Increasing solubility in H2O?
345145
O2 is the least soluble in water, followed by Br2.\n\nthe problem is that both LiCl and methanol dissolve completely in water. LiCl since it is an ionic compound, and methanol since it forms hydrogen bonding with water. Therefore any order for LiCl and methanol would be correct I presume.
964608
There's quite a few. Just google solubility chart. Here's one I found.
192522
You can\n1.) Boil the solution and reduce the solution, increasing saturation\n2.) Add more solute\n3.) reduce the temperature of the solution so it has a lower point of saturation.
809126
If the dissolving is endothermic, then increasing the temperature will increase its solubility, so the K+ concentration will increase.
781803
--->H2O  +  H+  +  Cl-
768648
Actually it is not a solubility problem, but a buffer.\n\nLet's use the Henderson-Hasselbalch equation for simplicity\n\npH=pKa+log[CH3COONa]/[CH3COOH]\n\nIf V is the volume of the solution in L then\n[CH3COOH]= mole/V= 0.010/V\n[CH3COONa]= mole/V =x/V\n\nso substitute in the equation and you get\n\n5.37=4.74+ log((x/V)/(0.01/V))=>\n0.63=log(x/0.01) =>\nx/0.01= 10^0.63 =>\nx =0.01*10^0.63= 0.0427 mole =42.7 mmole or since MW=82\nm= 0.0427*82= 3.50 g
797871
raising the temperature. like with a saturated solution of salt-water, if you reise the temperature the water will take in more salt. it's the same with other thing too.
811655
If by 'undissolved' you mean why are some salts insoluble in water, then I think we are in business.\n\nIn order for a salt to be soluble it has to be energetically favorable for the ions to separate. This requires a negative gibbs free energy. G = H - TS So you have to find the H and S values for the solid to aqueous solution and see if G will be negative for a given temperature.\n\nA method I recently learned to make insolube salts more soluble is to add other ions to the water...but you have to be careful to not add a common ion...or Le Chantliear will come into play.
712534
used to change colligative properties of solutions
423379
let me see if i can at least help you with part of your problem, solubility in chemistry is simple, basically the equation will always be:\n\n[Solubility equalibrium constant] = concentration of product divided by concentration of reagents (concentration of each reagents multiplying by each other).\n\nthe easiest way to understand this is to simplify things for you so as you know, a chemical equation always go like this:\n\nreagents <==> products\n\nJust know this from the above, the dynamic here is that when you have an excess of either the reagents or the product, the balance of the equation tend to shift in the opposite direction (ie excess concentration of product will cause more reagents to form and vice versa), in a chemical equation such as the one from above, the reaction can actually go either way, products or reagents can form.  What really matters is the equalibrium constant that I talked about in the first equation. This constant represent the concentration of the products and the reagents when they are in equailibrium or when they are balance, meaning that the rate at which products or reagent form stays the same and there are no change in the concentration of either the products or the reagents so the whole thing is static at the point.   Think about this, more concentration of products wouldn't form unless you have extra reagents to spare that can shift the balance.  When something is saturated it means that it can longer dissolve in a solvent (let's use water to simplify the explanation)so it just accumulates at the bottom, water molecules can only separate so many ions from each other (the reagents) and preventing them from forming a solid product since water itself has limited molecules (if you mix enough table salt in a glass of water, it'll get to the point where it'll accumulate at the bottom and won't dissolve).  If remember correctly, in a chemical equation, when there's different amount of reagents needed to form the product, there's a slight modification to the equation I gave you, for example:\n\n2 CaCO3 + 4 ZnO2 <----> 3 ZnCO3 + 3 CaO2\n\nthen  [equilibrium constant] = [Cao2]^3 x [ZnCO3]^3 divided by [CaCO3]^2 x [ZnO2]^2\n\nnote that you should balance the chemical equations first, I am just giving an example.\n\nhere's a summary of how to compare numbers, if you plug given concentrations of reagents and product into the equation and the number you get is larger than the real equilibrium constant, then the concentration of products is higher than it should be and the equation will shift towards forming more reagents (in the case of the solid product, it'll dissolve) and if there's more reagents, more solid product is gonna form.  Hope that helps. Remember the Chemistry textbook is still your best reference.
135415
That's a difficult question to answer because magnesium hydroxide is practically insoluble in water with a ksp value of like 1.8 x 10^-11 so very little magnesium hydroxide is dissolved in water.  But if you were to compare the solubility of magnesium hydroxide at 25C with 80C my guess is that it would be more soluble at a higher temperature but only by an almost insignificant amount.  \n\nIn short, Mg(OH)2 is pretty much insoluble in water so my direct answer after all the reasoning would probably be: no, Mg(OH)2 does not fit the pattern due to its extremely low solubility.
302581
add 2 parts HCl 1 part H20 to dilute properly
544916
Heat the solvent, then it can take more solute into solution than when it's cool.  Then when you cool it, it is supersaturated.
696007
Temperture and solubility are directly related. If you increase temperture, you increase solubility.
654039
its actually temperarure, pressure and polarity.\n\ntemperature increases then solubility increases\n\npressure increases then solubility decreases.\n\nand by the theory 'like dissolves like' polar is dissolved in polar solvents and vice versa.\n\n** though at room temperature, temp factor shall not have effect as temp remains constant
959876
a) the hydration energy is greater because the temp decreases. When you go from a low potential energy substance to a high potential energy substance the temp of solution will decrease.\n\nb) the solubilty will increase because energy is on the reactant side of the equation. (the reaction is endothermic and requires energy to occur)
955333
an easy example you can remember is H2O(water).\nif u want a more difficult one, u can use CH3COOH(acetic acid)!\n\nALL DE BEST!