Timeline for Dissolution of iron(III) hydroxide
Current License: CC BY-SA 4.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 17, 2020 at 8:07 | vote | accept | TheRelentlessNucleophile | ||
| Jul 17, 2020 at 8:07 | comment | added | TheRelentlessNucleophile | Excellent explanation! I now understand the point of further addition! So we only have to take into consideration the necessary condition for precipitation-dissolution, imposed by the constant. Now when I think about it, it has become obvious. Thank you very much! | |
| Jul 17, 2020 at 5:10 | comment | added | Mathew Mahindaratne | @M. Farooq: After dissolving all rust, the $[\ce{Fe^3+}]$ is large enough so that $Q_\mathrm{sp} = [\ce{Fe^3+}] \times (1.00 \times 10^{-7})^3 \gt K_\mathrm{sp}$, the condition to precipitate. So you add extra acid to get $Q_\mathrm{sp} = [\ce{Fe^3+}][\ce{OH}]^3 = K_\mathrm{sp}$, | |
| Jul 17, 2020 at 1:27 | comment | added | ACR | This is a tricky question. You have answered it well. I am also somewhat confused. I am wondering that $K_{sp}$ is an equilibrium value, calculated from the equilibrium concentrations of iron (III), and hydroxide ion. I am imagining that the aqueous solution is saturated with rust. How "valid" it is put the total dissolved iron concentration, after acid dissolution there in this expression: $K_\mathrm{sp} = [\ce{Fe^3+}][\ce{OH-}]^3$. There is no "rust" anymore after dissolution. | |
| Jul 17, 2020 at 1:01 | history | edited | Mathew Mahindaratne | CC BY-SA 4.0 |
added 3 characters in body
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| Jul 17, 2020 at 0:56 | history | answered | Mathew Mahindaratne | CC BY-SA 4.0 |