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Topic: Manipulating the Henderson-Hasselbalch Equation to Accommodate Dibasic compounds  (Read 5005 times)

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Offline paradoxicalpirate

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I am having trouble with some quantitative acid-base chemistry regarding a dibasic compound and the percentage to which the compound ionizes in three biological compartments. Hypothetical pH values for the three compounds are pH 1, pH 6, and pH 7.4. I am having trouble learning how to manipulate the Henderson-Hasselbalch equation (HHEQ) in a way to derive an equation that can be "flexibly used" to calculate the degree to which a compound ionizes in these biological compartments. The thread title specifies how to manipulate the equation to accommodate a dibasic compound, however I am also interested in learning how to accommodate diacidic compounds and mixed acid-base compounds. Insight as to setting up the HHEQ with hypothetical pKa values, as well as the origin of how the equation is derived is desired. If this thread is posted in the wrong forum please *Ignore me, I am impatient* it to the correct forum. Thank you for any and all responses. I have a pressing assignment containing problems of this nature due within 72 hours, all insight has potential positive impact. Thanks again.
« Last Edit: February 28, 2014, 11:11:45 AM by paradoxicalpirate »

Offline Babcock_Hall

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Forum rules require you to show an attempt before we can help you.

Offline paradoxicalpirate

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Forum rules require you to show an attempt before we can help you.


Interestingly enough, my problem with that is I cannot really attempt to compute anything until I derive the equation. The problem is with deriving an equation from the HHEQ ((10^(pH-pKa))=(b/bh+)) so that I can use it to compute the ionization ratio for diprotic compounds. I am not looking for a specific numerical answer, just with help deriving an equation. However if you want me to attempt this I will try. So ((10^(pH-pKa)))=(b/bh+]) then theoretically if the HHEQ is for a single area of ionization, to derive an equation for a compound with 2 specific points of ionization couldn't one simply substitute the HHEQ into the HHEQ equation? So that [BH+]*((10^(pH-pKa))=b and then plug that value back into the equation? I already understand that this is wrong, but you asked for an attempt so this is it. Sorry for breaking the rules on the first post. I assume that rule is to prevent this forum from becoming an answer tool, and not facilitating learning? I'm not looking for an answer tool, just a push or vicious shove in the right direction. Thanks for responding.

Offline Babcock_Hall

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Others with a greater knowledge of multiple equilibria are in a better position to help, but I may be able to start the process.  According to an old general chemistry textbook I have, a complete equation would be a quartic in [H+]  However, there is a simplification that can be applied when the two pKa values differ by several powers of ten.  Phosphoric acid conforms to this, inasmuch as the pKa values are separated by about 5 log units (although there are three pKa values, not two).  ChemBuddy is a program that can be used to make certain kinds of acid base calculations, but you may be more interested in the equation and the theory than a particular answer.  http://www.chembuddy.com/?left=BATE&right=example_of_pH_calculation

Offline paradoxicalpirate

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Thank you.
       However you are correct with regard that I am certainly more interested in the principle of how to arrive algebraically with an equation that can be conformed to several compounds as the molecules that I will be required to perform the calculations on the test are  specific pharmaceuticals. Generally I consider myself an exceptional student, however due to incompetence by the professor teaching my class (his 1st semester teaching, class has no text book to corroborate learned information with, few sample questions provided, vague non-informative slides) I am in quite the conundrum. The previous test he gave the class was related to this material. The class average was well below passing, he was ordered by his superiors to make the test null, and provide an alternate means to determine our grades. The class was issued a take home test identical to the test given in class and given 72 hours starting at noon today to complete the exam. We are allowed to use our notes and all material we can gather relating to the test, we just aren't allowed to copy other students. I am at a loss for information related to the content the professor wishes for us to learn, this is the primary reason I joined the forum. I have literally searched the internet frantically (for the past 24 hours) to try and obtain some information relative to the info he wishes us to learn. Consulting him about the information is not an option as he conveniently informed us he would not be in his office until monday afternoon after the exam was to be due. I have emailed him several times with no response. I am trying to give him the benefit of the doubt on this as I am highly aware of the degree of difficulty associated with his job, but I am struggling. Professional school is expensive, as is time, and being forced to repeat a class due to an instructors incompetence is not something I desire. I do not wish for this thread to turn into a rant about my situation however sometimes venting can help.

     I have the only set of practice problems he presented to the class related to the majority of what the test covers. For all relevant purposes the class is a 1st year Biopharmaceutical Science class. If there were any slides containing information related to this other than a slide with only the HHEQ listed and nothing else I would provide access to this.

     Once again I don't want answers. I want a "push or vicious shove" in the right direction. If there are any sources of information (i.e scientific journal entries) that would help I would go to about any length to obtain them. The professor does not own the material on the sample question page so I am in the clear to post it. And once again thank you to anyone who posts. 


"Ketoconazole is a weak dibasic compound with pKa values of 2.9 and 6.5. It is practically insoluble in
aqueous fluids unless both weak base groups are ionized.
1. Calculate the % of ketoconazole that is uncharged, singly charged, and doubly charged at pH 6
and pH 2.
2. Qualitatively describe how pH of the solution will influence the aqueous solubility of
ketoconazole. Will ketoconazole be more soluble in the stomach or in the small intestines?
3. Some elderly persons have achlorhydria, or insufficient production of gastric acids. Gastric pH in
these individuals may be as high as 5-6. Ketoconazole has been show to be poorly absorbed in
these patients. Explain.
4. The pH values of some carbonated beverages are given below. Drinking water has a pH of 6.5-7.0.
Explain why your patient’s physician recommended that she take ketoconazole with Coca Cola.
Which of the others may also work? How does the drink “make the drug work better”?"


The 1st question is my main concern. I understand questions 2-4. I understand that in order to receive help I am obligated to attempt the question so here goes:

1. pKa 2.9, pKa 6.5, pH 6
If the pKa were singly 6.5 then according to the HHEQ then {b}/{bH+} is 0.316. In order for ketoconazole to be diprotonated it must first be monoprotonated. This means that the ratio of diprotonated ketoconazole atoms must first be protonated to be diprotonated. This is where I run into trouble. I cannot factor in the third concentration {BH2+} in the equilibria. Any push from this point would be appreciated. Thanks.

Offline Yggdrasil

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You can get a good estimate of the answer by treating the two ionizable groups independently and interpreting the ratio of protonated species to unprotonated species as a probability of protonation.  For example, considering the group with pKa = 6.5 at pH 6.0, {b}/{bh+} = 0.316, so the probability that the group is protonated is 0.760.  Then if you find the probability that the group with pKa = 2.9 is protonated, you can figure out the probability that the entire molecule is uncharged, singly charged, or doubly charged.

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