What is the pH of a 0.115 M solution of Morphine if pKb is 5.79?
410035
good news...10.63 is correct. Now how to get there.  \nMorp + H20 ---> Morp+ + OH-\ninitial concentration of Morp=.115M  everything else is 0.\nEquilibrium concentration of Morp=.115M-X, H20 ignore, Morp+=X, OH-=X.   Set up your equilibrium equation.  Kb=[OH-][Morp+]/[Morp]  pKb=-log Kb  so Kb=1.62e-6.  1.62e-6=[X][X]/[.115-X]  We can ignore the X subtracted from .115 since it is very small when compared.  so solving for x we finally get X=4.32e-4 which is the OH- concentration.  pOH=-log OH-  pOH = 3.36.  pH = 14-pOH .....pH=10.63.
1009921
approximately 0.145mg/L
650541
pH = -log ([H3O+])\npOH = -log ([OH-])\n\npH + pOH = pKw = 14\npOH = 14 - pH\n\n[OH-] = 10^-(pOH)\n\nPlug in and solve.
652538
penicillin, morphine, codeine, insulin
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])
797573
pH 1.91 = .0123M\n\n     so, percent =  .0123M/.500M =.025 *100= 2.5%
378003
pH for a salt of strong base and weak acid is given by:\n         pH = 7 - (1/2)pKb - (1/2)log C\nwhere pKb = dissociation const of strong base\n       C  = concentration of Acid.\n\nKb = Kw / Ka = (10^-14) / (6.5 x 10^-5) = 1.53x10^-10\npKb = -log (1.53x10^-10) = 9.8154\nlog C = log 10^-1 = -1\n\npH = 7 - (1/2)x 9.8154 - (1/2)x-1\n   = 7 - 4.9077 + 0.5\n   = 2.5923
933864
to what?  Are you looking for the molecular weight of the monoprotic acid?  If so, remember M (molarity) is in moles/lter) and your colume of NaOH is in mL so you'ww want to convert that to liters.\n\n\n.01524 liters x .1578 moles/liter = 0.002405 moles\n\nso therefore you have 3.567  g / 0.002405 moles = 1483 gram/mole\n\nuse dimensional analysis and keep track pf the units to find the answer
950743
theoretical amount of product=10g lead acetate/ mw lead acetate x mw lead iodide\n\ntake the result of this formula as the denominator and divide 5.00 grams product/theoretical x 100%
560197
These are the two equations you need to use.\nBP Elevation: Delta T = i(Kb)m\nFP Depression: Delta T = i(Kf)m\n\nThe Molarity should be the moles of solute over L of solution.  Given room temperature the Density of ethanol is 0.789g/ml and the Density of benzene is .8786g/ml.\n\nOnce the L of the solvent are found, you can use the data given and the equations above to figure out the moles dissolved in each question.  Then convert the moles found to each of solute's mass using Molar masses of each solute.  \n\nNote that van 't Hoff factor is considered 1 (and omitted) in the calculations, since neither ethanol or pyrazole ionizes extensively.\n\nPROBLEM 1:\n\nSince: BP elevation = Delta T = i(Kb)m\n     : Density of Ethanol = 0.789g/ml\n\n78.1g(1ml/.0.789g)(1L/1000ml)= 0.5133L of ethanol.\n\n4.00 = [1.20(x)]/98.98\nx= 0.3299mol benzonic acid   \nMolar Mass benzonic acid = 122.12g/mol\n\nTherefore 0.3299mol(122.12g/mol) = 40.27g of Benzoic Acid must be added.\n\n\nPROBLEM 2:\n\nSince: FP Depression = Delta T = i(Kf)m\n     : Density of benzene = 8786g/ml\n451g(1ml/.8786g)(1L/1000ml)= 0.5133L of benzene.\n\n5.00 = [5.12(x)]/0.5133L\nx= 0.501mol pyrazole   \nMolar Mass pyrazole = 68.07g/mol\n\nTherefore 0.501mol(68.07g/mol) = 34.10g pyrazole must be added.
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.
866705
To solve this question we need to know the specific heat of the resulting KI solution.  To make things easier, we can probably assume that the specific heat of the solution is the same as the specific heat of water (c = 4.184 J/g degree C).\n\nThe energy (Q) released when a given mass, m, cools a given number of degrees C, (delta T), with a known specific heat, c, is:\nQ = mc(delta T)\n\nWe know the mass of the water, we know its specific heat, we know how many degrees its temperature changed, now just plug in and get an answer,\nQ = (60 g)(4.184 J/g degree C)(14 degree C)\nQ = 3514.56 Joules\nQ = 3.51 kJ\n\nBut this amount of heat energy was due to 28.7 grams of KI dissolving, and we want to know the heat given off by 1 mole, so we need to convert 28.7 grams of KI into moles of KI using the molar mass (166 g/mol).\n28.7 g KI / 166 g/mol KI = .173 moles KI\n\nSo if 3.51 kJ of energy was absorbed when .173 moles of KI dissolved, if one mole of KI dissolves, 20.3 kJ would be absorbed....the answer which you were looking for.\n\nBy the way....\nthe "(triangle) H" you are referring to is the change in the enthalpy of the reaction.  The "triangle" is the Greek letter "Delta".  It is more common to see it written out as "delta" or "change in" if one cannot use the actual Greek alphabet in typing, as opposed to "triangle".  Δ
986627
Divide the mass (in grams) by the molar mass of PbCrO4 (323.19 g/mol).
968320
here, you just use the ideal gas law, PV=nRT.  since molarity is equal to moles of solute per liters of solution, M=n/V, so P=MRT.\ntherefore,\nKp=Pco*Ph2o/(Ph2*Pco2)\n  =[CO]RT*[H2O]RT/ ([H2]RT*[CO2]RT)\nall the RT coefficients end up canceling out, so:\nKp=[CO][H2O]/[H2][CO2]\n  =Kc\nsince there were equal numbers of gas molecules on each side of the equation, Kp was equal to Kc.  as a general rule, Kp=Kc*(RT)^N, where N is the change in the number of gas molecules.
931815
Your KCl example is a good one.  A 0.5 molar would have 1/2 mole/liter or 37.75g KCl in a liter. A 5% w/v KCl would be 50 g/liter; so the 5% is more concentrated.\n\nThe 2.5% w/v sodium bisulfate is 25g/liter and the .025M is 3 g/liter.\n\nIt all depends on the molecular mass.
57350
Well for your information i would like to tell you that the pH value of lemonn is 2.5. Still if you wold like to find out, you could use either pH strips which indicate a pH  range to you or use a pH meter. This will give you the exact pH value. All the best!!
261269
The pH of Pepto Bismol is 5.81 .