I recently did a lab in Chemistry where we performed a titration between HCl and NaOH with Phenolphthalein as the indicator. We added DI H2O to the HCl before we began the titration to increase the volume. Why doesn't adding water affect the results?
204220
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.
71137
It's not C.  An increase in pH makes a solution more alkaline, meaning less H+ in the fluids.\n\nI'm going to say A here because a rise in pH is associated with increased O2 binding to hemoglobin.
269644
Acid +Alkali =Salt + Water\nHence,\n\nNaOH + HCl = NaCl + H2O\n\nWhen litmus paper is dipped in NaOH solution it turns blue it means that NaOH is basic in nature.When HCl is added to NaOH The mixture becomes acidic and so the litmus paper turns red.The product you get after mixing both solutions is common salt and water so when this is heated the water evaporates and the salt is left back.This is a small experiment which you can do by adding salt to water and then heating it till the water gets completely evaporated.
291349
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)
567547
I think you are leaving out the first step.\n\nRemember that 12.0 M means 12 moles of HCL per liter of aqueous solution.\n\nTo understand the equations so that you can use them correctly, first think that you will have to add some amount of water to some amount of the concentrated solution to make it the dilute solution.\n\nso, 12 mol/liter , assuming you had a liter of the solution, you keep the same number of moles and add water\n\nhow much water do you add to get to the weak, .6 mol/liter concentration\n\n12 mol/ (1+x) liters = .6 mol/liter\n\n12/(1+x)=.6\n12=.6(1+x)=.6+.6x\n11.4=.6x\n19 = x\n\nso if you had a liter of the solution at 12 M, you would have to add 19 liters of water and then you would have 20 liters of solution at the diluted .6 M\n\nbut you don't need that much of the final solution, so you use the same ratio of water addition, but only start with enough concentrated to end with 300 ml\n\nthe ratio is 19 to 1\n\nso 1/20 * 300 = 15   you need 15 ml of the conc. solution\n   19/20 * 300 = 285  you add 285 ml of water \n\nthe total is 300 ml of the .6 M solution\n\nnow look back at the equations and you will see where they came from and how to apply them properly
379839
Your question makes no sense.\nIf I understand correctly you are making artificial sea water by making a solution of KCl, HNO3,Hg(+2), then you are adding Pb(+2) and Cd(+2) to known concentrations... measuring potential (what method?) and you want to determine the concentration of...what? You know the concentrations of all species (unless a reaction is taking place)
479998
Get the chemical equation first:\nHCl + NaOH --> NaCl + H2O\n\nMoles of HCl used = (50/1000)dm3 X (1.0mol/dm3) = 0.05 mol\nMoles of NaOH used = (50/1000)dm3 x (1.0mol/dm3) = 0.05 mol\n\nFrom chemical equation, 1 mole of HCl needs 1 mole of NaOH for complete neutralization.\nFrom what you are given, you have the exact amount for complete neutralization (meaning, there is no excess reagents).\nFrom chemical equation again, 1 mole of HCl or NaOH produces 1 mole of water.\nThus you had exactly 0.05 mole of water produced here.\nHope this helps. :)
37018
Water is not a good conductor of electric current it conducts electric current because of the impurities present in it.\n\nWhy?\n\nWith this simple experiment you can verify that the distilled water presents a very low conductivity because the potential of the first circuit is much less than the developed by the battery. The small potential of this circuit is due to the presence of the H+ and OH- ions produced when some molecules of water are dissociated. However, the number of molecules of water that are dissociated is practically negligible, making the distilled water a weak electrons conductor. It is worth noting, that as the distance among electrodes increases, the resistance to the electrons transfer in the distilled water increases considerably. This fact evidences even more the low conductivity of the distilled water.\n\nWhen salt is added (NaCl) to the system, the conductivity of the system increases considerably. The water not only dissolves the salt, as well as it dissociates the salt molecule. The elementary sodium (Na) presents an excess electron and the chloride (Cl) has a strong affinity for electrons. Consequently, after dissociation, the sodium loses an electron to the chloride, forming Cl- and Na+ ions. The presence of these ions increases the conductivity of the water considerably. The positive ions (Na+) migrate to the negative electrode (linked to the pole - of the battery) and the negatives (Cl-) migrate to the positive electrodes (linked to the pole + of the pile). The conductivity of the aqueous solution water/salt is proportional to the ions concentration in the solution.\nWhen you try to measure the conductivity of tap water you can verify that it has higher conductivity in comparison with distilled water because this contains ions (less pure). The tap water has higher conductivity due to the presence of the ions: Ca2+ and Mg2+. And don´t forget, try the science at home!
