Doubt about calculating the solubility of lamivudine (3TC)

[Kathe]

Mentioned in the thread in the chemistry section,
https://www.physicsforums.com/threa...he-solubility-of-lamivudine-3tc.974052/page-2
I have calculated this curve (horizontal axis pH, vertical axis solubility) from the constants given by these authors.
View attachment 246759
It would be valuable and appropriate that we solve this here on the homework help thread, since this is a question arising from a textbook, and because this question involving solubility is a stage more complicated and difficult than the very frequent questions we get about pH etc. here.So a treatment could stand as a useful model.

For this, however, it is essential you use the site's conventions. For several reasons. Firstly, I'm convinced this problem would have been solved far faster if you had done so. Because setting out the problem according to theseConventions leads to a focus on the key points and where the Comprehension problems are.Explained also here:
https://www.physicsforums.com/help/noanswer/ Secondly if we continue to discuss this we have on the other thread, it will be practically a private discussion, not in the nature of this forum. Thirdly, be it a warning that, fairly unusually, your question got no answer in 10 days, although there are a number of competent chemists and chemical engineers here. That means your post was incomprehensible.

Fourthly, at least as important and useful as the theory of pH-dependent solubility for a degree and a career in pharmaceutical chemistry as well as for using this site in the future are

1. Being able to frame succinctly a scientific argument or question comprehensibly to readers
2. In English that does not need to be perfect (lots here isn't) but does need to be comprehensible
3. The comprehensibility helped by appropriate typography, e.g. LaTex

(1) I think can be done in about 10 lines or less, and that not having done so has caused you to lose more time than you have saved. For (2) you did not and you continue not to use even the suggestions for proper English words I made, again losing more time than you save. For (3) your refusal to use LaTex makes your texts look like a discouraging and impenetrable macchia Mediterranea, so again this is saving you no net time, whereas I think all you need for your text is the meaning and rules of $$, ##, and the codes for subscripts, superscripts and fractions. (A less satisfactory halfway house on my, but I'm told not all, versions of the site in the bar above the answer form about halfway along there is a thing that gives a drop down menu that also gives you subscripts and superscripts.)

I think attention to these points would give you an edge not only for this question, but for your studies in general.

Finally since you asked me about the link you could not open, I see this is the part of the site where we talk behind the students' backs and I can't resist giving you the quote

"
I think some of us have been developing remarkable deductive if not psychic powers in divining what a question would have been if the questioner had understood what it meant, like correcting three mistranscribed terms in a four-term equation and supplying the possible questions that could have been asked about it - this gift appears to be increasingly required. "
In this spirit, if we do nothing else, I would psych that since you likely understand the theory without the 2:1 Insoluble complex (?), chances are not using the solubility product equation for this complex is where you are missing something.

And really finally, I will come back, but understand I can't be in attendance all and every day OK?

 

[Kathe]

Now I can see the link! I will look bn latex writing! Anyway I was reflecting on your concept of [Cl-] constant and [ B ] decreasing. Perhaps from pH4.5 to pH4.2 [Cl -] = 0.081. The problem is that if it is true that salt starts to precipitate already at pH = 4.4 then the product [ B ] * [BH +] * [Cl-]> Kps 2: 1 for 4.5 <pH <4.2 in so that the bottom body is formed. Yet by doing the calculations the product is <Kps for 4.5 <pH <4.3 e = Kps for pH = 4.2 ... How is it possible ??[/B][/B]

 

[epenguin]

In brief, as for your various considerations which are hard to follow, if you have made some calculations they should be set out clearly as section 3 ("attempt at a solution") according to the homework template I linked to earlier, for just those reasons set out in the link. Please stop talking about "bottom body" etc. and find the right English words.

For some clarifications you ask: by "critical point" I mean a kink where the solubility curve changes slope.In both cases, the simple case of simple salt treated in the book, and the more complicated case with the 2:1 complex, I am saying that at pH above the high pH kink, [ B] is constant, below the low pH kink [ BH⁺] (and Cl^-) is constant.

