Hi, I have a doubt about the solubility / pH profile of 3tc. In practice, a solution of 3TC free base is prepared by adding an excess, which is why a bottom body will be formed which is in equilibrium with the aqueous concentration of B, which is precisely the intrinsic solubility S0.
. Then as I add HCl, BH + is formed. As the watery B is transformed into BH + by reaction with the acidulated water, another bottom body gradually dissolves to restore and maintain the concentration of B in water. Total solubility is ST = [B] [BH +] = S0 * S0 [H30 +] / Ka. Despite only for pH = pka = 4.2 the [B] = [BH +], already at pH = 4.5 (when [B]> [BH +]), partial salt 2: 1 ("partial") starts to form because the quantity of H + addition is not succulent to proton all B so that instead of having a B pronate and an B orphan of a proton, B and BH + share a proton leading to the formation of a hemichlorohydrate). What is the formula for calculating total solubility in the presence of a partial salt of the base? Graphically we note that solubility will increase up to pKa = 4.5 (which is likely as [BH +] grows, then partial salt is formed and solubility collapses to its minimum at pH = pKa (it is not clear to me why) and then the solubility returns to growth until all the salt 2: 1 is transformed into salt 1: 1 BH + Cl- whose total solubility is ST = (1 + Ka / [H3O +]) * (Kps) ^ 1 (2) To recap: how to calculate the solubility of a partial salt? It should be drawn as if it were a cationic dimer? On this case ST = [B] + [BH +] + [B2H +]. If you were to adjust for the degree of aggregation, since a dimer is formed, should be ST = [B] + [BH +] + 2 [B2H +]. The other hypothesis is that ST = (1 + Ka / [H3O +]) * (Kps) ^ 1/3 That is, since Kps of salt 2:1 is [B] * [BH +] * [Cl-] = s ^ 3, then s = (Kp of salt 2: 1) ^ 1 / 3. Help me to understand please. Good evening.
Caterina,
I think you have some errors in your equations. For example, your first equation is: ST = [B] [BH +] = S0 * S0 [H30 +] / Ka. If you mean to say that the total amount of 3tc in solution (total solubility) is the sum of the neutal and cationic forms (B + HB+), then ST = B + HB+ = B*(Ka*[H+] + 1), since Ka = [H+]*[B]/[HB+].
Please correct me if I am wrong.
From what you say, it seems that as the pH value is lowered with addition of HCl, the total amount of 3tc dissolved in the water, first goes up (until pH 4.5), then goes down (at pH 4.2), then increases again below pH 4.2, likely linear with a slope of 1 on plot of pH versus log concentration. Your hypothesis that it is due to a low solubility of the salt of the dimer (B2H+Cl-) might be correct (Ksp = [B][HB+][Cl-]) as this salt would have it's highest concentration at the pKa. One way to test this is to add NaCl or KCl and see if you can lower the concentration even further at the pH equal to the pKa value (salt it out by adding Cl-). From the full set of mass balance and equilibrium equations, you should be able to calculate the solubility product (Ksp) of the salt-dimer, and maybe the complexation constant for this species. Also, you can do the same (original) experiment by adding nitric acid (or sulfuric, or other) rather than HCl to see if this inflection goes away or is accentuated. This would show that the Cl- is the reason for the additional inflection point on the solubility curve.
Dear Mr. Chad T Jaftver, I have attached the 3tc plot so that you can view the trend of the curve. Yes, I was wrong to write the equations. The solubility of the free base (neutral form B) in solution represents the intrinsic solubility S0. The total solubility for pH> 4.5 is ST = [B] + [BH +] = S0 + S0 * [H3O +] / Ka = S0 * (1+ [H3O +] / Ka). Solubility for pH
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