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pisluca99

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- Thread starter pisluca99
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pisluca99

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- #2

Fra

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/Fredrik

- #3

pisluca99

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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?

/Fredrik

- #4

Fra

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/Fredrik

- #5

pisluca99

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The main question is this.

/Fredrik

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

DrJohn

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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.

- #7

pisluca99

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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?

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.

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

Fra

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Yesok, 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, 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+.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?

/Fredrik

- #9

pisluca99

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

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

Borek

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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".

- #11

pisluca99

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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 extentThat's dissociation equilibrium at work, not sure why "without".

- #12

Borek

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- #13

pisluca99

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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

Borek

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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.

- #15

Fra

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You mean without making it so complicating that you have to calculate the exact level of the balance?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..?

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

- #16

pisluca99

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Perfect. Thanks everyone!

- #17

pisluca99

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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

Borek

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- #19

pisluca99

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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

Borek

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"Reasonable ones" was intended to be the feedback you are looking for

- #21

pisluca99

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Thank you!"Reasonable ones" was intended to be the feedback you are looking for

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