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If I have a liter of water fully saturated with sucrose would it be possible to dissolve something like salt or any other substance in the water? Or when the solution is saturated, is it impossible to dissolve another solute in it?

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Saturating a liquid with one solute does not mean that the liquid will no longer dissolve another solute. However, you can expect the solubility of the second solute to be different, generally lower, than in the neat solvent.

One relevant concept here (though not specifically applicable to sucrose), in the case of ionic solutes, is the common-ion effect. According to this Wikipedia article:

The common ion effect is responsible for the reduction in the solubility of an ionic precipitate when a soluble compound containing one of the ions of the precipitate is added to the solution in equilibrium with the precipitate. It states that if the concentration of any one of the ions is increased, then, according to Le Chatelier's principle, some of the ions in excess should be removed from solution, by combining with the oppositely charged ions.

Regardless of whether the solute is ionic or not, when you are given a solubility value, for example, the solubility of some compound in water as $\pu{g solute/100mL water}$, this value is only relevant to the solubility in pure water. Once you have saturated water with sucrose, you then have a very different solvent system. There are no simple means of determining what the solubility of a particular solute will be in a solution saturated with another solute. However, there will, in general, be some solubility of additional solvents except as in the case described for the common ion effect.

Please don't hesitate to ask for clarifications in the comments below if I have misunderstood your question or left anything out.

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  • $\begingroup$ But can there be a limit? I mean in a particular temp, pressure can a specific volume of solvent dissolve limited amount of different solute alltogether? $\endgroup$ – Mockingbird May 13 '17 at 23:13
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    $\begingroup$ @Mockingbird , when we add solutes to a solution, we fundamentally create a different solution with different physical and chemical properties than the original solution. So, another way of asking "can there be a limit" is "can there be a solution that cannot dissolve anything". To my knowledge, there is no reason to believe that such a solution should exist. $\endgroup$ – airhuff May 14 '17 at 4:36
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There is no such thing as saturated solution in general. There is such thing as solution saturated with (for example) sucrose, but that doesn't prevent it from dissolving anything else.

Different compounds may indirectly affect each other's solubility via water activity, ionic strength, common ion effect, and/or complex formation, but that's another story.

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Yes, it is well possible. Try this at home: prepare an aqueous solution of table sugar (maybe not one litre, you cup of tea may be, in terms of volume offered, sufficient). Then look out for table salt, add some, and stir the liquid. Does this solution saturated (in terms of sugar) dissolve any salt?

After this experiment, do not drink this "beverage". Rinse your cup of tea in the sink. Besides, the taste is repulsive.

There are instances where the presence of one substance (your solute, here: sugar) in the solvent (here: water) does not influence the solvent capacity for another compound to be accommodated. It may happen that the presence of A aids the solution of B in your solvent, increasing the solvent capacity for B. Conversely, the presence of A may lower the solvent capacity for B, too -- the experiment will tell you the ultimate truth.