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.
339529
NaCl is not acidic. In the first reaction scheme above there is still effectively an OH- balancing the H+ so the addition of NaCl has no effect on pH. The tenuous clue might be the word "saline" which isn't necessarily pure NaCl. In the absence of any further information, I'd suggest that the most likely reason that NaCl dissolved in distilled water might have pH 5.5 in practice is because of dissolved CO2 from the air. CO2 forms carbonic acid when it dissolves in water:\n\nCO2 + H20 = H2CO3
176800
Since both HCl and KCl are strong elecrolyes, both should remain totally dissassocitated in solution.  I dont think any HCl will react with KCl in this case.\n\nHowever, have you ever heard the rule "Always Add Acid"?\nWhenever you dilute an acid with water, the resulting solution gives off energy (dissolving HCl in water is an exothermic reaction).  This when you dilsute 1 M HCl to .5 M HCl, energy will be released and will cause a temperature change in the solution.\n\na)  The number of moles of HCl used in this experiement is constant. This value will be equal to the number of moles of HCl present in the initial 100 ml of 1 M solution.\n1 M = x moles / .1 L, thus you have .1 moles of HCL.\n\nb)  The heat of the reaction will be the heat given off when the HCl solution is diluted.\n\nc)  Once you know the heat energy given off by the reaction, you can calculate the change in temperatutre if you know the specific heat of the final solution.  I am not sure if you are suppose to estimate the final solution's specific heat as being equal to that of water (4.186 J/g C), but that would be my guess.\nQ = m*c*(delta T)\n\n\nEDIT:\n"davidchingwong13" is correct in his application of the common ion effect.  By dissolving additional Cl- ions in the solution by adding HCl, the equilibrium between solid and aqeuous KCl is disturbed.  However, KCl is a strong electrolyte, it readily dissolves and breaks in to ions in water.  I doubt the addion of so few moles of Cl- ions to an already unsaturated KCl solution willl effect it that much.\n\nHowever....dissolving KCl in water is a endothermic reaction, so if, indeed, the KCl did precipitate out, it would be exothermic and give off additional energy.
175575
The different acids have a different pH so turn a differetn colour.\nThe concentrations are different. The HCl is more concentrated as it will fully dissociate.
357648
This sounds like a lab question and we don't have the details. Possible differences are lab versus standard conditions, quality of equipment, and impure reagents.
458466
Firstly, I would like to indicate that the bonds that chlorine forms with hydrogen are not hydrogen bonds at all. Hydrogen bonds are only present in compounds of hydrogen with the three most electronegative elements, i.e, fluorine, oxygen and nitrogen. Hence, hydrogen bonds may be found in water and ammonia, but not in hydrochloric acid.\n\nAlso you need to remember that the hydrogen bonds are intermolecular forces or attraction, that is the attraction between the hydrogen of one molecule of - taking the example of water - to the oxygen of another molecule of water. The bond within the molecule H - O - H is predominantly covalent, as it is formed by the sharing of electrons. The molecule might have an ionic tendency, how much depends on the electronegativity difference between the two elements sharing the electron.\n\nHence the bonds between the hydrogen and chlorine atoms sharing the electrons are predominantly covalent, with a slightly ionic tendency. Henceforth, water is very much an important factor in ionising an acid, and increase the hydrogen ion concentration in an acid.\n\nI hope this would be of good help to you.
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.
64742
The time necessary to diffuse H+ ions in the base. Thats it, when u add the necessary amount of acid at once.\n\nUsually when neutralization is carried out the conical flask (or whatever where the base is kept) is kept shaking to increase individual molecular motion. The more motion, the greater chance of H+ ion meeting OH- ion and neutralize each other in a given time interval\n\nH+(aq) + OH-(aq) --> H2O(l)
353813
The acid is used in a test for non reducing sugars. Without it, the test would come out negative, so you hydrolise the sugar produced during photosynthesis to reducing sugars (acid hydrolysis, followed by neutralisation), then perform Benedict's test. So basically the HCl hydrolises the sugar.