April 01, 2023, 07:43:57 PM
Forum Rules: Read This Before Posting

### Topic: pH and ionization of drugs  (Read 2804 times)

0 Members and 1 Guest are viewing this topic.

#### Luckenberg99

• Regular Member
• Posts: 14
• Mole Snacks: +0/-0
##### pH and ionization of drugs
« on: March 05, 2023, 12:02:20 PM »
Hi all, I have a little doubt.
Suppose we have a buffer solution HA/A- at pH=2 and we insert an acid drug HD with pKa = 5. From the Henderson-Hasselbalch it is possible to obtain all the relative quantities of the acid and conjugate base of the drug present in the buffer :
pH=pKa + log ([D-]\[HD]) <=> [HD] = 1000*[D-], so D- is practically negligible.

However, referring to the classic exercises on buffers, in theory adding the HD acid drug to the buffer it should react with A- of the buffer itself and, thinking in terms of moles, the HD acid drug is completely consumed and converted into the corresponding base -:
HD + A- --> HA + D-

So, in solution it would remain D-. So how does HD get 1000 times higher than D-?

#### Borek

• Mr. pH
• Deity Member
• Posts: 27304
• Mole Snacks: +1772/-408
• Gender:
• I am known to be occasionally wrong.
##### Re: pH and ionization of drugs
« Reply #1 on: March 05, 2023, 01:36:03 PM »
(sorry, misunderstood the question initially).

No, it is not like HD should react with the D-. Whether it does depends on the pKa values. Your example is a bit off, as at pH=2 typical buffer (one based on the pKa=2) won't work, still, it will need an acid much stronger than the drug is.

Plus, in typical situations therapeutic concentration of the drug is negligible compared to the buffer capacity.
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info, pH-meter.info

#### Luckenberg99

• Regular Member
• Posts: 14
• Mole Snacks: +0/-0
##### Re: pH and ionization of drugs
« Reply #2 on: March 05, 2023, 01:46:03 PM »
(sorry, misunderstood the question initially).

No, it is not like HD should react with the D-. Whether it does depends on the pKa values. Your example is a bit off, as at pH=2 typical buffer (one based on the pKa=2) won't work, still, it will need an acid much stronger than the drug is.

Plus, in typical situations therapeutic concentration of the drug is negligible compared to the buffer capacity.

so what you mean is that HD is 1000 times D- because it reacts minimally with A-? Is the reaction HD +A- --> HA + D- strongly shifted to the left according to pKas (2 and 5)?

Maybe a Better example:
buffer of acetic acid/acetate at pH=pKa= 4.75.
From Henderson Hasselbalch we obtain that [Asp-]=18[AspH]
Does this come from the fact that aspirin is a stronger acid than acetic acid, so the reaction AcO- + AspH <=> AcOH + Asp- is shifted to the right to some extent, generating these concentrations?
« Last Edit: March 05, 2023, 02:14:53 PM by Luckenberg99 »

#### Borek

• Mr. pH
• Deity Member
• Posts: 27304
• Mole Snacks: +1772/-408
• Gender:
• I am known to be occasionally wrong.
##### Re: pH and ionization of drugs
« Reply #3 on: March 06, 2023, 03:20:28 AM »
so what you mean is that HD is 1000 times D- because it reacts minimally with A-? Is the reaction HD +A- --> HA + D- strongly shifted to the left according to pKas (2 and 5)?

You can easily derive K for this reaction from both Ka values, and it will tell you what kind of equilibrium to expect.

Quote
Maybe a Better example:
buffer of acetic acid/acetate at pH=pKa= 4.75.
From Henderson Hasselbalch we obtain that [Asp-]=18[AspH]
Does this come from the fact that aspirin is a stronger acid than acetic acid, so the reaction AcO- + AspH <=> AcOH + Asp- is shifted to the right to some extent, generating these concentrations?

Again: you can't discuss what is happening in a meaningful way ignoring relative concentrations of the buffer and the aspirin.

Addition of every acid to the buffer lowers the pH. By how much depends on how strong the acid is and how much is added, relative to the amount of buffer present. In typical therapeutic situation amount of drug is negligible compared to the buffer concentration, so you can safely assume pH to be constant and be the driving force behind the HA/D- ratio.
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info, pH-meter.info

#### Luckenberg99

• Regular Member
• Posts: 14
• Mole Snacks: +0/-0
##### Re: pH and ionization of drugs
« Reply #4 on: March 06, 2023, 07:19:27 AM »
so what you mean is that HD is 1000 times D- because it reacts minimally with A-? Is the reaction HD +A- --> HA + D- strongly shifted to the left according to pKas (2 and 5)?

You can easily derive K for this reaction from both Ka values, and it will tell you what kind of equilibrium to expect.

Quote
Maybe a Better example:
buffer of acetic acid/acetate at pH=pKa= 4.75.
From Henderson Hasselbalch we obtain that [Asp-]=18[AspH]
Does this come from the fact that aspirin is a stronger acid than acetic acid, so the reaction AcO- + AspH <=> AcOH + Asp- is shifted to the right to some extent, generating these concentrations?

Again: you can't discuss what is happening in a meaningful way ignoring relative concentrations of the buffer and the aspirin.

Addition of every acid to the buffer lowers the pH. By how much depends on how strong the acid is and how much is added, relative to the amount of buffer present. In typical therapeutic situation amount of drug is negligible compared to the buffer concentration, so you can safely assume pH to be constant and be the driving force behind the HA/D- ratio.

so you mean that the degree of dissociation of HD doesn't solely depend on where the equilibrium is shifted?

#### Borek

• Mr. pH
• Deity Member
• Posts: 27304
• Mole Snacks: +1772/-408
• Gender:
• I am known to be occasionally wrong.
##### Re: pH and ionization of drugs
« Reply #5 on: March 06, 2023, 08:04:27 AM »
"Equilibrium shift" is just a rule of thumb, not something that can be quantitatively used to predict what will happen. Yes, just like le Chatelier's principle it can be often used to predict what to expect, but as every simplified rule they often fail if used blindly outside of the area where they are applicable.
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info, pH-meter.info

#### Luckenberg99

• Regular Member
• Posts: 14
• Mole Snacks: +0/-0
##### Re: pH and ionization of drugs
« Reply #6 on: March 06, 2023, 09:06:11 AM »
"Equilibrium shift" is just a rule of thumb, not something that can be quantitatively used to predict what will happen. Yes, just like le Chatelier's principle it can be often used to predict what to expect, but as every simplified rule they often fail if used blindly outside of the area where they are applicable.

Ok, thanks!