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!
1) Determining Ksp (constant of solubility) when given the concentration.
Ce(OH)3 solubility 5.2*10^(-6) mol/L
.......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..
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
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.
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.
Thank you so much for your help. I realized I had never responded, but thank you:)
Good evening! I'd be happy to help you with your questions regarding determining solubility constants and solubility in solutions.
1) To determine the solubility product constant (Ksp) when given the concentration, you can follow these steps:
Step 1: Write the balanced chemical equation for the dissolution of the compound. In this case, the compound is Ce(OH)3. The balanced equation will be:
Ce(OH)3(s) ⇌ Ce3+(aq) + 3OH-(aq)
Step 2: Write the expression for the solubility product constant (Ksp) using the concentrations of the ions. In this case, the Ksp expression will be:
Ksp = [Ce3+][OH-]^3
Step 3: Substitute the given concentration of Ce(OH)3 into the Ksp expression. In this case, you are given the solubility of Ce(OH)3 as 5.2*10^(-6) mol/L. Since Ce(OH)3 ionizes to one Ce3+ ion and three OH- ions, the concentration of Ce3+ will be equal to the solubility, and the concentration of OH- will be 3 times the solubility. Therefore:
Ksp = (5.2*10^(-6))(3(5.2*10^(-6)))^3
Step 4: Simplify and calculate the Ksp value. In this case, simplify and calculate the value to get the Ksp constant.
2) To determine the solubility in g/L for the CdS in a solution of CdCl2, you can use an ice table and the given information:
Step 1: Write the balanced chemical equation for the dissolution of CdS in CdCl2. The balanced equation will be:
CdCl2(aq) + H2S(g) ⇌ CdS(s) + 2HCl(aq)
Step 2: Set up an ice table to determine the equilibrium concentrations. In this case, you are given the molarity (4.25*10^(-3)M) of CdCl2 solution. Assume the initial concentration of CdS is x (unknown), and the equilibrium concentration of CdS is also x.
CdCl2 + H2S ⇌ CdS + 2HCl
Initial: 4.25*10^(-3)M ? M ? M 0 M
Change: -x M -1 M +x M +2x M
Equilibrium: 4.25*10^(-3)M - x M 1 M x M 2x M
Step 3: Write the expression for the solubility product constant (Ksp) using the concentrations of the ions. In this case, the Ksp expression will be:
Ksp = [Cd2+][S2-]
Step 4: Substitute the equilibrium concentration of CdS (x) into the Ksp expression. The concentration of Cd2+ will be equal to the initial concentration of CdCl2 (4.25*10^(-3)M - x), and the concentration of S2- will be equal to x. Therefore:
Ksp = (4.25*10^(-3)M - x)(x)
Step 5: Substitute the given value of Ksp (1.4*10^(-28)) into the Ksp expression and solve for x:
1.4*10^(-28) = (4.25*10^(-3)M - x)(x)
Solve quadratic equation above for x using the quadratic formula or other appropriate method to find the equilibrium concentration of CdS.
Step 6: After obtaining the equilibrium concentration of CdS, you can convert it to solubility in g/L using the molar mass of CdS. Multiply the equilibrium concentration (in moles/L) by the molar mass (g/mol) of CdS to get the solubility in g/L.
I hope this helps you understand how to approach these types of questions. Let me know if you have any further questions!