Just have a few questions on the experiment of partition coefficient of butanedioic acid between water and diethyl ether. I wonder why it is necessary to add about 10cm3 of distilled water to the 10cm3 of the ethereal solution before titration with the sodium hydroxide. Moreover, I would like to know why the separating funnel must be shaked after the water and diethylether were mixed. Thanks!
133051
Butanedioic acid is an organic compound.  It is somewhat polar (COOH acid groups), but has significant nonpolar portions.  These nonpolar portions make it harder for the compound to dissolve in water.\n\nEther is chemically similar in that it has polar and nonpolar portions.  Also the COOH groups can H-bond to the O in the ether (CH3CH2-O-CH2-CH3)\n\nWater is all polar.\n\nIf you mix water and ether, they form two phases, one water rich and one ether rich phase.  I would guess that the acid is more soluble in the ether than in the water.  However, IONIC compounds are much more soluble in water than in organic solvents, such as ether.  The addition of sodium hydroxide to the mixture causes the acid groups of the butanedioic acid to de-protonate, forming COO- ions.  This makes the butanedioic acid much more soluble in water than ether, and it will move to the water phase.\n\nIn order to move from one phase to another, the compound in solution must diffuse to the interface between the water and ether and then diffuse across the interface.  If you shake the funnel, you increase the surface area of the interface and also decrease the distance the compound must diffuse to get to the interface.  Shaking the funnel makes the system get to equilibrium faster if the initial state is not in equilibrium.
228320
LIQUID-VAPOR EQUILIBRIUM IN BINARY SYSTEMS\nApparatus\nBoiling apparatus with heating mantle and variac (S), 1/100o thermometer (S),\nrefractometer (L), glass beads or boiling chips (S), weighing bottles (D), small\nstoppered vials (D), 250 ml n-hexane (S) and 250 ml absolute ethanol (S).\nOther binary systems which may be investigated are (1) cyclohexane (S)-\nethanol (S); (2) ethyl acetate (S)-ethanol (S); (3) Methanol (S)-ethyl acetate (S);\n(4) isopropanol (S)-n-hexane (S); (5) isopropanol (S)-cyclohexane (S); (6)\nisopropanol (S)-ethylacetate (S); (7) Acetone (S)-cyclohexane (S); (8) Acetone\n(S)- hexane (S); and (9) cyclohexane (S)-2-propanol (S), (10) 1 Propanol-water.\nCONSULT YOUR LAB INSTRUCTOR TO FIND OUT WHICH SYSTEM YOU\nARE TO INVESTIGATE.\nCAUTION: The ORGANIC LIQUIDS are TOXIC and FLAMMABLE! Use all\nnecessary precautions to avoid accidents!\nCAUTION: DISPOSE OF THESE ORGANIC LIQUIDS by placing in the\nWaste Containers provided. DO NOT POUR DOWN THE DRAIN!\n\nAssemble the apparatus with care; consult instructor if you are at\nall doubtful how to put the pieces together. A 3-way stopcock (shown in Figure\n2) is provided as a means of withdrawing samples of distillate and residue at the\nappropriate time. Best procedure is to withdraw distillate first, then the residue.\n\nProcedure\nClean the apparatus by rinsing with a few milliliters of pure n-hexane. Make sure\nvials are clean and DRY and have labels for marking. Add glass beads to flask\n(carefully).\nThere are three particularly important requirements in this experiment.\n• Make sure that the condensed distillate and the liquid residue samples are\ncollected under equilibrium conditions\n• Read the equilibrium temperature accurately\n• Analyze the compositions accurately\nPlace the thermometer in a position near the top and center of the three legs of\nthe first insert as seen in Figure 3. The hole in the second insert should be\nplaced pointed towards the condenser\nPlace 50 ml n-hexane in a flask. Adjust the variac (not more than 100 -110 volts)\nso that liquid boils "reversibly" (very moderate boiling with only very tiny bubbles\nformed) and so that distillate comes over at a rate which will fill the sample tube\nonce every two or three minutes.\nDraw off the first fraction (containing impurities) which comes over and discard\n(use waste container). When a constant boiling mixture is obtained, record the\nboiling point and withdraw samples from distillate and residue into marked vials.\nAdd 1 ml ethanol to the distillate in the flask, obtain equilibrium-boiling, record the\nboiling point, and again withdraw samples of distillate and residue for analysis.\nBoiling points and refractive indices of the distillate and residue are then\ndetermined after successive additions of 0.2, 0.5, 1, 2, 5 and 5 ml of ethanol. The\nrefractive indices are used to obtain the mole fractions of ethanol in these\nsolutions.\nClean the apparatus by rinsing with a few milliliters of ethanol, and repeat the\nprocess starting with 50 ml of ethanol in the flask. Now, however, boiling points\nare to be determined after successive additions of 0.5, 1, 2, 5, 15 and 20 ml of nhexane\nto the flask.\nIn order to construct a plot of refractive index versus mole percent ethanol, the\nrefractive indices are determined for the pure hexane and ethanol and for a\nseries of solutions containing accurately known weights of hexane and ethanol.\nMixtures about 5 ml in volume containing approximately 1 volume of ethanol to 1,\n3, and 6 volumes of hexane are convenient. To obtain a decent standardization\ncurve of refractive index versus mole fraction, a series of 6 to 8 solutions\ncontaining accurately known weights of n-hexane and ethanol must be prepared\nand their refractive indices measured. Only a few milliliters of each of these\nsolutions are required, so do not waste chemicals when preparing them. The\nrequired data for the vapor pressures of the organic liquids are availab
455443
The previous answer is partially correct.\n\nWater reacts with the excess acetic anhydride which is converted to acetic acid.\n\nSince the product is only slightly soluble in water, this also makes it possible to collect the product as a solid.
