Given: A diprotic acid (H2A) has pKa1 = 2.8 and pKa2 = 8.4.Question: What is the equilibrium constant for the conversion ofH2A to A2−? The image depicts a chemical equilibrium equation:\[ \text{H}_2\text{A} \rightleftharpoons 2\text{H}^+ + \text{A}^{2-} \]This equation represents the dissociation of a generic diprotic acid (H₂A) into two hydrogen ions (H⁺) and a divalent anion (A²⁻) in an aqueous solution. The double arrow indicates that the reaction is in equilibrium, meaning it can proceed in both forward and reverse directions under certain conditions. Understanding this equation is essential in studying acid-base chemistry, particularly in analyzing how diprotic acids dissociate in solutions and affect pH levels.

Equilibrium constant is equal to the ratio of concentration of product to concentration of reactant. aA+bB⇌cC.K=[C]c[A]a[B]b Equilibrium constant for overall reaction is the product of the equilibrium constant in which reaction occurs in multiple steps. K=K1×K2×K3×K4×.......... In terms of pka value. pKa=pKa1+pKa2+pKa3+......... pKa=-logKa To determine the equilibrium constant for conversion of H2A to A2-. H2A⇌H++HA- pKa1=2.8.HA-⇌H++A2- pKa2=8.4. H2A⇌2H++A2- reaction-1. pKa3 for reaction 1=pKa1+pKa2.pKa3 for reaction 1=2.8+8.4=11.2. Equilibrium constant for reaction 1. pKa3=-logKa3.11.2=-logKa3.Ka3=6.30×10-12.Answer Equilibrium constant for conversion of H2A into A2- is 6.30 x 10-12.