Calcium sulfate is found in plaster.
a) At 25 °C, Ksp for CaSO4 is 2.4 × 10-5. Determine the molar solubility in water and in 0.15 M CaCl2.
b) Na2SO4 is added gradually to 100 cm3 of a solution that contains 0.15 M Ca2+ and 0.15 M Sr2+ ions (at 25 °C). Determine the Sr2+ concentration (in M), and hence the % Sr2+ precipitated, when CaSO4 just begins to precipitate. (Ksp SrSO4 = 3.2 x 10-7)
a) At 25 °C, Ksp for CaSO4 is 2.4 × 10-5. Determine the molar solubility in water and in 0.15 M CaCl2.
CaSO4 ionizes this way.
....................CaSO4 ==> Ca^2+ + [SO4]^2-
I.....................solid............0..................0
C....................solid............x..................x
E.....................solid............x.................x
Ksp = 2.4E-5 = (Ca^2+)(SO4^-2). Substitute the E line into the Ksp expression and solve for x = (Ca^2+) = (CaSO4) = solubility in mols/L in water.
In 0.15 M CaCl2 you have the added effect of Ca^2+ from CaCl2. This is the common ion effect.It ionizes this way.
.......................... CaCl2 ==> Ca^2+ + 2Cl^-
I...........................solid.............0...........0
E............................0.............0.15......0.30
Plug the E line from this plus the E line from the water solution into the Ksp expression.
Ksp = (Ca^2+)(SO4^2-) = 2.4E-5
(Ca^2+) = x from CaSO4 + 0.15 M from CaCl2
(SO4^2-) = x from CaSO4.
Ksp = (x + 0.15)(x) = 2.4E-5. Solve for x = (CaSO4) in mols/L in 0.15 M CaCl2 solution. Notice that the solubility of CaSO4 in CaCl2 is lower because of the common ion effect.
Na2SO4 is added gradually to 100 cm3 of a solution that contains 0.15 M Ca2+ and 0.15 M Sr2+ ions (at 25 °C). Determine the Sr2+ concentration (in M), and hence the % Sr2+ precipitated, when CaSO4 just begins to precipitate. (Ksp SrSO4 = 3.2 x 10-7)
Follow the steps above to calculate SO4 when SrSO4 and CaSO4 first begin to precipitate.
For SrSO4 Ksp = 3.2E-7 = (Sr^2+)(SO4^2-)
(SO4^2-) = Ksp/(Sr^2+) = 3.2E-7/0.15 = 2.13E-6
For CaSO4 Ksp = 2.4E-5 = (Ca^2+)(SO4^2-)
(SO4^2-) = Ksp/(Ca^2+) = 2.4E-5/0.15 = 1.6E-4
It is clear that when Na2SO4 is added dropwise to the mixture that SrSO4 will ppt first. The question is to calculate (Sr^2+) when CaSO4 just begins to ppt. It should be clear that Na2SO4 added dropwise ppts SrSO4 first. SrSO4 will continue to ppt until the [SO4]^2- reaches 1.6E-4. At that point CaSO4 will begin. So plug 1.6E-4 for SO4^2- and Ksp for SrSO4 and solve for (Sr^2+) at that point.
Post your work if you get stuck.
To determine the molar solubility of calcium sulfate (CaSO4) in water and in 0.15 M CaCl2, we can use the concept of solubility product (Ksp) and the common ion effect.
a) In water:
The solubility product (Ksp) expression for calcium sulfate (CaSO4) is given as:
Ksp = [Ca2+][SO42-]
Since calcium sulfate dissociates into one calcium ion (Ca2+) and one sulfate ion (SO42-), we can assume that the concentration of Ca2+ and SO42- ions in the saturated solution are both "x."
Therefore, the Ksp expression becomes:
Ksp = x * x = x^2
Given that Ksp for CaSO4 is 2.4 × 10^-5, we can set up the equation:
2.4 × 10^-5 = x^2
To solve for x, take the square root of both sides:
sqrt(2.4 × 10^-5) = x
x ≈ 4.9 × 10^-3 M
The molar solubility of calcium sulfate in water is approximately 4.9 × 10^-3 M.
In 0.15 M CaCl2:
When calcium sulfate is dissolved in a solution containing 0.15 M Ca2+ from CaCl2, we need to take into account the common ion effect, which will reduce the solubility of calcium sulfate.
The common ion effect states that the presence of a common ion (in this case, Ca2+) in the solution reduces the solubility of a salt.
Since the concentration of Ca2+ in the solution is 0.15 M, we can assume that the concentration of Ca2+ from CaSO4 is also "x." Thus, the remaining concentration of Ca2+ from CaCl2 is (0.15 - x) M.
The concentration of sulfate ions (SO42-) from CaSO4 will still be "x" M.
Now, the solubility product (Ksp) expression for calcium sulfate in the presence of CaCl2 becomes:
Ksp = [Ca2+][SO42-]
Substituting the concentrations, we have:
Ksp = (0.15 - x) * x
Again, the Ksp for CaSO4 is 2.4 × 10^-5. Thus, we can set up the equation:
2.4 × 10^-5 = (0.15 - x) * x
Solving this equation may require using an iterative or graphical method. By solving, we can find that the molar solubility (x) in 0.15 M CaCl2 is approximately 2.75 × 10^-5 M.
b) To determine the Sr2+ concentration (in M) and the % Sr2+ precipitated when CaSO4 just begins to precipitate in a solution containing 0.15 M Ca2+ and 0.15 M Sr2+, we need to consider the common ion effect again.
The solubility product (Ksp) expression for strontium sulfate (SrSO4) is given as:
Ksp = [Sr2+][SO42-]
Since we have 0.15 M Ca2+ and 0.15 M Sr2+ ions in the solution, the common ion (sulfate) will still be "x" M, and the concentration of Sr2+ will be (0.15 - x) M.
Using the Ksp expression for SrSO4:
Ksp = (0.15 - x) * x
Given that Ksp for SrSO4 is 3.2 × 10^-7, we can set up the equation:
3.2 × 10^-7 = (0.15 - x) * x
Solving this equation may again require an iterative or graphical method to find the value of x, which represents the concentration of sulfate ions (SO42-) and, as a consequence, the concentration of Sr2+ ions.
Once you find the value of x, the % Sr2+ precipitated can be calculated as:
% Sr2+ precipitated = (concentration of Sr2+ initially - concentration of Sr2+ at equilibrium) / concentration of Sr2+ initially * 100
Substitute the values obtained to calculate the % Sr2+ precipitated.