213879
Ummm...You will have to figure out which is faster... but I can give an idea as to proceedure...First, try to answer this question: What are all the colored species in grape juice? You need to know that first. Hopefully there is only one. That would simplfy things. Then you need an extinction coefficient for that compound, if you can find one. Using that, you can take several absorbance measures and use Beer's law to establish the concentration of that species in your stock solution. Beer's law states that Absorbance = extinction coefficient x concentration x path length. Your path length is the width of your cuvette, which is usually 1 cm. Beer's Law will not be valid if you use a solution that is so concentrated that it gives you an absorbance value greater than 2. At Absorbance = 2, well over 99% of light is being absorbed. A reading over 2 is thus considered analytically unusable. Your juice can be diluted, if necessary, with DEIONIZED (aka distilled) water. Once you do that, I would roughly time each reaction by just looking at it with my eye, adding just a drop or so of bleach/peroxide initially, and increasing the amount only if necessary. Make sure to add exactly the same amount of bleach/peroxide to exactly the same volume of grape juice. Once you have the reaction basically set up to your satisfaction, then place your juice sample in the colorimeter and get ready to add your reagent. Remember to take a measurement before you add any bleach/peroxide for EACH of the two tests. And take your measurements on the same time frame (ex: every 30 sec.)Good Luck!!! :-)
85408
wow, well put mrjeffy! I have one thing to add though. Because all acids and bases are electrolytes, you can use pH to determine what is and isn't an electrolyte in a solution. If it come out as acidic or basic, it is and electrolyte(the darker, the more basic or acidic). If it comes out neutral, like Distilled Water, it is not. Also don't forget that non-electrolytes will not completely dissolve into solution in water.
100911
Sorry, can't help. You use different units than I do.\n but concentration(mol per dm3) = no. of moles(mol) divided by volume(dm3)
184382
Osmolarity is in units of mol/L, while osmolality is in units of mol/kg. If the solvent for the solution is water, then there is no need for conversion, since 1 L of water weighs about 1 kg. So 0.5 molar solution of NaCl or any other substance will also be a 0.5 molal solution if the solvent is water.\n\nIf the solvent is not water (alcohol, benzene, etc) then you must convert the mol/L into mol/kg. The easiest way to do this is to find out what the solvent's density is.\n\nFor example: imagine you have a 0.500 molar solution of napthalene dissolved in benzene. That means that there is 0.500 moles of napthalene for every 1 liter of benzene (since we are talking about molarity). The density of benzene is 0.8765 g/ml, which is equivalent to 0.8675 kg/L. So use the density to help you solve for the molality:\n\nMOLARITY / DENSITY = MOLALITY\n\nIn our example:\n\n0.500 mol napthalene/L       \n------------------\n0.8675 kg/L\n\nThe liters cancel out and you are left with the units of mol/kg (molality). In this case, the answer is 0.576 mol/kg.
536609
We will need to find if there is a limiting material.\n\nCO + 2H2 ------> CH3OH\n\n(1.61 g H2)*(1 mole H2/2.016 g) = 0.7986 mole H2\n\n(0.7986 mole H2)*(0.5 mole CH3OH/mole H2) = 0.3993 mole CH3OH at most\n\n(15.95 g CO)*(1 mole CO/28g) = 0.5696 mole CO\n\n(0.5696 mole CO)*(1 mole CH3OH/1mole CO) = 0.5696 mole CH3OH at most\n\nH2 is the limiting reagent, therefore:\n0.3993 mole CH3OH * (32.032 g/mole CH3OH) = 12.79 g\n\nPlease check for errors, I went by memory on what the atomic and molecular weights were, you may want to check this.
