An acid dissociates sightly in solution. A 15.0M solution of this acid would be described as?
956301
If it dissociates slightly, it is still weak, even if it is 15M, which is very, very concentrated!
679263
I think that what is troubling you could be an exerecise involving "molarity" which uses two quantities: mass(moles)/volume(liter). \nBut you have to bear in mind that mass is referred to the solute, and volume refers to the solution. So, these quantities are not referring to ONE substance. \nIt would be better if you posted a question with the whole exercise, for clarity. \nGood luck.
982225
NOOOOOOOOOO\n\nFirst of all 5% means 5 g/100ml.\nThe molecular weights are different (60 for CH3COOH and 36.5 for HCl), so 5% HCl has actually approximately double concentration of molecules than 5% CH3COOH.\nAlso CH3COOH is a weak acid, so not all of it will dissociate.\nHCl is strong acid so all of it will dissociate.\n\nThis means that you will end up with a solution that is way much more acidic (approximately 2000 times more acidic if I remember the Ka of acetic acid correctly)\n\nThen you have the problem of taste and smell, which will be different and the danger of having impurities in the HCl which are toxic.\n\nAre these reasons enough?
1014284
2.70\n\nregard what ppl saying below, I can make this a little bit more clear.\n\nFirst of all, since the concentration of HCl is 0.00200M and HCl is a strong acid, so we can safely assume 100% dissociation. So, we have 0.02M hydronium ions [one to one ratio]; even if we consider the dissociation of water, it will not make any difference. the dissocation of water is depressed by the presence of "high" concentration of hydronium ions. \n\nyou can even calculate how much water actually dissociates into hydroxide and hydronium ions using the disociation constant of water, 10^-14. But I can guarantee you this will not make ANY difference in pH, say less than 1% for sure.\n\ndissociation of water need to be put into consideration ONLY IF the concentration of hydronium ions is VERY low, say 10^-6 -10^-7, or pH around 7, then it will make a difference. \n\nSo, pH = -log[H3O+]= -log[0.002]= 2.70
903570
Actually HCl is an acid even when it is in the gaseous state. The reaction of HCl gas with NH3 gas gives a cloud of the solid NH4Cl. In this reaction HCl, although it doesn't dissociate, it still donates its H+, so it is an acid according to the Bronsted-Lowry definition.\n\nIf the question is in the context of pH, then you should realize that pH and its 0-14 scale has a meaning only for aqueous solutions, so anyway it is not applicable to gases.\n\nYou should point out this to your teacher, since claiming that HCl gas is not acidic is wrong.
520869
A weak acid does not dissociate fully while a strong acid dissociates completely.\n\ne.g. HCl --> H+ + Cl-\nIf you have 0.1M of HCL, you will also have 0.1M of H+.\n\ne.g. CH3COOH <==> CH3COO- + H+\nIf you have 0.1M of CH3COOH, you may only get 0.01M of H+ as some of the CH3COOH does not dissociate to form H+.\n\nAn example of a weak organic acid is propanoic acid, C2H5COOH.
696206
the moles of NaOH = the moles of OH - = 0.125 * 0.606 = 0.07575\nthe moles of OH - in the new solution = 0.07575/(15 + 0.125 ) = 5*10^(-3) M\npOH = - log ( 5*10^ -3) = 2.3\n=> pH = 14 - pOH = 11.7
621533
NaOH ----------------> Na+   +   OH-\nFind the number of moles of NaOH\nMoles of NaOH = 0.10M * 0.150L = 0.015 moles\nSince the coefficients in fornt of NaOH, Na+ and OH- are all 1, therefore if 0.015 moles of NaOH disssociates, 0.015 moles of Na+ and 0.015 moles of OH- are produced.\n\nSimilarly we will work with HCl\nHCl --------------------> H+      +      Cl-\nMoles of HCL = 0.30M * 0.05L = 0.015 moles\nThis means the 0.015 moles of H+ will be produced\n\nSince H+ and OH- have equal amiunts, they will completely neutralize each other leaving the solution at neutral pH. Since the neutral pH is 7, therefore, the pH of this solution will be 7.
948423
You are going the wrong way.\nThe dissociation constant of a weak monoprotic acid is Ka. If you have a polyprotic acid then you have Ka1,Ka2,etc and more than one equivalence points.\n\nDo the plot, find the equivalence point, and then pKa=pH for 1/2 volume of base of that needed to reach the equivalence point.\nSince you have a weak acid you don't have the equivalence point at pH=7 but at higher pH (The equivalence point will be at the middle of the rising curve). \nThen Ka=10^-pKa\n\nYou can have a look at \n\nhttp://www.shodor.org/unchem/basic/ab/\n\nhttp://www.chemguide.co.uk/physical/acidbaseeqia/phcurves.html
739878
I am not sure what you are asking but the answer could be a monoatomic ion. The positive being a cation and the negative being an anion.\n\nYou might ask over at http://forums.mychemistrytutor.com though.
970448
N = Normality...in this case the concentration of Sulfuric Acid in a solution.\n\nThe concentration of Sulfuric Acid in the 1.00 N Solution is almost 4 times as concentrated as the 0.255 N Solution.\n\nIn practice Normality is "similar" to Molarity, except Normality takes into account the actual number of protons exchanged in a reaction.  Normality = Molarity times the number of protons exchanged.  In Sulfuric Acid, two protons are available for exchange...therefore, a one molar solution would be a two normal solution.  It is useful when doing titrations to determine concentrations.
772155
this is the electrolytic dissociation of acetic acid in water\nCH3COOH --> CH3COO- + H+
313595
Water... its got a pH=7
541583
pH from 0 to 7: Acid solution.\nPH from 7 to 14: Basic solution.
698593
You are not giving us all of the info.\nYes the total or initial concentration of the acid is C=(8.345*10^-3)/0.05 = 0.1669 M\n\nFor a monoprotic acid Ka=[H+][A-]/{HA]\n[A-]=[H+]=10^-pH=10^-4\n[HA]=C-[H+]= 0.1669-0.0001= 0.1668\nso you have calcualted Ka correctly.\n\nKa depends on the nature of the acid and temperature (for small molecules whereas for bigger molecules like proteins it is also affected by the nature of the surrounding chemical groups).\n \nThus addition of base doesn't affect Ka, only the pH.\nIs the base monoprotic or not? If not, what's the valency? Is it strong or weak? If it's weak what's the Kb? Is it the conjugate base of the weak acid? Without these info we can't calculate the pH.
445032
Click on "Add details," post the problem, and, as SMeghead said, someone might be able to help.\n\n(EDIT: You didn't provide too much information. From what you gave, about all we can say is that 12.5% of the solute - 1% from 8% - evaporated. If you need to know how many grams, milliliters, cubic centimeters, etc. evaporated, we'll need more information.)
711408
I am not a chemistry student but I make this research for you and maybe it helps you http://www.chem.latech.edu/~deddy/chem104/104Antacid.htm\nhttp://www.chemcool.com/regents/acidsbasessalts/aim3.htm\n\njust copy those links and you should be on your way to solve ur problem.