Consider the following buffer equilibrium:
HF (high concentration) + H2O <--> H3O+ (low concentration) + F- ( high concentration)
Using Le Chatelier's Principle, explain what happens to the pH of the buffer solution when a small amount of NaOH is added.
Wouldn't OH- react with HF to form water? But they're both on the reactants side.
I'm not sure of what your symbols mean. HF by itself is not a buffer; however, HF in combination with its salt (NaF) is a buffer and your designation of high concn and low concn may be intended to show that. If NaOH is added, it reacts with the H3O^+, and removing H3O^+ shifts the equilibrium to the right to create more H3O^+ to replace that which was removed. The pH will not change much.
Le Chatelier's Principle states that when a system at equilibrium is subjected to a stress, it will respond by shifting its equilibrium position to relieve the stress. In this case, the stress is the addition of a small amount of NaOH, which introduces hydroxide ions (OH-) into the buffer solution.
When NaOH is added, the hydroxide ions react with the hydrogen ions (H+) present in the solution to form water:
OH- + H+ -> H2O
Since there is already a high concentration of F- ions in the solution, the addition of OH- ions does not cause any significant change.
However, there is HF (a weak acid) present in the solution. When OH- ions react with HF, they produce water and form F- ions:
OH- + HF -> H2O + F-
This reaction consumes H+ ions and decreases the concentration of hydrogen ions in the solution. As a result, the equilibrium of the buffer system is disturbed, and to restore the balance, equilibrium shifts to the left, i.e., more HF dissociates to produce H+ ions.
To summarize, the addition of a small amount of NaOH disturbs the equilibrium of the buffer system by reacting with HF and consuming H+ ions. As a result, the concentration of H+ ions in the solution decreases, leading to an increase in pH (more basic) of the buffer solution.