Understanding the Effect of pH on Benzoic Acid Solubility

In summary, when benzoic acid is placed in a buffered solution at pH = 2, it will be mostly unionized and not very soluble. However, if it is placed in a basic environment, its solubility will increase. This is because the addition of a base, such as NaOH, will cause the formation of a highly soluble bensoate salt. When benzoic acid is added to a solution that is already acidic, the presence of H+ ions from another source will prevent the dissociation of benzoic acid. On the other hand, in a basic solution, the OH- ions will react with the H+ ions, shifting the equilibrium to the right and increasing the solubility of benzoic
  • #1
pisluca99
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Suppose we put benzoic acid in a buffered solution at pH=2. In this condition, it will be almost completely unionized and therefore not very soluble. Conversely, if it is placed in a basic environment, the solubility will be high. But why does this happen? Is there a shift in the acid equilibrium of benzoic acid? But shouldn't this move only if more H+ or more OH- are added to the benzoic acid already present in solution?
 
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  • #2
If you add a base such as NaOH, then you will get formation of the corresponding bensoate salt, which is very soluble, as compared to the nondissociated benzoic acid which has low solubility.

/Fredrik
 
  • #3
Fra said:
If you add a base such as NaOH, then you will get formation of the corresponding bensoate salt, which is very soluble, as compared to the nondissociated benzoic acid which has low solubility.

/Fredrik
yes, but I'm referring to a more upstream situation: the solution is already basic or acidic and benzoic acid is added. What happens to the acid dissociation equilibrium of benzoic acid in either situation when it is added?
 
  • #4
Not sure if I get the main question. In the acid-base balance, one refers to the soluble acid. The chemical activity of the not solved or precipitated compound is "constant" and doesnt enter the dissociation. So if you add more benzonic acid beyond saturation, it will just stay unsolved and not affect the dissociation of the acid. OTOH, if the solution is basic, when the acid is dissaociated, more acid is dissolved.

/Fredrik
 
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  • #5
Fra said:
Not sure if I get the main question. In the acid-base balance, one refers to the soluble acid. The chemical activity of the not solved or precipitated compound is "constant" and doesnt enter the dissociation. So if you add more benzonic acid beyond saturation, it will just stay unsolved and not affect the dissociation of the acid. OTOH, if the solution is basic, when the acid is dissaociated, more acid is dissolved.

/Fredrik
The main question is this.
Benzoic acid dissociates according to the following equilibrium:

Ar-COOH + H2O <=> Ar-COO- + H3O+

This said, let's suppose to insert this compound in a solution that is already acid. How does the acidity (H+) of this solution affect the dissociation equilibrium of benzoic acid?

Same question for a basic solution.
 
  • #6
The H3O+ from the acid solution will prevent the Ar-COOH reaction from producing many Ar-COO- ions, so it will not dissociate very much.
Think Le Chatelier's principle.
The presence of H+ ions from another source drives the disocciation of the benzoic acid to the left hand side.
 
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  • #7
DrJohn said:
The H3O+ from the acid solution will prevent the Ar-COOH reaction from producing many Ar-COO- ions, so it will not dissociate very much.
Think Le Chatelier's principle.
The presence of H+ ions from another source drives the disocciation of the benzoic acid to the left hand side.
ok, so placing it in a basic solution instead, the OH- reacts with the H+, removing them from equilibrium and shifting it to the right?

But shouldn't Le Chatelier's principle hold when we add H+ or OH- to a benzoic acid solution?
Or is it also valid in this case, that is by inserting the benzoic acid in an already buffered solution?
 
  • #8
pisluca99 said:
ok, so placing it in a basic solution instead, the OH- reacts with the H+, removing them from equilibrium and shifting it to the right?
Yes
pisluca99 said:
But shouldn't Le Chatelier's principle hold when we add H+ or OH- to a benzoic acid solution?
Or is it also valid in this case, that is by inserting the benzoic acid in an already buffered solution?
Yes, but to compute the exact balance when mixing things, one needs to consider the total volumes and the strength of all the acid and bases. There is always both H+ and OH-, it's just a question of balance. If you add benzoic acid in small amounts into a larger buffer, the pH of the buffer will determine the protolysis equilibrium. But If you add a stronger base or acid into a finite benzoate system you need to consider how the total amount shifts the pH in the total system. Multivariable systems where the different acids and bases are comparable in strength and approximations are diffucult, are higher order algebraic equations that are usually solved by numercially. It's the equations for H+ you get when all the reactions are coupled, and you solve for H+.

/Fredrik
 
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  • #9
Fra said:
Yes

Yes, but to compute the exact balance when mixing things, one needs to consider the total volumes and the strength of all the acid and bases. There is always both H+ and OH-, it's just a question of balance. If you add benzoic acid in small amounts into a larger buffer, the pH of the buffer will determine the protolysis equilibrium. But If you add a stronger base or acid into a finite benzoate system you need to consider how the total amount shifts the pH in the total system. Multivariable systems where the different acids and bases are comparable in strength and approximations are diffucult, are higher order algebraic equations that are usually solved by numercially. It's the equations for H+ you get when all the reactions are coupled, and you solve for H+.

