When performing a titration....?
453196
Actually let me explain a bit further.  You have added a know amount of one of the substances.  and you titrate the other against it.  In the presence of your indicator.\n\nNow it doesnt matter how much water you add you still have the same amount of the thing you measured out and so it will take the same amount of titrant to titrate it.\n\nhope that makes sense.
580895
This is a typical sort of problem found in real life laboratories- You have differences in concentrations because, in the real world, there are errors introduced at every step of the quantitation process..tiny differences in how much stays in the tube, in the pippette tip, the accuracy of the pippette itself, etc.  However, figuring out which is more accurate is fairly easy.  First, any value that actually falls *on* your standard curve is best-that way you don't have to extrapolate. Second, in general, the more you dilute, the less accurate the reading.  The first consideration is the more important.  In sample 1, your 1:50 dilution is too low to get an accurate reading, in unknown 2, your first dilution is too high to get an accurate reading.
610332
carwash's ideas are pretty good, particularly TLC and spectroscopy.\n\nI don't think I'd do a titration against a base, because esters will also react with bases (hydrolysis). For a synthesis, this doesn't matter. But for an analytical method, it invalidates the technique, unless you can account for it with a standard or something. You might end up getting a result higher than what it should be because of it. (It depends on the chemical environment of the ester)
1127796
Citric acid is triprotic so you have the ratio you need already written.  You last + sign should be an arrow and you are making salt and water from there.  To do the calculation you already have what you need from the reaction.  YOu need data as far as the amount of apple juice and concentration and amount of base.
73838
Its' probably a rounding error. It would depend on which masses for the elements you use, how precise you want to be.
754842
thats not so hard... all you have to do is titrate it against an acidic solution of known concentration that has a mole ratio of 1:1with sls. add a indicator ( phenolphthaleine) to find the endpoint and that should do it.
753594
It is perfectly permissible to write the products as Br2 + Br2.  It is probably easier to work this way.  After you balance the equations, just add the Br2's together to give the final reaction
249939
The three pKa's of phosphoric acid (2.12, 7.21, 12.38) are well separated from each other, considering they're on a log scale.  Generally, if they're separated by 3 or more, they don't interact significantly.  If you're calculating in the general pH range of pKa2 (say 5 - 9, which is the middle buffer region on your titration curve) the others won't affect it much.\n\nFor example: If you have a 1M "buffer" containing 99% H2PO4- and 1% HPO42- (the conjugates responsible for Ka2), the Henderson-Hasselbalch equation gives a pH = 5.2.\n\nSome of the H2PO4- there could pick up H+ from the solution to form H3PO4 (the conjugates reposnsible for Ka1).  The ratio of H2PO4-/H3PO4 also comes from the Henderson-Hasselbalch equation, but using pKa1:  pH = pKa1 + log(H2PO4-/H3PO4).  Plug in 5.2 = 2.12 + log(H2PO4-/H3PO4), and solve for H2PO4-/H3PO4 =~ 1000.  This means that out of the total amount of H2PO4- there, 1/1000 would try to pick up an H+.  The original concentration of H2PO4- would NOT drop significantly, so the first equation used to calculate the pH still holds.\n\nThe same goes for the base side.
495941
We have no idea what data you [we] have to work with here.\n\nWhat did you do in your lab?\n\nI assume you had a saturated solution of Ca(OH)2 which you then titrated with a stron acid like HCl.  You then should have recored the volume and concentration of the acid required to neutralize the Calcium Hydroxide.\n\nWrite out the chemical reaction between Ca(OH)2 and your acid, find the mole ratio between the moles of acid used and the moles of OH- in solution, and solve.
840473
Need more specifics as to your problem.\n\nHowever, I assume you are trying to do a quantitation on a UV or vis absorbing substance.\n\nMake up  a set of standards of known concentration of your analyte in the solvent you will be analyzing in.  Then do some scans over the whole wavelength range. \n\nUsually pick the wavelength at the top of the largest absorbance peak, unless that is a wavelength which will have interferences from other substances present.  If so, then pick a wavelength with good absorbance that is not subject to such interference.
