What causes alkalinity in Potassium chloride after electrolysis?
969279
You must be talking about Potassium Chloride solution and not Potassium Chloride in molten form. So it would contain K+, Cl-, H+, OH- ions, and once electricity is passed through, K+ and H+ migrate to the cathode. Hydrogen being lower in the electrochemical series get discharged first as Hydrogen gas, (2H+ + 2e- ----> H2) and both Cl- and OH- ions migrate to anode but because of concentration effect, Cl- loses electrons and get discharged as Cl2 gas. (2Cl- - 2e- ---> Cl2) Thus K+ and OH- ions are left behind. These form Potassium Hydroxide and all group I hydroxides are alkali hence the resulting solution is alkaline.
912330
Loss of potassium has to be balanced by some other ion, in order to maintian ionic charge balance.  If there were efflux of K with no corresponding change in anything else, then the loss of the + ion (K+) would mean the inside of the bacteria would become quite negatively charged.  However, what would tend to happen is an increase in influx of the other major + ion, sodium (Na+) and perhaps Calcium (Ca+2).  Though this would restore the electrical balance, you would now have a chemical imbalance- processes in the bacteria do not proceed exactly the same in the presence of different chemicals-so, there would be biochemical disruption.
874059
I think that isolating potassium is the wrong approach.  Potassium metal reacts with atmospheric water to form KOH and hydrogen gas.  If you accidentally get water on the metal the reaction is so vigourous that the metal will catch fire, which will be dangerous.  \n\nTry a different approach.  Dissolve KCl in water and add a molar equivalence of Silver iodide, AgI, also dissolved in water..  AgCl is nearly insoluble in water (~0.9 mg/L at 10C)  and will precipitate out, leaving KI in solution.  Filter out the AgCl and evaporate the water, leaving solid KI.  \n\nThe two biggest problems starting with KNO3 are:  It will be nearly impossible to salt out the nitrate because of its high solubility in salt form, and two, many nitrate salts, particularlyof metals, are highly explosive and dangerous.  KCl is cheap and plentiful, so why not just buy some?
396644
Like this?\nhttp://www.webelements.com/webelements/elements/media/pics/rainbow.jpg\n\nTo quote the web site,\n"The picture above shows the result of adding different metal salts to a burning reaction mixture of potassium chlorate and sucrose. The red colour originates from strontium sulphate. The orange/yellow colour originates from sodium chloride. The green colour originates from barium chlorate. The blue colour originates from copper (I) chloride. The lilac colour that should be evident from the potassium chlorate is washed out by the other colours, all of which are more intense."\n\nBasically, certain salts will color the flames various colors due to how the electrons are excited (more complicated than that, look into spectroscopy).
183538
I can't give you a sketch, but I can tell you how.\n\nThere are naturally occuring brines (very salty water if you aren't a specialist) which are mainly Potasium Bromide (KBr) in water. \n\nThis is put into the top of a tower and flows downwards over a series of internal plates or trays. \n\nAt the same time, Chlorine gas is sent upwards though the same tower so that the gas (Cl) and the salts (KBr) are in contact. The Cl replaces the Br in the salts and liberates free Bromine. \n\nThe liquid which come out the bottom of the tower is mainly KCl, potasium chloride solution.\n\nYou should find a typical type of tower if you do a search on Contactor Tower or Contactor Column on the web.
163806
It dissociates the bond between the sodium and chlorine atoms. Technically, it ceases to be a "salt" and becomes an electrolyte.
172666
When a salt derived from a strong acid and a weak base dissolves in water, the solution becomes acidic. For example, consider the process\n \nNH4Cl(s) ----H2O ---> NH4+(aq) + Cl-(aq)\n\nThe Cl- ion has no affinity for H+. The ammonium ion NH4+ is  a weak conjugate acid of the base NH4+(aq) and ionizes as follows \n\nNH4+(aq) + H2O(l) <=> NH3(aq) + H3O+(aq)\n\nor simply  \n\nNH4+(aq) <=> NH3(aq)  + H+(aq)\n\nThe equilibrium constant ( or ionization constant) for this process is given by \n\n\n\nKa = Kw / kb  =   1.0*10^-14 / 1.8*10^-5 = 5.6*10^-10\n\n\n      x^2 / .10 =  5.6*10^-10\n\n           x^2  =   5.6*10^-11\n\n             x  =    7.5*10^-6 M  =  [H+]\n\npH  =  -log [H+]\n    \n    =  5.12\n\n     my answer is  C , NH4Cl
266008
Read up on the common ion effect,\nhttp://en.wikipedia.org/wiki/Common-ion_effect\nhttp://dbhs.wvusd.k12.ca.us/webdocs/Equilibrium/Common-ion-effect.html\n\nSince both HCl and NaCl when dissolved in water produce Chloride (Cl-) ions insolution, when adding HCl to an NaCl solution, you are pushing the equilibrium of the dissolved to undissolved NaCl more towards the undissolved side.\nNaCl (s) + H2O <----> Na+ (aq) + Cl- (aq)\nSo by increasing the concentration of the Cl- in solution by means of HCl, you are shifting the NaCl(s)<-->NaCL(aq) reaction towards the "left".
