Say you add .050 mole of CaCl2 and .050 mole of Na2SO4 together in a solution and try to create a precip.  Why do you nee to know the ion concentrations when precipitation occurs?  Also why is it important to know the total volume of the solution?
217892
You need to know concentrations to determine the amount of substance that precipitates out by using solubility products. In this case the precipitate will be CaSO4, so the solubility product will be given by:\n\nk=[Ca(+2)]*[SO4(-2)]\n\nWhere k is the equilibrium constant (probably a given), and [Ca(+2)] & [SO4(-2)] are the concentrations of Ca(+2) and SO4(-2) in the solution. Any amounts greater than the equilbrium concentration will precipitate out as a solid.
625024
Your question is not very clear, however I am attempting it.\n1:  1 mol of Na3PO4 will have 3 mol of Na and 1 mol of PO4 ions.\n2:  2.5 mol of Na2SO4 will have 5 mol of Na and 2.5 mol of SO4 ions.\n3: 3 mol of CaSO4 will have 3 mol of Ca and 3 mol of SO4 ions.\n4: 4 mol of NH4NO3 will have 4 mol of NH4 and 4 mol of NO3 ions.
88853
For part a, I think you would need to know more information about the final buffer such as the final pH.\n\nFor part b, there are .020 mol of H+ ions in the solution because HCl is completely soluble. The molarity of the solution will be .020mol/1.00L = .020 M. To find pH, use -log(molarity). -log(.020) = 1.70
423162
is this your school homework,  are there suppossed to be other questions before these because my computer shows them starting from No. 4.\nnway,\n4.  filter the solution while still hot because as the soltion cools crystallization occurs, and you will nloose your product on the fiolter paper.\n5. crystallization can be induced by reduction in temperature i.e put in an ice bath or you can use a technique called seeding where you put a few crystals of the expected substance in the solution, the other method is to increase the concentration of the solution by evaporating it, to saturation point, ensure you do not evaporate it dry.\n6.i do not know what you started with so i can not tell.\n7.slow cooling will definately give bigger crystalls.\n8.sodium chloride reacts with silver nitrate to form a white precipitate, actually silver nitrate solution is used to analyse NaCl in industrial setting.\n\nope you will find these information sufficient.
362979
The response depends on what you call "not concentrated".  Any saturated solution, as you correctly indicate contains the maximum amount of solute at a specific temperature.  \n\nIf by "not concentrated" you mean that the maximum is low (compared to what?) then you can find sparingly soluble salts that reach saturation at relatively low concentrations (compared, for example, to NaCl).  The absolute value of these concentrations will be very much lower than the saturation level of NaCl.
180049
When performing a chemical reaction you want to use as little of each material to generate the quantity of material you need, with as little waste as possible.\n\nFor example:\n\nHCl + NaOH -> NaCl + H2O\n\nYou are using 1 mole of HCl and 1 mole of NaOH to generate 1 mole of salt and 1 mole of water, therefore stoichiometryt is 1:1.  This is the stoichiometry.  \n\nThe reaction:\n\nH2SO4 + 2NaOH -> Na2SO4 + 2H2O\n\nThe stoiciometry is 1 mole of H2SO4 and 2 moles of NaOH, therefore 1:2\n\nStoichiometry is the relative amounts of each reagent you use in a reaction.\n\nIf you were to use:\n\n2HCl + NaOH -> NaCl + H2O + HCl\n\nYou are using 2 moles of HCl and 1 mole of NaOH to generate the same thing, but you have an excess of HCl left over.  This is yield.  The yield of this experiment in relation to HCl would be 50%, since you have only consumed half of it, the rest is waste.  With respect to NaOH the yield is 100% since you have used all of it.  The yield of the reaction though is 100% as it is always based on the limiting reagent.\n\nThe product yield is the amount of product you recover based on a theoretical maximum.  In the first example above, and assuming you did the reaction on a 1 mole scale, you should recover 1 mole of NaCl (57.5g) and 1 mole of H2O (18g).  If you only recovered 30g of NaCl, then the yield of the reaction is 30 / 57.5 = 52.2%.\n\nNo chemical reaction is perfect so getting 100% return is rarely possible, although we do get very close occasionally in industry.\n\nEcologically the less EXCESS of reagent you can put into a reaction to maximise the recovery limits the amount of material going to waste, which has an impact on the environment.  Economically the less EXCESS of reagent you need, means it costs less to prepare a compound.\n\nWhen pharmaceutical companies make drugs they spend millions on materials.  Cutting down by 10% how much of everything they need can save hundreds of thousands of pounds per year / per drug on excess materials expenditure, increased environmental exposure etc.  Chemical companies spend 100's of thousands of pounds a year disposing of excess chemicals safely, recycling chemicals etc and are liable to very heavy fines if they don't.  Many savings can be made just by improving stoichiometries and product yields.\n\nLast year I spent over $1 million on chemicals, by improving stoichiometries and yields by 10% this alone saves an enormous amount of money.\n\nI hope this helps?
