[Araconan]

Let's say I have a buffer made of 0.5M acetic acid and 0.5M potassium acetate. I let this solution react until equilibrium, where I can calculate the pH, etc. Now, I add some NaOH to it. The method that my textbook uses to calculate the new pH, is to essentially separate the question into two parts. First considering the reaction between hydroxide and acetic acid, in which all the hydroxide is reacted. (Amount of acetic acid remaining being 0.5M - amount of hydroxide added) Then, considering the equilibrium reaction with the new values of concentration of acetic acid, and then calculating the Hydronium ion concentration, and therefore, pH.

What's confusing me, is that doesn't this in a way, not make sense? The method is essentially treating the problem as if I added sodium hydroxide along with the acetic acid and potassium acetate at the same time, as opposed to having the buffer first, then adding the sodium hydroxide. Wouldn't the hydroxide ions react with the Hydronium ions (in the buffer), instead of the acetic acid? And even even if the hydroxide ion reacts with the acetic acid instead, why is it using the original concentration of acetic acid in the calculation of the reaction with hydroxide? Wouldn't the actual concentration of acetic acid be smaller than the original, since the solution went to equilibrium first, before any sodium hydroxide was added?

[Hunter2]

No, there is no equilibrium. You have acetic acid and also acetate together.  The pH is given by the formula from Henderson Hasselbalch.  So if you add some hydroxide, some of the acetic acid will be neutralized and more acetate is produced. The same happens you add hydrochloric. Some of the Acetat will be covert to acetic acid. A new pH is the result.

[Borek]

What's confusing me, is that doesn't this in a way, not make sense? The method is essentially treating the problem as if I added sodium hydroxide along with the acetic acid and potassium acetate at the same time, as opposed to having the buffer first, then adding the sodium hydroxide. Wouldn't the hydroxide ions react with the Hydronium ions (in the buffer), instead of the acetic acid? And even even if the hydroxide ion reacts with the acetic acid instead, why is it using the original concentration of acetic acid in the calculation of the reaction with hydroxide? Wouldn't the actual concentration of acetic acid be smaller than the original, since the solution went to equilibrium first, before any sodium hydroxide was added?

You are right in your thinking - but these effects don't matter. Amount of acetic acid that dissociated is pretty small, so the concentration of acid in the solution is almost unchanged (please remember you already added a substantial amount of conjugate base, applying LeChatelier's principle you should see acid won't dissociate much). Hence error we are making assuming initial concentration of acetic acid and its direct reaction with added strong base is neglectable.

Let's use your numbers. 0.5M HAcetate and 0.5M KAcetate. After calculating equilibrium concentrations we get that solution is 0.499983 M in HAcetate and 0.500017 M in Acetate^- - note I had to use more than 4 significant digits to show the difference, as otherwise concentrations would be 0.5000 and 0.5000, as if there was no dissociation at all.

However, in some cases, dissociation of the acid can't be neglected. That's especially true for a stronger acids (like chloroacetic acid) and diluted solutions. See the very last paragraph on this page: http://www.chembuddy.com/?left=Buffer-Maker&right=buffer-calculation

[Borek]

No, there is no equilibrium.

Yes, there is an equilibrium. Just because we ignore it, doesn't mean it doesn't exist.

[Araconan]

What's confusing me, is that doesn't this in a way, not make sense? The method is essentially treating the problem as if I added sodium hydroxide along with the acetic acid and potassium acetate at the same time, as opposed to having the buffer first, then adding the sodium hydroxide. Wouldn't the hydroxide ions react with the Hydronium ions (in the buffer), instead of the acetic acid? And even even if the hydroxide ion reacts with the acetic acid instead, why is it using the original concentration of acetic acid in the calculation of the reaction with hydroxide? Wouldn't the actual concentration of acetic acid be smaller than the original, since the solution went to equilibrium first, before any sodium hydroxide was added?

You are right in your thinking - but these effects don't matter. Amount of acetic acid that dissociated is pretty small, so the concentration of acid in the solution is almost unchanged (please remember you already added a substantial amount of conjugate base, applying LeChatelier's principle you should see acid won't dissociate much). Hence error we are making assuming initial concentration of acetic acid and its direct reaction with added strong base is neglectable.

Let's use your numbers. 0.5M HAcetate and 0.5M KAcetate. After calculating equilibrium concentrations we get that solution is 0.499983 M in HAcetate and 0.500017 M in Acetate^- - note I had to use more than 4 significant digits to show the difference, as otherwise concentrations would be 0.5000 and 0.5000, as if there was no dissociation at all.

However, in some cases, dissociation of the acid can't be neglected. That's especially true for a stronger acids (like chloroacetic acid) and diluted solutions. See the very last paragraph on this page: http://www.chembuddy.com/?left=Buffer-Maker&right=buffer-calculation

Ahh, I see now. Thank you very much!