What happens when HCl is mixed with KCl 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.
270139
If you mean conductometric TITRATION, it won't work.  Assuming they're both at the same concentration, you'll get a flat line.  Conductance is a rough measure of the ion concetration.  The start (H+ + Cl-) has the same number as equivalence (Na+ + Cl-).  Adding more (Na+ + OH-) past equivalence will not change the concentration either.
804437
well I think no. because dilute HCl has a low concentration therefore it produces less hydrogen ions. so the less hydrogen ions it produces the less pH  it has
385728
mix potassium chloride with chlorine \n\nlol chemical weapons
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)
511983
Salts dissolve in water. The following reaction occurs in water.\n     NH4ClO4 + water ======> NH4+ (aq) and ClO4- (aq)\n\nCalculatin the pH is a little long, but, follow the followin steps\n       \n       ClO4-  +   H2O  ======>  HClO4  +   OH-\nNow write the Kb for this equation as\n              Kb = [HClO4] * [OH-] / [ClO4-]\nYou must be given the Kb value for this or you can calculate it as Kb = 1/Ka where Ka is for HClO4. Assume [HClO4] = [OH-] = x and [ClO4] = 0.4\nNow solve for x by plugging into the Kb equation. This will give you the [OH-]. From this value you can calculate pOH and then pH as pH = 14- pOH
386784
Well first of all you have to specify if the H's you want to account for are either comming from HCl alone or both HCl and some H's comming from the dissociation of water.\n\nI suppose that only the number of moles of H's comming from HCl is really worth while working the question for, since the H's of the dissociated water does not change the pH of the 0.10M solution.\n\nTherefore:\n\n0.10moles are in 1000mL\n? moles are in 50mL\n\n50 x 0.1 / 1000 = 0.005moles\n\nTherefore we can say that 0.005moles of HCl in 50mL of water results in a concentration of 0.10M
746627
Cl2 + 2KI --> 2KCl + I2\n\nsingle replacement
850929
The HCl will dissociate into its constituent ions H+ and Cl-  Zinc also  exists as ions Zn2+ or Zn3+ and it will combine with Cloride to make , Zinc Chloride.
82586
It has a cation or anion that isn't soluble. Stuff like NaCl won't work, because Na and Cl are always soluble. KBr has the same problem. But stuff like NaOH (the OH forms insoluble metal hydroxides) works well.\n\nBasically, you have to look at the solution and the precipitating agent and see if they'll form an insoluble compound.
231663
_KCl3 _KCL + _o2? K2Cl4o2 thats the answer i think.:)
355889
There are three species in attendance in this solution, H+ and ClO- ions (that come from the disassociation of hypochlorous acid) and hypochlorous acid itself. \n\nThe Ka value comes from the following equation: \nKa= [H+]*[ClO-]/[HClO]\n(the brackets stand for concentration)\n\nWe can set [H+] to the variable x, and because for each mole of H+ formed, one mole of ClO- is formed, [ClO-] is also x.  The concentration of HClO is initially .10M, and because x amount of HClO splits up into the two ions H+ and ClO-, the final concentration of HClO that we will be using is .10M-x.  \n\nThat gives us this following equation:  \nKa=x*x/(.10M-x) \nor \nKa=x^2/(.10M-x)\nand because Ka=3.5x10^-4, \n3.5x10^-4=x^2/(.10M-x)\n\nNow you can algebraically solve for the value of x, or use a graphing utility to do it for you.  \nCross-multiply:\n3.5x10^-4(.10M-x)=x^2\nNow distribute:\n3.5x10^-5 - 3.5x10^-4x = x^2\nand set it equal to zero:\n0=x^2 + 3.5x10^-4x - 3.5x10^-5\nand use the quadratic equation to solve for x:\nx= -3.5x10^-4 + sqrt((3.5x10^-4)^2 - (4*1*-3.5x10^-5)) all divided by 2*1\nx=.0057 \n\nSo the [H+] = [ClO-] = x = .0057 M.  \nThe HClO concentration is .10M-x=.094M.
573222
Once a solution of HCl hits pH 7, it is considered neutralized (or in other words it'll basically be water.) After a pH of 7, a solution has to be alkaline (or basic,) which an acid cannot be.
397908
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.
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
894546
You did not put the volume of HCl to neutralize so we will do the neutralization to pH = 7 only.\n\nmoles of HCl = 35 0 g/l / 36.46 g/mole = 9.60 mol/l\nmoles of NaOH = 30 g/l / 40 g/mol = 0.75 mol/l\nOne mole of HCl will neutralize one mole of NaOH.\n\n9.60 * Vol(HCl) = 0.75 * V(NaOH).\n\nOnce you know if you have any NaOH left you can calculate how much NaOH to add to bring it to pH = 8 or 9\n\npH = 8 =====> pOH = 6  (pOH = 14 - pH)\nthen 6 = - log [OH]\nCalculate OH from last equation and this will give you the moles of NaOH you need to add to bring to pH = 8.  Same for pH = 9.
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.