68419
When a salt derived from a strong acid and a weak base dissolves in water, the solution becomes acidic. For example, consider the process\n \nNH4Cl(s) ----H2O ---> NH4+(aq) + Cl-(aq)\n\nThe Cl- ion has no affinity for H+. The ammonium ion NH4+ is  a weak conjugate acid of the base NH4+(aq) and ionizes as follows \n\nNH4+(aq) + H2O(l) <=> NH3(aq) + H3O+(aq)\n\nor simply  \n\nNH4+(aq) <=> NH3(aq)  + H+(aq)\n\nThe equilibrium constant ( or ionization constant) for this process is given by \n\n\n\nKa = Kw / kb  =   1.0*10^-14 / 1.8*10^-5 = 5.6*10^-10\n\n\n      x^2 / .10 =  5.6*10^-10\n\n           x^2  =   5.6*10^-11\n\n             x  =    7.5*10^-6 M  =  [H+]\n\npH  =  -log [H+]\n    \n    =  5.12\n\n     my answer is  C , NH4Cl
378339
U mean how would it affect the molarity of :\n- Na2S2O3 solution;\n- iodine; or\n- the mixed soultion of iodine and Na2S2O3?\n\ni) Na2S2O3 solution conc will decrease slightly if left uncorked.\nii) lesser iodine will be required to neutralize Na2S2O3!\niii) pointless to calculate the conc of this soln.
361137
number of moles of NaCl = Mass/Mr\nalso\n\nNumber moles NaCl = molarity X vol(cm3)/1000\n\ntherefore\n\nMass/Mr = molarity X vol (cm3)/1000\n\nSo, if you rearrange the above:\n\nMass NaCl you need to dissolve = Mr X molarity X vol(cm3)/1000\n\nIn the question molarity is 2M and the volume is 300cm3.  I'll let you figure out the Mr of NaCl (just add together the relative atomic masses of Na [23] and Cl[35.5] and arrive at your answer.\n\nHope it helps.
448739
I will do the Na+ for you. All the others are similarly done.\n\nin 3 M of NaCl you have 3 mol/liter.  In 3 liters you have 9 mol.\nIf you mix both volumes you will have 3 + 2 = 5 liters\n\nso your Na+ concentration will be 9/5 = 1.8M
629476
HCl   +  NaOH   ===   NaCl   +   H2O\n0.0015 ,  0.0016 ,             0        ,       0        at start of reaction (moles)\n0,           0.0001 ,            0.0015 ,    0.0015  at end of reaction(moles)\n\nevery mole of water contributes about 10^-7(ten to the power -7)\nmoles of OH-  and H+ . thus their contribution to the pH is neglected when there is a considerable amount of acid or base present  i.e more than 10^-5 moles/litre (not moles but CONCENTRATION)\nThus concentration of NaOH = 0.0001/90*1000 moles/litre{total volume = 50+40 ml} = 0.001 .\nas NaOH is a strong electrolyte like any other sodium compound  we can assume complete ionisation . thus[OH-]=[NaOH] = 0.001. now pH = -log[H+]=14-(-log[OH-])= 14+log[OH-]  =14+log[0.001]=11\nthus the pH=11
371289
According to Sigma-Aldrich the solubility of MgCl2.6H2O is 1M at 20C (or 20.3030 g/100ml solution), so it shouldn't be possible to prepare 4.5 M MgCl2 in just water. If there are more solutes involved which might increase the solubility of MgCl2 try dissolving those first and then the MgCl2.\n\nYet I saw some papers using 4.5 or 2.5 M...\n\nI don't know why there is this discrepancy between sigma-aldrich and the solubility provided in the link in the above answer which would correspond to 8 M if you consider 100ml of solution and not water. In reality you would have a significant change in volume when you add so much salt and the solubility would be much less but I don't know exactly how much. If you consider 167g of MgCl2.6H2O then you are adding 88.8 g of water which means that the total volume is at least 189 ml and thus the maximum concentration is 4.3 M.\n\nWikipedia gives a different number; 54.2g/100ml at 20C, without specifying if it is for the anhydrous or hexahydrate... so it would be 5.6 or 2.6M respectively... \n\nWhat a mess !!!\n\nMaybe you should also post the question in biology, since someone who has been doing such purifications can clarify the issue.\n\nAnother idea would be to prepare a 100ml saturated solution according to the lowest reported solubility (that of sigma-aldrich) using a 250 ml volumetric cylinder, and then from a pre-weighed amount of MgCl2.6H2O keep adding powder slowly until it doesn't dissolve any more. Weigh the powder that is left in order to find how much you dissolved, measure the final volume and calculate the molarity. If you achieve higher than 4.5 M, dilute.\n\nOR\n\nprepare a 4 M if the amount of powder that you couldn't dissolve is very small.\n\nor \n\nprepare 450ml of 1M and evaporate to 100ml of solution while praying that there will be no precipitation...\nThat's very desperate but I am running out of ideas...\n\nI must say the whole story is very fishy (finding very different solubility values)\n\nI'd go mad too...
