Timeline for Algebraic treatment of equilibrium
Current License: CC BY-SA 4.0
17 events
| when toggle format | what | by | license | comment | |
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| May 4, 2020 at 6:26 | comment | added | Buck Thorn♦ | @AdnanAL-Amleh Yes, this is correct. | |
| May 3, 2020 at 19:53 | comment | added | Adnan AL-Amleh | @BuckThorn"or since $\ce{[M+]}=C_\ce{A-}$" Is it mean : $$\ce{[M+]}=C_\ce{A-} = [\ce{MA}]_0 $$ | |
| May 3, 2020 at 16:23 | comment | added | Buck Thorn♦ | @M.Farooq Repeating myself somewhat, the explanation in the text is straightforward (my opinion of course) if you have a solution prepared with only acid (no salt added). My intuition fails me when mixing salt and acid, but then if I strain my noggin a little I might reach clarity. And sometimes it's ok to let the math show you the way and not worry too much about intuition. | |
| May 3, 2020 at 13:58 | comment | added | ACR | It must be my own typo. | |
| May 3, 2020 at 9:49 | comment | added | Buck Thorn♦ | @M.Farooq Note I made a small amedment to the answer which I think catches an error in the textbook (or in your citation) | |
| May 3, 2020 at 9:46 | history | edited | Buck Thorn♦ | CC BY-SA 4.0 |
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| Apr 13, 2020 at 8:05 | comment | added | Buck Thorn♦ | @M.Farooq The analytical concentration of conjugate base is referred to as $c_{\ce{A−}}$, but for the purpose of keeping track of charge balance you have to consider both ions formed during conjugate base dissociation (A− and counterion), resulting in the "dummy" variable. In a "real" problem (creating a buffer say) you would specify the salt of the conjugate base that was used and things would be more explicit. In the OP it is ignored, which imho creates confusion. | |
| Apr 13, 2020 at 8:01 | comment | added | Buck Thorn♦ | There might be other ways to think about this that lead to a derivation without dummy variables (as in the OP). But if you understood such a derivation you would probably have the required background, ie you'd know the answer anyway. The point of the question seemed largely to come up with a clearer (more explicit) derivation. | |
| Apr 13, 2020 at 2:07 | comment | added | ACR | Thanks for detailed response. Why do we need to invoke another dummy variable [M$^+$] to derive equation (5) in the original post. | |
| Apr 12, 2020 at 18:30 | comment | added | Buck Thorn♦ | @MaxW I edited in an attempt to clarify | |
| Apr 12, 2020 at 18:29 | history | edited | Buck Thorn♦ | CC BY-SA 4.0 |
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| Apr 12, 2020 at 18:28 | comment | added | MaxW | I can't figure out $\ce{[H+] +[M+] =[H+]}+C_\ce{A-}$ since $\ce{[H+]}$ is on both sides. | |
| Apr 12, 2020 at 18:23 | comment | added | Buck Thorn♦ | @MaxW Maybe I should edit the equation as written but it is correct. The key point is that $[M^+] = C_{A-}$ | |
| Apr 12, 2020 at 18:04 | comment | added | MaxW | I'm confused. Shouldn't the second charge expression be $$\ce{[H+] + [M+]} = \ce{[OH-]} + C_\ce{A-}$$ | |
| Apr 12, 2020 at 15:06 | history | edited | Buck Thorn♦ | CC BY-SA 4.0 |
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| Apr 12, 2020 at 14:05 | vote | accept | ACR | ||
| Apr 12, 2020 at 17:27 | |||||
| Apr 12, 2020 at 12:48 | history | answered | Buck Thorn♦ | CC BY-SA 4.0 |