What is the pH of a 0.279M solution of hydrazine, N2H4, which has a Kb=1.3*10^-6?
832115
N2H4 + H2O -> N2H5+ + OH-\n\nKb = c[N2H5+] * c[OH-] / c[N2H4]\n\nAssuming that c[N2H5+] = c[OH-] and that c[N2H4] >> c[OH-] (so that we can use the starting concentration for c[N2H4]), we get approximately\n\nc[OH-] = sqrt( Kb * c[N2H4])\n\nor logarithmized\n\npOH = 1/2 ( pKb + 10^-(c[N2H4])\n\npH= 14 - pOH\n\n\ncalculator time.
858181
It would be 1.8N.  \n\nThree times in the case of Phosphoric Acid.
1122499
pH = pKa - log (.25/.4) ~= 4.3
1013801
Molar mass of NH4NO3 = 14+4+14+48 = 80\n\nSo, 1.4 g is equal to 1.4/80 = 0.0175 mol\n\nNow, mass of soln = 74+1.4 g = 75.4 g\n\nSo, heat absorbed = 75.4x4.18x(25-23.71) = 404.415 J\n\nSo, 0.0175 mol absorbs 404.415 J\n\nSo, 1 mol absorbs 23109.429 J\n\nSo, heat of dissolution of NH4NO3 is 23.109 kJ/mol (+ve as heat is absorbed)\n\nb) Lattice energy = 630+23.109 = 653.109 kJ/mol
651011
H2O2 + OC(NH2)2 -> HCOOH + NH4OH -> NH4(HCOO)
713454
The H3O+ concentration has to be the same as the acetate ion concentration, that is 0.425. Since you don't tell the units I cannot calculate the pH. Why do you need to know pKa of acid if you already know the concentration of the acetate ions? This is a chemistry question. Why don't you place it under chemistry?
663581
NH4NO2 --------------> N2 + 2H20 \n\nYOU NEED TO HEAT THE SOLUTION IN ORDER TO DECOMPOSE THE COMPOUND \n \nHOPE THIS HELPS
832121
Nope, :s sorry\n\nIts the same as before. The K(w) of water is 1.0 *10^-14\nK(w) = [H+][OH-]\n1.0*10^-14=[H+][2.8*10^-4]\n[H+]= 3.57*10^-11\n\npH=-log[H+]\n the pH is 10.4\n\nand pOH =-log[OH-] u have that in the ques\n\nHappy trails
692123
1. pH = -log [H+] which is the hydronium ion concentration.  [H+] = 10^-2.85 = 0.00141 M\n[OH-] = 1 x 10^-14/0.00141 = 7.09 x 10^-12\n\n2.  pH = -log 12 = -1.08\n\n3.  neutral water would be when [H+] = [OH-], so [H+] would be the square root of Kw, so [H+] = 1.58 x 10^-7.  pH = 6.80.
968707
NH3  -33.34 °C\nCH3F -78°C\nCH4 −161.6 °C\nHe -268.93 °C
452634
N2 +3H2 ---> 2NH3\n\nso, H2 Mwt is 2g/mol and NH3 Mwt is 34g/mol\nas per above eqn, 6g of H2 will give u 34g of NH3\nSo, 9.988g of H2 will give 56.6g of NH3\n\nYou got it?
