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

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

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Answer 1 · 2015-09-28T17:49:56

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 $A g C l$ $AgCl$ .

We write $A g C l (s) ⇌ A g^{+} + C l^{-}$ $AgCl(s) rightleftharpoons Ag^+ + Cl^-$ $(i)$ $(i)$

To describe this reaction, we write $K_{s p} = [A g^{+}] [C l^{-}]$ $K_(sp) = [Ag^+][Cl^-]$ . Should we add aqueous ammonia we will reduce the $[A g^{+}]$ $[Ag^+]$ value because of the following competing equiibrium:

$A g^{+} + 2 N H_{3} ⇌ {[A {g (N H_{3})}_{2}]}_{2}^{+}$ $Ag^+ + 2NH_3 rightleftharpoons [Ag(NH_3)_2]^+$ $(i i)$ $(ii)$

More silver ion, $[A g^{+}]$ $[Ag^+]$ , 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 $[A g^{+}]$ $[Ag^+]$ in equilibrium (i), we can drive the equilibrium to the right hand side.

On the other hand, for equilibrium (ii), if we reduce $[N H_{3}]$ $[NH_3]$ , say by increasing the $p H$ $pH$ to form $N H_{4}^{+}$ $NH_4^+$ , we would drive the equilibrium to the left. Of course, we are dealing with competing equilibria, that may be difficult to calculate.

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

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