Solubility (Questions)?
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1.  When you push down on a bicycle pump, you feel the pressure go up as the volume gets smaller.  If the volume gets smaller, the concentration of molecules per liter increases.  So you can easily observe there is a direct connection of pressure to change of concentrating.  Your experience with the bicycle pump or filling CO2 paintball tanks, or scuba tanks has made it obvious that you can achieve HUGE changes of pressure into a volume.  A little less obvious is the converse that you can have huge changes of volume from changes of pressure.  \nHowever, changing the pressure of a liquid makes very little change to its volume.  Hence, the concentration does not change with pressure.\n\n2.  Tiger is right that very many compounds will disolve readily in water and that is why people call in a universal solvent.  Inorganic salts have a great time in water, but not in organic solvents.  A really cool compound to look up is DMSO (Dimethylsulfoxide).  Because of the 2 methyl groups, it is an organic solvent.  However, the bond angle between those groups is very close to the bond angle between the 2 oxygens in water.  DMSO is a solvent for both organic AND inorganic compounds.\n\n3.  As mentioned in part 1, pressure and volume can change the concentration of a gas.  From the gas laws, you know that temperature is connected to changes in pressure.  So temperature as well as pressure will affect the solubility of gasses in liquids.\n\n4.  Raising the temperature.\n\n5.  CO2 is infinitely more soluble in air than it is in a soda.  Over time, the open glass of soda goes flat.  However, there are physical issues that affect the observation.  The larger bubbles tend to escape the sides of a glass and disappear into the air leaving the smaller ones behind.  When the glass is undisturbed and no new bubbles are being formed, the size of a bubble will decrease in relation to the concentration of CO2 (which is getting smaller as the soda goes flat).
735542
This a  difficult question to answer since the solubilty of these three liquids changes with changing temperature.  Water and acetic acid and miscible at all tempertures and chloroform and acetic acid are miscible at all tempertatures.  However, water and chloroform are only partially soluble in each other and their solubilty changes with changing temperatures.  \n\nPlease also refer to the reference that I have indicated below.  It discussed a point in the ternary system diagram called the Plait Point.  It is an example of a critical point, where the composition of the two phases is equal. At this point, the relative volumes of the two phases are equal and the boundary vanishes as the system becomes homogeneous.\n\nPlease refer to this document for more answers, sorry I could not have been more help.
459648
Themes > Science > Chemistry > Miscellenous > Help file Index > Precipitation Equilibrium > Ksp and Ion Concentration \n\nThe solubility product constant can be used to compute the equilibrium concentration of ions in solution in the same way that you use other equilibrium constant expressions to compute equilibrium concentrations. \n\nUse the Initial, Change, Equilibrium method to compute the concentrations: write the equilibrium expression, the initial amounts of each thing in the expression, how much they change by, then the equilibrium amounts. From that, you can compute the concentrations of all species. \n\nExample: BaF2 has a Ksp of 1.8*10-7. What is the concentration of barium in a saturated solution of BaF2? \n\nSolution:. First, write the balanced equation and the solubility constant expression: \n\nBaF2(s) < = > Ba+2(aq) + 2F-(aq) \nKsp = [Ba+2][F-]2\nWe start with no barium or fluoride ions in solution. If we set the amount of barium ions formed to x, then the number of flouride ions formed is just 2x \n\n [Ba+2] (M) [F-] (M) \nInitial 0 0 \nChange +x +2x \nEquilibrium x 2x \n\n\nNow plug in the equilibrium values shown above and solve for x \nKsp = [Ba+2][F-]2 \nKsp = (x)(2x)2 \nx = 3.6*10-3\nSince we defined the concentration of Ba+2 to be x, we've solved the problem \n[Ba+2] = 3.6*10-3 \n\nInformation provided by: http://learn.chem.vt.edu\n \n**try to visit the site below..** You can calculate the Ksp based on the above example..I'm sure the answer is in your books and notes..have fun learning..
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It will decrease if the solubilizing of the solid is endothermic, and increase if it is exothermic.\n\nIn most cases, solubilizing is endothermic (it takes heat to break up a solid and get the ions to stay in solution) so it will generally decrease, but this is not always the case.
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The solute (lesser component)is the substance which is dissolved in a solvent (greater component).\n\nHope this helps,\nUltimeaciax
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what affects buffers?\n\ndepends on which bufffers  -pH will with solubility, aas will temperature\n\nmaximum capacity depends on individual buffer - citric acid maxes at 3 base molecules, as does phosphates..\n\n\nnow to the question - what will you do for me?????   Good  thing I am a Christian - so i'll just take a thank you
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Water is polar and oil is nonpolar.  The general rule of thumb for solubility is that "like dissolves like," where "like" refers to polarity.  Polar solvents dissolve polar solutes, and nonpolar solvents dissolve nonpolar solutes.
