Recent content by Kathe

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...

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

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...

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...

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...

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...

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 =...

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 +]...

Hi Epiguin, I didn't make any big progress. 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 seems however that solubility decreases...

Hi Epiguin, I had misinterpreted myself. I thought you had to post the question in the "homework" section and I hadn't intended you to use the homework style or write formulas in full. I don't chew the "Latex" style but I send you the pages of the book where the formulas are. In any case I was...

I posted the exercise in the homework section as suvited by you. Meanwhile, I look at your new chart. Thanks for everything.

Hi, I have a doubt about the solubility calculation of lamivudine (3TC). Suppose we have an excess of free base lamivudine (B) in solution. An equilibrium is generated between B in the aqueous phase and B in the solid phase, ie between B in solution and B as the bottom body. As we acidify the pH...

However rereading I noticed that I was wrong in writing but I can't change the previous message ... By mistake I wrote "I don't want to learn" instead it's exactly the opposite. I want to learn and although everyone would like to be able to understand things on the fly, most of the time you have...

Hi, from your words it almost seems that I am not committing myself to solve this problem but I assure you that it is not so and I am sorry if you think this. If I wrote on this forum it is because I don't know how to deal with it and, as I have said many times, based on my studies (chemical...