Question

At ${{30}^{o}}C,$the solubility of $A{{g}_{2}}C{{O}_{3}}$ (${{K}_{sp}}=8X{{10}^{-12}}$ ) would be maximum in one litre of:
(A) 0.05 M $N{{a}_{2}}C{{O}_{3}}$
(B) 0.05 M $AgN{{O}_{3}}$
(C) pure water
(D) 0.05 M $N{{H}_{3}}$

Answer 1

The solubility product ${{K}_{sp}}$ describes the concentration of ions for a saturated solution, which means the solution at maximum equilibrium with molecules and ions of the solution. This equilibrium constant measures the maximum amount of ions and it changes with temperature.

Complete step by step answer:
However, the concentration of ions in a solution may be diluted, saturated, or supersaturated. so the ionic product which measures the maximum amount of ions in a solution and also varies on the concentration of ions dissolved.
If the solution is diluted, saturated, or supersaturated depends on the relation between ionic product and solubility product.
If the ionic product of solution is equal to the solubility product, then the solution is saturated without any precipitation.
If the ionic product is less than the solubility product, then the actual amount of ions is less than the maximum amount of ions and the solution is diluted but not precipitated.
If the ionic product is greater than the solubility product means that the actual amount of ions is more than the maximum amount of ions and the resulting solution is supersaturated.
Given the solubility product of $A{{g}_{2}}C{{O}_{3}}$ in one liter of solution ${{K}_{sp}}=8X{{10}^{-12}}$
If the 0.05 M $N{{a}_{2}}C{{O}_{3}}$, due to common ion effect of $C{{O}_{3}}^{-2}$ ion, the solubility of $A{{g}_{2}}C{{O}_{3}}$ will be less.
If pure water, $A{{g}_{2}}C{{O}_{3}}\rightleftharpoons 2A{{g}^{+}}+C{{O}_{3}}^{2-}$
In 0.05 M $N{{H}_{3}}$ ,
$A{{g}_{2}}C{{O}_{3}}\rightleftharpoons 2A{{g}^{+}}+C{{O}_{3}}^{2-}$
$2A{{g}^{+}}+4N{{H}_{3}}\rightleftharpoons 2Ag{{(N{{H}_{3}})}_{2}}^{+}$
Overall reaction: $A{{g}_{2}}C{{O}_{3}}+N{{H}_{3}}\Leftrightarrow 2Ag{{(N{{H}_{3}})}_{2}}^{+}+C{{O}_{3}}^{2-}$
Due to the complex formation between ammonium and silver ions shifts the solubility equilibrium in the forward reaction which increases the solubility of silver carbonate.
Therefore, $A{{g}_{2}}C{{O}_{3}}$ maximum solubility in 0.05 M $N{{H}_{3}}$ .

The correct answer is option D.

Note: In the case of polydentate ligands, coordination compounds stability depends upon chelate rings. This is known as the chelating effect. The stability of complex ions in this solution depends on the degree of dissociation between two species in the state of equilibrium.