"On the book it says that "for pH values <4.5 the solubility of the 3tc does not increase linearly with a slope equal to -1 as would be expected if the base body was constituted by the free base" ... To me it se.ems however that solubility decreases linearly with a +1 slope but is only a supposition". Increasing linearly with -1 slope and a decreasing linearly with +1 is the same thing. However the book is not right in saying that this is with pH, it is with [H⁺].

As for thinking about the 2:1 complex we solubilities and tendencies, this is quite difficult at first, so I didn't; I just calculated. One should try and make intuition and calculation coincide and reinforce each other. But sometimes one comes first and sometimes the other. In fact I found some qualitative insights that the calculation led me to only last night extremely interesting. This solubility curve has a minimum. And the calculation shows that this at this minimum [ B] = 2[Cl^-]. When you think of it this is obvious, or at least plausible-sounding. The composition of the precipitate is B₂HCl. So you could think that the solubility of a salt like this is minimum when the solution composition is exactly in the proportions of the precipitate, that solution contents to one side or the other of the proportion do not contribute to precipitation. I had not been aware of such a principle before now.Not completely sure of it yet, or how far it extends, I have a little problem with the H. But it seems we may have a principle for rationalising precipitates like this; is there anything in your book about it? Is any forum reader more familiar?

And still more. The above tells you that at this minimum [ B]/[BH⁺] = 2. Which gives us that
pH_Smin = pK_a - log 2
Which gives pH_Smin = 3.89. Look at the curve, we seem to have very good agreement between theoretical production and experimental results! A very interesting system after all I'm finding.

Anyway you should get on with setting things out according to the homework template.

 

[Kathe]

In brief, as for your various considerations which are hard to follow, if you have made some calculations they should be set out clearly as section 3 ("attempt at a solution") according to the homework template I linked to earlier, for just those reasons set out in the link. Please stop talking about "bottom body" etc. and find the right English words.

For some clarifications you ask: by "critical point" I mean a kink where the solubility curve changes slope.In both cases, the simple case of simple salt treated in the book, and the more complicated case with the 2:1 complex, I am saying that at pH above the high pH kink, [ B] is constant, below the low pH kink [ BH⁺] (and Cl^-) is constant.

"On the book it says that "for pH values <4.5 the solubility of the 3tc does not increase linearly with a slope equal to -1 as would be expected if the base body was constituted by the free base" ... To me it se.ems however that solubility decreases linearly with a +1 slope but is only a supposition". Increasing linearly with -1 slope and a decreasing linearly with +1 is the same thing. However the book is not right in saying that this is with pH, it is with [H⁺].

As for thinking about the 2:1 complex we solubilities and tendencies, this is quite difficult at first, so I didn't; I just calculated. One should try and make intuition and calculation coincide and reinforce each other. But sometimes one comes first and sometimes the other. In fact I found some qualitative insights that the calculation led me to only last night extremely interesting. This solubility curve has a minimum. And the calculation shows that this at this minimum [ B] = 2[Cl^-]. When you think of it this is obvious, or at least plausible-sounding. The composition of the precipitate is B₂HCl. So you could think that the solubility of a salt like this is minimum when the solution composition is exactly in the proportions of the precipitate, that solution contents to one side or the other of the proportion do not contribute to precipitation. I had not been aware of such a principle before now.Not completely sure of it yet, or how far it extends, I have a little problem with the H. But it seems we may have a principle for rationalising precipitates like this; is there anything in your book about it? Is any forum reader more familiar?

And still more. The above tells you that at this minimum [ B]/[BH⁺] = 2. Which gives us that
pH_Smin = pK_a - log 2
Which gives pH_Smin = 3.89. Look at the curve, we seem to have very good agreement between theoretical production and experimental results! A very interesting system after all I'm finding.

Anyway you should get on with setting things out according to the homework template.