324126
The salt that is formed is Barium sulphate, which simply has low solubility in water (or acid or bases - you need a chelating agent for the Barium to dissolve it again). There is  no other significant reaction (like redox or altering pH).
833753
sodium, potassium, chloride, trace amounts of other metal ions.
796075
At 10°C, saturated solution of potassium chloride is more concentrated than potassium nitrate. At this temperature, potassium chloride completely dissolves while potassium nitrate would form some precipitates. At 60°C, it would be the same since both would completely dissolve at this temperature.
76599
it depends on the state. if it has a (aq) after it, that mean's it's aqueous and it is hydrochloric acid. if it is not aqueous and has a (l) (s) or (g) meaning liquid, solid, or gas, respectively, then it is just hydrogen chloride. If it doesn't specify and it is for homework or something, you should put "hydrogen chloride, unless it's aqueous, then it's hydrochloric acid"  The reason for this is because in aqueous form, the compound dissociates, and there are extra H+ ions which is what forms an acid, so that makes it hydrochloric acid... the same thing happens with hydrogen flouride (HF) becoming hydrofloric acid, as well as HI and HBr.
846516
Yes, it actually is that simple. Potassium bonds very well to both chloride and bicarbonate. Potassium has a +1 charge, while both chloride and bicarbonate have a -1 charge.\n\nCheck out this site for a detailed description of these three electrolytes and their effect on the body:\n\nhttp://www.medicinenet.com/electrolytes/article.htm
808333
You have a common ion effect problem\n\nPbCl2 <=> Pb+2 +2Cl-\n\nKsp=[Pb+2][Cl-]^2\nYou know that [Pb+2]=5.5*10^-3\nAlso from the reaction you know that the amount of Cl- coming from the PbCl2 is double the amount of Pb+2.\nSo if C is the concentration of Cl- coming from HCl, the amount of Cl- at equilibrium is [Cl-]=C+2*[Pb+2] \n\nSo Ksp= [Pb+2]* (C+2*[Pb+2])^2=>\nif we assume that C>>2[Pb+2] then C+2*[Pb+2] =C and \nKsp=[Pb+2]C^2 =>1.7*10^-5= 5.5*10^-3 C^2 =>\nC=55.6 * 10^-3 but 2*[Pb+2]=11*10^-3, so we can't do the approximation...\n\nLet's name [Pb+2]=p for simplicity\n\nKsp=p(C+2p)^2 =>\nKsp/p= C^2+4pC+4p^2 =>\nC^2+4pC+4p^2-Ksp/p=0 =>\nC^2+2.2*10^-3 C -2.97*10^-3 =0\n\nC=0.5*(-2.2*10^-3 + SQRT( (2.2*10^-3)^2 +4*2.97*10^-3) ) =>\nC= 5.34*10^-2\n\nIf you added Va L of HCl, then\nC=mole HCl/Vtotal = Ca*Va/(Va+0.12) =>\nCVa+0.12C=CaVa =>\nVa=0.12C/(Ca-C)= 0.12*0.0534/(0.12-0.0534) = 0.0962L or 96.2 ml
726370
There should be no reaction.  Provided there is enough water to dissolve the CuCl, the Compounds would separate into their Ion counterparts.\n\nSo, youd end up with Cu+, K+, and Cl- in solution
400328
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.
1025655
Think of the net ionic equation which is occuring.\nSince both the HCl and Ca(OH)3 are dissolved in aqueous solution, broken into ions, just focus on the H+ and OH- part of it first.\n\nH+(aq) + OH-(aq) --> H2O(l)\nThe Hydronium ion and the Hydroxide ion react to form water.\nWhat is left over are the Ca+2 and Cl- ion which drop out of the net ionic eqaution since they form the very soluble compound, Calcium Chloride (CaCl2).\n\n2 HCl (aq) + Ca(OH)2 (aq) --> 2 H2O (l) + CaCl2 (aq)