Question

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)

Answer 1

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.

Answer 2

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.