Hi Epiguin, the Henderson Hasselbach equation to be applied in the case of a weak base salt is pH = pKa + log S0 / St-S0. In fact at pH 4.5 we have 4.5 = 4.2 + log (0.11 / x-0.11). So log 0.11 / (x-0.11) = 10 ^ 0.3. Then 0.11 / (x-0.11) = 10 ^ 0.3. Then 0.11 = 10 ^ 0.3 x-0.22. Then 10 ^ 0.3 x = 0.33. Thus x = 0.165. Unfortunately if I apply this equation in the minimum point I am not in the calculations, so I don't know if this equation is applicable to partial salts. If I apply the formula pH = pKa + log [ B ] / [BH +] at pH 4.2 we have that log [ B ] / [BH +] = 0. So [ B ] / [BH +] = 1 that is [ B ] = [BH +]. Unfortunately, if I apply the formula at pH 4.4 for example, we would have 4.4 = 4.2 + log 0.11 / x; log 0.11 / x = 0.2; 0.11 = 0.2 * 10 ^ x; x = 14408 ... Calculations are not found. I think I'm doing something wrong but I don't understand where exactly. Perhaps the equation should be modified but I don't know how or at what point.[/B][/B][/B][/B]

 

[epenguin]

Hi Epiguin, the Henderson Hasselbach equation to be applied in the case of a weak base salt is pH = pKa + log S0 / St-S0. In fact at pH 4.5 we have 4.5 = 4.2 + log (0.11 / x-0.11). So log 0.11 / (x-0.11) = 10 ^ 0.3. Then 0.11 / (x-0.11) = 10 ^ 0.3. Then 0.11 = 10 ^ 0.3 x-0.22. Then 10 ^ 0.3 x = 0.33. Thus x = 0.165. Unfortunately if I apply this equation in the minimum point I am not in the calculations, so I don't know if this equation is applicable to partial salts. If I apply the formula pH = pKa + log [ B ] / [BH +] at pH 4.2 we have that log [ B ] / [BH +] = 0. So [ B ] / [BH +] = 1 that is [ B ] = [BH +]. Unfortunately, if I apply the formula at pH 4.4 for example, we would have 4.4 = 4.2 + log 0.11 / x; log 0.11 / x = 0.2; 0.11 = 0.2 * 10 ^ x; x = 14408 ... Calculations are not found. I think I'm doing something wrong but I don't understand where exactly. Perhaps the equation should be modified but I don't know how or at what point.[/B][/B][/B][/B]

What you are doing wrong is resisting writing out just what the whole problem as in the homework template section 1, which could be done in about the same space and time as the above. Followed by sections 2 and 3. I do not know what is stopping you doing this. It would save me time and effort if I just did this for you, but it would be fairly unprecedented on this forum that the homework helper wrote the question in place of the student. I am only persisting because this problem is an interesting extension of ones we already frequently treat here, in which interesting things, challenges and even principles seem to emerge.

For what it is worth I calculate the pH at the high pH critical point to be 4.42. I am taking B₀ to be 0.115 from the graph in the publication. Above that pH things are exactly as in the theory and equations of your book for the simple case, so any strange result is no doubt due to some banal mistake in the calculation. Below that point [ B] is no longer equal to B₀.

 

[Kathe]

For a weak base and for pH>pKa +2 $$St=[ B ]+[BH^{+}];

[ B ]=S{_{0}};
[BH^{+}]=[ B ]\cdot[H{_{3}}O]/K {_a}=S{_{0}}\cdot[H{_{3}}O]/K {_a};

St= S{_{0}}+S {_{0}}\cdot[H{_{3}}O]/K {_a}=S{_{0}}\cdot (1+[H{_{3}}O]/K {_a});$$
For pH <pKa-2
$$St=[ B ]+[BH^{+}];

[BH^{+}]=\sqrt{Kps};

[ B ]=[BH^{+}]\cdot K{_a}/[H{_{3}}O]=\sqrt {Kps}\cdot K {_a}/[H{_{3}}O];

St=\sqrt {Kps} + \sqrt {Kps}\cdot K{_a}/[H{_{3}}O]=\sqrt {Kps}\cdot (1+K {_a}/[H{_{3}}O])$$;

 

[epenguin]

]

That is clearer, at least typographically. And some changes still to make there as well, haven't you noticed?

