concentration. Since Ca(OH)2 dissociates into 2 OH- ions, the concentration of OH- ions is:
2 * 0.00499 mol / 0.500 L = 0.01996 M
Now, we can calculate the ion product, Q:
Q = [Ca2+][OH-]² = (0.01996)(0.01996)² = 0.00000797
Comparing Q (0.00000797) to Ksp (6.5×10−6), we see that Q < Ksp. Therefore, the solution is unsaturated! It's not quite "salted" to perfection yet.
To determine if the solution will be saturated, we need to compare the ionic product (Qsp) with the solubility product constant (Ksp). If Qsp is equal to or greater than Ksp, the solution is saturated.
Let's calculate Qsp first:
- The molar mass of Ca(OH)2 is 74.093 g/mol.
- The number of moles of Ca(OH)2 added can be calculated using the formula: moles = mass / molar mass.
- The volume of the solution is 500 mL, which is equivalent to 0.5 L.
- The concentration of hydroxide ions (OH-) in the solution is given by the formula: concentration = moles / volume.
Now let's plug in the values:
moles of Ca(OH)2 = 0.370 g / 74.093 g/mol = 0.00499 mol
concentration of OH- ions = (0.00499 mol) / 0.5 L = 0.00998 M
Since Ca(OH)2 dissociates to form two hydroxide ions per formula unit, the concentration of OH- ions is doubled:
concentration of OH- ions = 2 * 0.00998 M = 0.01996 M
Now we can calculate Qsp:
Qsp = (concentration of Ca2+) * (concentration of OH-)²
Since Ca(OH)2 dissociates completely in water, the concentration of Ca2+ is equal to the concentration of OH-:
Qsp = (0.01996 M) * (0.01996 M)² = 7.984E-06
Comparing Qsp with Ksp:
Qsp = 7.984E-06
Ksp = 6.5E-06
Since Qsp (7.984E-06) is slightly greater than Ksp (6.5E-06), the solution is saturated.
To determine if the solution will be saturated, we need to compare the calculated ion product (Q) with the solubility product constant (Ksp).
First, let's calculate the amount of Ca(OH)2 that dissolves in the water.
The molar mass of Ca(OH)2 is:
Ca = 40.08 g/mol
O = 16.00 g/mol
H = 1.01 g/mol
Ca(OH)2 = (40.08 g/mol) + (2 * 1.01 g/mol) + (2 * 16.00 g/mol)
= 74.10 g/mol
To calculate the number of moles of Ca(OH)2, we divide the given mass (0.370 g) by the molar mass:
moles of Ca(OH)2 = 0.370 g / 74.10 g/mol
= 0.00499 mol
Now we need to find the concentration of Ca(OH)2 in the solution. Since we know that the solution is 500 mL, we can convert it to liters:
volume(L) = 500 mL = 0.5 L
Concentration (C) is calculated by dividing moles by volume:
C = moles / volume
= 0.00499 mol / 0.5 L
= 0.00998 mol/L
Next, we need to determine the concentration of Ca2+ and OH- ions in the solution. Since for every 1 mol of Ca(OH)2, we get 1 mol of Ca2+ and 2 mol of OH-, we multiply the concentration of Ca(OH)2 by the stoichiometric coefficients:
Ca2+ concentration = 0.00998 mol/L
OH- concentration = 2 * 0.00998 mol/L = 0.01996 mol/L
Now, let's calculate the ion product, Q, by multiplying the concentrations of Ca2+ and OH-:
Q = [Ca2+][OH-]
= 0.00998 mol/L * 0.01996 mol/L
= 1.99×10^-4
Finally, we can compare the ion product (Q) with the solubility product constant (Ksp) to determine if the solution is saturated.
If Q is greater than Ksp, the solution is supersaturated.
If Q is equal to Ksp, the solution is saturated.
If Q is less than Ksp, the solution is unsaturated.
In this case, Q (1.99×10^-4) is greater than Ksp (6.5×10−6), so the solution would be considered supersaturated.