Nicki Minaj
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Basically I have a molecule that is basic because it forms a salt. The pka is 12.4

I am confused when I did the henderson I got that the molecule will become more ionised in the small intestine compared to the stomach. Yet I thought because it is basic it would actually become more ionised in the stomach than small intestine ? The two conuteracts? Image
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TSR Jessica
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Sorry you've not had any responses about this. Are you sure you've posted in the right place? Here's a link to our subject forum which should help get you more responses if you post there.


Just quoting in Danny Dorito so she can move the thread if needed :wizard:
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h3rmit
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(Original post by Nicki Minaj)
Basically I have a molecule that is basic because it forms a salt. The pka is 12.4

I am confused when I did the henderson I got that the molecule will become more ionised in the small intestine compared to the stomach. Yet I thought because it is basic it would actually become more ionised in the stomach than small intestine ? The two conuteracts? Image
Hello again. If it's basic, then giving the pKa doesn't really make sense. Do you mean the pKa of the conjugate acid?
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Nicki Minaj
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(Original post by h3rmit)
Hello again. If it's basic, then giving the pKa doesn't really make sense. Do you mean the pKa of the conjugate acid?
Salts are ionic, i.e. they dissociation 100% in solution.

Your initial compound is a base if it forms a hydrochloride salt.

For example: RNH2 + HCl --> RNH3+Cl-

In solution:

RNH3+Cl- --> RNH3+ (aq) + Cl-(aq)

This reaction lies 100% to the RHS

So what you are dealing with is the hydrolysis of the cation (i.e. it's interaction with water at various pH levels)


RNH3+ (aq) + H2O <==> RNH2 + OH-

This is an equilibrium for which you have the ka value. The higher the acidity the greater the concentration of H+ ions, which will remove the OH- ions from the above equilibrium reducing the availability of the soluble form (the cation)

So the lower the pH, the less of the ionised form exists.

hi dear do you agree with this explanation
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h3rmit
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(Original post by Nicki Minaj)
Salts are ionic, i.e. they dissociation 100% in solution.

Your initial compound is a base if it forms a hydrochloride salt.

For example: RNH2 + HCl --> RNH3+Cl-

In solution:

RNH3+Cl- --> RNH3+ (aq) + Cl-(aq)

This reaction lies 100% to the RHS

So what you are dealing with is the hydrolysis of the cation (i.e. it's interaction with water at various pH levels)


RNH3+ (aq) + H2O <==> RNH2 + OH-

This is an equilibrium for which you have the ka value. The higher the acidity the greater the concentration of H+ ions, which will remove the OH- ions from the above equilibrium reducing the availability of the soluble form (the cation)

So the lower the pH, the less of the ionised form exists.

hi dear do you agree with this explanation
My mistake, I didn't know you actually knew the salt it formed, I thought you were saying it was basic from the pKa value like in the last thread, which didn't really make sense.

I'm not sure I agree with your last equation: "
RNH3+ (aq) + H2O <==> RNH2 + OH-"

It doesn't balance and I'd be expecting the amine ion to protonate the water. But in this equation, both species are acting as acids which is weird.
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Nicki Minaj
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(Original post by h3rmit)
My mistake, I didn't know you actually knew the salt it formed, I thought you were saying it was basic from the pKa value like in the last thread, which didn't really make sense.

I'm not sure I agree with your last equation: "
RNH3+ (aq) + H2O <==> RNH2 + OH-"

It doesn't balance and I'd be expecting the amine ion to protonate the water. But in this equation, both species are acting as acids which is weird.
RNH3+ (aq) + H2O <==> RNH2+ H3O+

does it make sense now
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Nicki Minaj
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(Original post by h3rmit)
My mistake, I didn't know you actually knew the salt it formed, I thought you were saying it was basic from the pKa value like in the last thread, which didn't really make sense.

I'm not sure I agree with your last equation: "
RNH3+ (aq) + H2O <==> RNH2 + OH-"

It doesn't balance and I'd be expecting the amine ion to protonate the water. But in this equation, both species are acting as acids which is weird.
dear can you explain me now with the corrected equation of what is going on? haha im confused
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h3rmit
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(Original post by Nicki Minaj)
RNH3+ (aq) + H2O <==> RNH2+ H3O+

does it make sense now
Following this equation, if you add more protons, you get more H3O+, so the equlibrium would shift to the left to form more RNH3+ ,so you'd expect more of the ionised form to exist in the stomach.

The RNH3+ reacting with hydroxides could work, since you'd then get an acid base neutralisation that gives you more of the unionised molecule in the most acidic environments.
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Nicki Minaj
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(Original post by h3rmit)
Following this equation, if you add more protons, you get more H3O+, so the equlibrium would shift to the left to form more RNH3+ ,so you'd expect more of the ionised form to exist in the stomach.

The RNH3+ reacting with hydroxides could work, since you'd then get an acid base neutralisation that gives you more of the unionised molecule in the most acidic environments.
thank you dear. will this be correct explanation in my own words. sorry i just want to see if i understand it.


Drug D is a salt as the drug is a base due to it forming hydrochloride salt. It is ionic and undergoes dissociation 100% in solution. For example: RNH2 + HCl --> RNH3+Cl-.


In solution:
RNH3+Cl- --> RNH3+ (aq) + Cl-(aq)
This reaction lies to the right hand side of the equation as it is dealing with the hydrolysis of the cation (i.e. it's interaction with water at various pH levels).
RNH3+ (aq) + H2O <==> RNH2+ H3O+
The higher the acidity (in stomach) the greater the concentration of H+ ions causing a higher concentration of H3O+. The equilibrium shifts to the left to form more RNH3+. Thus, more of the unionised form will exist in the stomach.6 This is proven by using the Henderson-Hasselbalch equation show below.


