How does Le Chatelier's principle explain the result you obtained after the addition of NaOH?
You need to get more explicit than that. What experiment did you do? To what did you add NaOH? What were the conditions?
Le Chatelier's principle is a fundamental concept in chemistry that states if a system in equilibrium is subjected to a change in conditions, the system will shift in a direction that minimizes the effect of that change. In the context of the addition of NaOH, let me explain how Le Chatelier's principle applies.
When NaOH is added to a solution, it acts as a strong base, increasing the concentration of hydroxide ions (OH-) in the solution. This change in concentration affects the equilibrium of any acid-base reactions in the system. Now, let's consider a specific example to understand the principle.
Suppose you have a solution in equilibrium, represented by the equation:
HA (acid) + H2O ⇌ H3O+ (hydronium ion) + A- (conjugate base)
In this equation, HA is the acid and A- is its conjugate base. When NaOH is added to the solution, the additional OH- ions will react with the hydronium ions (H3O+) to form water:
OH- + H3O+ ⇌ 2H2O
According to Le Chatelier's principle, the increased concentration of OH- ions will disrupt the equilibrium of the reaction. To counterbalance the increase in OH- ions, the reaction will shift to the left, consuming more hydronium ions and acid to minimize the effect of the added base. As a result, the concentration of HA (acid) and H3O+ (hydronium ions) will decrease, while the concentration of A- (conjugate base) will increase.
This shift in equilibrium favors the reactant side of the equation, reducing the acidity of the system and increasing the basicity. Hence, the result obtained after the addition of NaOH is a decrease in the acidity of the solution due to the shift in equilibrium towards the left to consume the added base.