I recognise that you are saying that for a weak base (with only one insoluble salt BHCl) total solubility S_t at lower pH is given by the last of the first set of equations, but at higher pH by the corresponding one in the second set. There is a sharp transition between the region where one or the other formula is applicable represented by a kink in the pH-solubility curves that we have seen. Whether anybody else who has not gone a bit into the theory could recognise that that is what you are saying, I have some doubts. So you are still scarcely beyond the 1st square for stating what the question is.

The point of transition between the two regimes is very sharp, given by a precise equation which you should give, rather than your very permissive inequalities which in fact I don't think even always hold.
It reads like notes towards section 1or 2 of homework help. Read a few other questions in the help forums, there is not this seeming reticence.

 

[epenguin]

To make it clear, the equations predict solubility curves like this.

 

[Kathe]

To make it clear, the equations predict solubility curves like this.

View attachment 246976

Exact! This is the typical solubility curve of a weak base. At very alkaline pH, at least 2 units higher than pKa, [ B ] >> [BH +] and [ B ] is about S0. At pH = pKa la [ B ] = [BH +] and at pH less than at least 2 units with respect to pKa we have that [BH +] >> [ B ]. In this case a further plateau is reached in which the solubility of BH + becomes equal to that of the BHCl salt. The graph of the 3tv is particular because it presents an anomaly in the behavior and consequently in the solubility curve. Moreover, it seems that this effect is due to the use of HCl (it is demonstrated by adding KCl and seeing that due to the presence of the common ion the solubility is further reduced) and not of other acids.

 

[epenguin]

Maybe I have been over-severe in trying to coax the student to set out what the question is, and maybe the question has been generally understood. In case any reader has not latched on, and to have an agreed clear starting point I will give how I would frame the question:

The total solubility S_t of a base to which a base to which acid is added, is, as a function of [H⁺] described by the equation

S_t = S₀(1 + [H⁺]/K_a). ...(1)

at high pH. There is a sharp break in the curve and the equation that describes solubility at lower pH is

S_t = √K_sp.(1 + K_a/[H⁺]). ...(2)

where K_sp is the solubility product of the (chloride – we are considering addition of HCl) salt of B

K_sp = [ BH⁺][Cl^-] ...(3)

(The student should describe exactly where this breakpoint is).

The corresponding curves have the general appearance as in #38 *

The substance (±)-3TC has the more complicated solubility curves shown in the attachment to #26 ( a calculated version is in #27). This is explained by the formation of a more complicated salt, termed the "2:1" complex, which precipitates at these intermediate pHs, has been isolated and proved to have the composition B.BH⁺Cl^- or B₂HCl.

The student has been unable to calculate the more complicated curves that would describe this situation and asks guidance.

She should now proceed to set up as per section 2 of the homework template the relevant equations, with sufficient indication of what they represent, and numbered to ease discussion. Including those already given (need typographical corrections). And then as per Section 3 of the template indicate attempts at the calculation or reasonings.

*,further commented later.

 

[epenguin]

Note on #40

I had to illustrate the general appearance of these curves in #38 because there are actually no adequate illustrations of it in the textbook extract given by the student. Such curves have a characteristic feature – the vertical distanceFrom the peak singularity to the height where the curve levels off at low pH is equal to B₀. Student should be able to work out why, and why this corresponds to the formula which may also not be evident at first sight. At first sight also the textbook curves 34.2, 34.3 seem to be missing this unavoidable feature - at the top they look just flat. But this is because the solubility is being expressed logarithmically – there are three orders of magnitude difference between B₀ and the concentration of a saturated salt solution, so the low-pH dip is practically not noticeable.The vertical coordinate in the case our problem (fig 34.4) is on the other hand linear.

