Answer to Question #97839 in Organic Chemistry for sagal

Question

if weak acids ionize only a few percent in aqueous solution, why is it possible to fully neutralize a weak acid by reacting with the stoichiometric equivalent of NaOH

Answer 1

It is because water is a weaker electrolyte than most of the weak acids. When we add weak acid to water, only a small amount of acid ionizes:

$HA = H^+ + A^-;$

The equilibrium of this process is greatly shifted to the left (the acid is weak, as we remember):

$K_a = \frac {[H^+][A^-]} {[HA]};$

$[HA]>>[H^+][A^-];$ (the denominator is much bigger than numerator)

When we add the sodium hydroxide to such a solution, all of the available protons react with sodium hydroxide, giving water:

$H^+ + NaOH = H_2O + Na^+;$

This process shifts equilibrium and the thermodynamic system (according to the Le-Chatelier principle) tries to offset the impact on the equilibrium by dissociating more molecules of acid. This action gives us more protons, which are immediately "consumed" by present OH^- ions from sodium hydroxide. So equilibrium shifts again and more acid molecules dissociate. And all protons react with existing hydroxo groups again. This process lasts as long, as we have acid molecules in the reaction mixture. As there is a stoichiometric amount of NaOH, at the end of this reaction we have only water and sodium salt of the acid. We can represent this by the equation:

$HA + NaOH = NaA + H_2O;$