Talk:Solubility equilibrium

message: Note to anybody editing the table: If you add a line to the table, or change a line, please be sure to cite your data source. Various sources disagree when it comes to solubility product data. Hence without citing a source, any data placed in the table is meaningless. The table has a data source column for that reason. Karlhahn 22:47, 13 December 2006 (UTC)
thread_title:  If you edit the table 

message: I think that solubility equilibrium should merge with equilibrium constant, and that page should have a solubility product constant inside the solubility equilibrium section. In fact, I was going to link to it in my post on the discussion of the redox  page. Kr5t 02:36, 29 March 2006 (UTC)
thread_title:  Merge with chemical equilibrium instead 

message: I believe the pages should be merged.  Solubility equilibrium is directly linked to chemical equilibrium and should be placed on the same page in its own sub-section. If, however, someone is to create a comprehensive solubility equilibrium constant table, the page should remain as it is. marcpatt14, May 2006
strongly oppose merger. these are both big topics and very different topics. if anything we need more articles such as total dissoved solids, a major topic in water quality analysis within environmental science Anlace 15:01, 10 June 2006 (UTC)
thread_title:  Merge unless... 

message: The following is a rough draft of a sol-prod table (from Lange's and CRC) that I will be adding to the article once I've finished proofreading it Karlhahn 01:34, 28 August 2006 (UTC)
replies: text: Doesn't give correct formulæ for hydrated compounds.  —DIV (128.250.80.15 (talk) 08:49, 5 April 2008 (UTC)
thread_title: Table

message: A casual observer may struggle to understand whether a compound with a high or low solubility constant was the most soluble.  Does a high solubity constant indicate the most soluble?  Maybe this should be clarified.King of Leon 16:57, 9 May 2007 (UTC)
thread_title:  Clarity 

message: umm, i was looking up the accepted Ksp value for Ca(OH)2 for a lab, and i noticed wikipedia gives 2 values for it, 4.68 × 10−6 on the Calcium Hydroxide page, and 8.0 × 10−6 on the chart on this page. which one would be correct? these things should be looked into in order to remove discrepencies.74.104.255.195 23:41, 17 May 2007 (UTC)
replies: text: The different numbers came from different sources, and different investigators have come up with different numbers. The solubility equilibrium table cites a source. The calcium hydroxide page does not.  and  both give still other values. I have added a line for one of the above sources. Karl Hahn (T) (C) 00:54, 18 May 2007 (UTC)
thread_title:  wikipedia contradicts itself 

message: it seems that the units are missed out throughout the context
shouldn't solubility products have the units like mol dm-3, mol2 dm-6, mol3 dm-9, etc (depending on the number of ions in the compound)218.215.27.145 13:19, 20 June 2007 (UTC)
thread_title: Units

message: I added a section on if the mole ratios are not even. it needs a little tidying up but the facts are solid hope someone finds it useful.  —Preceding unsigned comment added by 83.42.216.20 (talk) 17:42, 30 January 2008 (UTC)
replies: text: Thank you very much for that. I went through it and (hopefully) beautified it a bit. I also tried to clarify the units as much as I could. Hope I did not thoughtlessly affect your favourite bit in this nice and useful equation of yours.  Stan J. Klimas (talk) 12:41, 1 February 2008 (UTC)

text: How about explaining how the equation is derived? I can't arrive at this equation.220.255.4.133 (talk) 17:38, 13 August 2008 (UTC)
thread_title:  Complex Ksp 

message: I think that the general fomula
:\sqrt[n]{K_{\mathrm{sp}} \over {x^x \cdot y^y}} = {C \over M_M}
is wrong. I don't know where it comes from. 
The solubility product for a general binary compound ApBq is given by
:Ksp = [A]p[B]q
When the product dissociates the concentration of B is equal to q/p times the concentration of A. 
:[B] = q/p [A]
Therefore
:Ksp = [A]p (q/p)q [A]q
:= (q/p)q × [A]p+q
The solubility is 1/p [A]
:[A] = \sqrt[p+q]{K_{\mathrm{sp}} \over {(q/p)^q}}
:: The above does not incorporate the last step:  the solubility is the mass concentration in terms of the solid. One may incorporate 1/p and insert it under the root to obtain:
::S = {[A] \over p} = {[B] \over q} = \sqrt[p+q]{K_{\mathrm{sp}} \over {(q/p)^q} p^{p+q}}
::Then simplify the powers of p and q. Then convert the molar concentration to the mass concentration and you might find that the original formula is correct after all. Hope this helps. Cheers.  Stan J. Klimas (talk) 13:15, 2 July 2010 (UTC)
replies: text: Many thanks for this; is it worth putting in the main text? I want to go over this article. In particular, the treatment of activity is unsatisfactory. Petergans (talk) 06:26, 3 July 2010 (UTC)
replies: text:  I think that the current text could use improvements, including the addition of the derivation because this formula is not found in the English-language textbooks I am aware of. If German wikipedia is accurate, then the formula should be found in the following textbook, which I do not have: Jander, Blasius: Einführung in das anorganisch-Chemische Praktikum. S.Hirzel, Leipzig 1995 (14. Aufl.), . Cheers. Stan J. Klimas (talk) 14:01, 3 July 2010 (UTC)
thread_title:  general formula 

message: The text has been extensively revised and new sections added. There is overlap with the article solubility, but my feeling is that there is no need to merge these articles. Petergans (talk) 08:43, 7 July 2010 (UTC)
thread_title:  Extensive revision 

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thread_title:  External links modified 

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thread_title:  External links modified 

message: I think that some cases where activities are needed in calculations for the system solid salt-solvent should be mentioned by the article.--82.137.14.73 (talk) 21:05, 21 November 2017 (UTC)
thread_title:  Situations that require the use of activities 

message: I notice that this article does not include info re the solubility equilibrium
of substances in mixed solvents.--109.166.137.83 (talk) 23:36, 19 July 2020 (UTC)
thread_title:  Mixed solvents systems 

message: In the section regarding the solubility product, it is stated that, since in a calcium hydroxide solution the number of hydroxide anions is double the amount of calcium cations, the solubility product K_sp = [Ca2+][OH-]^2 reduces to K_sp = [Ca2+]^3.
How is this possible if the concentration of OH-, as previously stated, is double the concentration of Ca2+? Shouldn't the solubility product equal [Ca2+]{2*[Ca2+]}^2, and thus 4[Ca2+]^3?
This seems to be reinforced by the immediately following table.
Is this correct? 93.43.184.56 (talk) 13:11, 17 November 2021 (UTC)
replies: text: Yes.  It follows from   \mathrm{K_{sp} =  [Ca]  [OH]^2 }, since [OH] = 2 [Ca] Petergans (talk) 14:27, 17 November 2021 (UTC)
thread_title:  K_sp isn't equal to [Ca]^3 (?)