At 1.4 (stomach):

1.4 = 12.4 + log (A-/HA)

log (A- / HA) = -11

ratio of A- to HA: 10-11[HA] : [A-]

At 6.5 (small intestine, duodenum):

6.5 = 12.4 + log (A- / HA)

log (A- / HA) = -5.9

ratio of A- to HA = 1.26x10-6 [HA] : [A-]
According to the calculation, more of the drug will become unionised in the stomach and this needs to be considered for this formulation. os.c�ͱ� :
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h3rmit
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(Original post by Nicki Minaj)
chemistry
Yep, that works.
In your ratio, it might be a good idea to specify what HA and A- are, just to round out your explanation.

Ignore the last sentence of my previous post, by the way.

Edit: you say "
The equilibrium shifts to the left to form more RNH3+. Thus, more of the unionised form will exist in the stomach."
I'm assuming this is a typo and you meant more of the ionised form will exist in the stomach.
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Nicki Minaj
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(Original post by h3rmit)
Yep, that works.
In your ratio, it might be a good idea to specify what HA and A- are, just to round out your explanation.

Ignore the last sentence of my previous post, by the way.

Edit: you say "
The equilibrium shifts to the left to form more RNH3+. Thus, more of the unionised form will exist in the stomach."
I'm assuming this is a typo and you meant more of the ionised form will exist in the stomach.
thank you Image


Drug D is a salt as the drug is a basic. It has a hydrochloride present which is ionic and undergoes dissociation (100%) in solution. For example: RNH2 + HCl --> RNH3+Cl-.
RNH3+Cl- --> RNH3+ (aq) + Cl-(aq)
This reaction lies to the right hand side of the equation as it is 100% dissociated in aqueous solution. The reaction that forms an equilibrium is the hydrolysis of the cation:
RNH3+ (aq) + H2O <==> RNH2+ H3O+
The higher the acidity (in stomach) the greater the concentration of H+ ions causing a higher concentration of H3O+. The equilibrium shifts to the left to form more RNH3+. Thus, more of the ionised form will exist in the stomach.

sound better?
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h3rmit
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(Original post by Nicki Minaj)
thank you Image


Drug D is a salt as the drug is a basic. It has a hydrochloride present which is ionic and undergoes dissociation (100%) in solution. For example: RNH2 + HCl --> RNH3+Cl-.
RNH3+Cl- --> RNH3+ (aq) + Cl-(aq)
This reaction lies to the right hand side of the equation as it is 100% dissociated in aqueous solution. The reaction that forms an equilibrium is the hydrolysis of the cation:
RNH3+ (aq) + H2O <==> RNH2+ H3O+
The higher the acidity (in stomach) the greater the concentration of H+ ions causing a higher concentration of H3O+. The equilibrium shifts to the left to form more RNH3+. Thus, more of the ionised form will exist in the stomach.

sound better?
Perfect
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Nicki Minaj
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(Original post by h3rmit)
Perfect
thank you dear and i wrote HA: unionised and A- means ionised?
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h3rmit
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(Original post by Nicki Minaj)
thank you dear and i wrote HA: unionised and A- means ionised?
The species might be better, so HA=RNH3+ and A-=RNH2
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Nicki Minaj
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(Original post by h3rmit)
The species might be better, so HA=RNH3+ and A-=RNH2
but dear im confused henderson showed that more of the drug will remain unionised in the stomach though?
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h3rmit
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(Original post by Nicki Minaj)
but dear im confused henderson showed that more of the drug will remain unionised in the stomach though?
The A- is the unionised molecule, the HA is the ion. It makes more sense if you specify the species.
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Nicki Minaj
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(Original post by h3rmit)
The A- is the unionised molecule, the HA is the ion. It makes more sense if you specify the species.
At 1.4 (stomach):

1.4 = 12.4 + log (A-/HA)

log (A- / HA) = -11

ratio of A- to HA: 10-11[HA] : [A-]

At 6.5 (small intestine, duodenum):

6.5 = 12.4 + log (A- / HA)

log (A- / HA) = -5.9

ratio of A- to HA = 1.26x10-6 [HA] : [A-]
According to the calculation, more of the drug will become unionised in the stomach and this needs to be considered for this formulation.

this shows unionised in stomach

but in the explanation above it states that more will become ionised in stomach
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h3rmit
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(Original post by Nicki Minaj)
At 1.4 (stomach):

1.4 = 12.4 + log (A-/HA)

log (A- / HA) = -11

ratio of A- to HA: 10-11[HA] : [A-]

At 6.5 (small intestine, duodenum):

6.5 = 12.4 + log (A- / HA)

log (A- / HA) = -5.9

ratio of A- to HA = 1.26x10-6 [HA] : [A-]
According to the calculation, more of the drug will become unionised in the stomach and this needs to be considered for this formulation.

this shows unionised in stomach

but in the explanation above it states that more will become ionised in stomach
Don't use HA and A-, use the actual species the drug forms.
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Nicki Minaj
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(Original post by h3rmit)
Don't use HA and A-, use the actual species the drug forms.
dear do you agree henderson shows more unionised in the stomach right?

but when we did that equilibrium equation it showed more of the drug will become ionised in stomach?its contradicting
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h3rmit
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(Original post by Nicki Minaj)
dear do you agree henderson shows more unionised in the stomach right?

but when we did that equilibrium equation it showed more of the drug will become ionised in stomach?its contradicting
No, because you know HA is the ion (RNH3+) and A- is the unionised molecule (RNH2)
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