Question

Why does the conjugate base of a strong acid have no effect on solution `pH`?

Answer 1

Because the conjugate base of a strong acid DEMONSTRABLY does not compete strongly for the proton............

Explanation:

.......whereas the conjugate base of a weak acid DOES compete to some extent for the proton.......

And thus we can write the acid-base dissociation reaction:

`HA(aq) + H_2O(l)rightleftharpoonsH_3O^(+) + A^-`

And from this equilibrium expression we can derive an expression for `pH` of the buffer......

`K_a=([H_3O^+][A^-])/([HA])` (as you know `[H_2O]` is effectively a constant).

And if we take `log_10` of BOTH SIDES, we get........

`log_10K_a=log_10[H_3O^+]+log_10{([A^-])/([HA])}`

Which on rearrangement gives........

`underbrace(-log_10[H_3O^+])_"pH"=underbrace(-log_10K_a)_(pK_a)+log_10{([A^-])/([HA])}`

And thus we can write the so-called buffer equation............

`pH=pK_a+log_10{[[A^-]]/[[HA]]}`

And as we know, when `[A^-]=[HA]`, `pH=pK_a`. Do you agree?

`pH` could not be so moderated UNLESS `A^-` competed somewhat for the proton (and if it did not it would mean that the parent acid, `HA`, was a STRONG acid).