pKa for the molecular form and the anion are different

Marvin 5.6.0.1

My molecule has three hydroxyl groups. In calculating the pKa for the molecular form (particle charge 0) and fully deprotonated forms (particle charge -3) are obtained by varying the numbers:
7.58 and 7.53
12.69 and 12.21
8.90 and 8.76
Why? It's reversible equilibrium?

Hi,

Please, send your structure.

Thanks.

Jozsi

 

Hi, Jozsi !
I make the first steps in Marvin, so wrong ...
After checking these values ​​were the same ...
Excuse me ...

AlexandrI

hi there,

i donot know how to calculate the pKa of the structure properly by Marvin.. please guide anyone.. i am calculating pka for the phenobarbital drug and its 7.2pka at 298K temperature. but this value is not coming from the marvin sketcher... so guide anyone...

Hi,

..but this value is not coming from the marvin sketcher 

Please let me know what short of  pKa value(s)  were calculated by you with the Marvin's sketcher.

Jozsi

I am getting pKa 8.14 and 11.8 highlighted on  the phenobarbital molecule when I click on the pKa value button. What does that mean? How should i get 7.2pKa ( standardly calculated ) ?

I have attached the Pics of pKa value that I am getting and also of LogD value attached with this Message.. Please guide...

Hi,

By definition, pK_a is equal to  –log₁₀K_a, where K_a
is the acid dissociation constant, and it is constant at given temperature and
pressure.

1)     
K_a= ([A^-][H⁺])/[HA]

2)     
pK_a=-log₁₀K_a

3)     
pH=-log₁₀[H⁺]



Combine eq. 1 and eq. 2

4)     
pK_a= -log₁₀K_a = -log₁₀ (([A^-][H⁺])/[HA])

5)     
put eq.
3 to eq. 4

pK_a =  -log₁₀ ([A^-]/[HA]) + (-log₁₀ ) = -log₁₀
([A^-]/[HA]) + pH

pK_a=pH-log₁₀ ([A^-]/[HA])      or      pK_a=pH+log₁₀ ([HA]/[A^-])

 

6)     
Rearrange
eq.5

pH-pK_a= log₁₀ ([A^-]/[HA]) or pK_a-pH = log₁₀ ([HA]/[A^-])

It
can be seen from eq. 6 that the pH is equal to pK_a at half-neutralization, when [A⁻]/[HA]
= 1.

 

With the help of pK_a one can determine the amount
(concentration / or percentage) of the differently dissociated forms of the
molecule, that is the microspecies distribution, at a given pH. ChemAxon’s
pKa calculation shows the major microspecies distribution in two charts as well.

See attachement.

Since phenobarbital is a multi-protic drug, it has pKa1 and pKa2 values
as well. See more about pKa of multiprotic compunds:

http://www.chemaxon.com/marvin/help/calculations/pKa.html

http://en.wikipedia.org/wiki/Acid_dissociation_constant

 

In your case, our calculated  pK_a1 and pK_a2 of phenobarbital
at 298K is 8.14 and 11.80, respectively.

 

Best wishes,

Viktoria

Hi,

And one more note:

The pKa=8.14 is the first and the pKa=11.8 is the second ionization constants according to the pKa predictor.

The pKa=7.2 is an expermeintel value provided by you.

 It means that the error of the  pKa prediction is : 8.14 - 7.2 = 0.94 pKa unit.

 Jozsi

And one more note:

The calculation of property predictions (such as logP and pK_a ) can be enhanced when experimental data are available for molecules that are similar to the target. It will make the calculations more accurate.

http://www.chemaxon.com/marvin/help/calculations/cxtrain.html

Regards,

Viktoria