Why simple analytical concentrations can be used instead of equilibrium concentrations in the Henderson-Hasselbalch equation.
We've made an approximation in the calculation above. We kept things simple by neglecting the fact that some of the acetate will react with water to give us acetic acid, and some of the acetic acid will dissociate to give back acetate.
Consider the dissociation of acetic acid into acetate and hydrogen ions:
CH_{3}COOH = H^{+} + CH_{3}COO^{-}
All of that acetate you've produced in the titration will suppress this dissociation.
Consider the hydrolysis of acetate to form acetic acid and hydroxide ions:
CH_{3}COO^{-} + H_{2}O =
CH_{3}COOH + OH^{-}
All of the acetic acid you added after the titration will suppress this reaction.
Thus, changes due to BOTH of these reactions will be small compared to the total acetate and acetic acid concentrations themselves. That allowed us to use analytical and not equilibrium concentrations in the calculation of pK_{a} above.
Still not convinced? The approximation is easy to justify. We'll need to do a little bit of algebra, though.
Let the concentrations of CH_{3}COOH and CH_{3}COO^{-} before equilibration be C_{CH3COOH} and C_{CH3COO-}, and let the concentration after equilibration be [CH_{3}COOH] and [CH_{3}COO^{-}].
We want to show that
C_{CH3COOH} ~ [CH_{3}COOH]
C_{CH3COO-} ~ [CH_{3}COO^{-}]
We need to look for relationships that hold both before and after equilibration to relate these concentrations. Two things must be unchanged by the dissociation and hydrolysis reactions: the total moles of acetate in all its forms, and the total charge on the solution (which must be zero).
The total concentration of acetate in all its forms has to be the same before and after equilibration, if acetate isn't created or destroyed in the process. In your experiment, you have only 0.0025 + 0.0020 = 0.0045 moles of acetic acid and acetate to work with. Hydrolysis and dissociation will change the form of the acetate, but NOT the total amount of acetate.
This statement can be written as a mass balance equation:
C_{CH3COOH} + C_{CH3COO-} = [CH_{3}COOH] + [CH_{3}COO^{-}]
The solution must be electrically neutral; that is, the total molarity of positive charges has to be equal to the total molarity of negative charges. Hydrogen and sodium ions contribute to the positive charge, in this experiment; hydroxide and acetate contribute to the negative charge. This gives us a charge balance equation:
[H^{+}] + [Na^{+}] = [OH^{-}] + [CH_{3}COO^{-}]
The sodium ions come from the NaOH in the titration. Because one mole of acetate was produced for every mole of NaOH added,
[Na^{+}] = C_{CH3COO-}.
Combining the charge balance and mass balance equations gives us the relationships we're looking for:
[CH_{3}COOH] =
C_{CH3COOH}
- [H^{+}] + [OH^{-}]
[CH_{3}COO^{-}]
=
C_{CH3COO-}
+ [H^{+}] - [OH^{-}]
These equations make sense.
The first one says: The equilibrium
concentration of acetic acid is the analytical concentration minus the amount
that dissociates (equal to [H^{+}], plus the amount produced by hydrolysis of acetate (equal to [OH^{-}].
The second one says: The equilibrium
concentration of acetate is the analytical concentration plus the amount
produced by dissociation of acetic acid (equal to [H^{+}], minus the amount converted by hydrolysis back to acetic acid (equal to [OH^{-}].
Both equations say that the difference between the analytical and equilibrium acetic acid concentrations is equal to the difference between the hydrogen ion and hydroxide ion concentrations. If the difference is small relative to the analytical concentrations, the analytical and equilibrium concentrations are equal, and we've justified using analytical concentrations when we really should have used equilibrium concentrations in the Henderson-Hasselbalch equation.
Sometimes, the assumption is NOT justified. That will happen when the acid is really dissociates easily (K_{a} larger than about 10^{-3}, or when the concentration of acid or its conjugate base are very small. That tells us that your experimental determination of K_{a} will fail under those conditions!
Author: Fred Senese senese@antoine.frostburg.edu