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Let's say we have substance $\ce{A}$, which is mixed with substance $\ce{B}$ to improve shelf-life because $\mathrm{p}K_\mathrm{a}$ of the substance $\ce{A}$ is $7.9$ and in mix the $\mathrm{pH}$ is $5.2$.

Does this mean that in the solution with $\mathrm{pH}$ near $8$ the substance $\ce{A}$ has multiple molecules in neutral state and not dissociate, thus it forms precipitates?

Does $\mathrm{p}K_\mathrm{a}$ mean that most of the molecules of the substance $\ce{A}$ are in nondissociated state? Does lowering $\mathrm{pH}$ in this case causes substance $\ce{A}$ to dissociate?

Are bold parts right? Please explain in plain English without the use of external links.

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pKa and pH are related concepts but often confused. pKa is a property of a compound that tells us how acidic it is. The lower the pKa, the stronger the acid. pH is a property of a particular solution that depends on the concentrations and identities of the components.

For this discussion, I'm going to use the terms protonated and deprotonated to mean that a compound is associated or dissociated with a proton.

Based on the relationship between the pKa of a compound and the pH of a solution, we can predict whether a compound will be protonated or deprotonated. If the pH is lower than the pKa, then the compound will be protonated. If the pH is higher than the pKa, then the compound will be deprotonated.

A further consideration is the charge on the compound. Acids are neutral when protonated and negatively charged (ionized) when deprotonated. Bases are neutral when deprotonated and positively charged (ionized) when protonated.

Given the information you provided, if compound A (with pKa 7.9) is in a solution of pH 5.2, compound A will be in the protonated state.

Without knowing anything about the identities of A and B, the following is speculation. Most likely compound A is a base and compound B is an acid. The protonated version of A (its "conjugate acid") is more stable for long term storage than A itself.

Another possibility is that compound A is an acid and B is a stronger acid. If the deprotonated version of A is unstable, then compound B is added to ensure that if the mixture came in contact with a base, something more acidic than A would be present to react with the base.

EDIT BASED ON NEW INFORMATION: As Mateus B said, lidocaine will be protonated by HCl at the amine nitrogen. This is likely done for two reasons. First, as its hydrochloride, it will be more water soluble. In solution, the compound will dissociate into protonated lidocaine cation and chloride anion. Second, as the nitrogen is protonated, it is less reactive to the environment. Over time, amines can react with oxygen to form the corresponding N-oxide and/or can absorb carbon dioxide.

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  • $\begingroup$ From your answer I understand more, but If I provide the substances - can the answer be more precise and understandable? A- is lidocaine (local anaesthetic), B - hydrochloric acid. ?? Thanks! +1 $\endgroup$ – Ilan Oct 11 '14 at 18:50
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    $\begingroup$ The pKa of a compound also depends on the environment. Most published pKa values are an aqueous pKa, that is they tell you how acidic a compound is in water. However, that compound will likely have a different acidity and pKa, when they are an isolated molecule, or dissolved in some other solvent, not water, for example: ethanol, dichloromethane, or DMSO. This extra aspect to pKa that isn't necessarily relevant here, but good to keep in the back of your mind when thinking about the protonation state of a compound, you need to think about what it is dissolved in, too. $\endgroup$ – tmgriffiths May 12 '20 at 0:01
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Let's say we have substance A which was mixed with substance B to improve shelf-life because pKa of the substance A is 7.9 and in mix the pH is 5.2. Does this mean that in the solution with pH near 8 the substance A has multiple molecules in neutral state and not dissociate, thus it forms precipitates?

At all pH, the substance A will exist as a mixture of HA and A-. HA may or may not be soluble, so it may or may not precipitate.

Does pKa mean that most of the molecules of the substance A are in nondissociated state?

The higher the pKa, the lower the Ka, the lower the ratio of A- to HA, and thus the higher fraction of molecules of substance A will be in the non dissociated state.

Does lowering pH in this case causes substance A to dissociate?

No. Increasing pH causes substance A to dissociate.


Some additional information:

From the Henderson–Hasselbalch equation,

$\textrm{pH} = \textrm{pK}_{a}+ \log_{10} \left ( > \frac{[\textrm{A}^-]}{[\textrm{HA}]} \right )$

If pH = pKa, the concentration ratio of [A-] to [HA] is 1.

If your pH is higher than pKa, then [A-] must be greater than [HA].

If you lower your pH such that it is lower than pKa, then [HA] must be greater than [A-]

Therefore, by increasing the pH, you are dissociating HA into H+ and A-. Which makes sense according to the Le Chatelier principle, where H+ is released to counteract the increase in pH (or the decrease in [H+])

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  • $\begingroup$ Could you please answer in plain English the question as it was asked? $\endgroup$ – Ilan Oct 11 '14 at 7:51
  • $\begingroup$ @Ilan Explained in plain English, although I dare say the previous explanation was as simple as it could be, even to a non professional. $\endgroup$ – t.c Oct 11 '14 at 8:13
  • $\begingroup$ I'll include the original answer for completeness, under additional information. $\endgroup$ – t.c Oct 11 '14 at 8:28
  • $\begingroup$ If the substance A is lidocaine (base?), and B is hydrochloric acid, does this information help to focus the answer?? Thanks. $\endgroup$ – Ilan Oct 11 '14 at 18:52
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I think users @jerepierre and @t.c provided excellent answer above, but I will attempt a more simplified answer as requested.

First we need to understand what happens with an acid (with a specified pKa value), when a solution reaches certain pH values.

  1. At a pH above the pKa, more of the acid molecules are deprotonated.
  2. When the pH equals the pKa, the acid molecules are 50% protonated and 50% deprotonated.
  3. At a pH below the pKa, more acid is protonated.

Without knowing more about the specific chemical nature of the molecule, we cannot know more about what will happen. In general, in a water solution more deprotonated or protonated acids will be more soluable (more molecules stays in solution). This is a result of that part of the molecule gaining an electric charge when gaining or losing extra protons. This charge makes the molecule polar, and soluable in polar solutions (such as water).

Solubility of a molecule can be explained using another concept that relates to pKa. This is called the isoelectric point (pI). Sometimes molecules are large, and complex. Several different parts of such a large molecule can be either 'acidic' or 'basic', depending on their chemical nature. They then gain or lose charge differently as the pH of a solution changes. At a certain pH, all the charges even out and the molecule becomes neutrally charged (instead of positive, or negative). It then loses its polarity which makes it soluable in water and it precipitates.

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By specifying that A is lidocaine I could say that by adding HCl to it you'd be forming a diferent compound, lidocaine hydrocloride.

Which is more or less like this in the solution.

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    $\begingroup$ Can you elaborate a bit more? This doesn't really (directly) answer the question as-is. $\endgroup$ – jonsca Oct 11 '14 at 21:15
  • $\begingroup$ There isn't much more to elaborate on this. He asked about the influence of pKa on a generic case of two compounds, which has been answered already by the guys above, but then he asks about the specific case of lidocaine and HCl, to which I answered that it doesn't need to be directly related to pKa as the two compounds will react forming the compound shown in the image. In order to elaborate more I could only go toward reaction mechanism, which I don't think will interest a non-professional. $\endgroup$ – Mateus B. Oct 11 '14 at 22:34
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    $\begingroup$ @MateusB. I've read that Lidocaine hydrochloride has longer shelf life than just Lido as is... Could you please answer What are the causes for this extension? $\endgroup$ – Ilan Jun 24 '15 at 20:18

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