 

[Kathe]

Maybe I have been over-severe in trying to coax the student to set out what the question is, and maybe the question has been generally understood. In case any reader has not latched on, and to have an agreed clear starting point I will give how I would frame the question:

The total solubility S_t of a base to which a base to which acid is added, is, as a function of [H⁺] described by the equation

S_t = S₀(1 + [H⁺]/K_a). ...(1)

at high pH. There is a sharp break in the curve and the equation that describes solubility at lower pH is

S_t = √K_sp.(1 + K_a/[H⁺]). ...(2)

where K_sp is the solubility product of the (chloride – we are considering addition of HCl) salt of B

K_sp = [ BH⁺][Cl^-] ...(3)

(The student should describe exactly where this breakpoint is).

The corresponding curves have the general appearance as in #38 *

The substance (±)-3TC has the more complicated solubility curves shown in the attachment to #26 ( a calculated version is in #27). This is explained by the formation of a more complicated salt, termed the "2:1" complex, which precipitates at these intermediate pHs, has been isolated and proved to have the composition B.BH⁺Cl^- or B₂HCl.

The student has been unable to calculate the more complicated curves that would describe this situation and asks guidance.

She should now proceed to set up as per section 2 of the homework template the relevant equations, with sufficient indication of what they represent, and numbered to ease discussion. Including those already given (need typographical corrections). And then as per Section 3 of the template indicate attempts at the calculation or reasonings.

*,further commented later.

Thanks 1000 Epiguin ! In fact now the question is well formulated and I hope someone else can intervene to help us resolve the issue. My only doubt is that perhaps users could "snub" old posts in favor of new ones. Maybe, initially, having badly formulated the question, I involuntarily dismissed many users but I hope that they will reconsider and at least take a look at the question. Thank you so much for what you are doing!

 

[epenguin]

Nobody here is likely to help until they see you help yourself by setting out the necessary equations as per paragraph 2 of the homework help template, of which 3 are ready but not complete as I have pointed out, and others are necessary, then possibly a stab at para 3, Attempt at solution.

The problem is more challenging than average for this section of the site, but limited to calculating the curve in question is not as difficult as all that.

 

[jim mcnamara]

@Kathe
We cannot help you without your help. For me, most of your posts are exceedingly hard to decipher. I actually tried to decipher some broken latex and fixed it to give you some help on how to do that. PF helpers cannot go much further than that. Please note that the "ball is in your court" now, it is your turn to step up and help us to help you. We get that it is an interesting problem. Pharmacologists and chemists of all types need to know latex in order to communicate with peers, for example.

 

[Kathe]

@Kathe
We cannot help you without your help. For me, most of your posts are exceedingly hard to decipher. I actually tried to decipher some broken latex and fixed it to give you some help on how to do that. PF helpers cannot go much further than that. Please note that the "ball is in your court" now, it is your turn to step up and help us to help you. We get that it is an interesting problem. Pharmacologists and chemists of all types need to know latex in order to communicate with peers, for example.

Hello Jim Mcnamara, I tried to use Latex but it came out a very long formula that I had to break with ";" because I didn't know how to wrap up. I also didn't understand which formula to use for the division and that's why I kept using the "/". Unfortunately if I have been reluctant to use it it is because it is not a language familiar to me and using mostly the smartphone it became even more complex to write any formula. But I understand that it makes the reading of the formulas easier for this I will undertake to make the posts more understandable!

 

[Kathe]

Not knowing how to use the Latex well, I attached a word file where I reported the equations in an orderly manner.

 

[epenguin]

I don't know what you see on your iPhone but clicking on your attachment all I see on my iPad is

A (eg. by 2 units)

once repeated.I guess on i-phone you can see posts including your own as they are meant to be seen. Even before posting using the preview button bottom right. Re LaTex if you can do subscripts and superscripts for one thing that needs them, you can do them for another, therefore your #36 looks careless, especially after your attention has already been drawn to this.
Fractions: For the fractions in the equations so far it would be acceptable to keep them on one line as they are, but soon you will probably need the fraction thing. I do not know if you are using some other version of LaTex; if slightly tedious there is nothing enormously difficult about them in the site version https://www.physicsforums.com/help/latexhelp/. You can save some subscripts if you write [H⁺] as most people prefer instead of [H₃⁺O].