/Fredrik
yes, indeed the degree of ionization depends on the pH and pKa.

However, perhaps, without reasoning in terms of equilibrium, one could also simply say that in an acid environment the conjugate base (benzoate) tends to be reprotonated, so that the acid remains mostly undissociated, while in a basic environment the removal of the proton from benzoic acid Is favoured and it prevails in its undissociated form. I think this is also correct..?
 
  • #10
pisluca99 said:
However, perhaps, without reasoning in terms of equilibrium, one could also simply say that in an acid environment the conjugate base (benzoate) tends to be reprotonated, so that the acid remains mostly undissociated, while in a basic environment the removal of the proton from benzoic acid Is favoured and it prevails in its undissociated form.

That's dissociation equilibrium at work, not sure why "without".
 
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  • #11
Borek said:
That's dissociation equilibrium at work, not sure why "without".
Right. I meant not reasoning in terms of Le Chatelier's principle. In an acidic environment the many H+ reprotonate A-, while in a basic environment the many OH- deprotonate HA to a greater extent
 
  • #12
Le Chatelier's principle is - in principle - just a rule of thumb based on equilibrium considerations. These are not separate things.
 
  • #13
Borek said:
Le Chatelier's principle is - in principle - just a rule of thumb based on equilibrium considerations. These are not separate things.
so you mean that what I wrote just before is in any case connected to Le Chatelier's principle: if benzoic acid is placed in an acid solution, H+ shift the equilibrium to the left, so A- gets protonated. Conversely, if we place benzoic acid in a basic environment, the OH- consume the H+ which derive from the dissociation of the acid, so the equilibrium shifts to the right, so deprotonation is favoured. Simply this?
 
  • #14
Yes, that's the basic idea.

As another rule of thumb - at pH=pKa exactly half of the acid molecules of the acid in the solution are protonated. Go to higher pH - and dissociated form start to dominate. Go to lower pH - protonated form dominates. Ratio changes tenfold for each pH change by one. This can be easily derived just from the dissociation constant definition.

pKa for benzoic acid is 4.2. In a pH=2 buffer dissociated form you are around 2 pH units from pKa, that means amount of non-dissociated form is in a below single percent range.
 
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  • #15
pisluca99 said:
However, perhaps, without reasoning in terms of equilibrium, one could also simply say that in an acid environment the conjugate base (benzoate) tends to be reprotonated, so that the acid remains mostly undissociated, while in a basic environment the removal of the proton from benzoic acid Is favoured and it prevails in its undissociated form. I think this is also correct..?
You mean without making it so complicating that you have to calculate the exact level of the balance?

Then yes you are right. If the acidic environment and it's buffering capacity is large and strong enough, (and same for the basic environment) the equilibirum of the bensoate protolysis is either effectively fully to the right or to the left making the "calculation" easy, so simplifying it like you said is right.

/Fredrik
 
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  • #16
Perfect. Thanks everyone!
 
  • #17
@Borek @Fra
Sorry to reopen the thread, but I have another little doubt.
I have found that it is possible to evaluate the solubility of an acidic drug through the following equation:
Stot = So + So*10^(pH-pKa).
Plotting Stot vs pH then a sigmoid is obtained.

Is it correct to follow the following reasoning for the construction of the sigmoid? : when pH is at least two units lower than the pKa, then [A-] in solution can be neglected, so the So*10^(pH-pKa) term of the equation can be neglected and Stot = So = [HA], so in the plot we obtain a straight line parallel to the X axis.

However, when pH increases beyond this "condition", [A-] is no longer negligible and no term in the equation can be neglected, so the equation is used entirely and the curve begins to rise upwards.

At this point, when pH is at least two units higher than the pKa, the So term of the equation can be neglected, being [HA] negligible, so, in theory, as pH increases, the curve should rise towards infinity, but in reality this does not happen, as the [A-] does not increase to infinity.
Therefore, when pH is at least two units higher than the pKa, the curve again becomes parallel to the X axis, since all the HA can now be considered converted into A- and Stot no longer increases.

Is this reasoning correct? Thank you for your patience.
 
  • #18
These are all approximations. Reasonable ones, but whether you accept the reasoning and conclusion really depends on what level of accuracy you need.
 
  • #19
Borek said:
These are all approximations. Reasonable ones, but whether you accept the reasoning and conclusion really depends on what level of accuracy you need.

Oh sure, I really believe that these approximations are sufficient for my purpose. The important thing is that they are correct approximations/reasoning. I just needed some feedback on what I wrote.
 
  • #20
"Reasonable ones" was intended to be the feedback you are looking for :wink:
 
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  • #21
Borek said:
"Reasonable ones" was intended to be the feedback you are looking for :wink:
Thank you!
 

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