508655
Alright, there are two reasons \n\n1) when you add too much indicator to a titration, you are diluting the solution. Its just like if you were to add twenty drops of water to the solution, you would be changing the concentration. Changing the concentration means the titration might react differently, take much longer to complete, and the product will have a higher mass than what you would expect.\n\n2) SOME BUT NOT ALL indicators react with their solution. Phenolphthalein is an example of a chemical that does not react, (don't ask me to go over the indicators that do react, eg: adding an acid to a base shows you have a base...). If your indicator does react with your solution, then the titration will be less successful. This reason really depends what your indicator is. Bromothymol Blue is an example of an indicator that reactis with acid.
598714
Mg2+ < Na+ < Br-\n\nIon size depends first on the main quantum number. Na and Mg both strip down to the Ne configuration, while Br takes up the Xe configuration, two main quantum numbers higher.\n\nMg2+ is smaller than Na1+ because the remaining electrons are subject to a stronger electrostatic field.\n\nThe above is true for ions in crystal lattices.\n\nNote that in watery solutions ion size has to be calculated differently when you take the hydrate shell into account. Na+ becomes Na(H2O)6(1+), and Mg2+ becomes Mg(H2O)6(2+). Bromide won't take up a regular hydrate shell.
124024
If the freezing point of the solution had been incorrectly read .3̊ lower than the true freezing point, then the observed ΔTf would have been greater than it should have been. This would have caused the calculated molality to be too high. If the calculated molality were too high, then the number of moles present as calculated would be too high. An\nexaggerated number of moles would cause the calculated molar mass to be too low, as the same weight of the substance would be distributed against more moles of that substance.\nThe substance would seem lighter than it is.\n\nI'm sure this will help,\nUltimeaciax
405221
The precipitate is a solid, and it usually is in the form of a salt (i.e. NaCl)  If you are given an equation that begins with two liquids, for example, and one of the end products is solid (s), then you know that that is a precipitate.  \n\nPrecipitation can occur for a few different reasons.  One reason can be due to the insolubily of one of the reactants and wil result in precipitation  This can also be due to supersaturation of the mixture that causes the insoluble material to precipitate out of solution.\n\nWikipedia has a nice example using Silver Chloride (AgCl).
529024
I'm not going to solve these for you.  My homework days are far behind me.  I will try to teach you how to solve them, though.\n\n1.  Keep your eye on the moles and don't get confused by the changing volumes.  The final solution is 0.00436 M  and you know the final volume so calculate how much KNO3 is there (=0.00436 M x Vf, in liters).  That will tell you how much was in 25 mL of the second solution.\nSo how much was in the total second solution?  Multiply the moles above by five (25 mL goes into 125 mL five times.)  This is also how many moles was in original 20 mL solution.\nSo, to find the molarity of the original solution, divide the number of moles by the volume, in liters (20 mL = 0.02 L).\n\n2.  Same thing goes for this one.  Multiply 1.66 M by .037 L (remember that M is in terms of liters, not millilieters!) to get the moles in the first solution.  Do the same for the second.  Add the two volumes together to get the resulting volume (in liters!!!)  Divide the sum of the moles in both solutions by the total combined volume and you have your answer.\n\n3.  Same would go for this one if you hadn't left out what volume of solution the Pb++ was in.  Anyway, the way to do this one is to notice that the reaction of Pb+2 ion and SO4-2 ion is a one-to-one proposition.  So, if you figure out how many moles of Na2SO4 you have in 0.009 grams of the stuff, you will also calculate how many moles of Pb(NO3)2 you have.  Divide that by the volume in liters and you have your answer.  [Nice touch by Teach to throw in univalent species to cross you all up.  These are irrelevant to the reaction you're considering.  Still, I like that.  Makes you think.]\n\nGood luck.