Calculating solubility from Ksp
1. Calculate the solubility of calcium carbonate in water at 25 0 C. The Ksp of calcium carbonate is
4.8 x 10 5 .
2. What is the solubility of calcium hydroxide, in g/L if the solubility product constant for calcium O hydroxide is 4.0 x 10-6 ?
3. The Ksp for magnesium fluoride is 6.4 x 10-9 . What is the solubility in g/L?
4. The solubility product of copper (Il) sulfide, CuS, is 4 x 10 38 at 25 0 C. Calculate:
(a) the number of moles of Cus that will dissolve in 1.0 L of solution
(b) the number of grams of Cus that will dissolve in 1.0 L of solution
(c) the concentration of the sulfide-ion present
5. What is
(a) the solubility of calcium hydroxide, Ca(OH)2 in g/L?
(b) b) and the [OH-I ], if the solubility product constant for Ca(OH)2 is 4 x 10-6?
6. The solubility product constant, Ksp, of Ag2Cr04 in water is 5.00 x 10-1 at 25 0C, what is the concentration of Ag+l in a saturated solution of Ag2Cr04 at 25 0C?
7. The Ksp of Ag2C03 at 250C is 8.20 x 10-12. Calculate the mass of silver that may be recovered from a saturated solution of Ag2C03 at 25 0 C.
1. To calculate the solubility of calcium carbonate (CaCO3) in water at 25 0C, we need to find the square root of the Ksp value.
Ksp = [Ca2+][CO3 2-]
Since the stoichiometry of calcium carbonate is 1:1, the solubility of CaCO3 is equal to the concentration of the Ca2+ ions in solution.
√(4.8 x 10^5) = 691.96 g/L
The solubility of calcium carbonate is 691.96 g/L.
2. To find the solubility of calcium hydroxide (Ca(OH)2) in g/L, we need to find the square root of the Ksp value.
Ksp = [Ca2+][OH-]^2
Since the stoichiometry of calcium hydroxide is 1:2, the solubility of Ca(OH)2 is equal to twice the concentration of the Ca2+ ions in solution.
√(4.0 x 10^-6) = 0.002 g/L
The solubility of calcium hydroxide is 0.002 g/L.
3. To find the solubility of magnesium fluoride (MgF2) in g/L, we need to find the square root of the Ksp value.
Ksp = [Mg2+][F-]^2
Since the stoichiometry of magnesium fluoride is 1:2, the solubility of MgF2 is equal to twice the concentration of the Mg2+ ions in solution.
√(6.4 x 10^-9) = 0.00008 g/L
The solubility of magnesium fluoride is 0.00008 g/L.
4. (a) To find the number of moles of CuS that will dissolve in 1.0 L of solution, we need to divide the solubility product constant by the coefficient of the sulfide-ion (S2-).
Number of moles = Ksp / [S2-] = (4 x 10^38) / 1 = 4 x 10^38 moles
(b) To find the number of grams of CuS that will dissolve in 1.0 L of solution, we need to multiply the number of moles by the molar mass of CuS.
Mass = Number of moles x molar mass = (4 x 10^38) x (63.546 + 32.06) = 8.51 x 10^39 g
(c) Since the stoichiometry of copper (II) sulfide is 1:1 with sulfide-ion, the concentration of sulfide-ion is equal to the concentration of CuS.
Concentration of sulfide-ion = 4 x 10^38 moles/L
5. (a) The solubility of calcium hydroxide (Ca(OH)2) in g/L is equal to twice the concentration of the hydroxide-ion (OH-).
Solubility of Ca(OH)2 = 2 x [OH-]
(b) The concentration of OH- can be calculated from the square root of the Ksp value for Ca(OH)2.
√(4 x 10^-6) = 0.002 g/L
The solubility of calcium hydroxide is 0.002 g/L. The concentration of OH- is also 0.002 g/L.
6. The concentration of Ag+ in a saturated solution of Ag2CrO4 can be calculated using the Ksp value and the stoichiometry of the compound.
Ksp = [Ag+][CrO4 2-]2
Since the stoichiometry of silver chromate is 2:1, the concentration of Ag+ is equal to twice the square root of the Ksp value.
2 x √(5.00 x 10^-1) = 1.41 M
The concentration of Ag+ in a saturated solution of Ag2CrO4 is 1.41 M.
7. To calculate the mass of silver that may be recovered from a saturated solution of Ag2CO3, we can use the molar mass and stoichiometry of silver carbonate.
Ksp = [Ag+][CO3 2-]
Since the stoichiometry of silver carbonate is 2:1 with Ag+, the concentration of Ag+ is equal to half the square root of the Ksp value.
0.5 x √(8.20 x 10^-12) = 5.71 x 10^-6 g/L
To find the mass, we need to multiply the concentration by the volume of the solution in liters. Let's assume the volume is 1.0 L.
Mass of silver = 5.71 x 10^-6 g/L x 1.0 L = 5.71 x 10^-6 g
The mass of silver that may be recovered from a saturated solution of Ag2CO3 is 5.71 x 10^-6 g.