The two equilibria, AB ⇌ A+ + B- and AB + B- ⇌ AB2- are simultaneously maintained in a solution with — Chemical Equilibrium Chemistry Question

💡 Solution & Explanation

Step 1 - Express the equilibrium constants for both reactions We are given two chemical equilibria simultaneously maintained in an aqueous solution: 1. Dissociation of the neutral molecule \ce{AB}: \[\ce{AB <=> A+ + B-}\] According to the law of chemical equilibrium, the concentration-based equilibrium constant \(K_1\) is expressed as: \[K_1 = \frac{[\ce{A+}][\ce{B-}]}{[\ce{AB}]}\] 2. Formation of the complex ion \ce{AB2-}: \[\ce{AB + B- <=> AB2-}\] The equilibrium constant \(K_2\) for this reaction is expressed as: \[K_2 = \frac{[\ce{AB2-}]}{[\ce{AB}][\ce{B-}]}\] Step 2 - Isolate the concentration of the common species Since both equilibria exist simultaneously within the same solution, the concentration of \ce{AB} must be identical in both equilibrium expressions. From the first equilibrium expression: \[[\ce{AB}] = \frac{[\ce{A+}][\ce{B-}]}{K_1}\] From the second equilibrium expression: \[[\ce{AB}] = \frac{[\ce{AB2-}]}{K_2 [\ce{B-}]}\] Step 3 - Equate the expressions and solve for the ratio Setting both expressions for [AB] equal: \[\frac{[\ce{A+}][\ce{B-}]}{K_1} = \frac{[\ce{AB2-}]}{K_2 [\ce{B-}]}\] Rearranging to find the ratio: \[\frac{[\ce{A+}]}{[\ce{AB2-}]} = \frac{K_1}{K_2 [\ce{B-}]^2}\] Step 4 - Analyze the proportional relationship and evaluate the options At constant temperature, K1 and K2 are constants, so: \[\frac{[\ce{A+}]}{[\ce{AB2-}]} \propto \frac{1}{[\ce{B-}]^2}\] This shows the ratio is inversely proportional to the square of the concentration of \ce{B-}. * **Option (A) is incorrect:** Not directly proportional to [B-]. * **Option (B) is incorrect:** Inversely proportional to [B-]^2, not [B-]. * **Option (C) is incorrect:** Inversely, not directly, proportional to [B-]^2. * **Option (D) is correct:** The ratio is inversely proportional to the square of [B-]. \[\boxed{\text{D}}\]