If pK_a1 and pK_a2 for H₂CO₃ are 6.37 and 10.25,respectively,calculate the equilibrium constant for the reaction below: H₂CO₃(aq)+ 2H₂O(l)↔ 2H₃O⁺(aq)+ CO₃²^-(aq) A)4.1 * 10^-¹¹ B)4.3 * 10^-7 C)5.6 *10^-11 D)2.3 * 10^-8 E)2.4 * 10^-17

Question

Quizplus · Accepted Answer

The equilibrium constant for the reaction is the product of the two dissociation constants, K1 and K2. K1 = 10^(-pKa1) and K2 = 10^(-pKa2). Therefore, K = K1 * K2 = 10^(-6.37) * 10^(-10.25) = 2.4 * 10^(-17).

If PKa1 and PKa2 for H2CO3 Are 6

If PK_a1 and PK_a2 for H₂CO₃ Are 6