955885
In column chromatography, the elution of analytes is a function of both the stationary and mobile phases (among others things, like temperature).\n\nIf the analyte does not elute, then either the stationary or mobile phase was not chosen correctly.  K' might be very large without being infinite.  The compound may be moving through the column, but too slowly to elute entirely from the column.\n\nFor example:  if the stationary phase is silica gel and the compound does not elute with alcohol or water, then a basic or acidic aqueous solution might be needed to elute the compound.
53245
you must apply oxidation numbers to the equation
224478
The pH of a solution can be measured by titration, which consists of the neutralization of the acid (or base) by a measured quantity of base (or acid) of known concentration, in the presence of an indicator (a compound the color of which depends on the pH). The pH of a solution can also be determined directly by measuring the electric potential arising at special electrodes immersed in the solution.
758825
If you are trying to make a 1:80 dilution of some solution, by adding it to 3 L of water,  you need the 3 L of water to be 79 x the solution amount, S.\n\nThis will satisfy the need for S/(S+3) = 1/80, since 3=79S, and S+3=80S.\n\nTherefore, S=3/79, which rounds to 0.037975 L, or 37.975 mL.\n\nIf you are simply trying to combine some amount of solution so that 3 L is 80 times as much as the amount of the solution, then you need 3/80 or 0.0375L or 37.5 mL of the solution and the mixing ratio will be 1/80 or 1:80.  \n\nIt wasn't clear whether you wanted the mixing ratio or the dilution factor (also a ratio), so I've given you both.
327573
Let's see...it's desireable because it's likely to be a lot different than your starting material and your derivatizing agent - this gives you confidence that you actually made a derivative and are not just measuring the m.p. of either your unreacted agent or your starting material.  We used this a lot when I took a class called the Analysis of Organic Unkowns.  No instruments were allowed.
83196
I've usually made up a concentrated stock of various buffers and dilute it to the concentration I need.  From a 1M Tris stock it's a 1/100 dilution to 10mM and from a 1M MgCLs it's a 1/1000 dilution.  I would decide how many milliters I wanted to make, measure out the cesium powder then pipette out the amount of each Tris (1/100 of the total solution vulume) and Mg (1/1000 of the volume) then bring to the final volume with distilled water.  This works fine if the concentrated solutions don't interact and cause precipitates or something else like being nearly a saturated solution.
661226
It doesn't matter how much iron you happen to dissolve as long as you have the exact mass you dissolved for the later calculations. Use just enough sulphuric acid to dissolve all of the iron, then transfer the solution into a measuring bulb and fill up to the mark - the theoretical concentration of iron is the mass you used divided by the volume. \n\nTheoretically you need 1 mol of acid for each mol of iron, but you will want some excess acid to keep the ions in solution. Make sure that oxygen has no chance to react with your solution during dissolving and titration.\n\nTitration with potassium permanganate in acid solution will require 1 mol permanganate for each 5 moles of iron. Usual lab concentrations are 0.02 mol/l (1 ml of that solution detects 0.1 mmol iron). You will want to use roughly 20 ml permanganate solution per titration for a volume error <1%.\n\nNow we start calculating (using SI units). Iron has a molar mass of 55.8... g/mol. If you want to use about 20 ml, you will need about 0.002 mol of iron. \n\nYour iron solution will have a concentration of \n\nC = m / (M * V)\n\nm : mass of (impure) iron you dissolved, in grams\nM : molar mass of iron, in gram/mol\nV : volume of the volumetric bottle you filled your solution up with, in ml (for convenience)\n\nC : concentration of impure iron, in mol/ml\n\nDivide 0.002 by C to see (roughly) how many ml of your iron solution you want for your first titration. Take an appropriate full pipette which fits the order of magniture, or else dilute your original solution in a volumetric bottle so that you can use a full pipette you have at hand.\n\n\nNow to get percentage purity.\n\nI'll assume you can calculate a concentration from titration.\n\nWith your titration you will determine the measured concentration of iron (in small numbers of mol/ml). Divide that result by C (calculated above), and multiply by 100 to get percentiles.\n\nHave fun!