Calculate the pH of a 0.115 M solution of Morphine if pKb is 5.79?
443327
The p in front of the Kb stands for -log, so:\n-log Kb = 5.79\nlog Kb = -5.79\nRaise ten to both sides (to cancel out the log):\n10^log Kb = 10^(-5.79)\nKb= 1.62E-6\n\nNow we can set up a Kb expression:\nKb = [OH-][C+]/[morphine]\nor\n1.62E-6 = x^2/(.115 M - x)\nsolve for x (I just graphed it on my TI-83+ calculator, set it equal to zero, and used the zero function)\nx=4.31E-4 M = [OH-]\n \nNow we can calculate the pOH\npOH= -log[OH-] = -log 4.31E-4 = 3.37 \n\nUnder standard conditions, pOH and pH add up to 14.00.\npH + pOH = 14.00\npH + 3.37 = 14.00\npH = 14.00 - 3.37 = 10.63, the answer!
251343
(5.5 grams/1 Liter)x(molar mass of glucose, arrange it so the grams cancel) = moles/Liter of glucose
736124
The answer to the first part:\n1) You need to find the pKa ---> use pKa+pKb = 14\nso 14-pKb = pKa\n14-4.75 = 9.25\n\n2) work out the final Molarity of each\n[NH3] = 0.400M/0.480L = 0.833M\n[NH4Cl] = 0.100M/0.520L = 0.192M\n\n3) now use the henderson-hasselbach equation\npH = pKa + log([base]/[acid])\n= 9.25 + log([0.833]/[0.192])\n= 9.89
404275
divide 200g by the molecular weight of KCL and then divide that by .8
639146
It will be the pH of the sodium hypochlorite formed.  Determine that amount, look up the Ka of the acid and solve.
781593
I always have to derive the equations from scratch:\n\nKa = [H+][A-]/[HA]\n\nlog(Ka) = log(H+) + log(A-) - log(HA), multiple both sides by -1\n\npKa = pH + log(HA) - log(A)\n\npH = pKa + log(A-/HA)\n\n4.54 = 4.74 + log(A-/HA)\n\n-0.20 = log(A-/HA)\n\n10^(-0.20) = A-/HA = 0.631\n\nplease check for errors
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%
179716
Here are the equations you need to know:\n\nMolarity (M) = moles/Liters\n\nc=(t 1/2)(2k); \nor concentration  = half life (t 1/2) * 2 * rate constant (k)\n\nUsing this equation, you can solve for the rate constant, k.\n\nThen you can plug that value into the equation for a first order rate law:   \n\ncf=c1-kt; \nor final concentration = initial concentration - rate constant * time\n\nGood luck with the calculations!
695200
That's what I got (2.282e-5), so pKa = -logKa = 4.64.  The published value for pKa is 4.9:\nhttp://www.chem.ucalgary.ca/courses/351/Carey/Ch19/ch19-1.html
1073643
the easiest way is to use a pH paper to test the pH. and just compare the reading with the one given with the test kit. \n\nOr you may do some neutralisation. And then calculate with the  amount of alkali you use for neutralisation. By that you can calculate the concentration of hydroxide ions in the alkali that you use. Then calculate the pOH for the alkali you use. pH=14 - pOH.\neg. \n1mol of NaOH gives you 1 mole of OH ions.\nso the pOH = -log1 = 0\npH = 14 - 0 = 14\n\nJust some rough guide. Hopefully it helps...Good luck.
1140664
You're missing a unit. Do you mean 2.5 kg solution?
950979
pH = - log[H+], so [H+] = 10^(-pH)\n\nfor pH = 7, [H+] = 10^(-7)
759238
just look at your units..that normally hleps\n\nhf ...kj / mole\n3.22 kJ / 6.02 Kj/mol =    x moles\nxmoles * 18 gram / mol.....  Thats your answer
397741
by the simplified version of the Henderson hasselbach equation\n\npH=pka + log(A-/HA)\n\npH=-log(Ka) + log (1) = -log(Ka) + 0 = -log(1.8x10^-4)= 3.77\n\n- 2 sigfigs! Since when you take the log you should have as many decimal places as numbers you started with.\n\n-By consequence, this is at the point known as the Equivalence point.
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.
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.
105947
0.5%solution means 500mg.( 1/2gms )of solute in one liter of solvent.