If we have a buffer solution as follows where HA is a weak acid and $\ce{A-}$ is its conjugate base, $\ce{HA + H2O<=>H3O+ + A-}$, do we ever consider $\ce{A- + H2O<=>HA + OH-}$? This is because in all of the buffer system problems I've done we only use the first reaction to find pH and other values but, we never consider the reaction of the conjugate base with water to form hydroxide ions which should increase the pH. Why is this so?
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$\begingroup$ If you are in an acid media don't exist a reason for consider the second equilibrium. $\endgroup$– KobaDec 8, 2015 at 1:53
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Yes the equilibrium certainly exists. In fact, there exists a third equilibrium, the self ionisation of water. Considering any of the two equilibria gives the concentrations of all the species involved, there is no need to write down the third equilibrium as the third equation will be a combination of the first two. The problem arises only when you invoke stoichiometry. For example, if $0.1\ \mathrm{mol}$ of the acid dissociated, it would certainly give $0.1\ \mathrm{mol}\ \ce{A-}$, but $0.1\ \mathrm{mol}$ would not be the amount of $[\ce{H3O+}]$ as it is part of another equilibrium. However, if we do make this approximation, ignoring the hydrolysis of $\ce{A-} $, we can justify it in some cases. Say if $K_\mathrm{a}$ is around $10^{-5}$ then the equilibrium constant for the hydrolysis would be $K_\mathrm{w}/K_\mathrm{b}$ which would be around $10^{-9}$. Thus it would not be a bad approximation to ignore the second equilibrium for finding the pH.
