According to Le Chatelier's Principle: What happens when you add DIFFERENT species to the equilibrium?

Sep 28, 2015

Le Chatelier's principle applies to equilibrium concentrations in specific equilibria. Addition of different species that affect these equilibrium concentrations, will change the equilibria accordingly.

Explanation:

Consider the solution behaviour of a sparingly soluble or insoluble salt at equilibrium, say $A g C l$.

We write $A g C l \left(s\right) r i g h t \le f t h a r p \infty n s A {g}^{+} + C {l}^{-}$$\left(i\right)$

To describe this reaction, we write ${K}_{s p} = \left[A {g}^{+}\right] \left[C {l}^{-}\right]$. Should we add aqueous ammonia we will reduce the $\left[A {g}^{+}\right]$ value because of the following competing equiibrium:

$A {g}^{+} + 2 N {H}_{3} r i g h t \le f t h a r p \infty n s {\left[A g {\left(N {H}_{3}\right)}_{2}\right]}^{+}$$\left(i i\right)$

More silver ion, $\left[A {g}^{+}\right]$, will enter solution because it can be complexed by the ammonia. To give an answer in terms of Le Chatelier's principle, because we reduce $\left[A {g}^{+}\right]$ in equilibrium (i), we can drive the equilibrium to the right hand side.

On the other hand, for equilibrium (ii), if we reduce $\left[N {H}_{3}\right]$, say by increasing the $p H$ to form $N {H}_{4}^{+}$, we would drive the equilibrium to the left. Of course, we are dealing with competing equilibria, that may be difficult to calculate.