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Question: Two equilibria, are simultaneously maintained in a solution with equilibrium constants, \[{K_1}\] an...

Two equilibria, are simultaneously maintained in a solution with equilibrium constants, K1{K_1} and K2{K_2} respectively. The ratio of [[A+]  to  [AB2−]  [{A^ + }]\;to\;[A{B_2}^ - ]\;in the solution is:
A. Directly proportional to the concentration B−{B^ - }
B. Inversely proportional to the concentration of B−{B^ - }
C. Directly proportional to the square of the concentration of B−{B^ - }
D. Inversely proportional to the square of the concentration of B−{B^ - }

Explanation

Solution

The situation in which both reactants and products present in concentration do not change within the period of time in which a chemical reaction occurs, it means the solution is in chemical equilibrium. When the rate of both forward and backward reactions are the same, this means that the chemical reaction has achieved chemical equilibrium.

Complete step-by-step answer: There is no net change in the concentrations of the reactants and products as the rates of the reactions are equal. This state is referred to as a dynamic equilibrium and the rate constant is known as equilibrium constant.
When we define equilibrium constant, we can say that as the product of the molar concentration of products which is raised during a reaction is equal to tits stoichiometric coefficient and then divided by products of the molar concentrations of the reacts, each raised to power equal to its stoichiometric coefficients is constant at constant temperature.
It can be defined in terms of partial pressure of reactants and products. However, if it is expressed in terms of the partial pressure, it is denoted by Kp.{K_{p.}}The law of mass action influences the equilibrium. It states that the rate of a chemical reaction is directly proportional to product of the concentrations of the reactants with power to their respective stoichiometric coefficients.
Therefore, given the reaction –
thus K1=[A+][B−][AB]{K_1} = \frac{{[{A^ + }][{B^ - }]}}{{\left[ {AB} \right]}}
and
thus K2=[AB2−][AB][B−]{K_2} = \frac{{[A{B_2}^ - ]}}{{[AB][{B^ - }]}}

On dividing Eq. (i) from Eq. (ii),
[A+][B−][AB]X[AB2−][AB][B−]\frac{{[{A^ + }][{B^ - }]}}{{\left[ {AB} \right]}}X\frac{{[A{B_2}^ - ]}}{{[AB][{B^ - }]}} =K1/K2 = {K_1}/{K_2}
Or [A+][B−]2[AB2−]=K′\frac{{[{A^ + }]{{[{B^ - }]}^2}}}{{[A{B_2}^ - ]}} = K\prime

Where   K′=K1K2=constant\;K\prime = \frac{{{K_1}}}{{{K_2}}} = constant
or [A+][AB2−]=K′X1[B−1]2\frac{{[{A^ + }]}}{{[A{B_2}^ - ]}} = K'X\frac{1}{{{{[{B^{ - 1}}]}^2}}}

or   [A+]/[AB2−]∝1/[B−]2.\;[{A^ + }]/[A{B_2}^ - ] \propto 1/{[{B^ - }]^2}.

Thus the correct option is DD.

Note: Le Châtelier's principle is an important phenomenon which predicts the behavior of an equilibrium system when changes to its reaction conditions occur. It states that ”If a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium moves to partially reverse the change.”