In which solution is the [H+] in moles/liter greater than the [OH-]?
553835
Well to determine [H+] you use the equation that \npH=-log [H+] \nthus a solution with pH=3 will have a [H+] of 1x10^-3, \nand a solution with pH=8 will have a [H+] of 1x10^-8\nTo find [OH-] there are several equations you can use\n[OH-]=Kw/[H+], \nKw is the ionization constant of water which is equal to 1.0x10^-14\nThus you will have [OH-] of 1.0x10^-11 and 1.0x10^-6\nSo you answer will be that the solution with pH=3 will have a [H+]>[OH-]
802465
carboxylic acid, as represented by -(C=O)-OH. For more info, check the source.
952215
[H3O+][OH-] = 1.0 x 10^-14 M always at 25C\nso if [H3O+]= 5.0 x 10^-10M\n[OH-]= 2.0x10^-5
573821
I think this question is incomplete because the Ka of HF hasn't been provided.  The amount of H+ that is formed from HF can only be calculated from its Ka value.
679839
An acid has a higher concerntration of H+ ions in the solution. The PH scale is based on a logorithimic equation where a PH of 7 is neutral, andything below 7 is acidic and anything above 7 is basic. The conjugate acid is the H part of HX where the X- is the conjugate base. \n\nWikipedia has a great explination: http://en.wikipedia.org/wiki/Conjugate_acid
402204
H2SO4 - HCl - H3PO4 - HSO4(1-) - H2PO4(1-) - HPO4(2-)\n\nHCl is quite strong, but suffers the disadvantage at atmospheric pressure that it is volatile and will dissolve in water only up to 37%. It can be driven out of watery solutions by addition of less volatile but not necessarily stronger acids.\n\nH2CO3 would rank fairly high in this list too if you could prevent the CO2 from evaporating.\n\nWhen ranking H2SO4 and H3PO4 you'll have to state how far you want the protolysis to go. Taking the average of H2SO4 and HSO4(1-) makes it weaker than HCl, and worse so for H3PO4.
813100
ΔH =ΣΔHf°prod-ΣΔHf°react
48368
2HC2H3O2 + Ba(OH)2 --> Ba(C2H3O2)2 + 2H2O
651011
H2O2 + OC(NH2)2 -> HCOOH + NH4OH -> NH4(HCOO)
954955
HCl_aq + NaOH_aq -> NaCl_aq + H2O_l\n(H+)_aq + (Cl-)_aq + (Na+)_aq + (OH-)_aq -> (Na+)_aq + (Cl-)_aq + H2O_l\n(H+)_aq + (OH-)_aq -> H2O_l\n\nFor all the other equations, the (Cl-), (K+), (NO3-), (Na+) are spectator ions, and the n.i.e is the same:\n(H+)_aq + (OH-)_aq -> H2O_l
887006
pH= -log([H30+])\n\n\njust put the #'s in the above eqn.\n\nfor pOH, its pOH= -log([OH])\nand for pH given [OH], its pH= 14 - (-log([OH])
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.
262916
4 OH-(aq) --->  O2(g) + 2H2O(l) + 4e-
869137
2H3PO4 + 3Ba(OH)2 ---> 6H2O + Ba (PO4)2 +  2Ba
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.
290781
MgSO4(s) + 2HCL(aq) ---> H2SO4(aq) + MgCl2 (aq)
779566
17.13 mL = 0.01713 L\n\n0.01713 L x 0.3150 M = 0.005396 mol\n\nNaOH is a strong base, HCl is a strong acid.\nEach NaOH releases one hydroxide anion.\nEach HCl releases one proton.\n\nTherefore it takes 0.005396 mol of H+ to neutralize 0.005396 mol of -OH.\n\n0.005396 mol / 0.025 L = 0.2158 M\n\nThe concentration of the NaOH solution is 0.2158 M.