Question

Good evening!
I hope you are all well:)
I've been struggling through a full packet of these questions and was wondering if I could be taught how to do a few sample ones with all the details I was given.

1) Determining Ksp (constant of solubility) when given the concentration.
Ce(OH)3 solubility 5.2*10^(-6) mol/L

2) Determine the solubility in g/L for the following solutions. (Might have something to do with ice tables)
CdS in a solution of CdCl2 with a molarity of 4.25*10^(-3)M. Ksp=1.4*10^-28

Please help if you are able!

Answer 1

1) Determining Ksp (constant of solubility) when given the concentration.
Ce(OH)3 solubility 5.2*10^(-6) mol/L
<b> .......Ce(OH)3 ==> Ce^3+ + 3OH^-
I...............solid................0..............0
C..............solid................x..............3x
E...............solid................x...............3x
How did I get the table. Initially no Ce(OH)3 has dissolved; therefore, the products are zero. For the C line, for every 1 molecule of Ce(OH)3 that dissolves it will produce 1 of Ce^3+ and 3 of OH^-. I've called those x and 3x BUT you COULD put in 5.2E-6 instead of x and (3*5.2E-6)^3 for (3x)^3
The E line (E for equilibrium) is the sum of I and C. This set up will work any of the Ksp problems, sometimes with a little adjustments. So we write the Ksp expression = (Ce^3+)(OH^-)^3. Now let's plug in what we have above.
Ksp = (x)(3x)^3 = x*27x^3 = 27x^4. But notice that the problem TELLS you that x = 5.2E-6; therefore plug that in for x and solve for Ksp. You should get 1.97E-20 which I would round to 2.0E-20 to two significant figures in line with the 5.2 number..</b>

2) Determine the solubility in g/L for the following solutions. (Might have something to do with ice tables)
CdS in a solution of CdCl2 with a molarity of 4.25*10^(-3)M. Ksp=1.4*10^-28

<b>This is a problem involving the common effect. The common ion is Cd in this case; i.e., Cd from CdS and Cd from CdCl2. Notice that I start ALL of these problems the same way; some profs use different schemes and the two problems don't look alike.
..............CdS ==> Cd^2+ + S^2-
I..............solid.........0............0
C............solid.........x.............x
E............solid..........x.............x

In addition to the sparingly soluble CdS you have added 4.25E-3 M of CdCl2 so that looks like this. Remember that CdCl2 is a strong electrolyte; thus, it dissociates completely (that is a 100%).
...............CdCl2 ==> Cd^2+ + 2Cl^-
I..............4.25E-3.........0.............0
C............-4.25E-3......4.25E-3....2*4.25E-3

Write the Ksp expression.
Ksp = 1.4E-28 = (Cd^2+)(S^2-)
ng the two ice charts you substitute x for Cd^2 and 4.25E-3 for Cd. For (S) you have x. It looks like this.
Ksp = (x+4.25E-3)(x) and solve for x. I didn't solve this to get a number but it's straight forward. I expect you will need to use the quadratic formula but I'm not sure of that.
Hope this helps. I usually don't go into so much detail.

Answer 2

Sorry about that. I didn't complete the ICE table for the second problem.
...............CdCl2 ==> Cd^2+ + 2Cl^-
I..............4.25E-3.........0.............0
C............-4.25E-3......4.25E-3....2*4.25E-3
E...............0...............4.25E-3......8.50E-3

Also, on rereading the problem asks for solubility in g/L. Solving the equation I gave you x will be in M or mols/L.
To convert, use x = M CdS. grams = x mols/L CdS x molar mass CdS = grams CdS/L.

Answer 3

Thank you so much for your help. I realized I had never responded, but thank you:)