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  • $\begingroup$ Your "do not drink the experiment results" remark made me think I was on a cooking site for a moment. Also, your answer kind of reads: "I don't know what will happen, so why don't you try out?". $\endgroup$ – Dmitry Grigoryev Apr 18 '17 at 11:38
  • $\begingroup$ @DmitryGrigoryev One initial thought of mine was to find an example easy accessible and easy to perform; so far I found table sugar and salt in most kitchens. And it may be better not to keep this mixed solution around, in a place where in various extend people eat and drink, too. $\endgroup$ – Buttonwood Apr 18 '17 at 14:39
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One of the 'reasons' something dissolves is because the solution has a thermodynamically higher energy (more negative free energy) than the two separate compounds. Ionic compounds dissolve in water into two (or more) separate ions, so things are a bit more complicated than molecular dissolution like sugar in water. The simplest case is two chemicals which can be considered "non-polar". If you are familiar with the various van der Waals forces, then we can describe these as those chemicals which have only London dispersion interactions. These can be thought of as quantum mechanically induced transient (instantaneous) polarization of the electron clouds of two adjacent atoms. This would be true for two atoms with the same electronegativity, for example. Saturated hydrocarbons are in this category (methane, ethane, propane, etc.) since the H and the C atoms making them up have nearly the same electronegativity (but not quite), which suggest there aren't any permanent polarizations. So given two hydrocarbons, A and B, there won't be much stronger interaction between a molecule of A and another A molecule compared with a molecule of B (or the interaction between two Bs). This is only true up to a point. High molecular weight hydrocarbons aren't infinitely soluble in methane, for example, because of steric (geometric) influences. Big molecules (generally, polymers) are better able to align themselves with similar big molecules (compared to smaller ones which "bounce around" in random orientations. So you see (perhaps) that even the simplest case isn't subject to complexities. The next step towards complete charge separation is induced dipoles. It is possible for a ion or even a molecule with a polar group to change the charge distribution around a neighboring group or molecule or atom. For a given electric field, the induced dipole depends on how "loose" or "tight" the electrons are being held by that molecule (or group (molecular fragment) or atom). Explaining this in further detail is too complicated for this, I think. The next step (of course there aren't really steps, there's all sorts of "in-between" cases) is permanent dipoles. Now, depending on the geometry two dipoles can "cancel" and a molecule may have no net dipole. CO2 is an example. Although both C=O bonds are polar, the molecule is linear ← → so while the end oxygen atoms are more electronegative than the middle carbon both positive and negative center of charge is in the middle (centered at the C atom) so the dipole effects are much diminished compared to say H2O where the molecule is not linear, more of a V shape so that the partially negative charge on the central O atom is separated from the partial positive charge center (which is midway between the two H atoms, meaning on the geometric bisector of the central angle but away from the O atom. (the fact that water has a permanent dipole makes it a great material to use in a microwave oven, the microwaves make those little puppies dance like crazy (heats the heck out of them) and so makes microwave ovens effective at heating foods (the ones that contain water, anyway).) HCl, NH3 also have permanent dipoles (but ammonia does some funky stuff with its geometry). Materials with permanent dipoles tend to be soluble in water because the electric field orients the polar H2O molecule (negative electric field attracts the H atoms more than the O atom, and positive electric field the opposite). Water molecules surround and solvate polar and ionic chemicals. So, I've already mentioned the "last" step, which is full charge separation - ions. Of course there actually are ions which have non-integer charge, remember covalent bonds are basically shared electron pairs. There tend to be lots of water molecules oriented around each ion, which means there are fewer available to surround other ions, and other polar groups. So, that means more of one ionic solute will tend to reduce the solubility of another ionic solute. The important caveat here is the assumption that the ions (both cations and anions) aren't increasing the free energy (in a negative direction). If for some reason Na+ interacted with the -OH (hydroxy group) of sugar (and sugars have lots of hydroxy groups) and improved the thermodynamics then salt would increase the solubility of sugar and vice versa. I don't know of any cases where this happens. In fact, salts are frequently used to "salt out" organic compounds (meaning they reduce the solubility), sometimes very dramatically. So I've mentioned everything except what happens when you dissolve two polar chemicals in water. I've pretty strongly implied that they'll reduce each other's solubility. (BTW, you understand that when you've saturated a solution with one ionic species, and add another, that some of the first will preciptiate, as discussed above, right?) Well, that gets a little murky if one or both of the solutes are present in large enough concentrations to affect what the solvent is. Sucrose is soluble in water at 2100 grams per liter at 25°C. So, given a saturated solution, is the solvent the water or the sucrose. Which is solvent, which solute? I could argue it either way and in fact there's no one answer - it depends on the context. So, if you've got such a solution, what would happen to the solubility of some other polar compound? It turns out the answer is "it depends" there are examples of it decreasing solubility and examples where it increases solubility. No simple answer for highly concentrated solutions. (compounds where the solute approaches solvent concentration).

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    $\begingroup$ When you pasted this answer, your newlines failed to translate. $\endgroup$ – can-ned_food Apr 17 '17 at 17:03

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