no ,its pKa. kindly go through this paper,
http://pubs.rsc.org/en/content/articlelanding/1997/jm/a604311j/unauth#!divAbstract
Measure pH of one M pyridine solution.
It comes out to be= 9.17.
So [H+] = 6.67.10^-10 or [OH-1] =1.49.10^-5.
Py + H2O = PyH^+1 + OH^-1.
Kb = [PyH+] [OH-]/[Py].
(Neglet H2O)
[Py] = [PyH+1] = 1.
Kb = [OH-][OH-]=[OH-]^2
=(1.49.10^-5)2=0.171.10^-9 .
pKb = -log(Kb) = -log 0.171.10^-9.
which is =8.25.
The well known relation between pkb and the pka of conjugate acid –base pair is:
pk b + pka =14(pkW).
As pyridine and pyridinium ion form a conjugate Acid-Base pair.So:
pk (pyridine) + pk(pydinium ion)= 14 .
8.25+ pk (pyridinium ion) =14.
Therefore, pk (pyridinium ion) = 14-8.25=5.75.
What is the pKa (pK (pyridinium ion)) value of 2-Amino-6-methylpyridinium and 2-Amino-4-methylpyridinium
Dear Manohar Sehgal,
Could you please help me to calculate the pKa (pK (pyridinium ion)) value of 2-Amino-6-methylpyridinium and 2-Amino-4-methylpyridinium.
My young friend,
It is so simple but one needs to know the pH of the solution/s at any molar concentration (the simplest would be 1M).I am trying from the literature to know their pH values at any given molarity. But the easiest way will be that you prepare IM solution/s by the reported method/s and determine the pH value by a pH meter which is an ordinary instrument and should be available in your lab.
All you have to do is to simply write to me the value of pH of the compound/s.
Dear Sir
I will find the pH value of those two materials and then i will contact you. Thanks for your kindness
dear sir,
The pH of the both materials are measured for 1 mole solution. the pH value of 2-Amino-6-methylpyridine and 2-Amino-4-methylpyridine are 9.34 and 9.4 respectively. kindly help me to find the pK of pyridinium ion.
1. Pls. find below the values.
2. May I dare suggest you that in order to complete one series of your research problen, it will be in the fitness of the thing if you also find:
pkb values of 2 Amino,5methyl pyridine and 2,4; 2,5 and 2,6 dimethyl pyridines.
Then you have pkb values of six similarly substituted pyridine bases and study them wrt the unsubstituted pyridine. I can make some guess of your research ploblem.
[2-Amino-6-methylpyridine ]
pH= 9.34.
So [H+] = 4.571.10^-10 or [OH-1] =2.188.10^-5.
(2 A,6Mpy) + H2O = 2 A,6 Mpy H^+1 + OH^-1.
Kb = [2 A,6Mpy H+] [OH-]/[2 A,6Mpy].
(Neglet H2O)
[2 A,6 M py] = [2 A, 6 Mpy H+1] = 1.
Kb = [OH-][OH-]=[OH-]^2
=(2.188.10^-5)2=4.487.10^-10 .
pKb = -log(Kb) = -log 4.487.10^-10.
which is =9.448.
The well known relation between pkb and the pka of conjugate acid –base pair is:
pk b + pka =14(pkW).
As they form a conjugate Acid-Base pair.So:
pk (2 A,6Mpy) + pk(2 A,6MpyH+1)= 14 .
9.448+ pk (2 A, 6 MpyH+1) =14.
Therefore, pk (2 A,6 MpyH+1) = 14-9.448=4.652.
[2-Amino-4-methylpyridine ]
pH= 9.4.
So [H+] = 3.981.10^-10 or [OH-1] =2.512.10^-5.
(2 A,4Mpy) + H2O = 2 A,4 Mpy H^+1 + OH^-1.
Kb = [2 A, 4Mpy H+] [OH-]/ [2 A,4M py].
(Neglet H2O).
[2 A,4Mpy] = [2 A, 4 Mpy H+1] = 1.
Kb = [OH-][OH-]=[OH-]^2
=(2.512.10^-5)2=6.63.10^-10 .
pKb = -log(Kb) = -log 6.63.10^-10.
which is =9.178.
The well known relation between pkb and the pka of conjugate acid –base pair is:
pk b + pka =14(pkW).
As they form a conjugate Acid-Base pair.So:
pk (2 A,4Mpy) + pk(2 A,4MpyH+1)= 14 .
9.178+ pk (2 A,4MpyH+1) =14.
Therefore, pk (2 A,4Mpy H+1) = 14-9.178 =4.822.
Dear Sir,
Thanks for your kind help and also for your valuable guidance. Its very useful to carry out my research in the perfect direction.
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