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Why is $\ce{NaCl}$ more soluble in water than $\ce{NaOH}$? Well, this is a question from Chemical Equilibrium and the question hints for explanation in terms of Le Chatelier's principle.

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    $\begingroup$ This sounds like a homework question. Could you show us what you have tried? It will help you get a better answer. We like to see questions with a bit of your own effort. See the faq and How to Ask for more info. $\endgroup$ – Ben Norris Mar 5 '13 at 23:11
  • $\begingroup$ It probably has to do with the common ion effect. What equilibrium process is going on in plain old ordinary water? $\endgroup$ – Ben Norris Mar 6 '13 at 0:48
  • $\begingroup$ The solvation of a small amount of hydrogen cations to form hydronium, which would be disrupted by the hydroxide ion's higher affinity for these cations? $\endgroup$ – KeithS Mar 6 '13 at 22:30
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It is not. $\ce{NaOH}$ is more soluble ($>100~\mathrm{g}/100~\mathrm{ml}$) then $\ce{NaCl}$ ($<30 ~\mathrm{g}/100~\mathrm{ml}$). $\ce{NaOH}$ when dissolved produces $\ce{OH^-}$ anion, that has relatively small size and forms specific hydrogen bonds (where it is bonded to relatively small positively charged hydrogen) very well. In solid $\ce{NaOH}$ it has to form hydrogen bonds with quite big $\ce{Na^+}$ cation.

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The Water molecules move around the $\ce{Na}$ and $\ce{Cl}$ and break the salt apart into ions. The polarity of the water molecule allows this by the more negative oxygen being attracted to the positive Na ions and the positive hydrogen being attracted to the negative chlorine ions. In the case of NaOH, I am assume that bond between $\ce{Na+}$ and $\ce{OH-}$ is stronger than the bond in $\ce{NaCl}$. Also I don't see a case for the use of an equilibrium in this case because $\ce{NaOH}$ almost completely dissociates into its ions $\ce{Na+~~ OH-}$.

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Pure water, pH 7, has a concentration of roughly .1 ppm (100 ppb) of hydrogen cations (protons), in the form of hydronium ($\ce{H3O+}$). This is a natural state of equilibrium; the solution wants neither more nor fewer protons (if it did, it would take them or give them up, and so it would not be "neutral"; that's a bit circular, but it's a fundamental property of water).

Sodium hydroxide, $\ce{NaOH}$, is a strong base; it is a "proton acceptor", wanting desperately to share one of its lone pairs with a hydrogen cation. The sodium, which gave up the valence electron to the hydroxide, doesn't want it back, so the hydroxide effectively owns the electron. But, hydroxide isn't extremely electronegative, like say, the halogens; it doesn't want to own the electron outright and would rather share.

Hydrogen atoms, for their part, either want two electrons or none, and they aren't picky about which; again, this leads to a situation in which hydrogen would rather share. Hydrogen is the textbook definition of a neutral element in terms of electronegativity, and because of this it will readily form either ionic compounds with elements at the extremes, or covalent compounds with elements closer to the center. This makes combinations of hydrogen and halogens very acidic, because the hydrogen wants to get away from the electron-greedy halogen and form a more stable, equal covalent bond; the compound becomes a "proton donor".

Introduce $\ce{NaOH}$ to water, and the hydroxide ion will display its affinity for the protons of the hydronium. The water, however, was in equilibrium, and so the removal of the cations to form new water molecules, and the resulting release of un-paired sodium ions, disturbs this equilibrium. According to Le Chatelier's principle, this disruption is resisted with an opposing reaction (in this case the re-creation of sodium hydroxide from the sodium deprotonating water to form more hydroxide, releasing or re-releasing hydrogen cations). Because of this love-hate relationship between all of the component ions, the dissassociation of $\ce{NaOH}$ into water is resisted, but because of the extremely high solubility of Group I elements including sodium, it does eventually happen.

Sodium chloride, however, is a salt. It is quite literally the byproduct of an acid and base getting together to form water. The chlorine and sodium ions, abandoned by their respective proton and hydroxide when those two formed the much more stable polar covalent water molecule, form a marriage of convenience between them. The sodium, lacking the electron that the hydroxide now shares with the proton in the water molecule, is attracted to the chlorine, which had as an acid stripped a hydrogen of its electron to form a proton. The chlorine is very happy to "own" the extra electron that produces this charge difference, the sodium is equally happy to be rid of it, and as such, there is no "shock" to the system where the introduced components try to do something that disrupts the equilibrium of the water. Thus, there is no resistance to the natural tendency of the polar water molecules to dissociate the sodium and chloride ions (which happens readily, because the sodium and chloride are such opposites in terms of electron affinity that only their charge difference holds them together, and so any other charge difference, even a partial one, can pull them apart).

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  • $\begingroup$ This is oversimplified to the point of being almost untrue. As it stands, it currently has a red herring in argument. If the solubility of NaOH is being considered, is is pragamtically impossible to discuss its preference to take a proton from hydronium because very likely more than 100ppb of NaOH is added. So, to advance this argument you need to consider hydroxide attack of neutral water, which is an enthalpically neutral process. Also, the idea of 'disturbs the equilibrium' should be reconsidered as 'shits the equilibrium'. $\endgroup$ – Lighthart Mar 8 '13 at 18:02
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    $\begingroup$ 'Shits the equilibrium' - a colorful way to put it. $\endgroup$ – KeithS Mar 8 '13 at 19:02
  • $\begingroup$ shifts. Oops. Heh. $\endgroup$ – Lighthart Mar 8 '13 at 20:22

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