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Assumptions in buffer calculations help

I am confused on the assumptions you make in buffer calculations. I know that we assume that the acid has not dissociated, and that the salt has completely dissociated, but what does this actually mean? Surely if the salt had completely dissociated, then the concentration of the salt would be zero? Surely the concentration of the ions that the acid has dissociated too would also be zero if we assume it hasn't dissociated? Many thanks
Original post by Bertybassett
I am confused on the assumptions you make in buffer calculations. I know that we assume that the acid has not dissociated, and that the salt has completely dissociated, but what does this actually mean? Surely if the salt had completely dissociated, then the concentration of the salt would be zero? Surely the concentration of the ions that the acid has dissociated too would also be zero if we assume it hasn't dissociated? Many thanks

Let's say you have a weak acid HA, and the sodium salt of its conjugate base NaA.
HA + H2O <-> H+ + H3O+
NaA -> Na+ + A-

At equilibrium, we assume that the number of HA molecules dissociated is so small that inital [HA] = equilibrium [HA].
NaA ionises completely. [A-] is assumed to be only due to the NaA, as the contribution of A- from the acid is very small.
Original post by BobbJo
Let's say you have a weak acid HA, and the sodium salt of its conjugate base NaA.
HA + H2O <-> H+ + H3O+
NaA -> Na+ + A-

At equilibrium, we assume that the number of HA molecules dissociated is so small that inital [HA] = equilibrium [HA].
NaA ionises completely. [A-] is assumed to be only due to the NaA, as the contribution of A- from the acid is very small.


i am stuggling to understand what you mean. Can you use an example
Original post by Bertybassett
i am stuggling to understand what you mean. Can you use an example

A buffer is prepared containing 1.00 M CH3COOH and 1.00 M CH3COONa. What is its pH? [The Ka of acetic acid is 1.77 x 10^-5]

Ka = [H+][A-]/[HA]
1.77 x 10^-5 = [H+](1)/(1)
solve for [H+]
pH = -log [H+]
pH = 4.75

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