Calculate the concentrations of various species of .10M hypochlorous acid, for which Ka= 3.5x10^-4.?
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.
1185900
H2SO4 + 2KOH --> K2SO4 + 2H2O
864626
1) Calculate the initial concentration of the H+.\npH = - log [H+] = 2, therefore  [H+] = 10 ^ -2 mol / liter\n\n2) Calculate the dilution factor.\nWe want to change the concentration from 10 ^ -2 mol / liter to\n10 ^ -4 mol / liter. Therefore the dilution factor is 100.\n\n3) Calculate the amount of water to add.\nThe dilution factor is 100. We therefore want to have 100 x 14.5 ml = 1450 ml after dilution. We therefore have to add 1450 - 14.5 = 1435 ml of water
850425
KOH + HNO3 ---> KNO3+ H20\n\nmoles of KOH = 0.025L*0.2mol/L = 0.005 mol\nconcentration of nitric acid solution = 0.005mol/0.030L = 0.167M\nconcentration of KNO3 = 0.005mol/0.055L = 0.091M
160963
K2SO4 ---> 2K+  +  SO4-\n\nHCl ---> H+  +  Cl-\n\nNaHCO3 ---> Na+  +  HCO3-\n\nNH4OH ---> NH4+  +  OH-\n\nBaCl2 ---> Ba-2  +  2CL-\n\n(NH4)3PO4 ---> 3NH4+  +  PO4-3
737051
2HCl + Ca(OH)2 --> CaCl2 + 2H2O\n\n39.9 mL x L / 1000 mL x .466 mol / L Ca(OH)2 x 2 mol HCl / mol Ca(OH)2 x L / .127 mol x 1000 mL / L = 292.81 mL HCl
482555
AgBr-------->Ag+ + Br-\n(a) Ksp=[Ag][Br] solubility=s=[Ag]=[Br]=Ksp^0.5\ns=(6*10^-7)^0.5\n(b) [Br]=0.01 (because all of kBr was dissuciated)\ns=Ksp/[Br]=(6*10^-7)/0.01 mol/lit
231663
_KCl3 _KCL + _o2? K2Cl4o2 thats the answer i think.:)
971190
Molarity is defined as number of moles in 1 liter.\nYou have 0.01 mol and 10 liters\nM = 0.01 / 10 mol/liter - 0.001 mol/liter = 0.001 M.  That's it.  You did it.\n\nNow pH = - log[H+].\n\nIf you write the equation\nHCl ---> H+ + Cl-\nThis means that you have the same molarity for HCl and H(+).  For both of them the coefficient in the chemical equation is equal to 1.  For other problems the coefficients may be different so always look at the balanced equation.\n\nso M(HCl) = M(H+) = 0.001.\n\nPutting this value into the pH equation:\n\npH = - log 0.001 = 3\n\npH + pOH = 14.\nCalculate the pH according to the pH definition and then use the above equation to find the pOH.
295875
for each one you are given a ratio i am assuming you are combining them both, make sure you write out a balanced equation for each one to make sure, some may not be a 1 to 1 ratio, next Molarity=mol/L therfore 1M = 1 mol per liter, use your periodic table to figure out the molecular weight for each one, each one of these is a strong acid, strong base reaction, therefore they will create water as a product and a salt, in the first one you are given 1.0 mol of HCl which is reacting w/ NaOH 1 to 1, therefore you only need to convert to # of moles of NaOH i am assuming you can do this hopefully this helps
639844
[H3O+]=1 x 10^-4 => pH=-log[H+]=4 --- acidic\n[OH-]=1X10^-6 => pOH=-log[OH-]=6 => pH=14-6=8 --- basic\n\nHCl --> H+ + Cl-\n[H+]= 5.6 x 10^-4\npH = -log[H+] = -log[5.6 x 10^-4] = 3.25 --- acidic\npOH = 14 - 3.25 = 10.75\n-log[OH-] = 10.75 => [OH-]=1.78 x 10^-11\n\nSince HBr --> H+ + Br-\npH of 1 x 10^-4 of HBr = -log[1x10^-4] = 4 --- acidic\n\nSince Ca(OH)2 --> Ca2+ + 2OH-\npOH = -log[OH-] = -log(2 * 5 x 10^-4) = 3\npH = 14-3 = 11 --- basic
382458
H2 + Cl2 -> 2HCl\n2Ag + H2S -> Ag2S + H2 \nCH4 + 2O2 -> CO2 + 2H2O\n2HgO -> 2Hg + O2\n3Fe + 4H2O -> Fe3O4 + 4H2
490327
solution A has a hydrogen ion concentration of 10^-5\nsolution B has a hydrogen ion concentration of 10^-3\nAll you have to do is 10^-ph\n\nIs this homework?
883220
H2+O2-->H2O\nNaOH+HCl-->NaCl+H2O\nKOH-->(on heating,deomposition)K2O+H2O\ni guess this is what u want\nand i f u didn't get it i am sorry......
850392
You have to use the methods for equilibrium and pH of a salt to calculate the concentration of salt that would produce that pH.\nI cannot find my notes from chemistry right now,\nhowever you need to know the definition of Ka(or Kb)\nthen set up the equation as the salt reacts.\n\nnh4+ + h20 <--> nh3 + h30+\n\npH = -log[H30+}, so your [h30+] = 10^-pH\nyou can then calculate the concentration of nh4 using molar ratios and such, and then you will have molarity.
821226
1. sodium hypochlorite\n2. hydrogen chloride\n3. concentrated sulphuric acid\n4. oxidising it, for example with potassium manganate(VII).\n\nIf you've done the other 96, well done!
860365
use the equation  pH= -lgKa+ lg( salt/acid)\n(this equation only applies for buffer system)