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You cant.\nThe key word here being "completely".\n\nNo matter how insoluble a given precipitate is, it is always soluble to some, very minor, degree.\n\nHowever, by choosing to precipitate a compound with a very low, practically zero, solubility, the ion is (for all intensive purposes) removed from solution.\nWhen dealing with such low-low solubilities, the Ksp value for the compound becomes important...the lower the Ksp, the lower the solubility.
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well seeing that the molarity of the two solution is a direct ratio of how many molecules there are in the solution, i think that the ration would be 0.15/0.20\n\nso assuming that the reation is 1 to 1 (i cant remember so if it isnt 1 to 1, then i'm wrong) then you would need 18.75 mL of the 0.200 M HCl
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C because the solution is SATURATED with solute, meaning no more solute can be dissolved by the solvent in the solution.
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Hg2Br2, BaSO4, AgCl, PbCl2, Al(OH)3, are insoluble\nKBr is soluble.\nYou can refer to this link for details on solubility of salts.\nhttp://chemed.chem.purdue.edu/genchem/topicreview/bp/ch18/soluble.html
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Solute is anything tht is disolved in the liquid. If there was no solute, all you'd have is distilled water.
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First, you dissolve your compound in the solvent in which is is most soluable (perhaps THF or methylene chloride).  Then you very carefully use a pipette to layer the solvent in which it is not soluable on top (perhaps diethyl ether or pentane).  You need to add the second solvent slowly and carefully.  Let the solvent run down the side of the container.  Obviously the top solvent needs to be less dense than the bottom one.  You then let the container sit undisturbed on the bench or in the freezer away from any vibrations until crystals form.  As the upper solvent diffuses into the lower solvent, you compound will become less soluable until it falls out.  The first solution should be as close to saturated as possible.
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Click on "Add details," post the problem, and, as SMeghead said, someone might be able to help.\n\n(EDIT: You didn't provide too much information. From what you gave, about all we can say is that 12.5% of the solute - 1% from 8% - evaporated. If you need to know how many grams, milliliters, cubic centimeters, etc. evaporated, we'll need more information.)
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Uh, the rest of the question isn't appearing. But it's an equilibrium solubility problem. Write a reaction:\n\nMgF2 --> Mg+2 + 2F-\n\nKsp = 5.52e-9\n\nFill in the values on the ICE table (I don't know them, the numbers are missing), and set an equation up.\n\nKsp = [F-]^2 * [Mg+2]\n\nSolve with the numbers from the ICE table. Remember that those concentrations in the above equation are the equilibrated concentrations, not the initial.\n\n\nHope this helped.
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All nitrates are soluble.  You would need to consult a table of solubilities to see what other ions you could use to precipitate out the cations.\n\nEx:  If you had silver nitrate in a solution you could add something with a halide ion such as sodium chloride.  The silver ion would precipitate out with the chloride ion.  If no ppt forms with the sodium chloride you know it was not silver ions.\n\nThen look for other ions that you could test for.
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1.\nMn(s) + SnCl2(aq) --> MnCl2(aq) + Sn(s)\nCr(s) + 2HBr(aq) --> CrBr2(aq) + H2(g)\nCo(s) + CaBr2(aq) --> N.R\n2Al(s) + 3Ni(NO3)2(aq) --> 2Al(NO3)3(aq) + 3Ni(s)\n\n2.\nMg + 2AgNO3 --> Mg(NO3)2 + 2Ag\nmole of Ag = 3.580 / 108 = 0.033 mole\nmole of AgNO3 = 0.033 mole = 33 mmole\nmolarity of AgNO3 = 33 / 50 = 0.66 M\n\n3.\na.soluble\nb.soluble\nc.soluble\nd.insoluble\ne.soluble\nf.soluble\ng.insoluble in water, soluble in acid\nh.soluble\n\n4.\n3NaOH(aq) + Fe(ClO4)3(aq) --> Fe(OH)3(s) + 3NaClO4(aq)\n3Na+ + 3OH- + Fe3+ + 3ClO4- --> Fe(OH)3 + 3Na+ + 3ClO4-\nFe3+ + 3OH- --> Fe(OH)3\n(NH4)2SO4(aq) + Pb(NO3)2(aq) --> PbSO4(s) + 2NH4NO3(aq)\n2NH4+ + SO42- + Pb2+ + 2NO3- --> PbSO4 + 2NH4+ + 2NO3-\nPb2+ + SO42- --> PbSO4\n\nHope it helps\npeace\nvixklen