448049
Yes, it is true that the density of a solution is different of solvant but tjis difference is small ; So , i propose to dissove 40g of amyl acetate in a liter of water if that acetate is a solid. If that acetate is aliquid the total volume will be 1.04 l and you must use 47 cc tp water
254441
the OH group is para in the 4-Hydrobezoic acid, and the OH group is ortho in the salicylic, para bonds to a withdrawing group will cause greater stability in the ring due to resonance and therefore the boiling/melting points will be higher because greater stability in the ring will make it harder to break bonds. If you have more questions jsut get back to me :)
353203
There seems to be a missing link. It is in method 1 you added 100cm3 of water (in excess) to 0.10g of Li.\n\nIf that is true, then my method below works.\n\nNumber of moles of LiOH is 0.00368 (NOT 3.68) from your equation above.\n\nThis 0.00368mol of LiOH is found in 25cm3 (pipetted) of solution.\n\nSo number of moles of LiOH in 100cm3 of solution that you initially mixed is \n0.00368 x 4 (by ratio) = 0.01472mol\n\nSo 0.01472mol of LiOH was created in your initial reaction to find the amount of gas evolved.\n\nSo by the first equation number of moles of Li is also 0.01472mol.\n\nSo Ar of Li is 0.1 / 0.01472 = 6.79 (3 significant figures).
359653
The results may be affected by not cleaning the sides with distilled water.  Assume that your titrant has pH 10.  that will be 1000 times more concentrated that distilled water.  Losing some of this titrant to the wall may affect your result.  Putting some more water will not.
294956
your solutions looks strange to me, mixing DiNaEDTA with Tris-HCl may be difficult, usually EDTA needs very high pH for getting dissolved and and Tris-HCl solutions has pH around 6, so maybe your problem is pH. Sorry about amounts, can´t say if don´t have the molecular weights of the reagents you are using. Also check your NaCl, 2.5M is a bit more than saturated for NaCl at room temperature.\nAnd, just wondering, this doesn´t look like a lysis buffer... looks just like a buffer.
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9 weight percent\n\nhttp://en.wikipedia.org/wiki/Starting_fluid\nhttp://en.wikipedia.org/wiki/Diethyl_ether\n\nhttp://www.vs.ag/ida/index_dmefact.htm
513211
Because you have to know which is the top and which is the bottom layer. \n\nFor example, if you're using diethyl ether to extract something from water, the density of water is 1.00g/mL and of ether 0.71g/mL. So if you shake up a mixture of water and ether, the water will settle to the bottom and the ether will collect on top. You open a stopcock ("faucet") on the bottom of your vessel and let the water drain out and stop before the ether goes. \n\nIf you're using chloroform to extract from waster, the density of chloroform, is 1.48. So after shaking, it's the chloroform that winds up on the bottom.\n\nThe other solvent should not be too close to 1.00g/mL, because you'll get a liquid mush that doesn't separate into layers.
247613
benzioc acid in its neutral form is non-polar. So you cannot separate it from ether (also non-polar). you want to make benzoic acid polar by deprotonating it so you can separate it from ether.\n\nsodium bicarbonate is a weak base, so it would deprotonate the acidic hydrogen on benzoic acid and makes it polar. However, since sodium bicarbonate is a weak base, it does not deprotonate the benzoic acid very efficiently. To do that, you can you sodium hydroxide, which is a much stronger base and can deprotonoate benzoic acid almost completely. so sodium hydroxide is a better choice.
550727
You don't really say, but I'm assuming that you had a standardized solution of something like NaOH that you used in the buret, and you used that solution to determine the concentration of an acid by titration.  You also measured volumes of the acid for titration using a pipet.\n\nOK...So, for a).  If a bubble is present in the buret at the beginning of the experiment and it stays there the entire time, you shouldn't have any effect on your answer.\n\nFor b), if the bubble is not there at the end, then your buret reading will be off.  The readings from your buret would indicate that you put more sodium hydroxide into the flask of acid than you actually did...Does that make sense?  So, the concentration of your acid will actually be lower than your calculation says it is.\n\nc) is more difficult for me to answer, since I'm not really sure whether you determined the end-point of yoru titration by going to a specific pH, or whether you looked at a graph of pH and volumes of base added to dettermine the equivalence point.  If you did the first way, then your calculated result will be lower than your actual result.  If you did it the second way, then it really should affect your result.\n\nd) water in the pipet will dilute the acid as you dispense it, so that you are actually putting less acid in each titration than you should.  So, your calculated answer will be lower than the actual concentration....\n\nHope this helps.