Why are the results less accurate in a titration when too much indicator is added?
923865
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.
593819
It appears that the constants are calculated at one condition (dry,wet,etc) whereas the effective is closer to reality.  See\n\nhttp://www.pnas.org/cgi/content/full/102/45/16293
724819
To determine the concentration of a reagent.
186841
Volumetric Analysis also known as Titration is a standard laboratory method of quantitative/chemical analysis which can be used to determine the concentration of a known reactant. Because volume measurements play a key role in titration, it is also known as volumetric analysis. A reagent, called the titrant, of known concentration (a standard solution) and volume is used to react with a measured volume of reactant (Analyte). Using a calibrated burette to add the titrant, it is possible to determine the exact amount that has been consumed when the endpoint is reached. The endpoint is the point at which the titration is stopped. This is classically a point at which the number of moles of titrant is equal to the number of moles of analyte, or some multiple thereof (as in di- or tri- protic acids). In the classic strong acid-strong base titration the endpoint of a titration is when the pH of the reactant is just about equal to 7, and often when the solution permanently changes color due to an indicator. There are however many different types of titrations.\n\nMany methods can be used to indicate the endpoint of a reaction; titrations often use visual indicators (the reactant mixture changes colour). In simple acid-base titrations a pH indicator may be used, such as phenolphthalein, which turns (and stays) pink when a certain pH is reached or exceeded. Due to the logarithmic nature of the pH curve, the transitions are generally extremely sharp, and thus a single drop of titrant just before the endpoint can increase the pH by several points - leading to an immediate colour change in the indicator. That said, there is a slight difference between the change in indicator color and the actual equivalence point of the titration. This error is referred to as an indicator error, and it is indeterminate.\n\nEtymology\nThe word "titration" comes from the Latin "titalus," meaning inscription or title. The French word, titre, also comes from this origin, meaning rank. Titration is by definition the determination of rank or concentration of a solution, so it corresponds.\n\nPreparing a sample for titration\nTitrations are performed "in solution". If the sample isn't a liquid or solution the samples must be solved. If the analyte is very concentrated in the sample it might be useful to dilute the sample.\n\nAlthough the vast majority of titrations are carried out in aqueous solution, other solvents such as glacial acetic acid or ethanol (in petrochemistry) are used for special purposes.\n\nA measured amount of the sample can be given in the flask and then be solved or diluted. The mathematical result of the titration can be calculated directly with the measured amount. Sometimes the sample is solved or diluted beforehand and a measured amount of the solution is used for titration. In this case the solving or diluting must be done accurately with a known coefficient because the mathematical result of the titration must be multiplied with this factor.\n\nA lot of titrations require buffering to maintain a certain pH for the reaction. Therefore buffer solutions are added to the reactant solution in the flask.\n\nSome titrations require "masking" of a certain ion. This can be necessary when two reactants in the sample would react with the titrant and only one of them must be analysed. Or when the reaction would be disturbed or inhibited by this ion. In this case another solution is added to the sample which "masks" the unwanted ion (for instance by a weak binding with it or even forming a solid insoluble substance with it).\n\nSome redox reactions may require heating the solution with the sample and titration while the solution is still hot (to increase the reaction rate).\n\nProcedure\nN.B. Before starting, make sure that all of your glassware—especially the burette—is clean and dry. Alternatively, you can rinse out the glassware with the solution that is to be stored in the particular piece
616680
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.
1027970
Simply that during the induction period, the actual concentrations may not be 100% homogeneous and the rates will increase of decrease from theoretical.  This is much more important with macromolecules, but can be seen in other cases.  See\n\nwww.uem.mz/faculdade/ciencias/ ciencias_revista/textos/elalves.pdf\n\ndx.doi.org/10.1063/1.1954769\n\nwww.platinummetalsreview.com/ dynamic/article/view/50-2-54-63\n\nwww.citadel.edu/chemistry/scas2005.pdf
940302
you analyse the lowest first\na) to avoid contamination e.g. if you do highest first then, syringe will have traces of your original sample and thus giving you falsely higher result.\n\nb) using higher concentrations can sometimes "blind" the instrument to lower concentrations.
151863
Either:\n\n(a) you're not getting a very clean separation, as the peaks should be vivid\n\nor\n\n(b) your detector is insuitable for the acids you're studying.\n\nAtleast, that's my analysis. Any A-Chem experts here?
723747
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.
1171846
it depend on your solution. if your solvent was water you have some solids that disperse in it with different rate and different quality for more detect the solution(solvent) and contact me
490979
less elements, so less complicated maths.  \n\nYou get a less accurate result, which I don't consider an advantage.
870750
It sounds like there was a problem in making the dilution series. If the two samples are independent (If the 10^-4 sample was not made using the 10^-3 sample) then yes, I would average them (and it should be 70X10000- you have an extra zero). However, if they came from the same series I would use the sample that you have more confidence in. If the colonies were well spread out and not touching, I would use the 10^-3 sample. If there is a possibility of multiple colonies on top of each other when you counted the 10^-3 plate, then use the 10^-4 number.
245193
Use your buffering equation and plug in the numbers you know to get your answer.  You might have to do reverse calulations to find your answer.
965652
an extra zero after the decimal does not change the dose
892272
It tells you when you're at the end point, and the end point is where ever you want it to be, hopefully past but not too far from the equivalence point. \n\nThis isn't always used for pH. Let's say you wanted to know the concentration Sodium in a Na2C2O4 solution. You could use standardized KMnO4 (which is purple) and do a redox reaction which would make CO2 and a bunch of cations (K, Mn, Na). This would be a redox titration, which is different from acid base. This redox titration is actually pretty interesting because a reagent is the indicator, so the end point is when the solution changes to the color of the reactant (this particular reaction is also self catalyzing, which is kinda neat as well).
1181487
True the higher the pH the lower the concentrations
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.