a 1.00 M solutions of HCL dissociates 100%, what is the concentration of hyrdronium ion in the solution?
489184
I think they're both 1.ooM.  \n\n          HCL -> H+ + Cl-\n         1mole   1mole  1mole\nUsing mole ration, the hydronium and chloride ions will both be 1 M concentrations. :)\n\nI think that's right.
853370
The concept of 1.5M hydrogen gas is rather odd, but, yes, you use the Nernst Equation.  Don't forget to convert the pH to a molarity of H+ ions.
1136961
Are you preparing a buffer soln with NaHCO3 and NaCl ? Then take equimolar amounts of both and dissolve them in 100 ml of water and then dilute them to the desired strength.
9807
Well lots of chemistry questions lately....finals coming up? :)  Good Luck!!!  \n\nTo calculate this type of question first of all calculate the #of moles of HCl used.  So \nyou used 2100 mL = 2.1 L HCl \nThis HCl is 0.3 mol/L so you used:\n\n2.1 L HCl* (0.3 mol HCl/L) = 0.63 mol HCl\n\nnow use the ratio of Cl/HCl  (which is 1 to 1)\n\n0.63 mol HCL *(1mol Cl/1mol HCl)= 0.63 mol Cl ion\n\nSo far so good I hope.  \n\nNow do your solution diltion to calculate volume,\n2.1L + .2385L = 2.3385 L of mixed solution\n\nNow simply divide \n0.63 mol Cl/2.3385 L solution = 0.269 mol Cl ion in final solution.\n\nAgain these numbers are quick and dirty but the method of solving this type of problem is the same.
797573
pH 1.91 = .0123M\n\n     so, percent =  .0123M/.500M =.025 *100= 2.5%
967201
Write the dissociation reaction:\n\nAg2Co3(s) <--> 2Ag+1(aq) + CO3-2(aq)\n\nSince this is a solubility equilibrium, the solubility product expression is just the equilibrium expression.\n\nKsp=[Ag2]^2[CO3]\n\nTo find the actual value, you'll need to look it up in a table.\nBTW, it's 8.1*10^-12
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?
350659
Why do people answer complete and total bullcrap answers?\n\n\nThe only thing they were right about was it is a weak acid, and even that bears no relavance to its P.H only you use a diferent formula in calculating it\n\n\nthe question is incomplete, you need to know the molarity of the acid and then you get the square root of it disociation constant (Ka) multiplied by its molarity\n\nand then you -log(base10)the answer you got from the first part and that is the P.H\n\nin short  Sq root(Ka*molarity acid)= H+ concentration\n\nP.H= -logH+
1061977
(H+) + NH3 -> NH4+\n It makes an ammonium ion, if you are talking about a hydrogen ion and not H2.
757438
CaC2 + 2 H2O ---> Ca(OH)2 + HCCH
73319
the earlier equation is not balanced.\nCaO + 2HCl --------> CaCl2 + H2O
620217
HOCl <==========>  H+  + OCl-\n\nthe pKa of hypochlorous acid, HClO, is 7.4 at 25°C, so:\n\npKa = -log [Ka] = 7.4, therefore\n\nKa = 3.98 x 10^-8\n\nKa = [H+]*[OCl-]/[HOCl] = 3.98 x 10^-8\n\nfrom your problem, you state that you have 1.0M HOCl, and since the only source for ions is the HOCl dissociating;\n\n[H+] = [OCl-] = y, and [HOCl] = 1.0-y\n\nsubstituting this back into the Ka equation, we get:\n\n3.98 x 10^-8 = y^2/(1.0-y)\n\nwhich converts into the quadratic equation:\n\n0 = y^2 + (3.98x10^-8)y - 3.98x10^-8\n\nI leave it to you to solve the quadratic equation to get y = [H+], and then calculate the pH of the starting solution
16164
The instruction is not correct because the total volume yielded when you mix 100ml of concentrated solution and 900ml water will be somewhat less than 1000ml. The concentration will be slightly higher than 1.0M\n\nThe correct procedure is to take a 1l volumetric flask  and add the concentrated solution to about 800ml water, then fill up to the mark.
290781
MgSO4(s) + 2HCL(aq) ---> H2SO4(aq) + MgCl2 (aq)
162250
1. \nMoles of Ca(HCO3)2 produced = 3.05 x 10^10/Molecular Mass of Ca(HCO3)2\n=3.05 x 10^10/162\n=0.018 x 10^10\n\nFrom the equation we know on dissolving 1 mole of limestone we get 1 mole of Ca(HCO3)2 \nThus, if 0.018 x 10^10 moles of Ca(HCO3)2 are produced moles of limestone dissovled should be 0.018 x 10^10\nWeight of limestone dissolved = Moles x its Molecular Mass of limestone\n=(0.018 x 10^10) x(100)\n=1.8 x 10^10 gm.\n\n2.\na.) CO + 2H2 --> CH3OH\n\nb.)Moles of methanol to be produced = 45/Molecular Mass of CH3OH\n=45/32\n=1.40625\n\nFrom the equation we know 2 moles of H2 are required to produced 1 mole of methanol\nThus to produce 1.40625 moles of H2 required = 2*1.40625\nWeight of H2 required = Moles X Molecular mass of H2\n=(2 x 1.40625) x 2\n=5.625 gm
30238
No, HCL has nothing to do with the number of moles of CaCO3 that you used.  So you asume you used enough HCl so that all of your CaCO3 is dissolved.  All you need to do is devide 0.5215 g  with the molar weight of CaCO3 and you'll get your number of moles.\n\nHope this helps,\nUltimeaciax
916567
STP is 1 atm, 273 K.\n\nmoles of HCl = PV/RT = 1 atm * 1.1 L/ (0.08206 * 273K) = 0.049 moles\n\nSince HCl and NaOH is a 1:1 reaction, 1 mole HCl gives 1 mole NaCl, so you get 0.049 moles NaCl (MW 58.5), or 2.87g.