- #1
sneakycooky
- 13
- 3
- Homework Statement
- A saturated solution of cobalt(III) hydroxide (Ksp = 1.6 x 10^-44) is added to a saturated solution of thallium(III) hydroxide (Ksp = 6.3 x 10^-46). What is likely to occur?
a. both salts remain stable in solution
b. cobalt(III) hydroxide precipitates, thallium(III) hydroxide remains stable in solution
c. thallium(III) hydroxide precipitates, cobalt(III) hydroxide remains stable in solution
d. both salts precipitate
- Relevant Equations
- IP =[A^n+]^m * [B^m-]^n for degree of saturation, perhaps
At first I thought it was C. A few sources agree with me. The book says D is correct. Here was my reasoning for C:
With both salts having a formula of MX3, their [OH-]'s can be compared according to their Ksp values. Tl(OH)3 has a smaller Ksp, so it has a smaller [OH-] at saturation than Co(OH)3. I think mixing the two salts would make this be true:
[OH-] at Tl(OH)3 saturation < [OH-] of mixing both saturated salt solutions < [OH-] at Co(OH)3 saturation
With Tl(OH)3 being vastly supersaturated, I think it will all precipitate. I don't think Co(OH)3 would have any way to precipitate after this
The book says D is the answer. Here is its justification:
"When the solutions are mixed, [OH-] is above saturation levels for both the cobalt and the thallium in the solution. Since Tl(OH)3 has a smaller Ksp than Co(OH)3, it will react first. The ion product of the mixed solution is higher than the Ksp for Tl(OH)3, and the system will shift left to precipitate solid Tl(OH)3. After the Tl(OH)3 precipitates, a small excess of OH- will remain, which gives an ion product slightly above the Ksp of Co(OH)3. This will cause a small amount (1% - 3%) of Co(OH)3 to also precipitate."
The book's logic isn't making sense to me. Everything I underlined seems wrong to me; mixing the 2 solutions wouldn't make [OH-] be higher than in the Co(OH)3 solution as far as I can tell. I wonder if the "small excess of OH-" refers to the leftover 10^-7 M of OH- from the water of the Tl(OH)3 solution, but that seems like the wrong answer. If there were an excess, the precipitation of Co(OH)3 could make sense, but I guess that leads to the question: is there an excess of OH- after Tl(OH)3 precipitates?
Could someone tell me if the mistake is mine or the book's?
With both salts having a formula of MX3, their [OH-]'s can be compared according to their Ksp values. Tl(OH)3 has a smaller Ksp, so it has a smaller [OH-] at saturation than Co(OH)3. I think mixing the two salts would make this be true:
[OH-] at Tl(OH)3 saturation < [OH-] of mixing both saturated salt solutions < [OH-] at Co(OH)3 saturation
With Tl(OH)3 being vastly supersaturated, I think it will all precipitate. I don't think Co(OH)3 would have any way to precipitate after this
The book says D is the answer. Here is its justification:
"When the solutions are mixed, [OH-] is above saturation levels for both the cobalt and the thallium in the solution. Since Tl(OH)3 has a smaller Ksp than Co(OH)3, it will react first. The ion product of the mixed solution is higher than the Ksp for Tl(OH)3, and the system will shift left to precipitate solid Tl(OH)3. After the Tl(OH)3 precipitates, a small excess of OH- will remain, which gives an ion product slightly above the Ksp of Co(OH)3. This will cause a small amount (1% - 3%) of Co(OH)3 to also precipitate."
The book's logic isn't making sense to me. Everything I underlined seems wrong to me; mixing the 2 solutions wouldn't make [OH-] be higher than in the Co(OH)3 solution as far as I can tell. I wonder if the "small excess of OH-" refers to the leftover 10^-7 M of OH- from the water of the Tl(OH)3 solution, but that seems like the wrong answer. If there were an excess, the precipitation of Co(OH)3 could make sense, but I guess that leads to the question: is there an excess of OH- after Tl(OH)3 precipitates?
Could someone tell me if the mistake is mine or the book's?