Answer
This is the $NH_3$ ionization equation:
$NH_3(aq) + H_2O(l) \lt -- \gt N{H_4}^+(aq) + OH^-(aq)$
When we dilute the solution, we are adding more $H_2O$ to the reaction, and, according to Le Chatelier's principle, if we add more of one of the reactants, the equilibrium will shift to the right, increasing the direct reaction.
Since the equilibrium is favoring the direct reaction, the fraction of $NH_3$ that reacts with water will be increased.
Work Step by Step
- You can review this topic on section 12-6 of this book.
Le Chatelier's principle says that, if the equilibrium is affected and suffers a change, the reaction will readjust in a way that it partly counteracts the effects of this change.
So, if we dilute the solution of $NH_3$, the change is: The increase in water concentration.
Therefore, the reaction will readjust in a way that it consumes more $H_2O$, trying to get back to the concentration before the change.
Consumes more $H_2O --\gt$ Producing more $N{H_4}^+$ and $OH^-$.
So, the fraction of $NH_3$ that reacts with water is greater.