633715
5. KHCrO4 ----------------> K+   +   HCrO4-\nSince KHCrO4 dissociates complteley,\n the molarity of HCrO4- = molarity of KHCrO4\n the molarity of HCrO4- = 0.05M\n\nNow HCrO4 ------------------>    H+    +     (CrO4)2-\nCalculate the concentration of H+ from the pH value\n[H+] = 10^(-pH)   = 10^(-3.80)  = 1.58 * 10^(-4)\nIt is clear the only source of H+ is HCrO4-, therefore the concentration of (CrO4)2- is the same as concentration of H+\n[(CrO4)2-]   =   [H+]   = 1.58 * 10^-4\n\nKa = {H+} * [(CrO4)2-] / [HCrO4-]\nKa = [1.58 * 10^(-4)]^2/0.05\nKa = 5.02 * 10^ (-7)\n\n6.       KOH----------------> K+    +    OH-\n         HClO4 -------------> H+    +    ClO4-\n(a) \n    Moles of KOH = Molarity * Volume = 0.200M * 0.030L\n    Moles of KOH = 0.006 moles\n    From the equation above, \n    Moles of OH- = moles of KOH = 0.006 moles\n    \n    Moles of HClO4 = Molarity * Volume = 0.150M * 0.030L\n    Moles of HClO4 = 0.0045 moles\n    From the equation above,\n    Moles of H+ = moles of HClO4  = 0.0045 moles\n\n     H+    +    OH-    ------------>   H2O\nWhen this reaction is finished, 0.0015 moles of OH- are left\n[OH-]  =  moles of remaining OH- / Total Volume\n[OH-]  =  0.0015 moles/ 0.06L (30ml + 30ml)\n[OH-]  =  0.025M\npOH = -log [OH-]\npOH = -log (0.025) = 1.60\npH = 14 - pOH  = 14 - 1.60 = 12.40 \n\nDo the same for parts b, c, d, and e\n\n7. I will assume that \n   H2S --------------------------> 2H+    +    S2-\n   Ksp of HgS = 2.0 * 10^(-53)\n   Ksp of FeS = 6.3 * 10^(-18)\n\nKsp of HgS = [Hg2+] * [S2-]\n[Hg2+] = 0.01 M\n[S2-] = x\n2.0 * 10^(-53) = x*0.01\nx = 2.0 * 10^(-51)\nFrom the above equation, for every 1 mole of S2- there are 2 moles of H+\ntherefore,\n[H+] = 2 * 2.0*10^(-51) = 4*10^(-51)\npH = -log (4*10^(-51)) = 50.39\n\nKsp of FeS = [Fe2+] * [S2-]\n[Fe2+] = 0.01M\n[S2-] = x\n6.3* 10^(-18) = x* 0.01\nx = 6.3 * 10^(-18)\nFrom the above equation, for every 1 mole of S2- there are 2 moles of H+\ntherefore,\n[H+] = 2 * 6.3 * 10^(-18) = 1.26*10^(-17)\npH = -log (1.26*10^(-17)) = 16.90\n\nthe pH range is from 16.90-50.39\nThis is the best explanation I can gibe. By the way, pH level beyond 0 and 14 do exist, but, I'm not 100% about this answer.\nI fixed it, I think it is correct now.
358982
NaOH is completely dissociated when it is dissolved in water.  A .0025M KOH solution therefore produces a [OH-] equal to .0025M.  The ion product of [H30+] and [OH-] is a constant, 1.0 x 10^-14 M^2.  By substitution, the [H30+] can be determined.  The pH can then be calculated.\n\n[H30+] = 1.0 x 10^-14 M^2/[OH-] = 1.0 x 10^-14 M^2/2.5 x 10^-3 \n\naaaand all that equals .4 x 10^-11... and you put that in correct scientific notation... 4 x 10^-12 M for the concentration of [H30+].\n\nWe then plug that in...   \n\npH= -log[H30+] = -log (4 x 10^-12) = 11.4\n\nThe pH of a .0025M solution of KOH is 11.4, which is greater than 7, making it a basic solution.\n\nStrong base at that.\n\nP.S.- If it is supposed to have the correct number of significant figures/digits, the answer would then be 11.
411083
If the pH is 4, the pOH is 10 (pH + pOH = 14).  10^(-pOH) = [OH-].\n\nAnswer is 1E-10
272078
I take it you mean the concentration of hydronium ions (H3O+) at a pH of 6.2. ok, the equation you need is this:\npH = -log [H+]\n\nTo solve, do this:\npH = -log [H+]\n-pH = log [H+]\n10 ^ (-pH) = [H+]   (if you don't understand this step, look up logarithms. Basically, a= log b means exactly the same thing as 10^(a)=b )\n\n10^(-6.2) = [H+]\nYou have 10^(-6.2) mol/ L of hydronium ions.\n\nAnother useful equation is this:\n[H30+] * [OH-] = 10^-14\nwhich will give you the concentration of OH- ions when you have the concentration of hydronium ions, or vice versa.
557422
First, let's calculate the pOH of the solution:\npOH = -log [OH-] \nBecause KOH is a strong base (very soluble and complete dissociation), the concentration of KOH is the same as the concentration of the OH- because all of the KOH becomes K+ and OH- ions.  \npOH = -log [OH-] = -log 4.8 x 10^-2 M = 1.32\n\nNow, let's calculate the pH.  \npOH + pH always adds up to 14 under normal conditions.  \n1.32 + pH = 14\npH=12.68\n\nNow, we can calculate the H+ concentration. \npH=12.68=-log[H+]\n12.68 = -log[H+]\n-12.68 = log[H+]\nTake the antilog of both sides, which is the same as ten raised to the both sides. \n10^(-12.68)= 10^(log[H+])\n10^(-12.68)= [H+]\n2.1 x 10^-13 M= [H+]
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.