In the usual thread on this forum a student posts #1 consisting of standard homework template paras. 1, 2, 3 and then we can start to solve the problem collaboratively in as many posts as it takes – you can see what the sort of average of that is. Here we are on #40 and have not got as far as completing section 2 of #1. This is becoming unreasonable. I Will not answer your next post unless it is an attempt at a complete para 2, and a shot at 3 .

 

[Kathe]

I don't know what you see on your iPhone but clicking on your attachment all I see on my iPad is

A (eg. by 2 units)

once repeated.I guess on i-phone you can see posts including your own as they are meant to be seen. Even before posting using the preview button bottom right. Re LaTex if you can do subscripts and superscripts for one thing that needs them, you can do them for another, therefore your #36 looks careless, especially after your attention has already been drawn to this.
Fractions: For the fractions in the equations so far it would be acceptable to keep them on one line as they are, but soon you will probably need the fraction thing. I do not know if you are using some other version of LaTex; if slightly tedious there is nothing enormously difficult about them in the site version https://www.physicsforums.com/help/latexhelp/. You can save some subscripts if you write [H⁺] as most people prefer instead of [H₃⁺O].

In the usual thread on this forum a student posts #1 consisting of standard homework template paras. 1, 2, 3 and then we can start to solve the problem collaboratively in as many posts as it takes – you can see what the sort of average of that is. Here we are on #40 and have not got as far as completing section 2 of #1. This is becoming unreasonable. I Will not answer your next post unless it is an attempt at a complete para 2, and a shot at 3 .

I explain to you! As latex I used an automatic editor

https://www.codecogs.com/latex/eqneditor.php
But from the mobile it turns out to me however complicated because you click on a box and another one is selected.
With PC I used word and in fact I have attached a docx file to make the formulas understandable. I read from the phone but since you can't do it I thought I'd convert it to pdf. I also tried to turn word into latex but I couldn't. Let me know if you can open and view the pdf.

 

[epenguin]

OK I can read that now. Unfortunately I do not have the software or the skills to change that into something others can read here. Hopefully, someone else will volunteer to do this for you. Once you have one formula in LaTex code you can cannibalise it and get any other formula in latex code that needs only the same symbols. Awaiting this, I can give some positive advice.

About the first word. Please translate this into English.

We think order of things reflects comprehension as well as comprehensibility. So you should take out the equations that are always true and put them before the section which concerns things only sometimes, or conditionally, true e.g. under the different pH conditions. So generally true equations (I refer to them by line number not equation number) to take out and put first are 14, 15, 13, 2 (=11 and should have subscript). More equations than necessary will be unhelpful, we can safely eliminate Lines 4-6 even if they are true and stay part of your background knowledge. (12 is actually derived from 13, you could indicate this in the text with 12 ⇒ 13. The derivation depends on another very general equation that comes into all ionic equilibria that it is indispensable to include in the general equation section.) You could condense 8,9 and 17, 18 into single lines. Set out like this it will also be more helpful for your later revision than just a mass of formulae you might not easily see clear though any more. It would be convenient for others trying to follow if you list the numerical constants of the particular example we are treating. And number the equations.

For substance, you will not be able to understand the the complicated system of the problem unless you first understand the simpler case you are talking about above and that produces the behaviour in #38. Your comments so far do not give total confidence that you do understand this. You would have to be able to obtain the pH and other details of that maximum point, more than once asked. If your textbook seem a bit sketchy on that, the authors probably think it is the sort of thing pretty obvious to their readers. Make an attempt, this might go in either section 2 or 3. This is all about the background and a simple case. This will not suffice to to treat the complex case of the question because you have not yet included an equation specific to this question.

It begins to sound like quite a formidable lot of stuff to do but this would all be considered ordinary, even elementary routine in your field!